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he generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org Contents Projectile Motion Projectile Motion for an Object Launched Horizontally Projectile Motion for an Object Launched at an Angle Projectile Motion Problem Solving 1 4 7 11 1 2 3 4 iv www.ck12.org Chapter 1. Projectile Motion CHAPTER 1 Projectile Motion • Describe projectile motion and state when it occurs. • Give examples of projectile motion. The archer in the opening image is aiming his arrow a little bit above the bull’s eye of the target, rather than directly at it. Why doesn’t he aim at the bull’s eye instead? The answer is projectile motion. Combining Forces When the archer releases the bowstring, the arrow will be ﬂung forward toward the top of the target where she’s aiming. But another force will also act on the arrow in a different direction. The other force is gravity, and it will pull the arrow down toward Earth. The two forces combined will cause the arrow to move in the curved path shown in the Figure 1.1. This type of motion is called projectile motion. It occurs whenever an object curves down toward the ground because it has both a horizontal force and the downward force of gravity acting on it. Because of projectile motion, to hit the bull’s eye of a target with an arrow, you actually have to aim for a spot above the bull’s eye. You can see in theFigure 1.2 what happens if you aim at the bull’s eye instead of above it. 1 www.ck12.org FIGURE 1.1 FIGURE 1.2 Another Example of Projectile Motion You can probably think of other examples of projectile motion. One is shown in the Figure 1.3. The cannon shoots a ball straight ahead, giving it horizontal motion. At the same time, gravity pulls the ball down toward the ground. FIGURE 1.3 Q: How would you show the force of gravity on the cannon ball in the Figure 1.3? A: You would add a line pointing straight down from the cannon to the ground. To get a better feel for projectile motion, try these interactive animations: • http://phet.colorado.edu/en/simulation/projectile-motion • http://jersey.uoregon.edu/vlab/ (Click on the applet “Cannon.”) Summary • Projectile motion is movement of an object in a curved path toward the ground because it has both a horizontal force and the downward force of gravity acting on it. 2 www.ck12.org Chapter 1. Projectile Motion • Examples of objects that have projectile motion include arrows and cannon balls. Vocabulary • projectile motion: Motion of an object that has initial horizontal velocity but is also pulled down toward Earth by gravity. Practice Play the game at the following URL by shooting the cannon at a stationary target. Experiment with three variablespower, height of barrel, and angle of barrelyou ﬁnd at least three different combinations of variables that allow the cannon ball to hit the target. Record the values for the three combinations of variables. Then summarize what you learned by doing the activity. http://www.science-animations.com/support-ﬁles/projektielbeweging.swf Review 1. What is projectile motion? When does it occur? 2. How might knowledge of projectile motion help you shoot baskets in basketball? References 1. Laura Guerin. . CC BY-NC 3.0 2. Laura Guerin. . CC BY-NC 3.0 3. Christopher Auyeung. . CC-BY-NC-SA 3.0 3 CHAPTER 2 Projectile Motion for an Object Launched Horizontally www.ck12.org • State the relationship between the vertical and horizontal velocities of a projectile launched horizontally. • Find the time for a horizontally launched projectile to strike the ground. • Calculate the range of a horizontally launched projectile. The activity of bike jumping, like other sports that involve vector motions in perpendicular directions, requires more physical practice than mathematical analysis. The laws of physics apply to the activity, however, whether the biker is aware of them or not. Projectile Motion for an Object Launched Horizontally Objects that are launched into the air are called projectiles. The path followed by an object in projectile motion is called a trajectory. The motion of a projectile is described in terms of its position, velocity, and acceleration. Our knowledge that perpendicular components of vectors do not affect each other allow us to analyze the motion of projectiles. 4 www.ck12.org Chapter 2. Projectile Motion for an Object Launched Horizontally In the diagram, two balls (one red and one blue) are dropped at the same time. The red ball is released with no horizontal motion and the blue ball is dropped but also given a horizontal velocity of 10 m/s. As the balls fall to the ﬂoor, a photograph is taken every second so that in 5 seconds, we have 5 images of the two balls. Each vertical line on the diagram represents 5 m. Since the blue ball has a horizontal velocity of 10 m/s, you will see that for every second, the blue ball has moved horizontally 10 m. That is, in each second, the blue ball has increased its horizontal distance by 10 m. This horizontal motion is constant velocity motion. The red ball was dropped straight down with no horizontal velocity and therefore, in each succeeding second, the red ball falls straight down with no horizontal motion. The succeeding distances between seconds with the red ball motion indicates this motion is accelerated. A very important point here is that, the vertical motion of these two balls is identical. That is, they each fall exactly the same distance vertically in each succeeding second. The constant horizontal velocity of the blue ball has no effect on its accelerated vertical motion! Therefore, the vertical motion of the blue ball (the projectile) can be analyzed exactly the same as the vertical motion of the red ball. Example Problem: If an arrow if ﬁred from a bow with a perfectly horizontal velocity of 60.0 m/s and the arrow was 2.00 m above the ground when the it was released, how far will the arrow ﬂy horizontally before it strikes the ground? Solution: This problem is solved by determining how long it takes the arrow to fall to the ground in exactly the same manner as if the arrow was dropped with no horizontal velocity. Then the time required for the fall is multiplied by the horizontal velocity to get the horizontal distance. d = 1 2 at2 solved for t = s r 2d a = (2)(2:00 m) 9:80 m/s2 = 0:639 s The time required for the arrow to fall to the ground is the same time that the arrow ﬂies horizontally at 60.0 m/s, so dhorizontal = (vhorizontal)(time) = (60:0 m/s)(0:639 s) = 38:3 m Example Problem: A rock was thrown horizontally from a 100.0 m high cliff. It strikes the ground 90.0 m from the base of the cliff. At what speed was it thrown? 5 Solution: We can calculate how long it takes for a rock to free fall 100.0 m and then divide this time into the horizontal distance to get the horizontal velocity. www.ck12.org s r 2d a = d t = 90:0 m 4:52 s t = v = (2)(100:0 m) 9:80 m/s2 = 4:52 s = 19:9 m/s Summary • Perpendicular components of vectors do not inﬂuence each other. • The horizontal motion of a projectile does not inﬂuence its free fall. Practice The following video discusses projectile motion for projectiles launched horizontally. http://www.youtube.com/watch?v=-uUsUaPJUc0 1. Why does speedy need to drive a convertible? 2. What is speedy doing in this video? 3. How does the horizontal velocity change during the fall? 4. How does the vertical velocity change during the fall? Review 1. If a bullet is ﬁred from a high powered riﬂe at the exact time a duplicate bullet is dropped by hand near the barrel of the riﬂe, which bullet will hit the ground ﬁrst? (a) the one dropped straight down (b) the one ﬁred horizontally (c) both will hit the ground at the same time 2. A cannon is ﬁred from the edge of a small cliff. The height of the cliff is 80.0 m. The cannon ball is ﬁred with a perfectly horizontal velocity of 80.0 m/s. How far will the cannon ball ﬂy horizontally before it strikes the ground? 3. A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later. How high is the cliff and how far from the base of the cliff did the diver hit the water? • projectile motion: A form of motion where a particle (called a projectile) is thrown obliquely near the earth’s surface, it moves along a curved path under the action of gravity. The path followed by a projectile motion is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory after which there is no interference apart from gravity. • trajectory: The path followed by an object in projectile motion. References 1. Courtesy of PDPhoto.org. http://www.pdphoto.org/PictureDetail.php?mat=&pg=7672. Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 6 www.ck12.org Chapter 3. Projectile Motion for an Object Launched at an Angle CHAPTER 3 Pr
ojectile Motion for an Object Launched at an Angle • The student will calculate the maximum height and range of projectiles launched at an angle given the initial velocity and angle. In the case of the human cannonball shown, all the vector and gravitational calculations must be worked out perfectly before the ﬁrst practice session. With this activity, you cannot afford trial and error –the ﬁrst miss might be the last trial. Projectile Motion for an Object Launched at an Angle When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity. As usual, the analysis of the motion involves dealing with the two motions independently. 7 www.ck12.org Example Problem: A cannon ball is ﬁred with an initial velocity of 100. m/s at an angle of 45° above the horizontal. What maximum height will it reach and how far will it ﬂy horizontally? Solution: The ﬁrst step in the analysis of this motion is to resolve the initial velocity into its vertical and horizontal components. viup = (100: m/s)(sin 45) = (100: m/s)(0:707) = 70:7 m/s vihorizontal = (100: m/s)(cos 45) = (100: m/s)(0:707) = 70:7 m/s We will deal with the vertical motion ﬁrst. The vertical motion is symmetrical. The object will rise up to its highest point and then fall back. The distance it travels up will be the same as the distance it falls down. The time it takes to reach the top will be the same time it takes to fall back to its initial point. The initial velocity upward will be the same magnitude (opposite in direction) as the ﬁnal velocity when it returns to its original height. There are several ways we could approach the upward motion. We could calculate the time it would take gravity to bring the initial velocity to rest. Or, we could calculate the time it would take gravity to change the initial velocity from +70.7 m/s to -70.0 m/s. Yet another way would be to calculate the time for the height of the object to return to zero. v f = vi + at so t = v f vi a If we calculate the time required for the ball to rise up to its highest point and come to rest, the initial velocity is 70.7 m/s and the ﬁnal velocity is 0 m/s. Since we have called the upward velocity positive, then the acceleration must be negative or -9.80 m/s2. t = v f vi a = 0 m/s70:7 m/s 9:80 m/s2 = 7:21 s The maximum height reached can be calculated by multiplying the time for the upward trip by the average vertical velocity. The average upward velocity during the trip up is one-half the initial velocity. vupave = 1 height = (vupave)(tup) = (35:3 m/s)(7:21 s) = 255 m (70:7 m/s) = 35:3 m/s 2 Since this is the time required for the cannon ball to rise up to its highest point and come to rest, then the time required for the entire trip up and down would be double this value, or 14.42 s. The horizontal distance traveled during the ﬂight is calculated by multiplying the total time times the constant horizontal velocity. dhorizontal = (14:42 s)(70:7 m/s) = 1020 m Example Problem: A golf ball was knocked into the with an initial velocity of 4.47 m/s at an angle of 66° with the horizontal. How high did the ball go and how far did it ﬂy horizontally? Solution: viup = (4:47 m/s)(sin 66) = (4:47 m/s)(0:913) = 4:08 m/s vihor = (4:47 m/s)(cos 66) = (4:47 m/s)(0:407) = 1:82 m/s 8 www.ck12.org Chapter 3. Projectile Motion for an Object Launched at an Angle a = 0 m/s4:08 m/s 9:80 m/s2 = 0:416 s tup = v f vi vupave = 1 height = (vupave)(tup) = (2:04 m/s)(0:416 s) = 0:849 m (4:08 m/s) = 2:04 m/s 2 ttotal trip = (2)(0:416 s) = 0:832 s dhorizontal = (0:832 s)(1:82 m/s) = 1:51 m Example Problem: Suppose a cannon ball is ﬁred downward from a 50.0 m high cliff at an angle of 45° with an initial velocity of 80.0 m/s. How far horizontally will it land from the base of the cliff? Solution: In this case, the initial vertical velocity is downward and the acceleration due to gravity will increase this downward velocity. vidown = (80:0 m/s)(sin 45) = (80:0 m/s)(0:707) = 56:6 m/s vihor = (80:0 m/s)(cos 45) = (80:0 m/s)(0:707) = 56:6 m/s d = vidownt + 1 2 at2 50:0 = 56:6t + 4:9t2 Changing to standard quadratic form yields 4:9t2 + 56:6t 50:0 = 0 This equation can be solved with the quadratic formula. The quadratic formula will produce two possible solutions for t: p t = b+ 56:6+ t = p b2 4ac b2 4ac and t = b 2a 2a p(56:6)2 (4)(4:9)(50) (2)(4:9) = 0:816 s The other solution to the quadratic formula yields a negative number which is clearly not a reasonable solution for this problem. dhorizontal = (0:816 s)(56:6 m/s) = 46:2 m Summary • When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. • The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. • The horizontal motion is constant velocity motion and undergoes no changes due to gravity. • The analysis of the motion involves dealing with the two motions independently. Practice The following video shows a motion analysis for projectile motion launched upward. http://www.youtube.com/watch?v=rMVBc8cE5GU MEDIA Click image to the left for more content. 9 www.ck12.org The following video shows the famous "shoot the monkey" demonstration. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=cxvsHNRXLjw In this demonstration, a stuffed toy is hung from a high support and is attached to the support by an electric switch. A golf ball cannon is aimed up at the “monkey” while it is hanging on the support. The cannon is designed such that It when the golf ball projectile leaves the barrel, it triggers the switch and releases the toy monkey from its perch. would seem (to the non-physicist) that the projectile will miss the monkey because the monkey will fall under the line of ﬁre. The physicist knows, however, that the projectile falls from its line of ﬁre by exactly the same amount that the monkey falls and therefore, the projectile will hit the monkey every time . . . in fact, it cannot miss. 1. What is the cannon ball in this video? 2. What is used as the monkey in this video? Review 1. A player kicks a football from ground level with a velocity of magnitude 27.0 m/s at an angle of 30.0° above the horizontal. (a) Find the time the ball is in the air. (b) Find the maximum height of the ball. (c) Find the horizontal distance the ball travels. 2. A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land? 3. An arrow is ﬁred downward at an angle of 45 degrees from the top of a 200 m cliff with a velocity of 60.0 m/s. (a) How long will it take the arrow to hit the ground? (b) How far from the base of the cliff will the arrow land? • trajectory: The ballistic trajectory of a projectile is the path that a thrown or launched projectile will take under the action of gravity, neglecting all other forces, such as air resistance, without propulsion. References 1. Flickr: JoshBerglund19. http://www.flickr.com/photos/tyrian123/1539636464/. CC-BY 2.0 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 10 www.ck12.org Chapter 4. Projectile Motion Problem Solving CHAPTER 4 Projectile Motion Problem Solving Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components. Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components. Key Equations Break the Initial Velocity Vector into its Components Apply the Kinematics Equations y(t) = yi + viyt 1 Vertical Direction 2 gt2 vy(t) = viy gt 2 2g(Dy) vy ay = g = 9:8m=s2 10m=s2 2 = v0y Horizontal Direction x(t) = xi + vixt vx(t) = vix ax = 0 11 www.ck12.org Guidance • To work these problems, separate the “Big Three” equations into two sets: one for the vertical direction, and one for the horizontal. Keep them separate. • The only variable that can go into both sets of equations is time; use time to communicate between the x and y components of the object’s motion. Example 1 CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is ﬂat, i.e. the car came off the cliff horizontally). Question: v = ? [m=s] Given: h = Dy = 72 m d = Dx = 22 m g = 10:0 m=s2 Equation: h = viyt + 1 2 gt2 and d = vixt Plug n’ Chug: Step 1: Calculate the time required for the car to freefall from a height of 72 m. h = viyt + 1 s 2 gt2 but since viy = 0, the equation simpliﬁes to h = 1 s 2 gt2 rearranging for the unknown variable, t, yields t = 2h g = 2(72 m) 10:0 m=s2 = 3:79 s Step 2: Solve for initial velocity: vix = d 3:79 s = 5:80 m=s t = 22 m Answer: Example 2 5:80 m=s Question: A ball of mass m is moving horizontally with a speed of vi off a cliff of height h. How much time does it take the ball to travel from the edge of the cliff to the ground? Express your answer in terms of g (acceleration due to gravity) and h (height of the cliff). Solution: Since we are solving or how long it takes for the ball to reach ground, any motion in the x direction is not pertinent. To make this problem a little simpler, we will deﬁne down as the positive direction and the top of the cliff to be . In this solution we will use the equation y = 0 y(t) = yo + voyt + gt2 1 2 . 12 www.ck12.org Chapter 4. Projectile Mot
ion Problem Solving gt2 gt2 1 2 1 2 gt2 1 2 gt2 y(t) = yo + voyt + h = yo + voyt + h = 0 + voy + gt2 2 s 2h g start with the equation substitute h for y(t) because that’s the position of the ball when it hits the ground after time t substitute 0 for yobecause the ball starts at the top of the cliff substitute 0 for voy becauese the ball starts with no vertical component to it’s velocity simplify the equation solve for t Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A stone is thrown horizontally at a speed of 8:0 m=s from the edge of a cliff 80 m in height. How far from the base of the cliff will the stone strike the ground? 2. A toy truck moves off the edge of a table that is 1:25 m high and lands 0:40 m from the base of the table. a. How much time passed between the moment the car left the table and the moment it hit the ﬂoor? b. What was the horizontal velocity of the car when it hit the ground? 3. A hawk in level ﬂight 135 m above the ground drops the ﬁsh it caught. If the hawk’s horizontal speed is 20:0 m=s, how far ahead of the drop point will the ﬁsh land? 4. A pistol is ﬁred horizontally toward a target 120 m away, but at the same height. The bullet’s velocity is 200 m=s. How long does it take the bullet to get to the target? How far below the target does the bullet hit? 5. A bird, traveling at 20 m=s, wants to hit a waiter 10 m below with his dropping (see image). In order to hit the waiter, the bird must release his dropping some distance before he is directly overhead. What is this distance? 13 www.ck12.org 6. Joe Nedney of the San Francisco 49ers kicked a ﬁeld goal with an initial velocity of 20 m=s at an angle of 60. a. How long is the ball in the air? Hint: you may assume that the ball lands at same height as it starts at. b. What are the range and maximum height of the ball? 7. A racquetball thrown from the ground at an angle of 45 and with a speed of 22:5 m=s lands exactly 2:5 s later on the top of a nearby building. Calculate the horizontal distance it traveled and the height of the building. 8. Donovan McNabb throws a football. He throws it with an initial velocity of 30 m=s at an angle of 25. How much time passes until the ball travels 35 m horizontally? What is the height of the ball after 0:5 seconds? (Assume that, when thrown, the ball is 2 m above the ground.) 9. Pablo Sandoval throws a baseball with a horizontal component of velocity of 25 m=s. After 2 seconds, the ball is 40 m above the release point. Calculate the horizontal distance it has traveled by this time, its initial vertical component of velocity, and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time? 10. Barry Bonds hits a 125 m(4500) home run that lands in the stands at an altitude 30 m above its starting altitude. Assuming that the ball left the bat at an angle of 45 from the horizontal, calculate how long the ball was in the air. 11. A golfer can drive a ball with an initial speed of 40:0 m=s. If the tee and the green are separated by 100 m, but are on the same level, at what angle should the ball be driven? ( Hint: you should use 2 cos (x) sin (x) = sin (2x) at some point.) 12. How long will it take a bullet ﬁred from a cliff at an initial velocity of 700 m=s, at an angle 30 below the horizontal, to reach the ground 200 m below? 13. A diver in Hawaii is jumping off a cliff 45 m high, but she notices that there is an outcropping of rocks 7 m out at the base. So, she must clear a horizontal distance of 7 m during the dive in order to survive. Assuming the diver jumps horizontally, what is his/her minimum push-off speed? 14. If Monte Ellis can jump 1:0 m high on Earth, how high can he jump on the moon assuming same initial velocity that he had on Earth (where gravity is 1=6 that of Earth’s gravity)? 15. James Bond is trying to jump from a helicopter into a speeding Corvette to capture the bad guy. The car is going 30:0 m=s and the helicopter is ﬂying completely horizontally at 100 m=s. The helicopter is 120 m above the car and 440 m behind the car. How long must James Bond wait to jump in order to safely make it into the car? 14 www.ck12.org Chapter 4. Projectile Motion Problem Solving 16. A ﬁeld goal kicker lines up to kick a 44 yard (40 m) ﬁeld goal. He kicks it with an initial velocity of 22 m=s at an angle of 55. The ﬁeld goal posts are 3 meters high. a. Does he make the ﬁeld goal? b. What is the ball’s velocity and direction of motion just as it reaches the ﬁeld goal post (i.e., after it has traveled 40 m in the horizontal direction)? 17. In a football game a punter kicks the ball a horizontal distance of 43 yards (39 m). On TV, they track the hang time, which reads 3:9 seconds. From this information, calculate the angle and speed at which the ball was kicked. (Note for non-football watchers: the projectile starts and lands at the same height. It goes 43 yards horizontally in a time of 3:9 seconds) Answers to Selected Problems 1. 32 m 2. a. 0:5 s b. 0:8 m=s 3. 104 m 4. t = 0:60 s; 1:8 m below target 5. 28 m. 6. a. 3:5 s. b. 35 m; 15 m 7. 40 m; 8:5 m 8. 1:3 seconds, 7:1 meters 9. 50 m; v0y = 30 m=s; 500; on the way up 10. 4:4 s 11. 19 12. 0:5 s 13. 2:3 m=s 14. 6 m 15. 1:4 seconds 16. a. yes b. 14 m=s @ 23 degrees from horizontal 17. 22 m=s @ 62 degrees 15 Physics Unit 5 (Forces and Newton’s Laws) Patrick Marshall Ck12 Science Jean Brainard, Ph.D. James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science Jean Brainard, Ph.D. James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org 1 4 10 13 16 20 24 27 31 33 Contents 1 Newton’s First and Second Laws of Motion 2 Newton’s First Law 3 Newton’s Second Law 4 Newton’s Third Law of Motion 5 Types of Forces 6 Universal Law of Gravity 7 Mass versus Weight 8 9 Friction Free Body Diagrams 10 Problem Solving 1 iv www.ck12.org Chapter 1. Newton’s First and Second Laws of Motion CHAPTER 1 Newton’s First and Second Laws of Motion • Deﬁne force. • State the fundamental units for the Newton. • State Newton’s First Law of Motion. • Given two of the three values in F = ma, calculate the third. This image is of Buzz Aldrin, one of the ﬁrst men to walk on the moon. Apollo 11 was the spaceﬂight that landed the ﬁrst humans, Neil Armstrong and Buzz Aldrin, on the Moon on July 20, 1969. Armstrong became the ﬁrst to step onto the lunar surface 6 hours later on July 21. This accomplishment could not have occurred without a thorough understanding of forces and acceleration. Newton’s First and Second Laws of Motion What is a force? A force can be deﬁned as a push or pull. When you place a book on a table, the book pushes downward on the table and the table pushes upward on the book. The two forces are equal and there is no resulting If, on the other hand, you drop the book, it will fall to the ground pulled by a force called motion of the book. gravity. 1 www.ck12.org If you slide a book across the ﬂoor, it will experience a force of friction which acts in the opposite direction of the motion. This force will slow down the motion of the book and eventually bring it to rest. If the ﬂoor is smoother, If a perfectly smooth ﬂoor the force of friction will be less and the book will slide further before coming to rest. could be created, there would be no friction and the book would slide forever at constant speed. Newton’s First Law of Motion describes an object moving with constant speed in a straight line. In the absence of any force, the object will continue to move at the same constant speed and in the same straight line. If the object is at rest, in the absence of any force, it will remain at rest. Newton’s First Law states that an object with no force acting on it moves with constant velocity. (The constant velocity could, of course, be 0 m/s.) A more careful expression of Newton’s First Law is "an object with no net force acting on it remains at rest or moves with constant velocity in a straight line." The statement above is equivalent to a statement that "if there is no net force on an object, there will be no acceleration." In the absence of acceleration, an object will remain at rest or will move with constant velocity in a straight line. The acceleration of an object is the result of an unbalanced force. If an object suffers two forces, the
motion of the object is determined by the net force. The magnitude of the acceleration is directly proportional to the magnitude of the unbalanced force. The direction of the acceleration is the same direction as the direction of the unbalanced force. The magnitude of the acceleration is inversely proportional to the mass of the object. i.e. The more massive the object, the smaller will be the acceleration produced by the same force. These relationships are stated in Newton’s Second Law of Motion, "the acceleration of an object is directly proportional to the net force on the object and inversely proportional to the mass of the object." Newton’s Second Law can be summarized in an equation. a = F m or more commonly; F = ma According to Newton’s second law, a new force on an object causes it to accelerate and the larger the mass, the smaller the acceleration. Sometimes, the word inertia is used to express the resistance of an object to acceleration. Therefore, we say that a more massive object has greater inertia. The units for force are deﬁned by the equation for Newton’s second law. Suppose we wish to express the force that will give a 1.00 kg object an acceleration of 1.00 m/s2. F = ma = (1:00 kg)(1:00 m/s2) = 1:00 kg m/s2 This unit is deﬁned as 1.00 Newton or 1.00 N. kgm s2 = Newton Example Problem: What new force is required to accelerate a 2000. kg car at 2.000 m/s2? Solution: F = ma = (2000: kg)(2:000 m/s2) = 4000: N Example Problem: A net force of 150 N is exerted on a rock. The rock has an acceleration of 20. m/s2 due to this force. What is the mass of the rock? Solution: m = F = 7:5 kg a = (150 N) (20: m/s2 ) Example Problem: A net force of 100. N is exerted on a ball. If the ball has a mass of 0.72 kg, what acceleration will it undergo? Solution: a = F m = (100: N) (0:72 kg) = 140 m/s2 Summary • A force is a push or pull. • Newton’s First Law states that an object with no net force acting on it remains at rest or moves with constant velocity in a straight line. • Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the net 2 www.ck12.org Chapter 1. Newton’s First and Second Laws of Motion force on the object and inversely proportional to the mass of the object. Expressed as an equation, F = ma. Practice Professor Mac explains Newton’s Second Law of Motion. http://www.youtube.com/watch?v=-KxbIIw8hlc MEDIA Click image to the left for more content. Review 1. A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car’s brakes. What was the braking force? 2. Calculate the average force that must be exerted on a 0.145 kg baseball in order to give it an acceleration of 130 m/s2. 3. After a rocket ship going from the Earth to the Moon leaves the gravitational pull of the Earth, it can shut off its engine and the ship will continue on to the Moon due to the gravitational pull of the Moon. (a) True (b) False 4. If a space ship traveling at 1000 miles per hour enters an area free of gravitational forces, its engine must run at some minimum level in order to maintain the ships velocity. (a) True (b) False 5. Suppose a space ship traveling at 1000 miles per hour enters an area free of gravitational forces and free of air resistance. If the pilot wishes to slow the ship down, he can accomplish that by shutting off the engine for a while. (a) True (b) False • force: A push or pull. References 1. Courtesy of Neil A. Armstrong, NASA. http://spaceﬂight.nasa.gov/gallery/images/apollo/apollo11/html/as11 -40-5873.html. Public Domain 3 CHAPTER 2 Newton’s First Law www.ck12.org • Use skateboarding to explain Newton’s ﬁrst law of motion. There’s no doubt from Corey’s face that he loves skateboarding! Corey and his friends visit Newton’s Skate Park every chance they get. They may not know it, but while they’re having fun on their skateboards, they’re actually applying science concepts such as forces and motion. 4 www.ck12.org Starting and Stopping Chapter 2. Newton’s First Law Did you ever ride a skateboard? Even if you didn’t, you probably know that to start a skateboard rolling over a level surface, you need to push off with one foot against the ground. That’s what Corey’s friend Nina is doing in this picture 2.1. FIGURE 2.1 Do you know how to stop a skateboard once it starts rolling? Look how Nina’s friend Laura does it in the Figure 2.2. She steps down on the back of the skateboard so it scrapes on the pavement. This creates friction, which stops the skateboard. Even if Laura didn’t try to stop the skateboard, it would stop sooner or later. That’s because there’s also friction between the wheels and the pavement. Friction is a force that counters all kinds of motion. It occurs whenever two surfaces come into contact. Video Break Laura learned how to use forces to start and stop her skateboard by watching the videos below. Watch the video to see how the forces are applied. You can pick up some skateboarding tips at the same time! Starting: http://www.youtube.com/watch?v=OpZIVjbMAOU MEDIA Click image to the left for more content. Stopping: http://www.youtube.com/watch?v=6fuOwhx91zM 5 www.ck12.org FIGURE 2.2 MEDIA Click image to the left for more content. Laws of the Park: Newton’s First Law If you understand how a skateboard starts and stops, then you already know something about Newton’s ﬁrst law of motion. This law was developed by English scientist Isaac Newton around 1700. Newton was one of the greatest scientists of all time. He developed three laws of motion and the law of gravity, among many other contributions. Newton’s ﬁrst law of motion states that an object at rest will remain at rest and an object in motion will stay in motion unless it is acted on by an unbalanced force. Without an unbalanced force, a moving object will not only 6 www.ck12.org Chapter 2. Newton’s First Law keep moving, but its speed and direction will also remain the same. Newton’s ﬁrst law of motion is often called the law of inertia because inertia is the tendency of an object to resist a change in its motion. If an object is already at rest, inertia will keep it at rest. If an object is already in motion, inertia will keep it moving. Do You Get It? Q: How does Nina use Newton’s ﬁrst law to start her skateboard rolling? A: The skateboard won’t move unless Nina pushes off from the pavement with one foot. The force she applies when she pushes off is stronger than the force of friction that opposes the skateboard’s motion. As a result, the force on the skateboard is unbalanced, and the skateboard moves forward. Q: How does Nina use Newton’s ﬁrst law to stop her skateboard? A: Once the skateboard starts moving, it would keep moving at the same speed and in the same direction if not for another unbalanced force. That force is friction between the skateboard and the pavement. The force of friction is unbalanced because Nina is no longer pushing with her foot to keep the skateboard moving. That’s why the skateboard stops. Changing Direction Corey’s friend Jerod likes to skate on the ﬂat banks at Newton’s Skate Park. That’s Jerod in the picture above. As he reaches the top of a bank, he turns his skateboard to go back down. To change direction, he presses down with his heels on one edge of the skateboard. This causes the skateboard to turn in the opposite direction. Video Break Can you turn a skateboard like Jerod? To see how to apply forces to change the direction of a skateboard, watch this video: http://www.youtube.com/watch?v=iOnlcEk50CM MEDIA Click image to the left for more content. Do You Get It? Q: How does Jerod use Newton’s ﬁrst law of motion to change the direction of his skateboard? A: Pressing down on just one side of a skateboard creates an unbalanced force. The unbalanced force causes the skateboard to turn toward the other side. In the picture, Jerod is pressing down with his heels, so the skateboard turns toward his toes. Summary • Newton’s ﬁrst law of motion states that an object at rest will remain at rest and an object in motion will remain in motion unless it is acted on by an unbalanced force. 7 www.ck12.org FIGURE 2.3 • Using unbalanced forces to control the motion of a skateboard demonstrates Newton’s ﬁrst law of motion. Vocabulary • Newton’s ﬁrst law of motion: Law stating that an object’s motion will not change unless an unbalanced force acts on the object. Practice Do you think you understand Newton’s ﬁrst law? Go to the URL below to ﬁnd out. Review Newton’s law and watch what happens to the skateboarder in the animation. Then answer the questions at the bottom of the Web page. http://teachertech.rice.edu/Participants/louviere/Newton/law1.html 8 www.ck12.org Review Chapter 2. Newton’s First Law 1. State Newton’s ﬁrst law of motion. 2. You don’t need to push off with a foot against the ground to start a skateboard rolling down a bank. Does this violate Newton’s ﬁrst law of motion? Why or why not? FIGURE 2.4 3. Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the skateboard stopped suddenly and Nina ended up on the ground (see Figure above). Explain what happened. 4. Now that you know about Newton’s ﬁrst law of motion, how might you use it to ride a skateboard more safely? References 1. Image copyright DenisNata, 2012. . Used under license from Shutterstock.com 2. Image copyright DenisNata, 2012. . Used under license from Shutterstock.com 3. Image copyright Nikola Bilic, 2012. . Used under license from Shutterstock.com 4. Image copyright DenisNata, 2012. . Used under license from Shutterstock.com 9 CHAPTER 3 Newton’s Second Law www.ck12.org • State Newton’s second law of motion. • Compare and contrast the effects of force and mass on acceleration. These boys are racing around the track at Newton’s Skate Park. The boy who can increase his speed the most will win the race. Tony, who is closest to the camera in this picture, is bigger and stronger than the other two boys, so he can apply
greater force to his skates. Q: Does this mean that Tony will win the race? A: Not necessarily, because force isn’t the only factor that affects acceleration. Force, Mass, and Acceleration Whenever an object speeds up, slows down, or changes direction, it accelerates. Acceleration occurs whenever an unbalanced force acts on an object. Two factors affect the acceleration of an object: the net force acting on the object and the object’s mass. Newton’s second law of motion describes how force and mass affect acceleration. The law 10 www.ck12.org Chapter 3. Newton’s Second Law states that the acceleration of an object equals the net force acting on the object divided by the object’s mass. This can be represented by the equation: Acceleration = Net force Mass or a = F m Q: While Tony races along on his rollerblades, what net force is acting on the skates? A: Tony exerts a backward force against the ground, as you can see in the Figure 3.1, ﬁrst with one skate and then with the other. This force pushes him forward. Although friction partly counters the forward motion of the skates, it is weaker than the force Tony exerts. Therefore, there is a net forward force on the skates. FIGURE 3.1 Direct and Inverse Relationships Newton’s second law shows that there is a direct relationship between force and acceleration. The greater the force that is applied to an object of a given mass, the more the object will accelerate. For example, doubling the force on the object doubles its acceleration. The relationship between mass and acceleration is different. It is an inverse relationship. In an inverse relationship, when one variable increases, the other variable decreases. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force. Q: Tony has greater mass than the other two boys he is racing above. How will this affect his acceleration around the track? A: Tony’s greater mass will result in less acceleration for the same amount of force. Summary • Newton’s second law of motion states that the acceleration of an object equals the net force acting on the object divided by the object’s mass. • According to the second law, there is a direct relationship between force and acceleration and an inverse relationship between mass and acceleration. Vocabulary • Newton’s second law of motion: Law stating that the acceleration of an object equals the net force acting on the object divided by the object’s mass. 11 www.ck12.org Practice At the following URL, use the simulator to experiment with force, mass, and acceleration. First keep force constant at 1 N, and vary mass from 1–5 kg. Next keep mass constant at 1 kg, and vary force from 1–5 N. In each simulation, record the values you tested and the resulting acceleration. Finally, make two line graphs to plot your results. On one graph, show acceleration when force is constant and mass changes. On the other graph, show acceleration when mass is constant and force changes. Describe in words what the two graphs show. http://janggeng.com/newtons-second-law-of-motion/ Review 1. State Newton’s second law of motion. 2. How can Newton’s second law of motion be represented with an equation? 3. If the net force acting on an object doubles, how will the object’s acceleration be affected? 4. Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with the same force. Explain which boy will have greater acceleration. References 1. Uploaded by User:Shizhao/Wikimedia Commons. . CC BY 2.5 12 www.ck12.org Chapter 4. Newton’s Third Law of Motion CHAPTER 4 Newton’s Third Law of Motion • Deﬁne force. • State the fundamental units for the Newton. • State Newton’s First Law of Motion. • Given two of the three values in F = ma, calculate the third It was imagined as a geo-stationary transfer The image at above is a NASA artist’s concept of a space elevator. station for passengers and cargo between earth and space. This idea was not pursued but it began where all great ideas begin . . . in someone’s mind. Newton’s Third Law of Motion Where do forces come from? Observations suggest that a force applied to an object is always applied by another object. A hammer strikes a nail, a car pulls a trailer, and a person pushes a grocery cart. Newton realized that forces are not so one sided. When the hammer exerts a force on the nail, the nail also exerts a force on the hammer –after all, the hammer comes to rest after the interaction. This led to Newton’s Third Law of Motion, which states that whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction. 13 www.ck12.org This law is sometimes paraphrased as “for every action, there is an equal and opposite reaction.” A very important point to remember is that the two forces are on different objects –never on the same object. It is frequently the case that one of the objects moves as a result of the force applied but the motion of the other object in the opposite direction is not apparent. Consider the situation where an ice skater is standing at the edge of the skating rink holding on to the side rail. If the skater exerts a force on the rail, the rail is held in place with tremendous friction and therefore, will not move in any noticeable way. The skater, on the other hand, had little friction with the ice, and therefore will be accelerated in the direction opposite of his/her original push. This is the process people use to jump up into the air. The person’s feet exert force on the ground and the ground exerts an equal and opposite force on the person’s feet. The force on the feet is sufﬁcient to raise the person off the ground. The force on the ground has little effect because the earth is so large. One of the accelerations is visible but the other is not visible. A case where the reaction motion due to the reaction force is visible is the case of a person throwing a heavy object out of a boat. The object is accelerated in one direction and the boat is accelerated in the opposite direction. In this case, both the motion of the object is visible and the motion of the boat in the opposite direction is also visible. Rockets also work in this manner. It is a misconception that the rocket moves forward because the escaping gas pushes on the ground or the surrounding air to make the rocket go forward. Rockets work in outer space where there is no ground or surrounding air. The rocket exerts a force on the gases causing them to be expelled and the gases exert a force on the rocket causing it to be accelerated forward. Summary • A force applied to an object is always applied by another object. • Newton’s Third Law of Motion says, "whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction." Practice The following video contains a discussion and an example of Newton’s Third Law of Motion. http://www.youtube.com/watch?v=fKJDpPi-UN0 MEDIA Click image to the left for more content. Review 1. What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balances the original force and therefore, the ball will not accelerate in any direction. 2. When a bat strikes a ball, the force exerted can send the ball deep into the outﬁeld. Where is the equal and opposite force in this case? 3. Suppose you wish to jump horizontally and in order for you to jump a distance of 4 feet horizontally, you must exert a force of 200 N. When you are standing on the ground, you have no trouble jumping 4 feet horizontally. If you are standing in a canoe, however, and you need to jump 4 feet to reach the pier, you will surely fall into the lake. Why is it that you cannot jump 4 feet out of a canoe when you can easily do this when on land? 14 www.ck12.org Chapter 4. Newton’s Third Law of Motion • Newton’s Third Law of Motion: Whenever one object exerts a force on a second object, the second object also exerts a force on the ﬁrst object, equal in magnitude and opposite in direction. References 1. Courtesy of Pat Rawling, NASA. http://commons.wikimedia.org/wiki/File:Nasa_space_elev.jpg. Public Do- main 15 CHAPTER 5 The various types of common forces are discussed and analyzed. The various types of common forces are discussed and analyzed. www.ck12.org Types of Forces Key Equations Common Forces Guidance Normal Force 8 >>>>< >>>>: Fg = mg FN FT Fsp = kDx Force of spring stretched a distance Dx from equilibrium Gravity Normal force: acts perpendicular to surfaces Force of tension in strings and wires Often, objects experience a force that pushes them into another object, but once the objects are in contact they do not any move closer together. For instance, when you stand on the surface of the earth you are obviously not accelerating toward its center. According to Newton’s Laws, there must be a force opposing the earth’s gravity acting on you, so that the net force on you is zero. The same also applies for your gravity acting on the earth. We call such a force the Normal Force. The normal force acts between any two surfaces in contact, balancing what ever force is pushing the objects together. It is actually electromagnetic in nature (like other contact forces), and arises due to the repulsion of atoms in the two objects. Here is an illustration of the Normal force on a block sitting on earth: Tension Another force that often opposes gravity is known as tension. This force is provided by wires and strings when they hold objects above the earth. Like the Normal Force, it is electromagnetic in nature and arises due to the intermolecular bonds in the wire or string: 16 www.ck12.org Chapter 5. Types of Forces If the object is in equilibrium, tension mu
st be equal in magnitude and opposite in direction to gravity. This force transfers the gravity acting on the object to whatever the wire or string is attached to; in the end it is usually a Normal Force — between the earth and whatever the wire is attached to — that ends up balancing out the force of gravity on the object. Friction Friction is a force that opposes motion. Any two objects in contact have what is called a mutual coefﬁcient of friction. To ﬁnd the force of friction between them, we multiply the normal force by this coefﬁcient. Like the forces above, it arises due to electromagnetic interactions of atoms in two objects. There are actually two coefﬁcients of friction: static and kinetic. Static friction will oppose initial motion of two objects relative to each other. Once the objects are moving, however, kinetic friction will oppose their continuing motion. Kinetic friction is lower than static friction, so it is easier to keep an object in motion than to set it in motion. fs µsj ~FNj fk = µkj ~FNj [5] Static friction opposes potential motion of surfaces in contact [6] Kinetic frictions opposes motion of surfaces in contact There are some things about friction that are not very intuitive: • The magnitude of the friction force does not depend on the surface areas in contact. • The magnitude of kinetic friction does not depend on the relative velocity or acceleration of the two objects. • Friction always points in the direction opposing motion. If the net force (not counting friction) on an object is lower than the maximum possible value of static friction, friction will be equal to the net force in magnitude and opposite in direction. Spring Force Any spring has some equilibrium length, and if stretched in either direction it will push or pull with a force equal to: ~Fsp = k ~Dx [7] Force of spring ~Dx from equilibrium Example 1 Question: A woman of mass 70.0 kg weighs herself in an elevator. 17 www.ck12.org a) If she wants to weigh less, should she weigh herself when accelerating upward or downward? b) When the elevator is not accelerating, what does the scale read (i.e., what is the normal force that the scale exerts on the woman)? c) When the elevator is accelerating upward at 2.00 m/s2, what does the scale read? Answer a) If she wants to weigh less, she has to decrease her force (her weight is the force) on the scale. We will use the equation to determine in which situation she exerts less force on the scale. F = ma If the elevator is accelerating upward then the acceleration would be greater. She would be pushed toward the ﬂoor of the elevator making her weight increase. Therefore, she should weigh herself when the elevator is going down. b) When the elevator is not accelerating, the scale would read 70:0kg. c) If the elevator was accelerating upward at a speed of 2:00m=s2, then the scale would read F = ma = 70kg (9:8m=s2 + 2m=s2) = 826N which is 82:6kg. Example 2 Question: A spring with a spring constant of k = 400N=m has an uncompressed length of :23m and a fully compressed length of :15m. What is the force required to fully compress the spring? Solution: We will use the equation F = kx to solve this. We simply have to plug in the known value for the spring and the distance to solve for the force. F = kx = (400N=m)(:23m :15m) = 32N 18 www.ck12.org Watch this Explanation Chapter 5. Types of Forces MEDIA Click image to the left for more content. Time for Practice No Problems for this section. See Newton Law Problem Solving Concept. 19 CHAPTER 6 Universal Law of Gravity www.ck12.org • Describe and give the formula for Newton’s universal law of gravity. • Using Newton’s law of gravity Cavendish’s apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. Force of Gravity In the mid-1600’s, Newton wrote that the sight of a falling apple made him think of the problem of the motion of the planets. He recognized that the apple fell straight down because the earth attracted it and thought this same force of attraction might apply to the moon and that motion of the planets might be controlled by the gravity of the sun. He eventually proposed the universal law of gravitational attraction as 20 F = G m1m2 d2 www.ck12.org Chapter 6. Universal Law of Gravity where m1 and m2 are the masses being attracted, d is the distance between the centers of the masses, G is the universal gravitational constant, and F is the force of attraction. The formula for gravitational attraction applies equally to two rocks resting near each other on the earth and to the planets and the sun. The value for the universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N·m2/kg2. The moon is being pulled toward the earth and the earth toward the moon with the same force but in the opposite direction. The force of attraction between the two bodies produces a greater acceleration of the moon than the earth because the moon has smaller mass. Even though the moon is constantly falling toward the earth, it never gets any closer. This is because the velocity of the moon is perpendicular to the radius of the earth (as shown in the image above) and therefore the moon is moving away from the earth. The distance the moon moves away from the orbit line is exactly the same distance that the moon falls in the time period. This is true of all satellites and is the reason objects remain in orbit. Example Problem: Since we know the force of gravity on a 1.00 kg ball resting on the surface of the earth is 9.80 N and we know the radius of the earth is 6380 km, we can use the equation for gravitational force to calculate the mass of the earth. Solution: me = Fd2 Gm1 = 5:98 1024 kg )(6:38106 m)2 (9:80 m/s2 = (6:671011 Nm2=kg2 )(1:00 kg) Sample Problem: John and Jane step onto the dance ﬂoor about 20. M apart at the Junior Prom and they feel an attraction to each other. If John’s mass is 70. kg and Jane’s mass is 50. kg, assume the attraction is gravity and calculate its magnitude. Solution: Fg = Gm1m2 d2 = (6:671011 Nm2=kg2 (20: m)2 )(70: kg)(50: kg) = 1:2 108 N This is such an extremely weak force, it is probably not the force of attraction John and Jane felt. Summary • Newton proposed the universal law of gravitational attraction as F = G m1m2 d2 • The universal gravitational constant, G, was determined by Cavendish to be 6.67 1011 N·m2/kg2. • Even though satellites are constantly falling toward the object they circle, they do not get closer because their . straight line motion moves them away from the center at the same rate they fall. 21 Practice The following video is a lecture on universal gravitation. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=OZZGJfFf8XI www.ck12.org MEDIA Click image to the left for more content. 1. What factors determine the strength of the force of gravity? 2. Between which two points do we measure the distance between the earth and moon? The following video is The Mass vs. Weight Song. Use this resource to answer the questions that follow. http://w ww.youtube.com/watch?v=1whMAIGNq7E MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? The following website contains a series of solved practice problems on gravity. http://physics.info/gravitation/practice.shtml Review 1. The earth is attracted to the sun by the force of gravity. Why doesn’t the earth fall into the sun? 2. If the mass of the earth remained the same but the radius of the earth shrank to one-half its present distance, what would happen to the force of gravity on an object that was resting on the surface of the earth? 3. Lifting an object on the moon requires one-sixth the force that would be required to lift the same object on the earth because gravity on the moon is one-sixth that on earth. What about horizontal acceleration? If you threw a rock with enough force to accelerate it at 1.0 m/s2 horizontally on the moon, how would the required force compare to the force necessary to acceleration the rock in the same way on the earth? 4. The mass of the earth is 5.98 1024 kg and the mass of the moon is 7.35 1022 kg. If the distance between the earth and the moon is 384,000 km, what is the gravitational force on the moon? • gravity: A natural phenomenon by which physical bodies appear to attract each other with a force proportional to their masses and inversely proportional to the distance separating them. 22 www.ck12.org References Chapter 6. Universal Law of Gravity 1. Chris Burks (Wikimedia: Chetvorno). http://commons.wikimedia.org/wiki/File:Cavendish_Torsion_Balance _Diagram.svg. Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 23 www.ck12.org CHAPTER 7 Mass versus Weight • Distinguish between mass and weight. • Given the acceleration due to gravity and either the mass or the weight of an object, calculate the other one. “Space exploration is an international endeavor.” Three Japan Aerospace Exploration Agency astronauts – Akihiko Hoshide, Satoshi Furukawa and Naoko Yamazaki – are training with the 11-member astronaut candidate class of 2004. JAXA astronauts Satoshi Furukawa, Akihiko Hoshide and Naoko Yamazaki experience near-weightlessness on the KC-135 training airplane. Mass and Weight The mass of an object is deﬁned as the amount of matter in the object. The amount of mass in an object is measured
by comparing the object to known masses on an instrument called a balance. 24 www.ck12.org Chapter 7. Mass versus Weight Using the balance shown above, the object would be placed in one pan and known masses would be placed in the other pan until the pans were exactly balanced. When balanced, the mass of the object would be equal to the sum of the known masses in the other pan. The unit of measurement for mass is the kilogram. The mass of an object would be the same regardless of whether the object was on the earth or on the moon. The balance and known masses work exactly the same both places and would indicate the same mass for the same object as long as some gravitational force is present. The weight of an object is deﬁned as the force pulling the object downward. On the earth, this would be the gravitational force of the earth on the object. On the moon, this would be the gravitational force of the moon on the object. Weight is measured by a calibrated spring scale as shown here. Weight is measured in force units which is Newtons in the SI system. The weights measured for an object would not be the same on the earth and moon because the gravitational ﬁeld on the surface of the moon is one-sixth of the magnitude of the gravitational ﬁeld on the surface of the earth. The force of gravity is given by Newton’s Second Law, F = ma, when F is the force of gravity in Newtons, m is the mass of the object in kilograms, and a is the acceleration due to gravity, 9.80 m/s2. When the formula is used speciﬁcally for ﬁnding weight from mass or vice versa, it may appear as W = mg. Example Problem: What is the weight of an object sitting on the earth’s surface if the mass of the object is 43.7 kg? Solution: W = mg = (43:7 kg)(9:80 m/s2) = 428 N Example Problem: What is the mass of an object whose weight sitting on the earth is 2570 N? m = W a = 2570 N 9:80 m/s2 = 262 kg Summary • The mass of an object is measured in kilograms and is deﬁned as the amount of matter in an object. • The mass of an object is determined by comparing the mass to known masses on a balance. • The weight of an object on the earth is deﬁned as the force acting on the object by the earth’s gravity. • Weight is measured by a calibrated spring scale. • The formula relating mass and weight is W = mg. 25 Practice A song about the difference between mass and weight sung by Mr. Edmunds to the tune of Sweet Caroline. Remember to make allowances for the fact that he is a teacher, not a professional singer. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=1whMAIGNq7E www.ck12.org MEDIA Click image to the left for more content. 1. What is used to measure mass? 2. What is used to measure weight? 3. What units are used to measure mass? 4. What units are used to measure weight? This video shows what appears to be a magic trick but is actually a center of gravity demonstration. http://www.darktube.org/watch/simple-trick-magic-no-physics Review 1. The mass of an object on the earth is 100. kg. (a) What is the weight of the object on the earth? (b) What is the mass of the object on the moon? (c) Assuming the acceleration due to gravity on the moon is EXACTLY one-sixth of the acceleration due to gravity on earth, what is the weight of the object on the moon? 2. A man standing on the Earth can exert the same force with his legs as when he is standing on the moon. We know that the mass of the man is the same on the Earth and the Moon. We also know that F = ma is true on both the Earth and the Moon. Will the man be able to jump higher on the Moon than the Earth? Why or why not? • mass: The mass of an object is measured in kilograms and is deﬁned as the amount of matter in an object. • weight: The weight of an object on the earth is deﬁned as the force acting on the object by the earth’s gravity. References 1. Courtesy of NASA. http://spaceﬂight.nasa.gov/gallery/images/behindthescenes/training/html/jsc2004e45082.h tml. Public Domain 2. CK-12 Foundation - Christopher Auyeung. . CC BY-NC-SA 3.0 3. CK-12 Foundation - Christopher Auyeung. . CC-BY-NC-SA 3.0 26 www.ck12.org CHAPTER 8 • Deﬁne both static and sliding friction. • Explain what causes surface friction. • Deﬁne the coefﬁcient of friction. • Calculate frictional forces. • Calculate net forces when friction is involved. Chapter 8. Friction Friction Dealing with friction and a lack of friction becomes a very important part of the game in tennis played on a clay court. It’s necessary for this player to learn how to keep her shoes from sliding when she wants to run but also necessary to know how her shoes will slide when coming to a stop. Friction Most of the time, in beginning physics classes, friction is ignored. Concepts can be understood and calculations made assuming friction to be non-existent. Whenever physics intersects with the real world, however, friction must be taken into account. Friction exists between two touching surfaces because even the smoothest looking surface is quite rough on a microscopic scale. 27 www.ck12.org With the bumps, lumps, and imperfections emphasized as in the image above, it becomes more apparent that if we try to slide the top block over the lower block, there will be collisions involved when bumps impact on bumps. The forward motion causes the collisions with bumps which then exert a force in opposite way the block is moving. The force of friction always opposes whatever motion is causing the friction. The force of friction between these two blocks is related to two factors. The ﬁrst factor is the roughness of the surfaces that are interacting which is called the coefﬁcient of friction, µ (Greek letter mu). The second factor is the magnitude of the force pushing the top block down onto the lower block. It is reasonable that the more forcefully the blocks are pushed together, the more difﬁcult it will be for one to slide over the other. A very long time ago, when physics was young, the word “normal” was used in the same way that we use the word “perpendicular” today. The force pushing these blocks together is the perpendicular force pushing the top block down on the lower block and this force is called the normal force. Much of the time, this normal force is simply the weight of the top block but on some occasions, the weight of the top block has some added or reduced force so the normal force is not always the weight. The force of friction then, can be calculated by Ffriction = µ Fnormal This is an approximate but reasonably useful and accurate relationship. It is not exact because µ depends on whether the surfaces are wet or dry and so forth. The frictional force we have been discussing is referred to as sliding friction because it is involved when one surface is sliding over another. If you have ever tried to slide a heavy object across a rough surface, you may be aware that it is a great deal easier to keep an object sliding than it is to start the object sliding in the ﬁrst place. When the object to slide is resting on a surface with no movement, the force of friction is called static friction and it is somewhat greater than sliding friction. Surfaces that are to move against one another will have both a coefﬁcient of static friction and a coefﬁcient of sliding friction and the two values will NOT be the same. For example, the coefﬁcient of sliding friction for ice on ice is 0.03 whereas the coefﬁcient of static friction for ice on ice is 0.10 –more than three times as great. Example Problem: A box weighing 2000. N is sliding across a cement ﬂoor. The force pushing the box is 500. N and the coefﬁcient of sliding friction between the box and the ﬂoor is 0.20. What is the acceleration of the box. Solution: In this case, the normal force for the box is its weight. Using the normal force and the coefﬁcient of friction, we can ﬁnd the frictional force. We can also ﬁnd the mass of the box from its weight since we know the acceleration due to gravity. Then we can ﬁnd the net force and the acceleration. FF = µFN = (0:20)(2000: N) = 400: N weight g = 2000: N mass of box = 9:8 m=s2 = 204 g FNET = Pushing force frictional force = 500: N 400: N = 100: N a = FN = 0:49 m/s2 m = 100: N 204 kg Example Problem: Two boxes are connected by a rope running over a pulley (see image). The coefﬁcient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and velocity. Find the acceleration of the system. 28 www.ck12.org Chapter 8. Friction Solution: The force tending to move the system is the weight of box B and the force resisting the movement is the force of friction between the table and box A. The mass of the system would be the sum of the masses of both boxes. The acceleration of the system will be found by dividing the net force by the total mass. FN(box A) = mg = (5:0 kg)(9:8 m/s2) = 49 N Ffriction = µFN = (0:20)(49 N) = 9:8 N Weight of box B = mg = (2:0 kg)(9:8 m/s2) = 19:6 N FNET = 19:6 N 9:8 N = 9:8 N a = FNET = 1:4 m/s2 mass = 9:8 N 7:0 kg Summary • Friction is caused by bodies sliding over rough surfaces. • The degree of surface roughness is indicated by the coefﬁcient of friction, µ. • The force of friction is calculated by multiplying the coefﬁcient of friction by the normal force. • The frictional force always opposes motion. • Acceleration is caused by the net force which is found by subtracting the frictional force from the applied force. Practice A video explaining friction. http://www.youtube.com/watch?v=CkTCp7SZdYQ MEDIA Click image to the left for more content. Review 1. A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefﬁcient of sliding friction between the sidewalk and the metal runners of the sled? 2. If the coefﬁcient of sliding friction between a 25 kg crate
and the ﬂoor is 0.45, how much force is required to move the crate at a constant velocity across the ﬂoor? 29 www.ck12.org 3. A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coefﬁcient of sliding friction between the block and the table top? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity? • friction: A force that resists the relative motion or tendency to such motion of two bodies or substances in contact. • coefﬁcient of friction: The ratio of the force that maintains contact between an object and a surface (i.e. the normal force) and the frictional force that resists the motion of the object. • normal force: The perpendicular force one surface exerts on another surface when the surfaces are in contact. References 1. Courtesy of Eric Harris, U.S. Air Force. http://commons.wikimedia.org/wiki/File:Maria_Sharapova,_2008 _Family_Circle_Cup.JPG. Public Domain 2. CK-12 Foundation - Joy Sheng. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 30 www.ck12.org Chapter 9. Free Body Diagrams CHAPTER 9 Free Body Diagrams Students will learn how to draw a free-body diagram and apply it to the real world. Students will learn how to draw a free-body diagram and apply it to the real world. Guidance For every problem involving forces it is essential to draw a free body diagram (FBD) before proceeding to the problem solving stage. The FBD allows one to visualize the situation and also to make sure all the forces are accounted. In addition, a very solid understanding of the physics is gleaned and many questions can be answered solely from the FBD. Example 1 Watch this Explanation MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Time for Practice 1. Draw free body diagrams (FBDs) for all of the following objects involved (in bold) and label all the forces appropriately. Make sure the lengths of the vectors in your FBDs are proportional to the strength of the force: smaller forces get shorter arrows! a. A man stands in an elevator that is accelerating upward at 2 m=s2. a. A boy is dragging a sled at a constant speed. The boy is pulling the sled with a rope at a 30 angle. a. The picture shown here is attached to the ceiling by three wires. 31 www.ck12.org a. A bowling ball rolls down a lane at a constant velocity. a. A car accelerates down the road. There is friction f between the tires and the road. 2. For the following situation, identify the 3rd law force pairs on the associated free body diagrams. Label each member of one pair “A;00 each member of the next pair “B;00 and so on. The spring is stretched so that it is pulling the block of wood to the right. Draw free body diagrams for the situation below. Notice that we are pulling the bottom block out from beneath the top block. There is friction between the blocks! After you have drawn your FBDs, identify the 3rd law force pairs, as above. Answers Discuss in class 32 www.ck12.org Chapter 10. Problem Solving 1 CHAPTER 10 Problem Solving 1 In this lesson, students will learn how to solve difﬁcult problems using Newton’s 2nd law. In this lesson, students will learn how to solve difﬁcult problems using Newton’s 2nd law. Key Equations Common Forces 8 >>>>< >>>>: Fg = mg Gravity FN Normal force: acts perpendicular to surfaces FT Force of tension in strings and wires Fsp = kDx = Force of springDxfrom equilibrium Force Sums 8 >< >: Net force is the vector sum of all the forces Fnet = i Fi = ma Fnet,x = i Fix = max Horizontal components add also Fnet,y = i Fiy = may As do vertical ones Static and Kinetic Friction’ ( fs µsjFNj Opposes potential motion of surfaces in contact fk = µkjFNj Opposes motion of surfaces in contact Ultimately, many of these “contact” forces are due to attractive and repulsive electromagnetic forces between atoms in materials. Guidance Problem Solving for Newton’s Laws, Step-By-Step 1. Figure out which object is “of interest.” a. If you’re looking for the motion of a rolling cart, the cart is the object of interest. b. If the object of interest is not moving, that’s OK, don’t panic yet. c. Draw a sketch! This may help you sort out which object is which in your problem. 2. Identify all the forces acting on the object and draw them on object. (This is a free-body diagram –FBD) a. If the object has mass and is near the Earth, the easiest (and therefore, ﬁrst) force to write down is the force of gravity, pointing downward, with value mg. b. If the object is in contact with a ﬂat surface, it means there is a normal force acting on the object. This normal force points away from and is perpendicular to the surface. c. There may be more than one normal force acting on an object. For instance, if you have a bologna sandwich, remember that the slice of bologna feels normal forces from both the slices of bread! d. If a rope, wire, or cord is pulling on the object in question, you’ve found yourself a tension force. The direction of this force is in the same direction that the rope is pulling. e. Don’t worry about any forces acting on other objects. For instance, if you have a bologna sandwich as your object of interest, and you’re thinking about the forces acting on the slice of bologna, don’t worry about the force of gravity acting on either piece of bread. 33 www.ck12.org f. Remember that Newton’s 3rd Law, calling for “equal and opposite forces,” does not apply to a single object. None of your forces should be “equal and opposite” on the same object in the sense of Newton’s 3rd Law. Third law pairs act on two different objects. g. Recall that scales (like a bathroom scale you weigh yourself on) read out the normal force acting on you, not your weight. If you are at rest on the scale, the normal force equals your weight. If you are accelerating up or down, the normal force had better be higher or lower than your weight, or you won’t have an unbalanced force to accelerate you. h. Never include “ma” as a force acting on an object. “ma” is the result of the net force Fnet which is found by summing all the forces acting on your object of interest. 3. Determine how to orient your axes a. A good rule to generally follow is that you want one axis (usually the x-axis) to be parallel to the surface your object of interest is sitting on. b. If your object is on a ramp, tilt your axes so that the x-axis is parallel to the incline and the y-axis is perpendicular. In this case, this will force you to break the force of gravity on the object into its components. But by tilting your axes, you will generally have to break up fewer vectors, making the whole problem simpler. 4. Identify which forces are in the x direction, which are in the y direction, and which are at an angle. a. If a force is upward, make it in the ydirection and give it a positive sign. If it is downward, make it in the ydirection and give it a negative sign. b. Same thing applies for right vs. left in the xdirection. Make rightward forces positive. c. If forces are at an angle, draw them at an angle. A great example is that when a dog on a leash runs ahead, pulling you along, it’s pulling both forward and down on your hand. d. Draw the free body diagram (FBD). e. Remember that the FBD is supposed to be helping you with your problem. For instance, if you forget a force, it’ll be really obvious on your FBD. 5. Break the forces that are at angles into their x and y components a. Use right triangle trigonometry b. Remember that these components aren’t new forces, but are just what makes up the forces you’ve already identiﬁed. c. Consider making a second FBD to do this component work, so that your ﬁrst FBD doesn’t get too messy. 6. Add up all the x forces and x components. a. Remember that all the rightward forces add with a plus (+) sign, and that all the leftward forces add with a minus () sign. b. Don’t forget about the xcomponents of any forces that are at an angle! c. When you’ve added them all up, call this "the sum of all x forces" or "the net force in the xdirection." 7. Add up all the y forces and y components. a. Remember that all the upward forces add with a (+) sign, all the downward forces add with a () sign. b. Don’t forget about the ycomponents of any forces that are at an angle! c. When you’ve added them all up, call this "the sum of all y forces" or "net force in the ydirection." 8. Use Newton’s Laws twice. a. The sum of all xforces, divided by the mass, is the object’s acceleration in the xdirection. b. The sum of all yforces, divided by the mass, is the object’s acceleration in the ydirection. c. If you happen to know that the acceleration in the xdirection or ydirection is zero (say the object is just sitting on a table), then you can plug this in to Newton’s 2nd Law directly. d. If you happen to know the acceleration, you can plug this in directly too. 9. Each body should have a FBD. a. Draw a separate FBD for each body. 34 www.ck12.org Chapter 10. Problem Solving 1 b. Set up a sum of forces equation based on the FBD for each body. c. Newton’s Third Law will tell you which forces on different bodies are the same in magnitude. d. Your equations should equal your unknown variables at this point. Example 1 Question: Using the diagram below, ﬁnd the net force on the block. The block weighs 3kg and the inclined plane has a coefﬁcient of friction of :6. Answer: The ﬁrst step to solving a Newton’s Laws problem is to identify the object in question. In our case, the block on the slope is the object of interest. Next, we need to draw a free-body diagram. To do this, we need to identify all of the forces acting on the block and their direction. The forces are friction, which acts in the negative x direction, the normal force, which acts in the positive y direction, and gravity, which acts in a combination of the negative y direction and the
positive x direction. Notice that we have rotated the picture so that the majority of the forces acting on the block are along the y or x axis. This does not change the answer to the problem because the direction of the forces is still the same relative to each other. When we have determined our answer, we can simply rotate it back to the original position. Now we need to break down gravity (the only force not along one of the axises) into its component vectors (which do follow the axises). The x component of gravity : 9:8m=s2 cos60 = 4:9m=s2 The y component of gravity : 9:8m=s2 sin60 = 8:5m=s2 Yet these are only the acceleration of gravity so we need to multiply them by the weight of the block to get the force. F = ma = 3kg 4:9m=s2 = 14:7NF = ma = 3kg 8:5m=s2 = 25:5N 35 Now that we have solved for the force of the y-component of gravity we know the normal force (they are equal). Therefore the normal force is 25:5N. Now that we have the normal force and the coefﬁcient of static friction, we can ﬁnd the force of friction. www.ck12.org Fs = µsFN = :6 25:5N = 15:3N The force of static friction is greater than the component of gravity that is forcing the block down the inclined plane. Therefore the force of friction will match the force of the x-component of gravity. So the net force on the block is net force in the x direction : xcomponent o f gravity z { }\| 14:7N f orce o f f riction z { }\| 14:7N net force in the y direction : 25:5N } {z \| Normal Force 25:5N {z } \| ycomponent o f gravity = 0N = 0N Therefore the net force on the block is 0N. Example 2 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Watch this Explanation 36 www.ck12.org Simulation Chapter 10. Problem Solving 1 • http://simulations.ck12.org/FreeBody/ Time for Practice 1. Find the mass of the painting. The tension in the leftmost rope is 7:2 N, in the middle rope it is 16 N, and in the rightmost rope it is 16 N. 2. Find Brittany’s acceleration down the frictionless waterslide in terms of her mass m, the angle q of the incline, and the acceleration of gravity g. 37 www.ck12.org 3. The physics professor holds an eraser up against a wall by pushing it directly against the wall with a completely horizontal force of 20 N. The eraser has a mass of 0:5 kg. The wall has coefﬁcients of friction µS = 0:8 and µK = 0:6: a. Draw a free body diagram for the eraser. b. What is the normal force FN acting on the eraser? c. What is the frictional force FS equal to? d. What is the maximum mass m the eraser could have and still not fall down? e. What would happen if the wall and eraser were both frictionless? 4. A tractor of mass 580 kg accelerates up a 10 incline from rest to a speed of 10 m=s in 4 s. For all of answers below, provide a magnitude and a direction. 38 www.ck12.org Chapter 10. Problem Solving 1 a. What net force Fnet has been applied to the tractor? b. What is the normal force, FN on the tractor? c. What is the force of gravity Fg on the tractor? d. What force has been applied to the tractor so that it moves uphill? e. What is the source of this force? 5. A heavy box (mass 25 kg) is dragged along the ﬂoor by a kid at a 30 angle to the horizontal with a force of 80 N (which is the maximum force the kid can apply). a. Draw the free body diagram. b. What is the normal force FN? c. Does the normal force decrease or increase as the angle of pull increases? Explain. d. Assuming no friction, what is the acceleration of the box? e. Assuming it begins at rest, what is its speed after ten seconds? f. Is it possible for the kid to lift the box by pulling straight up on the rope? g. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 30 angle? h. In the absence of friction, what is the net force in the xdirection if the kid pulls at a 45 angle? i. In the absence of friction, what is the net force in the xdirection if the kid pulls at a60 angle? j. The kid pulls the box at constant velocity at an angle of 30. What is the coefﬁcient of kinetic friction µK between the box and the ﬂoor? k. The kid pulls the box at an angle of 30, producing an acceleration of 2 m=s2. What is the coefﬁcient of kinetic friction µK between the box and the ﬂoor? 6. Spinal implant problem —this is a real life bio-med engineering problem! 39 www.ck12.org Here’s the situation: both springs are compressed by an amount xo. The rod of length L is ﬁxed to both the top plate and the bottom plate. The two springs, each with spring constant k, are wrapped around the rod on both sides of the middle plate, but are free to move because they are not attached to the rod or the plates. The middle plate has negligible mass, and is constrained in its motion by the compression forces of the top and bottom springs. The medical implementation of this device is to screw the top plate to one vertebrae and the middle plate to the vertebrae directly below. The bottom plate is suspended in space. Instead of fusing broken vertebrates together, this implant allows movement somewhat analogous to the natural movement of functioning vertebrae. Below you will do the exact calculations that an engineer did to get this device patented and available for use at hospitals. a. Find the force, F, on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region both springs are providing opposite compression forces.) b. Find the force, F, on the middle plate for the region of its movement 4x xo. Give your answer in terms of the constants given. ( Hint: In this region, only one spring is in contact with the middle plate.) c. Graph F vs. x. Label the values for force for the transition region in terms of the constants given. 7. You design a mechanism for lifting boxes up an inclined plane by using a vertically hanging mass to pull them, as shown in the ﬁgure below. The pulley at the top of the incline is massless and frictionless. The larger mass, M, is accelerating downward with a measured acceleration a. The smaller masses are mA and mB ; the angle of the incline is q, and the coefﬁcient of kinetic friction between each of the masses and the incline has been measured and determined to be µK. a. Draw free body diagrams for each of the three masses. b. Calculate the magnitude of the frictional force on each of the smaller masses in terms of the given quantities. c. Calculate the net force on the hanging mass in terms of the given quantities. d. Calculate the magnitudes of the two tension forces TA and TB in terms of the given quantities. e. Design and state a strategy for solving for how long it will take the larger mass to hit the ground, assuming at this moment it is at a height h above the ground. Do not attempt to solve this: simply state the strategy for solving it. 40 www.ck12.org Chapter 10. Problem Solving 1 8. You build a device for lifting objects, as shown below. A rope is attached to the ceiling and two masses are allowed to hang from it. The end of the rope passes around a pulley (right) where you can pull it downward to lift the two objects upward. The angles of the ropes, measured with respect to the vertical, are shown. Assume the bodies are at rest initially. a. Suppose you are able to measure the masses m1 and m2 of the two hanging objects as well as the tension TC. Do you then have enough information to determine the other two tensions, TA and TB? Explain your reasoning. b. If you only knew the tensions TA and TC, would you have enough information to determine the masses m1 and m2? If so, write m1 and m2 in terms of TA and TC. If not, what further information would you require? 9. A stunt driver is approaching a cliff at very high speed. Sensors in his car have measured the acceleration and velocity of the car, as well as all forces acting on it, for various times. The driver’s motion can be broken down into the following steps: Step 1: The driver, beginning at rest, accelerates his car on a horizontal road for ten seconds. Sensors show that there is a force in the direction of motion of 6000 N, but additional forces acting in the opposite direction with magnitude 1000 N. The mass of the car is 1250 kg. Step 2: Approaching the cliff, the driver takes his foot off of the gas pedal (There is no further force in the direction of motion.) and brakes, increasing the force opposing motion from 1000 N to 2500 N. This continues for ﬁve seconds until he reaches the cliff. Step 3: The driver ﬂies off the cliff, which is 44:1 m high and begins projectile motion. (a) Ignoring air resistance, how long is the stunt driver in the air? (b) For Step 1: i. Draw a free body diagram, naming all the forces on the car. ii. Calculate the magnitude of the net force. iii. Find the change in velocity over the stated time period. iv. Make a graph of velocity in the xdirection vs. time over the stated time period. v. Calculate the distance the driver covered in the stated time period. Do this by ﬁnding the area under the curve in your graph of (iv). Then, check your result by using the equations for kinematics. (c) Repeat (b) for Step 2. (d) Calculate the distance that the stunt driver should land from the bottom of the cliff. Answers 1. 3:6 kg 41 www.ck12.org 2. g sin q 3. b.20 N c. 4:9 N d. 1:63 kg e. Eraser would slip down the wall 4. a. 1450 N b. 5600 N c. 5700 N d. Friction between the tires and the ground e. Fuel, engine, or equal and opposite reaction 5. b. 210 N c. no, the box is ﬂat so the normal force doesn’t change d. 2:8 m=s2 e.28 m=s f. no g. 69 N h. 57 N i. 40 N j. 0:33 k. 0:09 6. a. zero b. kx0 7. b. f1 = µkm1g cos q; f2 = µkm2g cos q c. Ma d. TA = (m1 + m2)(a + µ cos q) and TB = m2a + µm2 cos q e. Solve by using d = 1=2at2 and substituting h for d 8. a. Yes, because it is static and you know the angle and m1 b. Yes, TA and the angle gives you m1 and the angle and TC gives you m2; m1 = TA cos 25=g and m2 = TC cos 30=g 9. a. 3 seconds d. 90 m 42 Physics Unit 6 (Energy) Patric
k Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: February 4, 2014 iii Contents www.ck12.org 1 4 9 12 15 Contents 1 Kinetic Energy 2 Potential Energy 3 Conservation of Energy 4 Energy Problem Solving 5 Springs iv www.ck12.org Chapter 1. Kinetic Energy CHAPTER 1 Kinetic Energy • Deﬁne energy. • Deﬁne kinetic energy. • Given the mass and speed of an object, calculate its kinetic energy. • Solve problems involving kinetic energy. This military jet requires a tremendous amount of work done on it to get its speed up to takeoff speed in the short distance available on the deck of an aircraft carrier. Some of the work is done by the plane’s own jet engines but work from a catapult is also necessary for takeoff. Kinetic Energy Energy is the ability to change an object’s motion or position. Energy comes in many forms and each of those forms can be converted into any other form. A moving object has the ability to change another object’s motion or position simply by colliding with it and this form of energy is called kinetic energy. The kinetic energy of an object can be calculated by the equation KE = 1 2 mv2, where m is the mass of the object and v is its velocity. The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy. The SI unit for kinetic energy (and all forms of energy) is kg m2 s2 which is equivalent to Joules, the same unit we use for work. The kinetic energy of an object can be changed by doing work on the object. The work done on an object equals the kinetic energy gain or loss by the object. This relationship is expressed in the work-energy theorem WNET = DKE. Example Problem: A farmer heaves a 7.56 kg bale of hay with a ﬁnal velocity of 4.75 m/s. 1 (a) What is the kinetic energy of the bale? (b) The bale was originally at rest. How much work was done on the bale to give it this kinetic energy? www.ck12.org Solution: (a) KE = 1 2 mv2 = 1 2 (7:56 kg)(4:75)2 = 85:3 Joules (b) Work done = DKE = 85:3 Joules Example Problem: What is the kinetic energy of a 750. kg car moving at 50.0 km/h? Solution: 50:0 km h 1000 m km 1 h 3600 s = 13:9 m/s KE = 1 2 mv2 = 1 2 (750: kg)(13:9 m/s)2 = 72; 300 Joules Example Problem: How much work must be done on a 750. kg car to slow it from 100. km/h to 50.0 km/h? Solution: From the previous example problem, we know that the KE of this car when it is moving at 50.0 km/h If the same car is going twice as fast, its KE will be four times as great because KE is is 72,300 Joules. proportional to the square of the velocity. Therefore, when this same car is moving at 100. km/h, its KE is 289,200 Joules. Therefore, the work done to slow the car from 100. km/h to 50.0 km/h is (289; 200 Joules) (72; 300 Joules) = 217; 000 Joules. Summary • Energy is the ability to change an object’s motion or position. • There are many forms of energy. • The energy of motion is called kinetic energy. • The formula for kinetic energy is KE = 1 2 mv2. • The work done on an object equals the kinetic energy gain or loss by the object, WNET = DKE. Practice The following video discusses kinetic energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=g157qwT1918 MEDIA Click image to the left for more content. 1. Potential energy is present in objects that are ______________. 2. Kinetic energy is present in objects that are ______________. 3. What formula is given for kinetic energy? Practice problems involving kinetic energy: http://www.physicsclassroom.com/Class/energy/u5l1c.cfm 2 www.ck12.org Review Chapter 1. Kinetic Energy 1. A comet with a mass of 7.85 1011 kg is moving with a velocity of 25,000 m/s. Calculate its kinetic energy. 2. A riﬂe can shoot a 4.00 g bullet at a speed of 998 m/s. (a) Find the kinetic energy of the bullet. (b) What work is done on the bullet if it starts from rest? (c) If the work is done over a distance of 0.75 m, what is the average force on the bullet? (d) If the bullet comes to rest after penetrating 1.50 cm into a piece of metal, what is the magnitude of the force bringing it to rest? • energy: An indirectly observed quantity that is often understood as the ability of a physical system to do work. • kinetic energy: The energy an object has due to its motion. References 1. Courtesy of Mass Communication Specialist 3rd Class Torrey W. Lee, U.S. Navy. Jet Takeoff. Public Domain 3 CHAPTER 2 www.ck12.org Potential Energy • Deﬁne potential energy. • Solve problems involving gravitational potential energy. • Solve problems involving the conversion of potential energy to kinetic energy and vice versa. Shooting an arrow from a bow requires work done on the bow by the shooter’s arm to bend the bow and thus produce potential energy. The release of the bow converts the potential energy of the bent bow into the kinetic energy of the ﬂying arrow. Potential Energy When an object is held above the earth, it has the ability to make matter move because all you have to do is let go of the object and it will fall of its own accord. Since energy is deﬁned as the ability to make matter move, this object has energy. This type of energy is stored energy and is called potential energy. An object held in a stretched If the stretched rubber band is released, the object will move. A rubber band also contains this stored energy. pebble on a ﬂexed ruler has potential energy because if the ruler is released, the pebble will ﬂy. If you hold two positive charges near each other, they have potential energy because they will move if you release them. Potential If the chemical bonds are broken and allowed to form lower energy is also the energy stored in chemical bonds. 4 www.ck12.org Chapter 2. Potential Energy potential energy chemical bonds, the excess energy is released and can make matter move. This is frequently seen as increased molecular motion or heat. If a cannon ball is ﬁred straight up into the air, it begins with a high kinetic energy. As the cannon ball rises, it slows down due to the force of gravity pulling it toward the earth. As the ball rises, its gravitational potential energy is increasing and its kinetic energy is decreasing. When the cannon ball reaches the top of its arc, it kinetic energy is zero and its potential energy is maximum. Then gravity continues to pull the cannon ball toward the earth and the ball will fall toward the earth. As it falls, its speed increases and its height decreases. Therefore, its kinetic energy increases as it falls and its potential energy decreases. When the ball returns to its original height, its kinetic energy will be the same as when it started upward. When work is done on an object, the work may be converted into either kinetic or potential energy. If the work results in motion, the work was converted into kinetic energy and if the work done resulted in change in position, the work was converted into potential energy. Work could also be spent overcoming friction and that work would be converted into heat but usually we will consider frictionless systems. If we consider the potential energy of a bent stick or a stretched rubber band, the potential energy can be calculated by multiplying the force exerted by the stick or rubber band times the distance over which the force will be exerted. The formula for calculating this potential energy looks exactly like the formula for calculating work done, that is W = Fd. The only difference is that work is calculated when the object actually moves under the force and potential energy is calculated when the system is at rest before any motion actually occurs. In the case of gravitational potential energy, the force exerted by the object is its weight and the distance it could travel would be its height above the earth. Since the weight of an object is calculated by W = mg, then gravitational potential energy can be calculate by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height the object will fall. 5 Example Problem: A 3.00 kg object is lifted from the ﬂo
or and placed on a shelf that is 2.50 m above the ﬂoor. (a) What was the work done in lifting the object? (b) What is the gravitational potential energy of the object sitting on the shelf? (c) If the object falls off the shelf and free falls to the ﬂoor, what will its velocity be when it hits the ﬂoor? Solution: weight of the object = mg = (3:00 kg)(9:80 m=s2) = 29:4 N www.ck12.org (a) W = Fd = (29:4 N)(2:50 m) = 73:5 J (b) PE = mgh = (3:00 kg)(9:80 m=s2)(2:50 m) = 73:5 J (c) KE = PE so 1 2 mv2 = 73:5 J s v = (2)(73:5 J) 3:00 kg = 7:00 m=s Example Problem: A pendulum is constructed from a 7.58 kg bowling ball hanging on a 3.00 m long rope. The ball is pulled back until the rope makes an angle of 45 with the vertical. (a) What is the potential energy of the ball? (b) If the ball is released, how fast will it be traveling at the bottom of its arc? Solution: You can use trigonometry to ﬁnd the vertical height of the ball in the pulled back position. This vertical height is found to be 0.877 m. PE = mgh = (7:58 kg)(9:80 m=s2)(0:877 m) = 65:1 J When the ball is released, the PE will be converted into KE as the ball swings through the arc. KE = 1 s 2 mv2 = 65:1 J v = (2)(65:1 kg m2=s2) 7:58 kg = 4:14 m=s Summary • Stored energy is called potential energy. 6 www.ck12.org Chapter 2. Potential Energy • Energy may be stored by holding an object elevated in a gravitational ﬁeld or by holding it while a force is attempting to move it. • Potential energy may be converted to kinetic energy. • The formula for gravitational potential energy is PE = mgh. • In the absence of friction or bending, work done on an object must become either potential energy or kinetic energy or both. Practice The following video discusses types of energy. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=nC6tT1wkXEc MEDIA Click image to the left for more content. 1. What is the deﬁnition of energy? 2. Name two types of potential energy. 3. How is energy transferred from one object to another? Potential and kinetic energy practice problems with solutions: http://www.physicsclassroom.com/Class/energy/U5L2bc.cfm Review 1. A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top? 2. A 50.0 kg shell was ﬁred from a cannon at earth’s surface to a maximum height of 400. m. What is the potential energy at maximum height? 3. If the shell in problem #3 then fell to a height of 100. m, what was the loss of PE? 4. A person weighing 645 N climbs up a ladder to a height of 4.55 m. (a) What work does the person do? (b) What is the increase in gravitational potential energy? (c) Where does the energy come from to cause this increase in PE? • potential energy: Otherwise known as stored energy, is the ability of a system to do work due to its position or internal structure. For example, gravitational potential energy is a stored energy determined by an object’s position in a gravitational ﬁeld while elastic potential energy is the energy stored in a spring. References 1. Image copyright Skynavin, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 7 2. Image copyright Tribalium, 2013; modiﬁed by CK-12 Foundation - Samantha Bacic. http://www.shutterstock. com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 www.ck12.org 8 www.ck12.org Chapter 3. Conservation of Energy CHAPTER 3 Conservation of Energy • State the law of conservation of energy. • Describe a closed system. • Use the law of conservation of energy to solve problems. There are many energy conversions between potential and kinetic energy as the cars travel around this double looping roller coaster. Conservation of Energy The law of conservation of energy states that within a closed system, energy can change form, but the total amount of energy is constant. Another way of expressing the law of conservation of energy is to say that energy can neither be created nor destroyed. An important part of using the conservation of energy is selecting the system. In the In a closed system, conservation of energy just as in the conservation of momentum, the system must be closed. objects may not enter or leave, and it is isolated from external forces so that no work can be done on the system. In the analysis of the behavior of an object, you must make sure you have included everything in the system that is involved in the motion. For example, if you are considering a ball that is acted on by gravity, you must include the earth in your system. The kinetic energy of the ball considered by itself may increase and only when the earth is included in the system can you see that the increasing kinetic energy is balanced by an equivalent loss of potential energy. The sum of the kinetic energy and the potential energy of an object is often called the mechanical energy. Consider a box with a weight of 20.0 N sitting at rest on a shelf that is 2.00 m above the earth. The box has zero kinetic energy but it has potential energy related to its weight and the distance to the earth’s surface. PE = mgh = (20:0 N)(2:00 m) = 40:0 J 9 www.ck12.org If the box slides off the shelf, the only force acting on the box is the force of gravity and so the box falls. We can calculate the speed of the box when it strikes the ground by several methods. We can calculate the speed directly 2 = 2ad. We can also ﬁnd the ﬁnal velocity by setting the kinetic energy at the bottom of the using the formula v f fall equal to the potential energy at the top, KE = PE and, 1 2 = 2gh. You may note these formulas 2 mv2 = mgh, so v f are essentially the same. q (2)(9:80 m=s2)(2:00 m) = 6:26 m=s v = Example Problem: Suppose a cannon is sitting on top of a 50.0 m high hill and a 5.00 kg cannon ball is ﬁred with a velocity of 30.0 m/s at some unknown angle. What will be the velocity of the cannon ball when it strikes the earth? Solution: Since the angle at which the cannon ball is ﬁred is unknown, we cannot use the usual equations from projectile motion. However, at the moment the cannon ball is ﬁred, it has a certain KE due to the mass of the ball and its speed and it has a certain PE due to its mass and it height above the earth. Those two quantities of energy can be calculated. When the ball returns to the earth, its PE will be zero and therefore, its KE at that point must account for the total of its original KE + PE. This gives us a method of solving the problem. ETOTAL = KE + PE = 1 ETOTAL = 2250 J + 2450 J = 4700 J (5:00 kg)(30:0 m=s)2 + (5:00 kg)(9:80 m=s2)(50:0 m) 2 mv2 + mgh = 1 2 1 2 mv f 2 = 4700 J so v f = s (2)(4700 J) 5:00 kg = 43:4 m=s Example Problem: A 2.00 g bullet moving at 705 m/s strikes a 0.250 kg block of wood at rest on a frictionless surface. The bullet sticks in the wood and the combined mass moves slowly down the table. (a) What is the KE of the bullet before the collision? (b) What is the speed of the combination after the collision? (c) How much KE was lost in the collision? Solution: (a) KEBULLET = 1 (b) mBvB + mW vW = (mB+W )(vB+W ) 2 mv2 = 1 2 (0:00200 kg)(705 m=s)2 = 497 J (0:00200 kg)(705 m=s) + (0:250 kg)(0 m=s) = (0:252 kg)(V ) (1:41 kg m=s) = (0:252 kg)(V ) V = 5:60 m=s 2 mv2 = 1 (c) KECOMBINATION = 1 KELOST = KEBEFORE KEAFTER = 497 J 4 J = 493 J 2 (0:252 kg)(5:60 m=s)2 = 3:95 J Summary • In a closed system, energy may change forms but the total amount of energy is constant. Practice The following video demonstrates Newton Ball tricks. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=JadO3RuOJGU 10 www.ck12.org Chapter 3. Conservation of Energy MEDIA Click image to the left for more content. 1. What happens when one ball is pulled up to one side and released? 2. What happens when three balls are pulled up to one side and released? 3. What happens when two balls are pulled out from each side and released? Practice problems with answers for the law of conservation of energy: http://www.physicsclassroom.com/class/energy/u5l2bc.cfm Review 1. A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00 m to the surface of the water, (a) what is the kinetic energy of the chunk of ice when its hits the water, and (b) what is its velocity? 2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill. (a) Assuming the movement is frictionless, at what vertical height will the cart come to rest? (b) Do you need to know the mass of the cart to solve this problem? 3. A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer’s feet touch the ground 9.00 m below where the rope is tied. How fast is the performer moving at the bottom of the arc? 4. A skier starts from rest at the top of a 45.0 m hill, coasts down a 30slope into a valley, and continues up to the top of a 40.0 m hill. Both hill heights are measured from the valley ﬂoor. Assume the skier puts no effort into the motion (always coasting) and there is no friction. (a) How fast will the skier be moving on the valley ﬂoor? (b) How fast will the skier be moving on the top of the 40.0 m hill? 5. A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitutde of the ﬁnal velocity when it strikes the ground? Ignore air resistance. 6. If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum height will it reach? Neglect air resistance. • conservation of energy: An empirical law of physics (meaning it cannot be derived), states that the total amount of energy within an isolated system is constant. Although energy can be transformed from one form into another, energy cannot be created or destroyed • closed system: Means it cannot exchange any of heat, work, or matter with the surroundings. • mechanical energy: The s
um of potential energy and kinetic energy. References 1. User:Zonk43/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Teststrecke_Roller_Coaster3.J PG. Public Domain 11 CHAPTER 4 Energy Problem Solving Students will learn how to analyze and solve more complicated problems involving energy conservation. Students will learn how to analyze and solve more complicated problems involving energy conservation. www.ck12.org Key Equations Einitial = E ﬁnal ; T hetotalenergydoesnotchangeinclosedsystems KE = 1 2 mv2 ; Kinetic energy PEg = mgh ; Potential energy of gravity PEsp = 1 2 kx2; Potential energy of a spring W = FxDx = Fd cos q ; Work is equal to the distance multiplied by the component of the force in the direction it is moving. Guidance The main thing to always keep prescient in your mind is that the total energy before must equal the total energy after. If some energy has transferred out of or into the system via work, you calculate that work done and include it in the energy sum equation. Generally work done by friction is listed on the ’after’ side and work put into the system, via a jet pack for example, goes on the ’before’ side. Another important point is that on turns or going over hills or in rollercoaster loops, one must include the centripetal motion equations -for example to insure that you have enough speed to make the loop. Example 1 Watch this Explanation 12 MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. www.ck12.org Time for Practice Chapter 4. Energy Problem Solving 1. A rock with mass m is dropped from a cliff of height h. What is its speed when it gets to the bottom of the cliff? a. b. p r p mg 2g h 2gh c. d. gh e. None of the above 2. In the picture above, a 9:0 kg baby on a skateboard is about to be launched horizontally. The spring constant is 300 N=m and the spring is compressed 0:4 m. For the following questions, ignore the small energy loss due to the friction in the wheels of the skateboard and the rotational energy used up to make the wheels spin. a. What is the speed of the baby after the spring has reached its uncompressed length? b. After being launched, the baby encounters a hill 7 m high. Will the baby make it to the top? If so, what is his speed at the top? If not, how high does he make it? c. Are you ﬁnally convinced that your authors have lost their minds? Look at that picture! 3. When the biker is at the top of the ramp shown above, he has a speed of 10 m=s and is at a height of 25 m. The bike and person have a total mass of 100 kg. He speeds into the contraption at the end of the ramp, which slows him to a stop. a. What is his initial total energy? (Hint: Set Ug = 0 at the very bottom of the ramp.) b. What is the length of the spring when it is maximally compressed by the biker? (Hint: The spring does not compress all the way to the ground so there is still some gravitational potential energy. It will help to draw some triangles.) 4. An elevator in an old apartment building in Switzerland has four huge springs at the bottom of the shaft to cushion its fall in case the cable breaks. The springs have an uncompressed height of about 1 meter. Estimate the spring constant necessary to stop this elevator, following these steps: a. First, guesstimate the mass of the elevator with a few passengers inside. b. Now, estimate the height of a ﬁve-story building. c. Lastly, use conservation of energy to estimate the spring constant. 13 5. You are skiing down a hill. You start at rest at a height 120 m above the bottom. The slope has a 10:0 grade. Assume the total mass of skier and equipment is 75:0 kg. www.ck12.org a. Ignore all energy losses due to friction. What is your speed at the bottom? b. If, however, you just make it to the bottom with zero speed what would be the average force of friction, including air resistance? 6. Two horriﬁc contraptions on frictionless wheels are compressing a spring (k = 400 N=m) by 0:5 m compared to its uncompressed (equilibrium) length. Each of the 500 kg vehicles is stationary and they are connected by a string. The string is cut! Find the speeds of the vehicles once they lose contact with the spring. 7. A roller coaster begins at rest 120 m above the ground, as shown. Assume no friction from the wheels and air, and that no energy is lost to heat, sound, and so on. The radius of the loop is 40 m. a. If the height at point G is 76 m, then how fast is the coaster going at point G? b. Does the coaster actually make it through the loop without falling? (Hint: You might review the material from centripetal motion lessons to answer this part.) Answers to Selected Problems 1. . 2. a. 2:3 m=s c. No, the baby will not clear the hill. 3. a. 29; 500 J b. Spring has maximum compressed length of 13 m 4. . 5. a. 48:5 m=s b. 128 N 6. 0:32 m=s each 7. a.29 m/s b. just barely, aC = 9:8 m=s2 14 www.ck12.org CHAPTER 5 Chapter 5. Springs Springs Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion. Key Equations T = 1 f ; Period is the inverse of frequency Tspring = 2p r m k ; Period of mass m on a spring with constant k Fsp = kx ; the force of a spring equals the spring constant multiplied by the amount the spring is stretched or compressed from its equilibrium point. The negative sign indicates it is a restoring force (i.e. direction of the force is opposite its displacement from equilibrium position. Usp = 1 that it is stretched or compressed from equilibrium 2 kx2 ; the potential energy of a spring is equal to one half times the spring constant times the distance squared Guidance • The oscillating object does not lose any energy in SHM. Friction is assumed to be zero. • In harmonic motion there is always a restorative force, which attempts to restore the oscillating object to its equilibrium position. The restorative force changes during an oscillation and depends on the position of the object. In a spring the force is given by Hooke’s Law: F = kx • The period, T , is the amount of time needed for the harmonic motion to repeat itself, or for the object to go one full cycle. In SHM, T is the time it takes the object to return to its exact starting point and starting direction. • The frequency, f ; is the number of cycles an object goes through in 1 second. Frequency is measured in Hertz (Hz). 1 Hz = 1 cycle per sec. • The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object. The amplitude can vary in harmonic motion, but is constant in SHM. Example 1 MEDIA Click image to the left for more content. 15 www.ck12.org Watch this Explanation Simulation MEDIA Click image to the left for more content. Mass & Springs (PhET Simulation) Time for Practice 1. A rope can be considered as a spring with a very high spring constant k; so high, in fact, that you don’t notice the rope stretch at all before it “pulls back.” a. What is the k of a rope that stretches by 1 mm when a 100 kg weight hangs from it? b. If a boy of 50 kg hangs from the rope, how far will it stretch? c. If the boy kicks himself up a bit, and then is bouncing up and down ever so slightly, what is his frequency of oscillation? Would he notice this oscillation? If so, how? If not, why not? 2. If a 5:0 kg mass attached to a spring oscillates 4.0 times every second, what is the spring constant k of the spring? 3. A horizontal spring attached to the wall is attached to a block of wood on the other end. All this is sitting on a frictionless surface. The spring is compressed 0:3 m. Due to the compression there is 5:0 J of energy stored in the spring. The spring is then released. The block of wood experiences a maximum speed of 25 m=s. a. Find the value of the spring constant. b. Find the mass of the block of wood. c. What is the equation that describes the position of the mass? d. What is the equation that describes the speed of the mass? e. Draw three complete cycles of the block’s oscillatory motion on an x vs. t graph. 16 www.ck12.org Chapter 5. Springs 4. A spider of 0:5 g walks to the middle of her web. The web sinks by 1:0 mm due to her weight. You may assume the mass of the web is negligible. a. If a small burst of wind sets her in motion, with what frequency will she oscillate? b. How many times will she go up and down in one s? In 20 s? c. How long is each cycle? d. Draw the x vs t graph of three cycles, assuming the spider is at its highest point in the cycle at t = 0 s. Answers to Selected Problems 1. a. 9:8 105 N=m b. 0:5 mm c. 22 Hz 2. 3:2 103 N=m 3. a. 110 N=m d. v(t) = (25) cos(83t) 4. a. 16 Hz b. 16 complete cycles but 32 times up and down, 315 complete cycles but 630 times up and down c. 0:063 s Investigation 1. Your task: Match the period of the circular motion system with that of the spring system. You are only allowed to change the velocity involved in the circular motion system. Consider the effective distance between the block and the pivot to be to be ﬁxed at 1m. The spring constant(13.5N/m) is also ﬁxed. You should view the charts to check whether you have succeeded. Instructions: To alter the velocity, simply click on the Select Tool, and select the pivot . The Position tab below will allow you to numerically adjust the rotational speed using the Motor ﬁeld. To view the graphs of their respective motion in order to determine if they are in sync, click on Chart tab below. 2. MEDIA Click image to the left for more content. 3. Now the mass on the spring has been replaced by a mass that is twice the rotating mass. Also, the distance between the rotating mass and the pivot has been changed to 1.5 m. What velocity will keep the period the same now? 4. MEDIA Click image to the left for more content. 17 Physics Unit 7 (Work, Power, Kinetic Energy
Theorem) Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Work 2 Power 3 Work-Energy Principle 1 4 7 iv www.ck12.org CONCEPT 1 Concept 1. Work Work • Deﬁne work. • Identify forces that are doing work. • Given two of the three variables in the equation, W = Fd , calculate the third. In order for the roller coaster to run down the incline by gravitational attraction, it ﬁrst must have work done on it towing it up to the top of the hill. The work done on the coaster towing it to the top of the hill becomes potential energy stored in the coaster and that potential energy is converted to kinetic energy as the coaster runs down from the top of the hill to the bottom. Work The word work has both an everyday meaning and a speciﬁc scientiﬁc meaning. In the everyday use of the word, work would refer to anything which required a person to make an effort. In physics, however, work is deﬁned as the force exerted on an object multiplied by the distance the object moves due to that force. W = Fd In the scientiﬁc deﬁnition of the word, if you push against an automobile with a force of 200 N for 3 minutes but the automobile does not move, then you have done NO work. Multiplying 200 N times 0 meters yields zero work. If you are holding an object in your arms, the upward force you are exerting is equal to the object’s weight. If 1 www.ck12.org you hold the object until your arms become very tired, you have still done no work because you did not move the object in the direction of the force. When you lift an object, you exert a force equal to the object’s weight and the If an object weighs 200. N and you lift it 1.50 meters, then your work is, object moves due to that lifting force. W = Fd = (200: N)(1:50 m) = 300: N m . One of the units you will see for work is the Newton·meter but since a Newton is also a kilogram·m/s 2 , then a Newton·meter is also kg·m 2 /s 2 . This unit has also been named the Joule (pronounced Jool) in honor of James Prescott Joule, a nineteenth century English physicist. Example Problem: A boy lifts a box of apples that weighs 185 N. The box is lifted a height of 0.800 m. How much work did the boy do? Solution: W = Fd = (185 N)(0:800 m) = 148 N m = 148 Joules Work is done only if a force is exerted in the direction of motion. If the force and motion are perpendicular to each other, no work is done because there is no motion in the direction of the force. If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Example Problem: Suppose a 125 N force is applied to a lawnmower handle at an angle of 25° with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done? Solution: The solution requires that we determine the component of the force that was in the direction of the motion of the lawnmower. The component of the force that was pushing down on the ground does not contribute to the work done. Fparallel = (Force)(cos 25) = (125 N)(0:906) = 113 N W = Fparalleld = (113 N)(56 m) = 630 J Summary • In physics, work is deﬁned as the force exerted on an object multiplied by the distance the object moves due to that force. • The unit for work is called the joule. • If the force is at an angle to the motion, then the component of the force in the direction of the motion is used to determine the work done. Practice The following video introduces energy and work. Use this resource to answer the questions that follow. http://wn.com/Work_physics_#/videos 2 www.ck12.org Concept 1. Work MEDIA Click image to the left for more content. 1. What deﬁnition is given in the video for energy? 2. What is the deﬁnition given in the video for work? 3. What unit is used in the video for work? The following website contains practice questions with answers on the topic of work. http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section6.rhtml Review 1. How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m? 2. A workman carries some lumber up a staircase. The workman moves 9.6 m vertically and 22 m horizontally. If the lumber weighs 45 N, how much work was done by the workman? 3. A barge is pulled down a canal by a horse walking beside the canal. exerted is 400. N, and the barge is pulled 100. M, how much work did the horse do? If the angle of the rope is 60.0°, the force • work: A force is said to do work when it acts on a body so that there is a displacement of the point of application, however small, in the direction of the force. Thus a force does work when it results in movement. The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product, W = Fd . • joule: The SI unit of work or energy, equal to the work done by a force of one Newton when its point of application moves through a distance of one meter in the direction of the force. References 1. Image copyright Paul Brennan, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic. . CC BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3 CONCEPT 2 • Deﬁne power. • Given two of the three variables in P = W t , calculate the third. www.ck12.org Power Typical Pressurized Water Reactors (PWR) reactors built in the 1970’s produce about 1100 megawatts, whilst the latest designs range up to around 1500 megawatts. That is 1,500,000,000 Joules/second. A windmill farm, by comparison, using hundreds of individual windmills produces about 5 megawatts. That is 5,000,000 Joules/second (assuming the wind is blowing). Power Power is deﬁned as the rate at which work is done or the rate at which energy is transformed. In SI units, power is measured in Joules per second which is given a special name, the watt , W . Power = Work Time 1.00 watt = 1.00 J/s 4 www.ck12.org Concept 2. Power Another unit for power that is fairly common is horsepower. 1.00 horsepower = 746 watts Example Problem: A 70.0 kg man runs up a long ﬂight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. Calculate the power output of the man in watts and horsepower. Solution: The force exerted must be equal to the weight of the man = mg = (70:0 kg)(9:80 m/s2) = 686 N W = Fd = (686 N)(4:5 m) = 3090 N m = 3090 J 4:0 s = 770 J/s = 770 W P = W t = 3090 J P = 770 W = 1:03 hp Since P = W that is produced by the power. t = F d P = W t = Fd t = Fv t and W = Fd , we can use these formulas to derive a formula relating power to the speed of the object The velocity in this formula is the average speed of the object during the time interval. Example Problem: Calculate the power required of a 1400 kg car if the car climbs a 10° hill at a steady 80. km/h. Solution: 80. km/h = 22.2 m/s In 1.00 s, the car would travel 22.2 m on the road surface but the distance traveled upward would be (22.2 m)(sin 10°) = (22.2 m)(0.174) = 3.86 m. The force in the direction of the upward motion is the weight of the car = (1400 kg)(9.80 m/s 2 ) = 13720 N. W = Fd = (13720 N)(3:86 m) = 53; 000 J Since this work was done in 1.00 second, the power would be 53,000 W. If we calculated the upward component of the velocity of the car, we would divide the distance traveled in one second by one second and get an average vertical speed of 3.86 m/s. So we could use the formula relating power to average speed to calculate power. P = Fv = (13720 N)(3:86 m/s) = 53; 000 W Summary • Power is deﬁned as the rate at which work is done or the rate at which energy is transformed. • Power = Work Time • Power = Force velocity Practice In the following video, Mr. Edmond sings about work and power. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=5EsMmdaYClQ 5 www.ck12.org MEDIA Click image to the left for more content. 1. What units are used for work? 2. What units are used for power? The following website has practice problems on work and power. http://www.angelﬁre.com/sciﬁ/dschlott/workpp.html Review 1. If the circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N, how much work does the centripetal force do on the toy when it follows its orbit for one cycle? 2. A 50.0 kg woman climb
s a ﬂight of stairs 6.00 m high in 15.0 s. How much power does she use? 3. (a) Assuming no friction, what is the minimum work needed to push a 1000. kg car 45.0 m up a 12.5° incline? (b) What power would be needed for the same problem if friction is considered and the coefﬁcient of friction for the car is 0.30? • power: The rate at which this work is performed. References 1. Courtesy of Ryan Hagerty, U.S. Fish and Wildlife Service. http://digitalmedia.fws.gov/cdm/singleitem/coll ection/natdiglib/id/13455/rec/2 . Public Domain 6 www.ck12.org Concept 3. Work-Energy Principle CONCEPT 3 Work-Energy Principle The reason the concept of work is so useful is because of a theorem, called the work-energy principle , which states that the change in an object’s kinetic energy is equal to the net work done on it : DKe = Wnet [2] Although we cannot derive this principle in general, we can do it for the case that interests us most: constant acceleration. In the following derivation, we assume that the force is along motion. This doesn’t reduce the generality of the result, but makes the derivation more tractable because we don’t need to worry about vectors or angles. Recall that an object’s kinetic energy is given by the formula: Ke = 1 2 mv2 [3] Consider an object of mass m accelerated from a velocity vi to v f under a constant force. The change in kinetic energy, according to [2], is equal to: DKe = Kei Ke f = 1 2 mv2 f 1 2 mv2 i = 1 2 m(v2 f v2 i ) [4] Now let’s see how much work this took. To ﬁnd this, we need to ﬁnd the distance such an object will travel under these conditions. We can do this by using the third of our ’Big three’ equations, namely: alternatively, v f 2 = vi 2 + 2aDx [5] Dx = v f 2 2 vi 2a [6] Plugging in [6] and Newton’s Third Law, F = ma , into [2], we ﬁnd: W = FDx = ma v f 2 2 vi 2a which was our result in [4]. Using the Work-Energy Principle = 1 2 m(v2 f v2 i ) [7], The Work-Energy Principle can be used to derive a variety of useful results. Consider, for instance, an object dropped a height Dh under the inﬂuence of gravity. This object will experience constant acceleration. Therefore, we can again use equation [6], substituting gravity for acceleration and Dh for distance: 7 www.ck12.org multiplying both sides by $mg$, we ﬁnd: Dh = v f 2 2 vi 2g mgDh = mg v f 2 2 vi 2g = DKe [8] In other words, the work performed on the object by gravity in this case is mgDh . We refer to this quantity as gravitational potential energy; here, we have derived it as a function of height. For most forces (exceptions are friction, air resistance, and other forces that convert energy into heat), potential energy can be understood as the ability to perform work. Spring Force A spring with spring constant k a distance Dx from equilibrium experiences a restorative force equal to: This is a force that can change an object’s kinetic energy, and therefore do work. So, it has a potential energy associated with it as well. This quantity is given by: Fs = kDx [9] Esp = 1 2 kDx2 [10] Spring Potential Energy The derivation of [10] is left to the reader. Hint: ﬁnd the average force an object experiences while moving from x = 0 to x = Dx while attached to a spring. The net work is then this force times the displacement. Since this quantity (work) must equal to the change in the object’s kinetic energy, it is also equal to the potential energy of the spring. This derivation is very similar to the derivation of the kinematics equations — look those up. This applet may be useful in reviewing Spring Potential Energy: http://phet.colorado.edu/en/simulation/mass-spring-lab 8 Physics Unit 8 Momentum Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. James H Dann, Ph.D. Ck12 Science CONTRIBUTORS Chris Addiego Alexander Katsis Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Momentum 2 Momentum 3 Law of Conservation of Momentum 4 Conservation of Momentum in One Dimension 5 Conservation of Momentum in Two Dimensions 6 Inelastic Collisions 7 Elastic Collisions 8 Momentum and Impulse 1 4 7 10 14 18 22 26 iv www.ck12.org Concept 1. Momentum CONCEPT 1 • Deﬁne momentum. • Relate momentum to mass and velocity. • Calculate momentum from mass and velocity. Momentum Cody seems a little reluctant to launch himself down this ramp at Newton’s Skate Park. It will be his ﬁrst time down the ramp, and he knows from watching his older brother Jerod that he’ll be moving fast by the time he gets to the bottom. The faster he goes, the harder it will be to stop. That’s because of momentum. What Is Momentum? Momentum is a property of a moving object that makes it hard to stop. The more mass it has or the faster it’s moving, the greater its momentum. Momentum equals mass times velocity and is represented by the equation: 1 www.ck12.org Momentum = Mass x Velocity Q : What is Cody’s momentum as he stands at the top of the ramp? A : Cody has no momentum as he stands there because he isn’t moving. However, Cody will gain momentum as he starts moving down the ramp and picks up speed. In other words, his velocity is zero. Q : Cody’s older brother Jerod is pictured 1.1 . If Jerod were to travel down the ramp at the same velocity as Cody, who would have greater momentum? Who would be harder to stop? A : Jerod obviously has greater mass than Cody, so he would have greater momentum. He would also be harder to stop. FIGURE 1.1 You can see an animation demonstrating the role of mass and velocity in the momentum of moving objects at this URL: http://www.science-animations.com/support-files/momentum.swf Calculating Momentum To calculate momentum with the equation above, mass is measured in (kg), and velocity is measured in meters per second (m/s). For example, Cody and his skateboard have a combined mass of 40 kg. If Cody is traveling at a velocity of 1.1 m/s by the time he reaches the bottom of the ramp, then his momentum is: Momentum = 40 kg x 1.1 m/s = 44 kg m/s Note that the SI unit for momentum is kg m/s. Q : The combined mass of Jerod and his skateboard is 68 kg. If Jerod goes down the ramp at the same velocity as Cody, what is his momentum at the bottom of the ramp? A : His momentum is: Momentum = 68 kg x 1.1 m/s = 75 kg m/s Summary • Momentum is a property of a moving object that makes it hard to stop. It equals the object’s mass times its velocity. 2 www.ck12.org Concept 1. Momentum • To calculate the momentum of a moving object, multiply its mass in kilograms (kg) by its velocity in meters per second (m/s). The SI unit of momentum is kg m/s. Vocabulary • momentum : Property of a moving object that makes it hard to stop; equal to the object’s mass times its velocity. Practice At the following URL, review how to calculate momentum, and then solve the problems at the bottom of the Web page. http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienceIntegation-848.htm Review 1. Deﬁne momentum. 2. Write the equation for calculating momentum from mass and velocity. 3. What is the SI unit for momentum? 4. Which skateboarder has greater momentum? a. Skateboarder A: mass = 60 kg; velocity = 1.5 m/s b. Skateboarder B: mass = 50 kg; velocity = 2.0 m/s References 1. . . used under license from Shutterstock 3 CONCEPT 2 www.ck12.org Momentum Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Students will learn what momentum is and how to calculate momentum of objects. In addition, students will learn how to use conservation of momentum to solve basic problems. Key Equations p = mv Momentum is equal to the objects mass multiplied by its velocity pinitial = p ﬁnal The total momentum does not change in closed systems Example 1 A truck with mass 500 kg and originally carrying 200 kg of dirt is rolling forward with the transmission in neutral and shooting out the dirt backwards at 2 m/s (so that the dirt is at relative speed of zero compared with the ground). If the truck is originally moving at 2 m/s, how fast will it be moving after it has shot out all the dirt. You may ignore the effects of friction. To solve this problem we will apply conservation of momentum to the truck when it is full of dirt and when it has dumped all the dirt. Solution mivi = m f v f
(mt + md)vi = mtv f start by setting the initial momentum equal to the ﬁnal momentum substitute the mass of the truck plus the mass of the dirt in the truck at the initial and ﬁnal states v f = v f = (mt + md)vi mt (500 kg + 200 kg) 2 m/s 500 kg v f = 2:8 m/s solve for the ﬁnal velocity plug in the numerical values Example 2 John and Bob are standing at rest in middle of a frozen lake so there is no friction between their feet and the ice. Both of them want to get to shore so they simultaneously push off each other in opposite directions. If John’s mass is 50 kg and Bob’s mass is 40 kg and John moving at 5 m/s after pushing off Bob, how fast is Bob moving? 4 www.ck12.org Concept 2. Momentum Solution For this problem, we will apply conservation of momentum to the whole system that includes both John and Bob. Since both of them are at rest to start, we know that the total momentum of the whole system must always be zero. Therefore, we know that the sum of John’s and Bob’s momentum after they push off each other is also zero. We can use this to solve for Bob’s velocity. 0 = m jv j + mbvb mbvb = m jv j vb = m jv j mb 50 kg 5 m/s 40 kg vb = 6:25 m/s vb = The answer is negative because Bob is traveling in the opposite direction to John. Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. You ﬁnd yourself in the middle of a frozen lake. There is no friction between your feet and the ice of the lake. You need to get home for dinner. Which strategy will work best? a. Press down harder with your shoes as you walk to shore. b. Take off your jacket. Then, throw it in the direction opposite to the shore. c. Wiggle your butt until you start to move in the direction of the shore. d. Call for help from the great Greek god Poseidon. 2. You and your sister are riding skateboards side by side at the same speed. You are holding one end of a rope and she is holding the other. Assume there is no friction between the wheels and the ground. If your sister lets go of the rope, how does your speed change? a. It stays the same. b. It doubles. c. It reduces by half. 5 www.ck12.org 3. You and your sister are riding skateboards (see Problem 3), but now she is riding behind you. You are holding one end of a meter stick and she is holding the other. At an agreed time, you push back on the stick hard enough to get her to stop. What happens to your speed? Choose one. (For the purposes of this problem pretend you and your sister weigh the same amount.) a. It stays the same. b. It doubles. c. It reduces by half. 4. An astronaut is using a drill to ﬁx the gyroscopes on the Hubble telescope. Suddenly, she loses her footing and ﬂoats away from the telescope. What should she do to save herself? 5. A 5:00 kg ﬁrecracker explodes into two parts: one part has a mass of 3:00 kg and moves at a velocity of 25:0 m=s towards the west. The other part has a mass of 2:00 kg . What is the velocity of the second piece as a result of the explosion? 6. A ﬁrecracker lying on the ground explodes, breaking into two pieces. One piece has twice the mass of the other. What is the ratio of their speeds? 7. While driving in your pickup truck down Highway 280 between San Francisco and Palo Alto, an asteroid lands in your truck bed! Despite its 220 kg mass, the asteroid does not destroy your 1200 kg truck. In fact, it landed perfectly vertically. Before the asteroid hit, you were going 25 m=s . After it hit, how fast were you going? 8. An astronaut is 100 m away from her spaceship doing repairs with a 10:0 kg wrench. The astronaut’s total mass is 90:0 kg and the ship has a mass of 1:00 104 kg . If she throws the wrench in the opposite direction of the spaceship at 10:0 m=s how long would it take for her to reach the ship? Answers to Selected Problems 1. . 2. . 3. . 4. . 5. 37:5 m=s 6. v1 = 2v2 7. 21 m=s 8. a. 90 sec 6 www.ck12.org Concept 3. Law of Conservation of Momentum CONCEPT 3 Law of Conservation of Momentum • State the law of conservation of momentum. • Describe an example of momentum being transferred and conserved. These skaters are racing each other at Newton’s Skate Park. The ﬁrst skater in line, the one on the left, is distracted by something he sees. He starts to slow down without realizing it. The skater behind him isn’t paying attention and keeps skating at the same speed. Q : Can you guess what happens next? A : Skater 2 runs into skater 1. Conserving Momentum When skater 2 runs into skater 1, he’s going faster than skater 1 so he has more momentum. Momentum is a property of a moving object that makes it hard to stop. It’s a product of the object’s mass and velocity. At the moment of the collision, skater 2 transfers some of his momentum to skater 1, who shoots forward when skater 2 runs into him. Whenever an action and reaction such as this occur, momentum is transferred from one object to the other. However, the combined momentum of the objects remains the same. In other words, momentum is conserved. This is the law of conservation of momentum . 7 Modeling Momentum The Figure 3.1 shows how momentum is conserved in the two colliding skaters. The total momentum is the same after the collision as it was before. However, after the collision, skater 1 has more momentum and skater 2 has less momentum than before. www.ck12.org FIGURE 3.1 Q : What if two skaters have a head-on collision? Do you think momentum is conserved then? A : As in all actions and reactions, momentum is also conserved in a head-on collision. You can see how at this URL: http://www.physicsclassroom.com/mmedia/momentum/cthoi.cfm Summary • Whenever an action and reaction occur, momentum is transferred from one object to the other. However, total momentum is conserved. This is the law of conservation of momentum. 8 www.ck12.org Vocabulary Concept 3. Law of Conservation of Momentum • momentum : Property of a moving object that makes it hard to stop; equal to the object’s mass times its velocity. • law of conservation of momentum : Law stating that, when an action and reaction occur, the combined momentum of the objects remains the same. Practice Watch the astropitch animation at the following URL. Experiment with different velocities. Then take the quiz and check your answers. http://www.phys.utb.edu/~pdukes/standard/PhysApplets/AstroPitch/TabbedastroPitch2.html Review 1. State the law of conservation of momentum. 2. Fill in the missing velocity (x) in the diagram of a vehicle collision seen in the Figure 3.2 so that momentum is conserved. FIGURE 3.2 Solve for x. References 1. Laura Guerin. . CC BY-NC 3.0 2. Laura Guerin. . CC BY-NC 3.0 9 CONCEPT 4 Conservation of Momentum in One Dimension www.ck12.org • State the law of conservation of momentum. • Use the conservation of momentum to solve one-dimensional collision problems. For this whale to leap out of the water, something underwater must be moving in the opposite direction, and intuition tells us it must be moving with relatively high velocity. The water that moves downward is pushed downward by the whale’s tail, and that allows the whale to rise up. Conservation of Momentum in One Dimension When impulse and momentum were introduced, we used an example of a batted ball to discuss the impulse and momentum change that occurred with the ball. At the time, we did not consider what had happened to the bat. According to Newton’s third law, however, when the bat exerted a force on the ball, the ball also exerted an equal and opposite force on the bat. Since the time of the collision between bat and ball is the same for the bat and for the ball, then we have equal forces (in opposite directions) exerted for equal times on the ball AND the bat. That means that the impulse exerted on the bat is equal and opposite (-Ft) to the impulse on the ball (Ft) and that also means that there was a change in momentum of the bat [D(mv)BAT] that was equal and opposite to the change in momentum of the ball [D(mv)BALL] . The change in momentum of the ball is quite obvious because it changes direction and ﬂies off at greater speed. However, the change in momentum of the bat is not obvious at all. This occurs primarily because the bat is more massive than the ball. Additionally, the bat is held ﬁrmly by the batter, so the batter’s mass can be combined with the mass of the bat. Since the bat’s mass is so much greater than that of the ball, but they have equal and opposite forces, the bat’s ﬁnal velocity is signiﬁcantly smaller than that of the ball. Consider another system: that of two ice skaters. If we have one of the ice skaters exert a force on the other skater, the force is called an internal force because both the object exerting the force and the object receiving the force 10 www.ck12.org Concept 4. Conservation of Momentum in One Dimension are inside the system. In a closed system such as this, momentum is always conserved. The total ﬁnal momentum always equals the total initial momentum in a closed system. Conversely, if we deﬁned a system to contain just one ice skater, putting the other skater outside the system, this is not a closed system. If one skater pushes the other, the force is an external force because the receiver of the force is outside the system. Momentum is not guaranteed to be conserved unless the system is closed. In a closed system, momentum is always conserved. Take another example: if we consider two billiard balls colliding on a billiard table and ignore friction, we are dealing with a closed system. The momentum of ball A before the collision plus the momentum of ball B before collision will equal the momentum of ball A after collision plus the momentum of ball B after collision. This is called the law of conservation of momentum and is given by the equation pAbefore + pBbefore = pAafter + pBafter Example Problem: Ball A has a mass of 2.0 kg and is moving due west with a velocity of 2.0 m/s while ball B has a mass of 4.0 kg and is moving west with a velocity of 1.0 m/s. Ball A overtakes ball B and co
llides with it from behind. After the collision, ball A is moving westward with a velocity of 1.0 m/s. What is the velocity of ball B after the collision? Solution: Because of the law of conservation of momentum, we know that pAbefore + pBbefore = pAafter + pBafter . mAvA + mBvB = mAv0 A + mBv0 B (2:0 kg)(2:0 m/s) + (4:0 kg)(1:0 m/s) = (2:0 kg)(1:0 m/s) + (4:0 kg)(vB 0 m/s) 4:0 kg m/s + 4:0 kg m/s = 2:0 kg m/s + 4vB 0 kg m/s 4vB 0 = 8:0 2:0 = 6:0 0 = 1:5 m/s vB After the collision, ball B is moving westward at 1.5 m/s. Example Problem: A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars? Solution: Note that since the two trains stick together, the ﬁnal mass is m A +m B , and the ﬁnal velocity for each object is the same. Thus the conservation of momentum equation, mAvA + mBvB = mAv0 B , can be A + mBv0 rewritten mAvA + mBvB = (mA + mB) v0 (30; 000: kg)(2:2 m/s) + (30; 000: kg)(0 m/s) = (60; 000: kg)(v0 m/s) 66000 + 0 = 60000v0 v0 = 66000 60000 = 1:1 m/s After the collision, the two cars move off together toward the east with a velocity of 1.1 m/s. Summary • A closed system is one in which both the object exerting a force and the object receiving the force are inside the system. • In a closed system, momentum is always conserved. 11 Practice www.ck12.org The following video shows the Mythbusters building and using various sizes of a toy called "Newton’s Cradle." Use this resource to answer the question that follows. https://www.youtube.com/watch?v=BiLq5Gnpo8Q MEDIA Click image to the left for more content. 1. What is Newton’s Cradle? 2. How does Newton’s Cradle work? 3. How does a Newton’s Cradle show conservation of momentum? If you are interested in learning more about Newton’s Cradles, visit this site: http://science.howstuffworks.com/ne wtons-cradle.htm Review 1. A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. With what speed does the goalie slide on the (frictionless) ice? 2. A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet ﬂy off together at 9.0 m/s. What was the original velocity of the bullet? 3. A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s. After the collision, the 0.50 kg ball moves away at -14 m/s. Find the velocity of the second ball after the collision. 4. Two carts are stationary with a compressed spring between them and held together by a thread. When the thread is cut, the two carts move apart. After the spring is released, one cart m = 3:00 kg has a velocity of 0.82 m/s east. What is the magnitude of the velocity of the second cart (m = 1:70 kg) after the spring is released? 5. Compared to falling on a tile ﬂoor, a glass may not break if it falls onto a carpeted ﬂoor. This is because a. less impulse in stopping. b. longer time to stop. c. both of these d. neither of these. 6. A butterﬂy is hit by a garbage truck on the highway. The force of the impact is greater on the 12 www.ck12.org Concept 4. Conservation of Momentum in One Dimension a. garbage truck. b. butterﬂy. c. it is the same for both. 7. A riﬂe recoils from ﬁring a bullet. The speed of the riﬂe’s recoil is small compared to the speed of the bullet because a. the force on the riﬂe is small. b. the riﬂe has a great deal more mass than the bullet. c. the momentum of the riﬂe is unchanged. d. the impulse on the riﬂe is less than the impulse on the bullet. e. none of these. • Law of Conservation of Momentum: The total linear momentum of an isolated system remains constant regardless of changes within the system. References 1. Courtesy of NOAA. http://commons.wikimedia.org/wiki/File:Humpback_whale_noaa.jpg . Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 13 CONCEPT 5 Conservation of Momentum in Two Dimensions www.ck12.org • Use the conservation of momentum and vector analysis to solve two-dimensional collision problems. • Review vector components. In a game of billiards, it is important to be able to visualize collisions in two dimensions – the best players not only know where the target ball is going but also where the cue ball will end up. Conservation of Momentum in Two Dimensions Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide. Most objects are not conﬁned to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated; momentum is a vector and collisions of objects in two dimensions can be represented by axial vector components. To review axial components, revisit Vectors: Resolving Vectors into Axial Components and Vectors: Vector Addition. Example Problem: A 2.0 kg ball, A , is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B , also with a mass of 2.0 kg. After the collision, ball A moves off at 30° south of west while ball B moves off at 60° north of west. Find the velocities of both balls after the collision. Solution: Since ball B is stationary before the collision, then the total momentum before the collision is equal to momentum of ball A . The momentum of ball A before collision is shown in red below, and can be calculated to be p = mv = (2:00 kg)(5:00 m/s) = 10:0 kg m/s west 14 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions Since momentum is conserved in this collision, the sum of the momenta of balls A and B after collsion must be 10.0 kg m/s west. pAafter = (10:0 kg m/s)(cos 30) = (10:0 kg m/s)(0:866) = 8:66 kg m/s pBafter = (10:0 kg m/s)(cos 60) = (10:0 kg m/s)(0:500) = 5:00 kg m/s To ﬁnd the ﬁnal velocities of the two balls, we divide the momentum of each by its mass. Therefore, vA = 4:3 m/s and vB = 2:5 m/s . Example Problem: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. The two cars stick together after the collision. What is the speed and direction of the two cars after the collision? Solution: Northward momentum = (1325 kg)(27:0 m/s) = 35800 kg m/s Eastward momentum = (2165 kg)(17:0 m/s) = 36800 kg m/s q (35800)2 + (36800)2 = 51400 kg m/s R = 15 www.ck12.org q = sin1 35800 m = 51400 kgm/s 3490 kg velocity = p 51400 = 44 north of east = 14:7 m/s @ 44 N of E Example Problem: A 6.00 kg ball, A , moving at velocity 3.00 m/s due east collides with a 6.00 kg ball, B , at rest. After the collision, A moves off at 40.0° N of E and ball B moves off at 50.0° S of E. a. What is the momentum of A after the collision? b. What is the momentum of B after the collision? c. What are the velocities of the two balls after the collision? Solution: pinitial = mv = (6:00 kg)(3:00 m/s) = 18:0 kg m/s This is a right triangle in which the initial momentum is the length of the hypotenuse and the two momenta after the collision are the legs of the triangle. a. pA = (18:0 kg m/s)(cos 40:0) = (18:0 kg m/s)(0:766) = 13:8 kg m/s b. pB = (18:0 kg m/s)(cos 50:0) = (18:0 kg m/s)(0:643) = 11:6 kg m/s c. vA = 2:30 m/s vB = 1:93 m/s • The conservation of momentum law holds for all closed systems regardless of the directions of the objects before and after they collide. • Momentum is a vector; collisions in two dimensions can be represented by axial vector components. Summary Practice This video shows circus performers using conservation of momentum. Use this resource to answer the questions that follow. http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ 16 www.ck12.org Concept 5. Conservation of Momentum in Two Dimensions MEDIA Click image to the left for more content. 1. Why do the ﬂiers scrunch up in the air while spinning and twisting? 2. What happens to the rate at which they spin when they change shape in the air? Review 1. Billiard ball A , mass 0.17 kg, moving due east with a velocity of 4.0 m/s, strikes stationary billiard ball B , also mass of 0.17 kg. After the collision, ball A moves off at an angle of 30° north of east with a velocity of 3.5 m/s, and ball B moves off at an angle of 60 ° south of east . What is the speed of ball B ? 2. A bomb, originally sitting at rest, explodes and during the explosion breaks into four pieces of exactly 0.25 kg each. One piece ﬂies due south at 10 m/s while another pieces ﬂies due north at 10 m/s. (a) What do we know about the directions of the other two pieces and how do we know it? (b) What do we know about the speeds of the other two pieces and how do we know it? 3. In a head-on collision between protons in a particle accelerator, three resultant particles were observed. All three of the resultant particles were moving to the right from the point of collision. The physicists conducting the experiment concluded there was at least one unseen particle moving to the left after the collision. Why did they conclude this? References 1. Image copyright VitCOM Photo, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 17 CONCEPT 6 Inelastic Collisions Students will learn how to solve problems involving inelastic collisions. www.ck12.org Key Equations pinitial = p ﬁnal The total momentum does not change in closed systems Example 1 Question : Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed of 2:5m=s . Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial speed of block B? Solution : To ﬁnd mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg and the mass
of block A is 8.0kg. 10kg 8:0kg = 2:0kg Now that we know the mass of both blocks we can ﬁnd the speed of block B. We will use conservation of momentum. This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This problem is one dimensional, because all motion happens along the same line. Thus we will use the equation (mA + mB)v f = mA vA + mB vB and solve for the velocity of block B. (mA + mB)v f = mA vA + mBvB ) (mA + mB)(v f ) (mA)(vA) mB = vB Watch this Explanation 18 www.ck12.org Concept 6. Inelastic Collisions MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 0% for perfectly inelastic collisions. Collision Lab (PhET Simulation) 19 www.ck12.org Car Collision (CK-12 Simulation) Time for Practice 2. Two blocks collide on a frictionless surface, as shown. Afterwards, they have a combined mass of 10 kg and a speed of 2:5 m=s . Before the collision, one of the blocks was at rest. This block had a mass of 8:0 kg . What was the mass and initial speed of the second block? 1. 3. 4. In the above picture, the carts are moving on a level, frictionless track. After the collision all three carts stick together. Determine the direction and speed of the combined carts after the collision. 5. The train engine and its four boxcars are coasting at 40 m=s . The engine train has mass of 5; 500 kg and the boxcars have masses, from left to right, of 1; 000 kg , 1; 500 kg , 2; 000 kg; and 3; 000 kg . (For this problem, you may neglect the small external forces of friction and air resistance.) 20 www.ck12.org Concept 6. Inelastic Collisions a. What happens to the speed of the train when it releases the last boxcar? ( Hint: Think before you blindly calculate. ) b. If the train can shoot boxcars backwards at 30 m=s relative to the train’s speed, how many boxcars does the train need to shoot out in order to obtain a speed of 58:75 m=s ? 6. In Sacramento a 4000 kg SUV is traveling 30 m=s south on Truxel crashes into an empty school bus, 7000 kg traveling east on San Juan. The collision is perfectly inelastic. a. Find the velocity of the wreck just after collision b. Find the direction in which the wreck initially moves 7. Manrico (80:0 kg) and Leonora (60:0 kg) are ﬁgure skaters. They are moving toward each other. Manrico’s speed is 2:00 m=s ; Leonora’s speed is 4:00 m=s . When they meet, Leonora ﬂies into Manrico’s arms. a. With what speed does the entwined couple move? b. In which direction are they moving? c. How much kinetic energy is lost in the collision? Answers to Selected Problems 1. 2:0 kg; 12:5 m=s 2. 0:13 m=s to the left 3. a. no change b. the last two cars 4. a. 15 m=s b. 49 S of E 5. a. 0.57 m/s b. the direction Leonora was originally travelling c. 297.26 J 21 CONCEPT 7 www.ck12.org Elastic Collisions Students will learn how to solve problems involving elastic collisions Key Equations pinitial = p ﬁnal The total momentum does not change in closed systems KEinitial = KE ﬁnal The total kinetic energy does not change in elastic collisions Example 1 MEDIA Click image to the left for more content. Example 2 Question : Chris and Ashley are playing pool. Ashley hits the cue ball into the 8 ball with a velocity of 1:2m=s . The cue ball (c) and the 8 ball ( e ) react as shown in the diagram. The 8 ball and the cue ball both have a mass of :17kg . What is the velocity of the cue ball? What is the direction (the angle) of the cue ball? Answer : We know the equation for conservation of momentum, along with the masses of the objects in question as well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the velocity of the cue ball. mcvic + mevie = mcv f c + mev f e v f c = v f c = mcvic + mevie mev f e mc :17kg 2:0m=s + :17kg 0m=s :17kg 1:2m=s :17kg v f c = :80m=s Now we want to ﬁnd the direction of the cue ball. To do this we will use the diagram below. 22 www.ck12.org Concept 7. Elastic Collisions We know that the momentum in the y direction of the two balls is equal. Therefore we can say that the velocity in the y direction is also equal because the masses of the two balls are equal. mcvcy = mevey ! vcy = vey Given this and the diagram, we can ﬁnd the direction of the cue ball. After 1 second, the 8 ball will have traveled 1:2m . Therefore we can ﬁnd the distance it has traveled in the y direction. sin25o = opposite hypotenuse = x 1:2m ! x = sin25 1:2m = :51m Therefore, in one second the cue ball will have traveled :51m in the y direction as well. We also know how far in total the cue ball travels in one second (:80m) . Thus we can ﬁnd the direction of the cue ball. sin1 opposite hypotenuse = sin1 :51m :80m = 40o Watch this Explanation MEDIA Click image to the left for more content. Simulation Note: move the elasticity meter to 100% for perfectly elastic collisions. 23 www.ck12.org Collision Lab (PhET Simulation) Time for Practice 1. You are playing pool and you hit the cue ball with a speed of 2 m=s at the 8 -ball (which is stationary). Assume an elastic collision and that both balls are the same mass. Find the speed and direction of both balls after the collision, assuming neither ﬂies off at any angle. 2. A 0:045 kg golf ball with a speed of 42:0 m=s collides elastically head-on with a 0:17 kg pool ball at rest. Find the speed and direction of both balls after the collision. 3. Ball A is traveling along a ﬂat table with a speed of 5:0 m=s , as shown below. Ball B , which has the same mass, is initially at rest, but is knocked off the table in an elastic collision with Ball A . Find the horizontal distance that Ball B travels before hitting the ﬂoor. 4. Students are doing an experiment on the lab table. A steel ball is rolled down a small ramp and allowed to hit the ﬂoor. Its impact point is carefully marked. Next a second ball of the same mass is put upon a set screw and a collision takes place such that both balls go off at an angle and hit the ﬂoor. All measurements are taken with a meter stick on the ﬂoor with a co-ordinate system such that just below the impact point is the origin. The following data is collected: (a) no collision: 41:2 cm (b) target ball: 37:3 cm in the direction of motion and 14:1 cm perpendicular to the direction of motion 24 www.ck12.org Concept 7. Elastic Collisions i. From this data predict the impact position of the other ball. ii. One of the lab groups declares that the data on the ﬂoor alone demonstrate to a 2 % accuracy that the collision was elastic. Show their reasoning. iii. Another lab group says they can’t make that determination without knowing the velocity the balls have on impact. They ask for a timer. The instructor says you don’t need one; use your meter stick. Explain. iv. Design an experiment to prove momentum conservation with balls of different masses, giving apparatus, procedure and design. Give some sample numbers. 5. A 3 kg ball is moving 2 m/s in the positive x direction when it is struck dead center by a 2 kg ball moving in the positive y direction at 1 m/s. After collision the 3 kg ball moves at 3 m/s 30 degrees from the positive x axis. Find the velocity and direction of the 2 kg ball. Answers to Selected Problems 1. 8 m=s same direction as the cue ball and 0 m=s 2. vgol f = 24:5 m=s; vpool = 17:6 m=s 3. 2:8 m 4. . 5. 1:5 m=s 54 25 CONCEPT 8 Momentum and Impulse www.ck12.org • Deﬁne momentum. • Deﬁne impulse. • Given mass and velocity of an object, calculate momentum. • Calculate the change in momentum of an object. • State the relationship that exists between the change in momentum and impulse. • Using the momentum-impulse theorem and given three of the four variables, calculate the fourth. Rachel Flatt performs a layback spin at the 2 011 Rostelecom Cup in Moscow, Russia. When an ice skater spins, angular momentum must be conserved. When her arms or feet are far away from her body, her spin slows; when she brings her arms and feet close in to her body, she spins faster. Momentum and Impulse If a bowling ball and a ping-pong ball are each moving with a velocity of 5 mph, you intuitively understand that it will require more effort to stop the bowling ball than the ping pong ball because of the greater mass of the bowling ball. Similarly, if you have two bowling balls, one moving at 5 mph and the other moving at 10 mph, you know it 26 www.ck12.org Concept 8. Momentum and Impulse will take more effort to stop the ball with the greater speed. It is clear that both the mass and the velocity of a moving object contribute to what is necessary to change the motion of the moving object. The product of the mass and velocity of an object is called its momentum . Momentum is a vector quantity that has the same direction as the velocity of the object and is represented by a lowercase letter p . The momentum of a 0.500 kg ball moving with a velocity of 15.0 m/s will be p = mv p = mv = (0:500 kg)(15:0 m/s) = 7:50 kg m/s You should note that the units for momentum are kg·m/s. According to Newton’s ﬁrst law, the velocity of an object cannot change unless a force is applied. If we wish to change the momentum of a body, we must apply a force. The longer the force is applied, the greater the change in momentum. The impulse is the quantity deﬁned as the force multiplied by the time it is applied. It is a vector quantity that has the same direction as the force. The units for impulse are N·s but we know that Newtons are also kg·m/s 2 and so N·s = (kg·m/s 2 )(s) = kg·m/s. Impulse and momentum have the same units; when an impulse is applied to an object, the momentum of the object changes and the change of momentum is equal to the impulse. Ft = Dmv Example Problem: Calculating Momentum A 0.15 kg ball is moving with a velocity of 35 m/s. Find the momentum of the ball. Solution: p = m
v = (0:15 kg)(35 m/s) = 5:25 kg m/s Example Problem: If a ball with mass 5.00 kg has a momentum of 5:25 kg m/s , what is its velocity? m = 5:25 kgm/s 5:00 kg It should be clear from the equation relating impulse to change in momentum, Ft = Dmv , that any amount of force would (eventually) bring a moving object to rest. If the force is very small, it must be applied for a long time, but a greater force can bring the object to rest in a shorter period of time. Solution: v = p = 1:05 m/s If you jump off a porch and land on your feet with your knees locked in the straight position, your motion would be brought to rest in a very short period of time and thus the force would need to be very large – large enough, perhaps, to damage your joints or bones. Suppose that when you hit the ground, your velocity was 7.0 m/s and that velocity was brought to rest in 0.05 seconds. If your mass is 100. kg, what force was required to bring you to rest? 0:050 s If, on the other hand, when your feet ﬁrst touched the ground, you allowed your knees to ﬂex so that the period of time over which your body was brought to rest is increased, then the force on your body would be smaller and it would be less likely that you would damage your legs. F = Dmv t = (100: kg)(7:0 m/s) = 14; 000 N Suppose that when you ﬁrst touch the ground, you allow your knees to bend and extend the stopping time to 0.50 seconds. What force would be required to bring you to rest this time? t = (100: kg)(7:0 m/s) 0:50 s With the longer period of time for the force to act, the necessary force is reduced to one-tenth of what was needed before. = 1400 N F = Dmv Extending the period of time over which a force acts in order to lessen the force is a common practice in design. Padding in shoes and seats allows the time to increase. The front of automobiles are designed to crumple in an accident; this increases the time the car takes to stop. Similarly, barrels of water or sand in front of abutments on 27 www.ck12.org the highway and airbags serve to slow down the stoppage time. These changes all serve to decrease the amount of force it takes to stop the momentum in a car crash, which consequently saves lives. Example Problem: An 0.15 kg baseball is thrown horizontally at 40. m/s and after it is struck by a bat, it is traveling at -40. m/s. (a) What impulse did the bat deliver to the ball? (b) If the contact time of the bat and bat was 0.00080 seconds, what was the average force the bat exerted on the ball? (c) Calculate the average acceleration of the ball during the time it was in contact with the bat. Solution: We can calculate the change in momentum and give the answer as impulse because we know that the impulse is equal to the change in momentum. (a) p = mDv = (0:15 kg)(40: m/s 40: m/s) = (0:15 kg)(80: m/s) = 12 kg m/s The minus sign indicates that the impulse was in the opposite direction of the original throw. t = 12 kgm/s Again, the negative sign indicates the force was in the opposite direction of the original throw. 0:00080 s = 15000 N (b) F = Dmv (c) a = F = 100; 000 m/s2 m = 15000 N 0:15 kg Summary • The product of the mass and velocity of an object is called momentum, given by the equation r = mv . • Momentum is a vector quantity that has the same direction as the velocity of the object. • The quantity of force multiplied by the time it is applied is called impulse. • Impulse is a vector quantity that has the same direction as the force. • Momentum and impulse have the same units: kg·m/s. • The change of momentum of an object is equal to the impulse. Ft = Dmv Practice Use this resource to answer the question that follows. https://www.youtube.com/watch?v=3g4v8x7xggU MEDIA Click image to the left for more content. 1. Why don’t the glasses of water spill when the tablecloth is pulled out from under them? 2. How does the video get from momentum to impulse? 28 www.ck12.org Concept 8. Momentum and Impulse Review 1. A small car with a mass of 800. kg is moving with a velocity of 27.8 m/s. (a) What is the momentum of the car? (b) What velocity is needed for a 2400. kg car in order to have the same momentum? 2. A scooter has a mass of 250. kg. A constant force is exerted on it for 60.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 28.0 m/s. (a) What is the change in momentum? (b) What is the magnitude of the force exerted on the scooter? 3. The brakes on a 15,680 N car exert a stopping force of 640. N. The car’s velocity changes from 20.0 m/s to 0 m/s. (a) What is the car’s mass? (b) What was its initial momentum? (c) What was the change in momentum for the car? (d) How long does it take the braking force to bring the car to rest? • momentum: A measure of the motion of a body equal to the product of its mass and velocity. Also called linear momentum . • impulse: The product obtained by multiplying the average value of a force by the time during which it acts. The impulse equals the change in momentum produced by the force in this time interval. References 1. User:deerstop/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Flatt-3.jpg . Public Domain 29 Physics Unit 9: Circular Motion Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 5 10 Contents 1 Circular Motion 2 Circular Motion 3 Centripetal Force iv www.ck12.org Concept 1. Circular Motion CONCEPT 1 Circular Motion • Deﬁne centripetal acceleration. • Understand the theory of the centripetal acceleration equation. • Use the centripetal acceleration equation. • Understand the angular relationship between velocity and centripetal acceleration. • Use the equations for motion in two directions and Newton’s Laws to analyze circular motion. Weather satellites, like the one shown above, are found miles above the earth’s surface. Satellites can be polar orbiting, meaning they cover the entire Earth asynchronously, or geostationary, in which they hover over the same spot on the equator. Circular Motion The earth is a sphere. If you draw a horizontal straight line from a point on the surface of the earth, the surface of the earth drops away from the line. The distance that the earth drops away from the horizontal line is very small – so small, in fact, that we cannot represent it well in a drawing. In the sketch below, if the blue line is 1600 m, the amount of drop (the red line) would be 0.20 m. If the sketch were drawn to scale, the red line would be too short to see. 1 www.ck12.org When an object is launched exactly horizontally in projectile motion, it travels some distance horizontally before it strikes the ground. In the present discussion, we wish to imagine a projectile ﬁred horizontally on the surface of the earth such that while traveling 1600 m horizontally, the object would fall exactly 0.20 m. If this could occur, then the object would fall exactly the amount necessary during its horizontal motion to remain at the surface of the earth, but not touching it. In such a case, the object would travel all the way around the earth continuously and circle the earth, assuming there were no obstacles, such as mountains. What initial horizontal velocity would be necessary for this to occur? We ﬁrst calculate the time to fall the 0.20 m: s t = r 2d a = (2)(0:20 m) 9:80 m/s2 = 0:20 s The horizontal velocity necessary to travel 1600 m in 0.20 s is 8000 m/s. Thus, the necessary initial horizontal velocity is 8000 m/s. In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration. In the case of satellites orbiting the earth, the centripetal acceleration is caused by gravity. If you were swinging an object around your head on a string, the centripetal acceleration would be caused by your hand pulling on the string toward the center of the circle. It is important to note that the object traveling in a circle has a constant speed but does not have a constant velocity. This is because direction is part of velocity; when an object changes its direction, it is changing its velocity. Hence the object’s acceleration. The acceleration in the case of uniform circular motion is the change in the direction of the velocity, but not its magnitude. For an object traveling in a circular path, the centripetal acceleration is
directly related to the square of the velocity of the object and inversely related to the radius of the circle. ac = v2 r Taking a moment to consider the validity of this equation can help to clarify what it means. Imagine a yo-yo. Instead of using it normally, let it fall to the end of the string, and then spin it around above your head. If we were to increase the speed at which we rotate our hand, we increase the velocity of the yo-yo - it is spinning faster. As it spins faster, it also changes direction faster. The acceleration increases. Now let’s think about the bottom of the equation: the radius. If we halve the length of the yo-yo string (bring the yo-yo closer to us), we make the yo-yo’s velocity greater. Again, it moves faster, which increases the acceleration. If we make the string longer again, this decreases the acceleration. We now understand why the relationship between the radius and the acceleration is an inverse relationship - as we decrease the radius, the acceleration increases, and visa versa. 2 www.ck12.org Concept 1. Circular Motion Example Problem: A ball at the end of a string is swinging in a horizontal circle of radius 1.15 m. The ball makes exactly 2.00 revolutions per second. What is its centripetal acceleration? Solution: We ﬁrst determine the velocity of the ball using the facts that the circumference of the circle is 2pr and the ball goes around exactly twice per second. v = (2)(2pr) = 14:4 m/s t = (2)(2)(3:14)(1:15 m) 1:00 s r = (14:4 m/s)2 We then use the velocity and radius in the centripetal acceleration equation. ac = v2 Example Problem: The moon’s nearly circular orbit around the earth has a radius of about 385,000 km and a period of 27.3 days. Calculate the acceleration of the moon toward the earth. 1:15 m = 180: m/s2 Solution: v = 2pr T = (2)(3:14)(3:85108 m) r = (1020 m/s)2 (27:3 d)(24:0 h/d)(3600 s/h) 3:85108 m = 0:00273 m/s2 = 1020 m/s ac = v2 As shown in the previous example, the velocity of an object traveling in a circle can be calculated by v = 2pr T 3 Where r is the radius of the circle and T is the period (time required for one revolution). This equation can be incorporated into the equation for centripetal acceleration as shown below. www.ck12.org ac = v2 r = ( 2pr T )2 r = 4p2r T 2 Summary • In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration . • The acceleration in the case of uniform circular motion changes the direction of the velocity but not its magnitude. • Formulas for centripetal acceleration are ac = v2 r and ac = 4p2r T 2 . Practice This video is a demonstration of centripetal force using balloons and trays of water. Use this resource to answer the questions that follow. https://www.youtube.com/watch?v=EX5DZ2MHlV4 MEDIA Click image to the left for more content. 1. What does centripetal mean? 2. What is uniform circular motion? 3. Why is centripetal acceleration always towards the center? Review 1. An automobile rounds a curve of radius 50.0 m on a ﬂat road at a speed of 14 m/s. What centripetal acceleration is necessary to keep the car on the curve? 2. An object is swung in a horizontal circle on a length of string that is 0.93 m long. If the object goes around once in 1.18 s, what is the centripetal acceleration? • circular motion: A movement of an object along the circumference of a circle or rotation along a circular path. • centripetal acceleration: The acceleration toward the center that keeps an object following a circular path. References 1. Greg Goebel . http://www.public-domain-image.com/space-public-domain-images-pictures/weather-satellit e.jpg.html . Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 4 www.ck12.org Concept 2. Circular Motion CONCEPT 2 Circular Motion Students will learn that in circular motion there is always an acceleration (and hence a force) that points to the center of the circle deﬁned by the objects motion. This force changes the direction of the velocity vector of the object but not the speed. Students will also learn how to calculate that speed using the period of motion and the distance of its path (circumference of the circle it traces out). Vocabulary • centripetal acceleration: The inward acceleration that keeps an object in circular motion. • centripetal force: The inward force that keeps an object in circular motion. Introduction A satellite orbits around the Earth in Figure below . A car travels around a curve in Figure below . All of these objects are engaged in circular motion. Let us consider the satellite ﬁrst. The satellite is held in place by the Earth’s gravity. The gravity holds the satellite in its orbit. In what direction does this force act? If the earth were “magically” gone, the satellite would ﬂy off tangent to its motion at the instant gravity no longer held it. The force preventing this from happening must keep pulling the satellite toward the center of the circle to maintain circular motion. FIGURE 2.1 What is the force in Figure above that prevents the car from skidding off the road? If you guessed “the friction between the tires and the road” you’d be correct. But is it static or kinetic friction? Unless the tires skid, there can be no kinetic friction. It is static friction that prevents the tires from skidding, just as it is static friction that permits you to walk without slipping. In Figure below , you can see the foot of a person who walks toward the right by pushing their foot backward with a horizontal component of force F . They move forward because the ground exerts a horizontal component force fs in the opposite direction. (Note that vertical forces are ignored.) The force the ground exerts on the person’s foot is a static friction force. Because the foot does not slide, we know that F and fs are equal opposed forces. We can easily see which direction the static friction force must act when we walk, but what about a car performing circular motion? In what direction does the static friction act on the car in Figure above ? 5 www.ck12.org FIGURE 2.2 FIGURE 2.3 Figure below shows the top view of a car moving around a circular track with a constant speed. Since acceleration is deﬁned as a = Dv , you may be tempted to say that since the speed remains constant, Dv = 0 , the acceleration Dt must also be zero. But that conclusion would be incorrect because Dv represents a change in velocity, not a change in speed. The velocity of the car is not constant since it is continuously changing its direction. How then do we ﬁnd the acceleration of the car? Figure below shows the instantaneous velocity vectors for the car in two different positions a very small time apart. Notice that the vector DV points toward the center of the circle. (Recall that DV can be thought of as the sum of the vectors V2 + (V1) .) The direction of the acceleration points in the direction of DV since acceleration is deﬁned as a = D~v Dt . This is reasonable, since if there were no force directed toward the center of the circle, the car would move off tangent to the circle. We call the inward force that keeps an object in circular motion a “center seeking”, or centripetal force and the acceleration, centripetal acceleration. The centripetal acceleration is often denoted as ac In order to ﬁnd the correct expression for the magnitude of the centripetal acceleration we’ll need to use a little geometric reasoning. Figure below and Figure below show two “almost” similar triangles. The magnitudes of V1 and V2 are equal, and the change in location of the car occurs over a very small increment in time, Dt . The velocities are tangent to the circle and therefore perpendicular to the radius of the circle. As such, the “radius” triangle and the “velocity” triangle are approximately similar (see the ﬁgures above). We construct an approximate ratio between the two triangles by assuming that during the time Dt , the car has traveled a distance Ds 6 www.ck12.org Concept 2. Circular Motion FIGURE 2.4 FIGURE 2.5 along the circle. The ratio below is constructed in order to determine the acceleration. : = Dv v , which leads to, vDs Ds r Dividing both sides of the equation by Dt , we have: v Ds Dt : = rDv . : = r Dv Dt . But Ds Dt is the speed v of the car and Dv Dt is the acceleration of the car. 7 www.ck12.org If we allow the time to become inﬁnitesimally small, then the approximation becomes exact and we have: v2 = ra; a = v2 r . Thus, the magnitude of the centripetal acceleration for an object moving with constant speed in circular motion is ac = v2 r , and its direction is toward the center of the circle. Illustrative Examples using Centripetal Acceleration and Force Example 1A: A 1000 kg car moves with a constant speed 13.0 m/s around a ﬂat circular track of radius 40.0 m. What is the magnitude and direction of the centripetal acceleration? Answer: The magnitude of the car’s acceleration is ac = v2 r = 132 acceleration is toward the center of the track. 40 = 4:225 = 4:23 m=s2 and the direction of its Example 1b: Determine the force of static friction that acts upon the car in Figure below FIGURE 2.6 Answer: Using Newton’s Second Law: F = fs = ma = 1000(4:225) = 4225 = 4230 N Example 1c: Determine the minimum necessary coefﬁcient of static friction between the tires and the road. Answer: y F = FN Mg = 0; FN = Mg but fs = µsFN = ma Thus, µsMg = Ma; µsg = a; µs = a g = 4:225 9:8 = 0:431 = 0:43 Check Your Understanding True or False? 1. Kinetic friction is responsible for the traction (friction) between the tires and the road. Answer : False. As long as the car does not skid, there is no relative motion between the instantaneous contact area of the tire and the road. 2. True or False? The force of static friction upon an object can vary. Answer : True. In attempting to move an object, a range of forces of different magnitudes can be applied unt
il the maximum static friction between the object and the surface it rests upon is overcome and the object is set into motion. Recall that the magnitude of static friction is represented by the inequality: fs µsFN 8 www.ck12.org Concept 2. Circular Motion 3. The greater the mass of the car, the greater the coefﬁcient of friction. Answer : False. The coefﬁcient of friction is independent of the mass of an object. Recall that it is the ratio of the friction force to the normal force. As such, it is a pure number dependent only upon the nature of the materials in contact with each other- in this case rubber and asphalt. References 1. Image copyright Paul Fleet, 2012. http://www.shutterstock.com . Used under license from Shutterstock 2. Tim White (Flickr: TWHITE87). http://www.ﬂickr.com/photos/tjwhite87/8102931300/ 3. Image copyright Andre Adams, 2012; modiﬁed by CK-12 Foundation - Raymond Chou . http://www.shut . CC-BY 2.0 terstock.com . Used under license from Shutterstock.com 4. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 5. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 9 CONCEPT 3 www.ck12.org Centripetal Force • Deﬁne centripetal force. • Solve problems involving centripetal force. • Explain the difference between centripetal and centrifugal forces. Jupiter’s moons and ring materials follow all the laws of physics, including centripetal force and centripetal acceleration. Centripetal Force Centripetal force is, simply, the force that causes centripetal acceleration. Objects that move in uniform circular motion all have an acceleration toward the center of the circle and therefore, they must also suffer a force toward the center of the circle. That force is the centripetal force. For orbiting satellites, such as the moon orbiting the earth or the earth orbiting the sun, the centripetal force is produced by gravity. When an Olympic hammer thrower whirls a massive ball on a chain, the centripetal force is created by the athlete and transmitted by the chain. Newton’s second law shows the relationship between force and acceleration, F = ma . Since we have formulas expressing the relationships for centripetal acceleration, they can easily be altered to show the relationships for centripetal force. ac = v2 and ac = 4p2r r and F = ma so Fc = mv2 r T 2 so Fc = 4p2rm T 2 10 www.ck12.org Concept 3. Centripetal Force Common Misconceptions Many people incorrectly use the term centrifugal force instead of centripetal force . Often, you will hear the term centrifugal force used to describe the outward force pushing an object away from the center of a circle. In reality, however, centrifugal forces are inertial, or ﬁctional, forces. They only exist in the frame of reference of the object that is moving and, even then, are theoretical. Physicists dealing in a moving frame of reference use centrifugal forces to ease calculations. For a great explanation of the difference between centrifugal and centripetal force, see this video: https://www.youtube.com/watch?v=DLgy6rVV-08 MEDIA Click image to the left for more content. Summary • Centripetal force is the force that causes centripetal acceleration. • Equations for centripetal force are Fc = mv2 r and Fc = 4p2rm . T 2 Practice A video of physics students riding a roller coaster. Use this resource to answer the questions that follow. http://www.teachersdomain.org/asset/phy03_vid_roller/ 1. Does the roller coaster in the video have a complete circle as part of its path? 2. What is it that keeps the glass of water on the tray as it swings over the student’s head? Review 1. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. What is the centripetal force needed to keep this runner on the curve and what supplies this force? 2. A 1000. kg car rounds a curve of 50.0 m radius on a ﬂat road with a speed of 14.0 m/s. (a) Will the car make the turn successfully if the pavement is dry and the coefﬁcient of friction is 0.60? (b) Will the car make the turn successfully if the pavement is wet and the coefﬁcient of friction is 0.20? 3. An 0.500 kg object tied to a string is swung around a person’s head in a horizontal circle. The length of the string is 1.00 m and the maximum force the string can withstand without breaking is 25.0 N. What is the maximum speed the object may be swung without breaking the string? • centripetal force: The component of force acting on a body in curvilinear motion that is directed toward the center of curvature or axis of rotation. References 1. Courtesy of NASA/JPL/Cornell University. http://www.nasa.gov/centers/goddard/multimedia/largest/EduI mageGallery.html . Public Domain 11 Physics Unit 10: Waves Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. CONTRIBUTOR Catherine Pavlov www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: April 6, 2014 iii Contents www.ck12.org 1 7 10 14 17 20 24 28 32 35 39 42 45 56 Contents 1 Measuring Waves 2 Mechanical Wave 3 Transverse Waves 4 Longitudinal Waves 5 Reﬂection of Mechanical Waves 6 Refraction of Mechanical Waves 7 Wave Interactions 8 Wave Interference 9 Wave Speed 10 Sound Waves 11 Frequency and Pitch of Sound 12 Speed of Sound 13 Resonance with Sound Waves 14 Sound in a Tube iv www.ck12.org Concept 1. Measuring Waves CONCEPT 1 Measuring Waves Lesson Objectives • Deﬁne wave amplitude and wavelength. • Relate wave speed to wave frequency and wavelength. Lesson Vocabulary • hertz (Hz) • wave amplitude • wave frequency • wavelength • wave speed Introduction Tsunamis, or the waves caused by earthquakes, are unusually large ocean waves. You can see an example of a tsunami in Figure 1.1 . Because tsunamis are so big, they can cause incredible destruction and loss of life. The tsunami in the ﬁgure crashed into Thailand, sending people close to shore running for their lives. The height of a tsunami or other wave is just one way of measuring its size. You’ll learn about this and other ways of measuring waves in this lesson. FIGURE 1.1 This tsunami occurred in Thailand on De- cember 26, 2004. 1 www.ck12.org Wave Amplitude and Wavelength The height of a wave is its amplitude. Another measure of wave size is wavelength. Both wave amplitude and wavelength are described in detail below. Figure 1.2 shows these wave measures for both transverse and longitudinal waves. You can also simulate waves with different amplitudes and wavelengths by doing the interactive animation at this URL: http://sci-culture.com/advancedpoll/GCSE/sine%20wave%20simulator.html . FIGURE 1.2 Wave amplitude and wavelength are two important measures of wave size. Wave Amplitude Wave amplitude is the maximum distance the particles of a medium move from their resting position when a wave passes through. The resting position is where the particles would be in the absence of a wave. • In a transverse wave, wave amplitude is the height of each crest above the resting position. The higher the crests are, the greater the amplitude. • In a longitudinal wave, amplitude is a measure of how compressed particles of the medium become when the wave passes through. The closer together the particles are, the greater the amplitude. 2 www.ck12.org Concept 1. Measuring Waves What determines a wave’s amplitude? It depends on the energy of the disturbance that causes the wave. A wave caused by a disturbance with more energy has greater amplitude. Imagine dropping a small pebble into a pond of still water. Tiny ripples will move out from the disturbance in concentric circles, like those in Figure above . The ripples are low-amplitude waves. Now imagine throwing a big boulder into the pond. Very large waves will be generated by the disturbance. These waves are high-amplitude waves. Wavelength Another important measure of wave size is wavelength. Wavelength is the distance between two corresponding points on adjacent waves (see Figure 1.2 ). Wavelength can be measured as the distance between two adjacent crests of a transverse wave or two adjacent compressions of a longitudinal wave. It is usually measured in meters. Wavelength is related to the energy of a wave. Short-wavelength waves have more energy than long-wavelength waves of the same amplitude. You can see examples of waves with shorter and longer wavelengths in Figure 1.3 . FIGURE 1.3 Both of these wa
ves have the same ampli- tude, but they differ in wavelength. Which wave has more energy? Wave Frequency and Speed Imagine making transverse waves in a rope, like the waves in Figure above . You tie one end of the rope to a doorknob or other ﬁxed point and move the other end up and down with your hand. You can move the rope up and down slowly or quickly. How quickly you move the rope determines the frequency of the waves. Wave Frequency The number of waves that pass a ﬁxed point in a given amount of time is wave frequency . Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. The higher the number is, the greater is the frequency of the wave. The SI unit for wave frequency is the hertz (Hz) , where 1 hertz equals 1 wave passing a ﬁxed point in 1 second. Figure 1.4 shows high-frequency and low-frequency transverse waves. You can simulate transverse waves with different frequencies at this URL: http://zonalandeduc ation.com/mstm/physics/waves/partsOfAWave/waveParts.htm . 3 www.ck12.org FIGURE 1.4 A transverse wave with a higher fre- quency has crests that are closer to- gether. The frequency of a wave is the same as the frequency of the vibrations that caused the wave. For example, to generate a higher-frequency wave in a rope, you must move the rope up and down more quickly. This takes more energy, so a higher-frequency wave has more energy than a lower-frequency wave with the same amplitude. Wave Speed Assume that you move one end of a rope up and down just once. How long will take the wave to travel down the rope to the other end? This depends on the speed of the wave. Wave speed is how far the wave travels in a given amount of time, such as how many meters it travels per second. Wave speed is not the same thing as wave frequency, but it is related to frequency and also to wavelength. This equation shows how the three factors are related: Speed = Wavelength Frequency In this equation, wavelength is measured in meters and frequency is measured in hertz, or number of waves per second. Therefore, wave speed is given in meters per second. The equation for wave speed can be used to calculate the speed of a wave when both wavelength and wave frequency are known. Consider an ocean wave with a wavelength of 3 meters and a frequency of 1 hertz. The speed of the wave is: Speed = 3 m 1 wave/s = 3 m/s You Try It! Problem: Jera made a wave in a spring by pushing and pulling on one end. The wavelength is 0.1 m, and the wave frequency is 0.2 m/s. What is the speed of the wave? If you want more practice calculating wave speed from wavelength and frequency, try the problems at this URL: htt p://www.physicsclassroom.com/class/waves/u10l2e.cfm . The equation for wave speed (above) can be rewritten as: 4 www.ck12.org Concept 1. Measuring Waves Frequency = Speed Wavelength or Wavelength = Speed Frequency Therefore, if you know the speed of a wave and either the wavelength or wave frequency, you can calculate the missing value. For example, suppose that a wave is traveling at a speed of 2 meters per second and has a wavelength of 1 meter. Then the frequency of the wave is: Frequency = 2 m/s 1 m = 2 waves/s, or 2 Hz You Try It! Problem: A wave is traveling at a speed of 2 m/s and has a frequency of 2 Hz. What is its wavelength? The Medium Matters The speed of most waves depends on the medium through which they are traveling. Generally, waves travel fastest through solids and slowest through gases. That’s because particles are closest together in solids and farthest apart in gases. When particles are farther apart, it takes longer for the energy of the disturbance to pass from particle to particle. Lesson Summary • Wave amplitude is the maximum distance the particles of a medium move from their resting positions as a wave passes through. Wavelength is the distance between two corresponding points of adjacent waves. Waves with greater amplitudes or shorter wavelengths have more energy. • Wave frequency is the number of waves that pass a ﬁxed point in a given amount of time. Higher frequency waves have more energy. Wave speed is calculated as wavelength multiplied by wave frequency. Wave speed is affected by the medium through which a wave travels. Lesson Review Questions Recall 1. How is wave amplitude measured in a transverse wave? 2. Describe the wavelength of a longitudinal wave. 3. Deﬁne wave frequency. Apply Concepts 4. All of the waves in the sketch below have the same amplitude and speed. Which wave has the longest wavelength? Which has the highest frequency? Which has the greatest energy? 5 www.ck12.org 5. A wave has a wavelength of 0.5 m/s and a frequency of 2 Hz. What is its speed? Think Critically 6. Relate wave amplitude, wavelength, and wave frequency to wave energy. 7. Waves A and B have the same speed, but wave A has a shorter wavelength. Which wave has the higher frequency? Explain how you know. Points to Consider You read in this lesson that waves travel at different speeds in different media. • When a wave enters a new medium, it may speed up or slow down. What other properties of the wave do you think might change when it enters a new medium? • What if a wave reaches a type of matter it cannot pass through? Does it just stop moving? If not, where does it go? References 1. David Rydevik. http://commons.wikimedia.org/wiki/File:2004-tsunami.jpg . Public Domain 2. Christopher Auyeung. CK-12 Foundation . 3. Christopher Auyeung. CK-12 Foundation . 4. Christopher Auyeung. CK-12 Foundation . 6 www.ck12.org Concept 2. Mechanical Wave CONCEPT 2 • Describe mechanical waves. • Deﬁne the medium of a mechanical wave. • Identify three types of mechanical waves. Mechanical Wave No doubt you’ve seen this happen. Droplets of water fall into a body of water, and concentric circles spread out through the water around the droplets. The concentric circles are waves moving through the water. Waves in Matter The waves in the picture above are examples of mechanical waves. A mechanical wave is a disturbance in matter that transfers energy through the matter. A mechanical wave starts when matter is disturbed. A source of energy is needed to disturb matter and start a mechanical wave. Q: Where does the energy come from in the water wave pictured above? A: The energy comes from the falling droplets of water, which have kinetic energy because of their motion. The Medium The energy of a mechanical wave can travel only through matter. The matter through which the wave travels is called the medium ( plural , media). The medium in the water wave pictured above is water, a liquid. But the medium of a mechanical wave can be any state of matter, even a solid. Q: How do the particles of the medium move when a wave passes through them? A: The particles of the medium just vibrate in place. As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don’t actually travel along with the wave. Only the energy of the wave travels through the medium. 7 Types of Mechanical Waves There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move. You can see this in the Figure 2.1 and in the animation at the following URL. http://www. acs.psu.edu/drussell/Demos/waves/wavemotion.html www.ck12.org FIGURE 2.1 • In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. • In a longitudinal wave, particles of the medium vibrate back and forth parallel to the direction of the wave. • In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle. Q: How do you think surface waves are related to transverse and longitudinal waves? A: A surface wave is combination of a transverse wave and a longitudinal wave. Summary • A mechanical wave is a disturbance in matter that transfers energy through the matter. • The matter through which a mechanical wave travels is called the medium ( plural , media). • There are three types of mechanical waves: transverse, longitudinal, and surface waves. They differ in how particles of the medium move when the energy of the wave passes through. Vocabulary • mechanical wave : Disturbance in matter that transfers energy from one place to another. • medium (plural, media ): Matter through which a mechanical wave moves. Practice At the following URL, read the short introduction to waves and watch the animations. Then answer the questions below. http://www.acs.psu.edu/drussell/Demos/waves-intro/waves-intro.html 8 www.ck12.org Concept 2. Mechanical Wave 1. The article gives a dictionary deﬁnition of wave. What is the most important part of this deﬁnition? 2. What happens to particles of the medium when a wave passes? 3. How is “doing the wave” in a football stadium like a mechanical wave? Review 1. Deﬁne mechanical wave. 2. What is the medium of a mechanical wave? 3. List three types of mechanical waves. 4. If you shake one end of a rope up and down, a wave passes through the rope. Which type of wave is it? References 1. Zachary Wilson. . CC BY-NC 3.0 9 CONCEPT 3 www.ck12.org Transverse Waves • Describe transverse waves. • Explain how waves transfer energy without transferring matter. • Deﬁne wavelength, frequency, and period of a transverse wave. • State the relationship between speed, wavelength, and frequency. • Solve problems using the relationships between speed, wavelength, frequency, and period. Professional surfer Marcio Freire rides a giant wave at the legendary big wave surf break known as as "Jaws" during one the largest swells of the winter March 13, 2011 in Maui, HI. Massive waves, such as this one, transfer huge amounts of energy. Transverse Waves Types of Waves Water waves, sound waves, and the waves that travel along a rope are mechanical waves . Mechanical wa
ves require a material medium such as water, air, or rope. Light waves, however, are electromagnetic waves and travel without a material medium. They are not mechanical waves. In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. One type of mechanical wave is the transverse wave . In the case of transverse waves, the movement of the medium is perpendicular to the direction of the energy movement. 10 www.ck12.org Concept 3. Transverse Waves In the sketch above, consider the transverse wave produced when the boy jerks one end of a rope up and down while the other end is tied to a tree. The energy spent by the boy transfers permanently down the rope to the tree. The rope, however, only moves up and down. If we stuck a piece of tape somewhere on the rope, we would see that the particles of medium do not travel with the energy. After the wave has passed by, the piece of tape would still be in the same place it was before the wave approached. In all transverse waves, the movement media vibrates perpendicularly to the direction of wave motion, and the medium is not permanently moved from one place to another. Frequency, Wavelength, and Velocity Waves are identiﬁed by several characteristics. There is a center line where the medium would be if there were no wave, which is sometimes describes as the undisturbed position. The displacement of the medium above this undisturbed position is called a crest and the displacement below the undisturbed position is called a trough . The maximums of the crest and trough are equal and are called the amplitude . The distance between equivalent positions on succeeding waves is called the wavelength . The wavelength could be measured from a crest to the next crest or from a trough to the next trough, and is commonly represented with the Greek letter lambda, l . The time interval required for one complete wave to pass a point is called the period . During the period of the wave, an entire wavelength from one crest to the next crest passes a position. The number of waves that pass a single position in one second is called the frequency . The period of a wave and its frequency are reciprocals of each other. f = 1 T The units for the period are seconds and the units for frequency are s 1 or 1 name Hertz (Hz). s . This unit has also been given the 11 Another important characteristic of a wave is its velocity. The wave velocity is different from the velocity of the medium; the wave velocity is the velocity of the linearly transferred energy. Since the energy travels one wavelength, l , in one period, T , the velocity can be expressed as distance over time: www.ck12.org v = l T : Since period and frequency are reciprocals, the speed of the wave could also be expressed as v = l f . Example Problem: A sound wave has a frequency of 262 Hz. What is the time lapse between successive wave crests? Solution: The time lapse between successive crests would be the period and the period is the reciprocal of the frequency. T = 1 f = 1 262 s1 = 0:00382 s Example Problem: A sound wave has a frequency of 262 Hz has a wavelength of 1.29 m. What is the velocity of the wave? Solution: v = l f = (1:29 m)(262 s1) = 338 m/s Summary • Mechanical waves require a material medium such as water, air, or rope. • In all types of mechanical waves, energy moves from one place to another while the media carrying the wave only vibrates back and forth in position. • One type of mechanical wave is the transverse wave, in which the movement of the medium is perpendicular to the direction of the energy propagation. • The maximum displacement of the medium is the distance from the undisturbed position to the top of a crest, or the amplitude. • The distance along the line of motion of the wave from equivalent positions on succeeding waves is the wavelength. • The time interval required for one entire wave to pass a point is the period. • The number of periods per second is the wave’s frequency. • The period of a wave and its frequency are reciprocals of each other. • The velocity of the wave’s energy transfer is given by v = l f or v = l T : Practice The following video explains wave characteristics. Pause the video before each practice question and try to solve it yourself before moving on. http://www.youtube.com/watch?v=5ENLxaPiJJI 12 www.ck12.org Concept 3. Transverse Waves MEDIA Click image to the left for more content. 1. What is the distance between the base line and crest called? 2. What symbol is used for wavelength? 3. What is the relationship between period and frequency? Review 1. A sound wave produced by a chime 515 m away is heard 1.50 s later. (a) What is the speed of sound in air? (b) The sound wave has a frequency of 436 Hz. What is its period? (c) What is the wavelength of the sound? 2. A hiker shouts toward a vertical cliff 685 m away. The echo is heard 4.00 s later. (a) What is the speed of sound in air? (b) Why is this speed of sound slightly different from the previous answer? (c) The wavelength of the sound is 0.750 m. What is the frequency? (d) What is the period of the wave? 3. The speed of light in air is 3.00 10 8 m/s. If a light wave has a wavelength of 5.80 10 7 m, what is its frequency? • transverse wave: A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer • undisturbed position: The equilibrium or rest position of the medium in a wave. • crest: A crest is the point on a wave with the maximum value or upward displacement within a cycle • trough: A trough is the opposite of a crest, so the minimum or lowest point in a cycle • amplitude: The maximum displacement from a zero value during one period of an oscillation. • wavelength: The distance between corresponding points of two consecutive waves. • frequency: The number of waves that pass a ﬁxed point in unit time. • period: A period is the time required for one complete cycle of vibration to pass a given point. References 1. Image copyright EpicStockMedia, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. Kid: Image copyright Richcat, 2013; Tree: Image copyright AlexeyZet, 2013; Composite created by CK-12 . Used under licenses from Shutterstock.com Foundation - Samantha Bacic. http://www.shutterstock.com/ 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 13 CONCEPT 4 • Describe longitudinal waves. www.ck12.org Longitudinal Waves https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=6e9QZ8b6JzZZRM&tbnid=CAVrIJRLaBDfM:&ved=0CAUQjRw&url=http%3A%2F%2Fsirius.ucsc.edu%2Fdemoweb%2Fcgi-bin%2F%3Fwavesvisible-slinky&ei=kE4FUpSDN4HXygHR84HwDg&bvm=bv.50500085,d.b2I&psig=AFQjCNEJ23OS_x3Ga2tbC3Vi2VVUfbPmQ&ust=1376165898023972 Playing with a Slinky is a childhood tradition, but few children realize they are actually playing with physics. Longitudinal Waves Like transverse waves, longitudinal waves are mechanical waves, which means they transfer energy through a medium. Unlike transverse waves, longitudinal waves cause the particles of medium to move parallel to the direction of the wave. They are most common in springs, where they are caused by the pushing an pulling of the spring. Although the surface waves on water are transverse waves, ﬂuids (liquids, gases, and plasmas) usually transmit longitudinal waves. As shown in the image below, longitudinal waves are a series of compressions and rarefactions , or expansions. The wavelength of longitudinal waves is measured by the distance separating the densest compressions. The amplitude of longitudinal waves is the difference in media density between the undisturbed density to the highest density in a compression. Example Problem: A sonar signal (sonar is sound waves traveling through water) of 1:00 106 Hz frequency has a wavelength of 1.50 mm in water. What is the speed of sound in water? Solution: v = l f = (0:00150 m)(1:00 106 s1) = 1500 m/s Example Problem: A sound wave of wavelength 0.70 m and velocity 330 m/s is produced for 0.50 s. 14 www.ck12.org Concept 4. Longitudinal Waves a. What is the frequency of the wave? b. How many complete waves are emitted in this time interval? c. After 0.50 s, how far is the wave front from the source of the sound? Solution: f = v l = 330 m/s 0:70 m = 470 s1 a. b. complete waves = (470 cycles/s)(0:50 s) = 235 cycles c. distance = (330 m/s)(0:50 s) = 115 m Summary • Longitudinal waves cause the particles of medium to move parallel to the direction of the wave. Practice The following video explains how a tuning fork creates sound . Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=bomzzHC-59k MEDIA Click image to the left for more content. 1. In your own words, how are compressions and rarefactions produced by the tuning fork? 2. Make a guess why sound can easily travel around corners (Hint: think of its medium). Review 1. Some giant ocean waves have a wavelength of 25 m long, and travel at speeds of 6.5 m/s. Determine the frequency and period of such a wave. 2. Bats use sound echoes to navigate and hunt. They emit pulses of high frequency sound waves which reﬂect off obstacles in the surroundings. By detecting the time delay between the emission and return of a pulse, a bat can determine the location of the object. What is the time delay between the sending and return of a pulse from an object located 12.5 m away? The approximate speed of sound is 340 m/s. 3. Sachi is listening to her favorite radio station which broadcasts radio signals with a frequency of 1:023 108 Hz . If the speed of the signals in air is 2:997 108 m/s , what is the wavelength of these radio signals? 4. A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to move past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave. 5. A sona
r signal leaves a submarine, travels through the water to another submarine and reﬂects back to the original submarine in 4.00 s. If the frequency of the signal was 512 cycles per second and the wavelength of the signal was 2.93 m, how far away is the second submarine? • longitudinal wave: A wave in which the direction of media displacement is the same as the direction of wave propagation. 15 References 1. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 www.ck12.org 16 www.ck12.org Concept 5. Reﬂection of Mechanical Waves CONCEPT 5 Reﬂection of Mechanical Waves • State the law of reﬂection. • Solve problems using the law of reﬂection. • Given data about the media on either side of a barrier, determine whether the reﬂected wave will be upright or inverted. When mechanical waves strike a barrier, at least part of the energy of the waves will be reﬂected back into the media from which they came. You experience this every single day, when you look in the mirror and see your own reﬂection. Reﬂection of Mechanical Waves When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some portion of the wave is reﬂected back into the original medium. It reﬂects back at an equal angle that it came in. These angles are called the angle of incidence and the angle of reﬂection . The normal line, the incident and reﬂected rays, and the angles of incidence and reﬂection are all shown in the diagram sketched above. The law of reﬂection states that the angle of incidence equals the angle of reﬂection. These rules of reﬂection apply in the cases of water waves bouncing off the side of a pool, sound waves echoing off a distant cliff, or wave pulses traveling down a rope or a slinky. Consider the change that would occur with a light rope joined to a heavier rope. When a wave pulse travels down the rope and encounters the media change, a reﬂection will occur. Look at the image below. In the top sketch, we see a lightweight (black) rope attached to a heavier rope (red). There is a wave pulse traveling down the rope from left to right. When the wave pulse encounters the barrier (the change in rope weight), part of the wave moves into the new medium and part of the wave is reﬂected back into the old medium. As you can see in the bottom half of the diagram, the transmitted portion of the wave continues into the new medium right side up. The transmitted wave is somewhat diminished because some of the energy of the wave was reﬂected and also because the rope to be lifted is heavier. The reﬂected wave is also diminished because some of the energy was transmitted through the barrier. The reﬂected wave is also inverted (upside down). This is a general rule for mechanical waves passing from a less dense medium into a more dense medium, that is, the reﬂected wave will be inverted. 17 www.ck12.org The situation changes when the wave is passing from a more dense medium into a less dense medium. As you can see in the sketch below, when a wave pulse moving in denser medium encounters a media interface to a medium of less density, the reﬂected wave is upright rather than inverted. It is also possible for a mechanical wave to encounter an impenetrable barrier, that is, a barrier which does not allow In such a case, the complete wave pulse will be reﬂected and the reﬂected wave will be any transmission at all. inverted. Summary • When a wave strikes an obstacle or comes to the end of the medium it is traveling in, some part of the wave is reﬂected back into the original medium. • The law of reﬂection states that the angle of incidence equals the angle of reﬂection. • The general rule, for mechanical waves passing from a less dense medium into a more dense medium, the reﬂected wave will be inverted. • When a wave pulse moving in denser medium encounters a media interface to a medium of lesser density, the reﬂected wave is upright rather than inverted. • When a mechanical wave encounters an impenetrable barrier, the complete wave pulse will be reﬂected and the reﬂected wave will be inverted. Practice The following video shows a wave machine in action. Use this video to answer the questions that follow. http://www.youtube.com/watch?v=YQHbRw_hyz4 MEDIA Click image to the left for more content. 1. What happens to the wave when it is reﬂected from an open end? 2. What happens to the wave when it is reﬂected from a ﬁxed end? 18 www.ck12.org Review Concept 5. Reﬂection of Mechanical Waves 1. Draw a diagram showing a surface with a normal line. On the diagram, show a wave ray striking the surface with an angle of incidence of 60 . Draw the reﬂection ray on the diagram and label the angle of reﬂection. 2. Light strikes a mirror’s surface at 30 to the normal. What will the angle of reﬂection be? 3. If the angle between the incident ray and the reﬂected ray is 90 , what is the angle of incidence? 4. When a water wave is reﬂected from a concrete wall, will the reﬂected wave be inverted or upright? 5. If you tie a heavy spring to a light spring and send a wave pulse down the heavy spring, some of the wave will be reﬂected when the wave passes into the lighter spring. Will the reﬂected pulse be upright or inverted? • reﬂection: The change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. • law of reﬂection: Says the angle at which the wave is incident on the surface equals the angle at which it is reﬂected. • angle of incidence: The angle formed by a ray incident on a surface and a perpendicular to the surface at the point of incidence. • angle of reﬂection: The angle formed by a reﬂected ray and a perpendicular to the surface at the point of reﬂection. References 1. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 19 CONCEPT 6 Refraction of Mechanical Waves www.ck12.org • Deﬁne and describe refraction of mechanical waves. • State and use the law of refraction. A straw in a glass of water seen from the side often appears broken, even though it is not. The apparent break is due to the bending of light rays leaving the straw; as the light passes from the water to the glass and from the glass to the air, the light rays are bent. Nonetheless, your eye traces the light ray backward as if the light has followed a straight path from its origin at the straw. Since the light appears to have come from a different place, your eye sees the straw as being broken. Refraction of Mechanical Waves When any wave strikes a boundary between media, some of the energy is reﬂected and some is transmitted. When the wave strikes the media interface at an angle, the transmitted wave will move in a slightly different direction than the incident wave. This phenomenon is known as refraction . 20 www.ck12.org Concept 6. Refraction of Mechanical Waves Consider the image sketched above. Suppose that the waves represented here are water waves. The wave crests are represented by the black lines in the image. As such, the distance between two consecutive black lines is the wavelength. Let the red line represent a transition from deep to shallow water. This transition is called the media interface . As the waves hit the boundary, the waves slow down. The right side of the wave reaches the boundary before the left side of the wave, causing the left side to catch up and the angle of propagation to change slightly. This change in direction can be seen in the yellow line, which is slightly angled at the boundary. The refraction of waves across boundaries operates similarly to the method by which tanks are steered. Tanks do not have a steering wheel. Instead, they have an accelerator to produce forward motion and separate brakes on each tread. The operator uses brakes on both treads at the same time in order to stop, but brakes on only one tread to turn the tank. By braking one side, the operator causes that side to slow down or stop while the other side continues at the previous speed, causing the tank to turn towards the slower tread. This sketch shows a wave ray striking an interface between old medium and new medium. A normal line has been drawn as a dotted line perpendicular to the interface. The angle between the incident ray and the normal line is called the angle of incidence , shown as q i , and the angle between the refracted ray and the normal line is called the angle of refraction , q r . 21 www.ck12.org We already understand that the change in the wave direction at the border depends on the difference between the two velocities. This relationship is conveniently expressed in a mathematical relationship: sin qr sin qi = vr vi = lr li The ratio of the sine of the angle of refraction to the sine of the angle of incidence is the same as the ratio of the velocity of the wave in the new medium to the velocity of the wave in the old medium and equal to the ratio of wavelength (l) in the old medium to the wavelength in the new medium. Example Problem: A water wave with a wavelength of 3.00 m is traveling in deep water at 16.0 m/s. The wave strikes a sharp interface with shallow water with an angle of incidence of 53:0 . The wave refracts into the shallow water with an angle of refraction of 30:0 . What is the velocity of the wave in shallow water and what is its wavelength in the new medium? so sin 30 sin 53 = vr 16:0 m=s and vr = 10:0 m=s . Solution: = vr vi sin qr sin qi so 10:0 m=s 16:0 m=s = lr = lr li vr 3:00 m and lr = 1:88 m . vi Example Problem: The ratio of the sin qr to sin qi is 0.769 . 5:00 109 m , what is its wavelength in the original medium? If the wavelength of a wave in a new medium is Solution: Summary 0:769 = lr li so li = 5:00 109 m 0:769 = 6:50 109 m • When any wave strikes a boundary between media, some of the energy is reﬂected and some is transmitted. • When a wave strikes the media i
nterface at an angle, the transmitted wave will move in a different direction than the incident wave. This phenomenon is known as refraction. = vr vi • At any media interface, sin qr sin qi = lr li Practice http://www.youtube.com/watch?v=mH9VwivqjmE Follow up questions. 22 www.ck12.org Concept 6. Refraction of Mechanical Waves 1. What causes refraction? 2. What doesn’t change during refraction? Review 1. A laser beam passes through water and enters a glass block at an angle. The ratio of the speed of the wave in glass to the speed in water is 0.866 . If the angle of incidence to the interface is 60 , what is the angle of refraction? 2. A ray of light is traveling from air into glass at an angle of 30:0 to the normal line. The speed of the light in air is 3:00 108 m=s and in glass the speed drops to 2:00 108 m=s . What is the angle of refraction? 3. Which of the following change when a water wave moves across a boundary at an angle between deep water and shallow water? a. frequency b. wavelength c. speed d. wave direction e. period 4. Which of the following change when a water wave moves across a boundary exactly along the media interface between deep water and shallow water? a. frequency b. wavelength c. speed d. wave direction e. period 5. The speed of sound is 340 m/s. A particular sound wave has a frequency of 320. Hz. a. What is the wavelength of this sound in air? b. If this sound refracts into water where the speed of sound is 4 times faster, what will be the new wavelength? c. What will be the new frequency? 6. When a light ray passes from air into diamond, the angle of incidence is 45:0 and the angle of refraction is 16:7 . If the speed of light in air is 3:00 108 m=s , what is the speed of light in diamond? • refraction: The turning or bending of a wave direction when it passes from one medium to another of different density. References 1. Image copyright cheyennezj, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. Greg Goebel. http://www.public-domain-image.com/full-image/transportation-vehicles-public-domain-ima ges-pictures/tanks-public-domain-images-pictures/m7-priest-self-propelled-105-millimeter-howitzer-tank.jpgcopyright-friendly-image.html . Public Domain 4. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 23 CONCEPT 7 www.ck12.org Wave Interactions • Identify ways that waves can interact with matter. • Deﬁne and give examples of wave reﬂection, refraction, and diffraction. Did you ever hear an echo of your own voice? An echo occurs when sound waves bounce back from a surface that they can’t pass through. The woman pictured here is trying to create an echo by shouting toward a large building. When the sound waves strike the wall of the building, most of them bounce back toward the woman, and she hears an echo of her voice. An echo is just one example of how waves interact with matter. How Waves Interact with Matter Waves interact with matter in several ways. The interactions occur when waves pass from one medium to another. The types of interactions are reﬂection, refraction, and diffraction. Each type of interaction is described in detail below. You can see animations of the three types at this URL: http://www.acoustics.salford.ac.uk/schools/teacher/ lesson3/ﬂash/whiteboardcomplete.swf Reﬂection An echo is an example of wave reﬂection. Reﬂection occurs when waves bounce back from a surface they cannot pass through. Reﬂection can happen with any type of waves, not just sound waves. For example, light waves can 24 www.ck12.org Concept 7. Wave Interactions also be reﬂected. In fact, that’s how we see most objects. Light from a light source, such as the sun or a light bulb, shines on the object and some of the light is reﬂected. When the reﬂected light enters our eyes, we can see the object. Reﬂected waves have the same speed and frequency as the original waves before they were reﬂected. However, the direction of the reﬂected waves is different. When waves strike an obstacle head on, the reﬂected waves bounce straight back in the direction they came from. When waves strike an obstacle at any other angle, they bounce back at the same angle but in a different direction. This is illustrated in diagram below. In this diagram, waves strike a wall at an angle, called the angle of incidence. The waves are reﬂected at the same angle, called the angle of reﬂection, but in a different direction. Notice that both angles are measured relative to a line that is perpendicular to the wall. FIGURE 7.1 Refraction Refraction is another way that waves interact with matter. Refraction occurs when waves bend as they enter a new medium at an angle. You can see an example of refraction in the picture below. Light bends when it passes from air to water or from water to air. The bending of the light traveling from the ﬁsh to the man’s eyes causes the ﬁsh to appear to be in a different place from where it actually is. FIGURE 7.2 Waves bend as they enter a new medium because they start traveling at a different speed in the new medium. For 25 example, light travels more slowly in water than in air. This causes it to refract when it passes from air to water or from water to air. Q: Where would the ﬁsh appear to be if the man looked down at it from straight above its actual location? A: The ﬁsh would appear to be where it actually is because refraction occurs only when waves (in this case light waves from the ﬁsh) enter a new medium at an angle other than 90°. www.ck12.org Diffraction Did you ever notice that you can hear sounds around the corners of buildings even though you can’t see around them? The Figure 7.3 shows why this happens. As you can see from the ﬁgure, sound waves spread out and travel around obstacles. This is called diffraction . It also occurs when waves pass through an opening in an obstacle. All waves may be diffracted, but it is more pronounced in some types of waves than others. For example, sound waves bend around corners much more than light does. That’s why you can hear but not see around corners. FIGURE 7.3 For a given type of waves, such as sound waves, how much the waves diffract depends on the size of the obstacle (or opening in the obstacle) and the wavelength of the waves. The Figure 7.4 shows how the amount of diffraction is affected by the size of the opening in a barrier. Note that the wavelength of the wave is the distance between the vertical lines. FIGURE 7.4 26 www.ck12.org Summary Concept 7. Wave Interactions • Three ways that waves may interact with matter are reﬂection, refraction, and diffraction. • Reﬂection occurs when waves bounce back from a surface that they cannot pass through. • Refraction occurs when waves bend as they enter a new medium at an angle and start traveling at a different speed. • Diffraction occurs when waves spread out as they travel around obstacles or through openings in obstacles. Vocabulary • diffraction : Bending of a wave around an obstacle or through an opening in an obstacle. • reﬂection : Bouncing back of waves from a barrier they cannot pass through. • refraction : Bending of waves as they enter a new medium at an angle and change speed. Practice Make a crossword puzzle of terms relating to wave interactions. Include at least seven different terms. You can use the puzzle maker at the following URL. Then exchange and solve puzzles with a classmate. http://puzzlemaker.disc overyeducation.com/CrissCrossSetupForm.asp Review 1. What is reﬂection? What happens if waves strike a reﬂective surface at an angle other than 90°? 2. Deﬁne refraction. Why does refraction occur? 3. When does diffraction occur? How is wavelength related to diffraction? References 1. Zachary Wilson. . CC BY-NC 3.0 2. Zachary Wilson. . CC BY-NC 3.0 3. Student: Flickr:MaxTorrt; Radio: Flickr:Kansir. . CC BY 2.0 4. Zachary Wilson. . CC BY-NC 3.0 27 CONCEPT 8 www.ck12.org Wave Interference • Deﬁne wave interference. • Compare and contrast constructive and destructive interference. • Explain how standing waves occur. When raindrops fall into still water, they create tiny waves that spread out in all directions away from the drops. What happens when the waves from two different raindrops meet? They interfere with each other. When Waves Meet When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference . Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude. Amplitude is the maximum distance the particles of the medium move from their resting positions when a wave passes through. How amplitude is affected by wave interference depends on the type of interference. Interference can be constructive or destructive. Constructive Interference Constructive interference occurs when the crests, or highest points, of one wave overlap the crests of the other wave. You can see this in the Figure 8.1 . As the waves pass through each other, the crests combine to produce a wave with greater amplitude. You can see an animation of constructive interference at this URL: http://phys23p.sl.psu.e du/phys_anim/waves/embederQ1.20100.html Destructive Interference Destructive interference occurs when the crests of one wave overlap the troughs, or lowest points, of another wave. The Figure 8.2 shows what happens. As the waves pass through each other, the crests and troughs cancel each other out to produce a wave with zero amplitude. You can see an animation of destructive interference at this URL: htt p://phys23p.sl.psu.edu/phys_anim/waves/embederQ1.20200.html 28 www.ck12.org Concept 8. Wave Interference FIGURE 8.1 Standing Waves Waves may reﬂect off an obstacle that they are unable to pass through. When waves are reﬂected straight back from an obstacle, the reﬂected waves interfere with the original waves and create standing waves . These are waves that
appear to be standing still. Standing waves occur because of a combination of constructive and destructive interference. You can see animations of standing waves at the URLs below. http://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/ http://www.physicsc lassroom.com/mmedia/waves/swf.cfm Q : How could you use a rope to produce standing waves? A : You could tie one end of the rope to a ﬁxed object, such as doorknob, and move the other end up and down to generate waves in the rope. When the waves reach the ﬁxed object, they are reﬂected back. The original waves and the reﬂected waves interfere to produce a standing wave. Try it yourself and see if the waves appear to stand still. Summary • Wave interference is the interaction of waves with other waves. • Constructive interference occurs when the crests of one wave overlap the crests of the other wave, causing an increase in wave amplitude. • Destructive interference occurs when the crests of one wave overlap the troughs of the other wave, causing a decrease in wave amplitude. 29 www.ck12.org FIGURE 8.2 • When waves are reﬂected straight back from an obstacle, the reﬂected waves interfere with the original waves and create standing waves. Vocabulary • standing wave : Wave appearing to stand still that forms when a wave and its reﬂected wave interfere. • wave interference : Interaction of waves with other waves. Practice Review wave interference at the following URL. Then do the Check Your Understanding problem at the bottom of the Web page. Be sure to check your answers. http://www.physicsclassroom.com/class/waves/u10l3c.cfm Review 1. What is wave interference? 2. Create a table comparing and contrasting constructive and destructive interference. 30 www.ck12.org Concept 8. Wave Interference 3. What are standing waves? How do they form? References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Christopher Auyeung. . CC BY-NC 3.0 31 www.ck12.org Wave Speed CONCEPT 9 Objective The student will: • Solve problems involving wavelength, wave speed, and frequency. Vocabulary • wave equation: Relates wavelength and wave speed. Distance is speed times time, x = vt , so wavelength is wave speed times period l = vT . • wave speed: How quickly the peak of each wave is moving forward. The Wave Equation Simple harmonic motion, moving back and forth in place, has amplitude along with period and frequency. Wave motion means that the back-and-forth change is also moving through space. This means that there are two further qualities of a wave. • The wavelength is the distance between two compressions in the direction of motion of the wave, and is represented by the Greek letter lambda, l . For an ocean wave, it would be the distance in meters from the top of one wave to the next. • The wave speed is how quickly the peak of each wave is moving forward, and is represented by v (for velocity, although for our current purpose the direction is not important). These two are related by the period T of the wave. The period is the time it takes for a wave to complete one cycle, which is the time it takes for a peak to move forward one wavelength. The wave equation expresses this. Distance is speed times time, x = vt , so wavelength is wave speed times period l = vT . This can alternately be expressed in terms of frequency. Suppose there are three waves every second. This is frequency f = 3:0 Hz , equivalent to period T = 1 3 s . This means that during one second, three waves come out from the source. Since each peak is one wavelength, l , ahead of the other, this means that during that one second, the lead wave has gone ahead three wavelengths. The distance the wave goes in one second is the wave speed. So, the wave speed is equal to the frequency times the wavelength, v = l f . Because f = 1 l = vT ! v = l T , these are mathematically the same: T = l 1 T = l f The wave equation says distance wavelength is equal to wave speed multiplied by period T . http://demonstrations.wolfram.com/SpeedOfSound/ 32 www.ck12.org Concept 9. Wave Speed FIGURE 9.1 Illustrative Example 1 a. The ripple tank arm in Figure 9.1 has a period of 0.25 s and the length of the arrow in the ﬁgure is 3.76 cm. What is the wavelength of the water waves if the picture is to scale? Answer: There are four wavelengths from the start to the end of the arrow. Wavelength is: l = 3:76cm 4 = 0:94cm b. What is the velocity of the wave? Answer: Since the period is T = 0:25 s , the frequency is , velocity is v = l f = (0:94 cm)(4:0 Hz) = 3:76 cm s . Check Your Understanding f = 1 T = 1 0:25 s = 4:0 Hz or four cycles per second. So the 1. A sound wave travels at the speed of 343:0 m/s through the air at room temperature, 20:0C(68:0F) . If the frequency of the sound is 261.6 Hz (a middle-C note), what is the wavelength of the note? Answer = 343:0 m/s 261:6 Hz = 1:311 m 2. X-rays are electromagnetic waves. A particular type of x-ray has a frequency of 3:0 1017 Hz and a wavelength of 1:0 109 m . What is the velocity of this type of x-ray? = 3:0 108 m/s Answer: v = l f = 1:0 109 m)(3:0 1017 1 s This result is true for any electromagnetic radiation traveling in a vacuum, including visible light. 3. Compared to the speed of sound in air at a temperature of 20C , how many times faster is the speed of light through the air at the same temperature? Answer: We’ll assume that the velocity of light vL = 3:0 108 m s is approximately the same through the air as it is 33 www.ck12.org through a vacuum. Number 1 above gave the velocity of sound for a temperature of 20C as vs = 343:0 m s . Answer: VL = 874; 635:57 ! 8:7 105 . The velocity of light is indeed a good deal greater than the velocity of Vs sound in air. In fact, the velocity of light in vacuum is the greatest velocity that exists. No material object can travel at this velocity! We will discuss these ideas in Chapter 23 (The Special Theory of Relativity). 4. At the moment a lightning ﬂash is seen, a person begins to count off seconds. If he hears the thunder after seven seconds, approximately how far away from the person did the lightning ﬂash originate? Answer : We’ll assume that the sound travels at a velocity of 343 m/s. In seven seconds the sound has traveled x = vt ! x = 343 m s (7s) = 2401 ! 2400 m . 5. Sound travels about 1,500 m/s in water. A destroyer ship locates an enemy submarine using a sonar signal which takes a 4.3 s to travel to, reﬂect, and return to the destroyer. How far is the sub from the destroyer? Answer : The time for the signal to reach the sub is half of the total time of travel, ! 4:3 s x = vt ! x = 1; 500 m s (2:15 s) = 3225 ! 3200 m or 3.2 km. 2 = 2:15 ! 2:2 s . Using References 1. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 34 www.ck12.org Concept 10. Sound Waves CONCEPT 10 • Deﬁne sound. • Describe sound waves and how they are generated. • Identify media through which sound waves can travel. Sound Waves Crack! Crash! Thud! That’s what you’d hear if you were in the forest when this old tree cracked and came crashing down to the ground. But what if there was nobody there to hear the tree fall? Would it still make these sounds? This is an old riddle. To answer the riddle correctly, you need to know the scientiﬁc deﬁnition of sound. Deﬁning Sound In science, sound is deﬁned as the transfer of energy from a vibrating object in waves that travel through matter. Most people commonly use the term sound to mean what they hear when sound waves enter their ears. The tree above generated sound waves when it fell to the ground, so it made sound according to the scientiﬁc deﬁnition. But the sound wasn’t detected by a person’s ears if there was nobody in the forest. So the answer to the riddle is both yes and no! How Sound Waves Begin All sound waves begin with vibrating matter. Look at the ﬁrst guitar string on the left in the Figure 10.1 . Plucking the string makes it vibrate. The diagram below the ﬁgure shows the wave generated by the vibrating string. The moving string repeatedly pushes against the air particles next to it, which causes the air particles to vibrate. The vibrations spread through the air in all directions away from the guitar string as longitudinal waves. In longitudinal waves, particles of the medium vibrate back and forth parallel to the direction that the waves travel. You can see an animation of sound waves traveling through air at this URL: http://www.mediacollege.com/audio/01/sound-wave s.html Q: If there were no air particles to carry the vibrations away from the guitar string, how would sound reach the ear? A: It wouldn’t unless the vibrations were carried by another medium. Sound waves are mechanical waves, so they can travel only though matter and not through empty space. 35 www.ck12.org FIGURE 10.1 A Ticking Clock The fact that sound cannot travel through empty space was ﬁrst demonstrated in the 1600s by a scientist named Robert Boyle. Boyle placed a ticking clock in a sealed glass jar. The clock could be heard ticking through the air and glass of the jar. Then Boyle pumped the air out of the jar. The clock was still ticking, but the ticking sound could no longer be heard. That’s because the sound couldn’t travel away from the clock without air particles to pass the sound energy along. You can see an online demonstration of the same experiment—with a modern twist—at this URL: http://www.youtube.com/watch?v=b0JQt4u6-XI MEDIA Click image to the left for more content. Sound Waves and Matter Most of the sounds we hear reach our ears through the air, but sounds can also travel through liquids and solids. If you swim underwater—or even submerge your ears in bathwater—any sounds you hear have traveled to your ears through the water. Some solids, including glass and metals, are very good at transmitting sounds. Foam rubber and heavy fabrics, on the other hand, tend to mufﬂe sounds. They absorb rather than pass on the sound energy. Q: How can you tell that sounds travel through solids? A: One way is that you can hea
r loud outdoor sounds such as sirens through closed windows and doors. You can also hear sounds through the inside walls of a house. For example, if you put your ear against a wall, you may be able to eavesdrop on a conversation in the next room—not that you would, of course. 36 www.ck12.org Summary Concept 10. Sound Waves • In science, sound is deﬁned as the transfer of energy from a vibrating object in waves that travel through matter. • All sound waves begin with vibrating matter. The vibrations generate longitudinal waves that travel through matter in all directions. • Most sounds we hear travel through air, but sounds can also travel through liquids and solids. Vocabulary • sound : Transfer of energy from a vibrating object in longitudinal waves that travel through matter. Practice Watch the video “How Sound Waves Travel” at the following URL. Then explain how sound waves begin and how they travel, using the human voice as an example. http://www.youtube.com/watch?v=_vYYqRVi8vY MEDIA Click image to the left for more content. Review 1. How is sound deﬁned in science? How does this deﬁnition differ from the common meaning of the word? 2. Hitting a drum, as shown in the Figure 10.2 , generates sound waves. Create a diagram to show how the sound waves begin and how they reach a person’s ears. FIGURE 10.2 3. How do you think earplugs work? 37 References 1. Guitar string photo by Flickr:jar(); illustration by Christopher Auyeung (CK-12 Foundation). . CC BY 2.0 2. S.L. Ratigan. . CC BY 2.0 www.ck12.org 38 www.ck12.org Concept 11. Frequency and Pitch of Sound CONCEPT 11 Frequency and Pitch of Sound • Deﬁne the pitch of sound. • Relate the pitch of sound to the frequency of sound waves. • Identify infrasound and ultrasound. A marching band passes you as it parades down the street. You heard it coming from several blocks away. Now that the different instruments have ﬁnally reached you, their distinctive sounds can be heard. The tiny piccolos trill their bird-like high notes, and the big tubas rumble out their booming bass notes. Clearly, some sounds are higher or lower than others. High or Low How high or low a sound seems to a listener is its pitch . Pitch, in turn, depends on the frequency of sound waves. Wave frequency is the number of waves that pass a ﬁxed point in a given amount of time. High-pitched sounds, like the sounds of the piccolo in the Figure 11.1 , have high-frequency waves. Low-pitched sounds, like the sounds of the tuba Figure 11.1 , have low-frequency waves. For a video demonstration of frequency and pitch, go to this URL: http://www.youtube.com/watch?v=irqfGYD2UKw Can You Hear It? The frequency of sound waves is measured in hertz (Hz), or the number of waves that pass a ﬁxed point in a second. Human beings can normally hear sounds with a frequency between about 20 Hz and 20,000 Hz. Sounds with frequencies below 20 hertz are called infrasound . Infrasound is too low-pitched for humans to hear. Sounds with frequencies above 20,000 hertz are called ultrasound . Ultrasound is too high-pitched for humans to hear. 39 www.ck12.org FIGURE 11.1 Some other animals can hear sounds in the ultrasound range. For example, dogs can hear sounds with frequencies as high as 50,000 Hz. You may have seen special whistles that dogs—but not people—can hear. The whistles produce sounds with frequencies too high for the human ear to detect. Other animals can hear even higher-frequency sounds. Bats, like the one pictured in the Figure 11.2 , can hear sounds with frequencies higher than 100,000 Hz! FIGURE 11.2 Q: Bats use ultrasound to navigate in the dark. Can you explain how? A: Bats send out ultrasound waves, which reﬂect back from objects ahead of them. They sense the reﬂected sound waves and use the information to detect objects they can’t see in the dark. This is how they avoid ﬂying into walls and trees and also how they ﬁnd ﬂying insects to eat. Summary • How high or low a sound seems to a listener is its pitch. Pitch, in turn, depends on the frequency of sound waves. • High-frequency sound waves produce high-pitched sounds, and low-frequency sound waves produce low- pitched sounds. 40 www.ck12.org Concept 11. Frequency and Pitch of Sound • Infrasound has wave frequencies too low for humans to hear. Ultrasound has wave frequencies too high for humans to hear. Vocabulary • infrasound : Sound with a frequency below the range of human hearing (less than 20 hertz). • pitch : How high or low a sound seems to a listener. • ultrasound : Sound with a frequency above the range of human hearing (greater than 20,000 hertz). • wave frequency : Number of waves that pass a ﬁxed point in a given amount of time. Practice At the following URL, complete the interactive module to review and test your knowledge of the frequency and pitch of sound. http://www.engineeringinteract.org/resources/oceanodyssey/flash/concepts/pitch.htm Review 1. What is the pitch of sound? 2. How is the pitch of sound related to the frequency of sound waves? 3. Deﬁne infrasound and ultrasound. References 1. Piccolo: U.S. Navy photo by Chief Mass Communication Specialist David Rush; Tuba: Bob Fishbeck. . Public Domain 2. . . Public Domain 41 CONCEPT 12 • Give the speed of sound in dry air at 20 °C. • Describe variation in the speed of sound in different media. • Explain the effect of temperature on the speed of sound. www.ck12.org Speed of Sound Has this ever happened to you? You see a ﬂash of lightning on the horizon, but several seconds pass before you hear the rumble of thunder. The reason? The speed of light is much faster than the speed of sound. What Is the Speed of Sound? The speed of sound is the distance that sound waves travel in a given amount of time. You’ll often see the speed of sound given as 343 meters per second. But that’s just the speed of sound under a certain set of conditions, speciﬁcally, through dry air at 20 °C. The speed of sound may be very different through other matter or at other temperatures. Speed of Sound in Different Media Sound waves are mechanical waves, and mechanical waves can only travel through matter. The matter through which the waves travel is called the medium (plural, media). The Table 12.1 gives the speed of sound in several different media. Generally, sound waves travel most quickly through solids, followed by liquids, and then by gases. Particles of matter are closest together in solids and farthest apart in gases. When particles are closer together, they can more quickly pass the energy of vibrations to nearby particles. You can explore the speed of sound in different media at this URL: http://www.ltscotland.org.uk/resources/s/sound/speedofsound.asp?strReferringChannel=resources&strReferringPageI D=tcm:4-248291-64 42 www.ck12.org Concept 12. Speed of Sound Medium (20 °C) Dry Air Water Wood Glass Aluminum TABLE 12.1: speed of sound Speed of Sound Waves (m/s) 343 1437 3850 4540 6320 Q: The table gives the speed of sound in dry air. Do you think that sound travels more or less quickly through air that contains water vapor? (Hint: Compare the speed of sound in water and air in the table.) A: Sound travels at a higher speed through water than air, so it travels more quickly through air that contains water vapor than it does through dry air. Temperature and Speed of Sound The speed of sound also depends on the temperature of the medium. For a given medium, sound has a slower speed at lower temperatures. You can compare the speed of sound in dry air at different temperatures in the following Table 12.2 . At a lower temperature, particles of the medium are moving more slowly, so it takes them longer to transfer the energy of the sound waves. Temperature of Air 0 °C 20 °C 100 °C TABLE 12.2: speed of sound Speed of Sound Waves (m/s) 331 343 386 Q: What do you think the speed of sound might be in dry air at a temperature of -20 °C? A: For each 1 degree Celsius that temperature decreases, the speed of sound decreases by 0.6 m/s. So sound travels through dry, -20 °C air at a speed of 319 m/s. Summary • The speed of sound is the distance that sound waves travel in a given amount of time. The speed of sound in dry air at 20 °C is 343 meters per second. • Generally, sound waves travel most quickly through solids, followed by liquids, and then by gases. • For a given medium, sound waves travel more slowly at lower temperatures. Vocabulary • speed of sound : Speed at which sound waves travel, which is 343 m/s in dry air at 20 °C. Practice At the following URL, read about the speed of sound in different materials. Be sure to play the animation. Then answer the questions below. http://www.ndt-ed.org/EducationResources/HighSchool/Sound/speedinmaterials.htm 43 1. Describe what you hear when you play the animation. Explain your observations. 2. Name two properties of materials that affect the speed of sound waves. How do they affect the speed of sound? 3. Explain why sound waves moves more quickly through warmer air than cooler air. www.ck12.org Review 1. What is the speed of sound in dry air at 20 °C? 2. Describe variation in the speed of sound through various media. 3. Explain how temperature affects the speed of sound. 44 www.ck12.org Concept 13. Resonance with Sound Waves CONCEPT 13 Resonance with Sound Waves Objectives The student will: • Understand the conditions for resonance. • Solve problems with strings and pipes using the condition for resonance. Vocabulary • beat frequency • natural frequency: The frequency at which a system vibrates normally when given energy without outside interference. • resonance: Timing force to be the same as natural frequency. • sympathetic vibrations Introduction Many systems have a tendency to vibrate. When the forced vibration frequency is the same as the natural frequency, the amplitude of vibration can increase tremendously. A well-known example of this situation is pushing a person on a swing, Figure below . We know from study of simple pendulums that without being pushed, the person in th
e swing rocks back and forth with a frequency that depends on gravity and the length of the chain. f = 1 2p r g L This is one example of a natural frequency – the frequency at which a system vibrates normally when given energy without outside interference. Pushing on the person in the swing will affect the amplitude of the swinging. This is called forced vibration – when a periodic force from one object (the person pushing) affects the vibration of another object (the person swinging). To get the most effect, the person pushing will start just at the very back of the swing. In other words, the frequency of how often they push is exactly the same as the frequency of the swing. Suppose they do not push at the right time, but instead push at some other frequency. That would mean that sometimes they are pushing forward when the swing is still going backward. In that case, the swing would slow down – i.e. the amplitude of the swing will be reduced. Timing the pushes to be the same as the natural frequency is called resonance . For this reason, the natural frequency is also known as the resonant frequency. If the pushes are timed just right, then even if each individual push is small, the vibration will get larger with each push. 45 www.ck12.org FIGURE 13.1 A classic example of an unfortunate consequence of a forced vibration at resonant frequency is what happened to the Tacoma Narrows Bridge, in 1940. See the link below. http://www.youtube.com/watch?v=3mclp9QmCGs In Figure below , the bridge is beginning to resonate, in part, due to the frequency of vibration of the wind gusts. In Figure below , we see that the bridge is no longer able to respond elastically to the tremendous amplitude of vibration from the forced vibration of wind energy (at its resonant frequency), and it is torn apart. FIGURE 13.2 Modern bridges are built to avoid this effect, but through history there are a number of documented situations where a forced vibration at resonance had dire results. The Broughton Suspension Bridge (1831) and the Angers Bridge (1850) are two examples of bridges believed to have collapsed due to the effect of soldiers marching at a regular pace that caused resonance. The Albert Bridge in West London, England has been nicknamed The Trembling Lady because it has been set into resonance so often by marching soldiers. Though soldiers no longer march across the bridge, there still remains a sign of concern as shown in Figure below . 46 www.ck12.org Concept 13. Resonance with Sound Waves FIGURE 13.3 FIGURE 13.4 Sympathetic Vibrations There is a typical classroom physics demonstration where one tuning fork is set into motion and an identical tuning fork, if placed closed enough, will also be set vibrating, though with smaller amplitude. The same effect occurs when tuning a guitar. One string is plucked and another, whose length is shortened by holding it down some distance from the neck of the guitar, will also be set into vibration. When this condition is met, both strings are vibrating with the same frequency. We call this phenomenon sympathetic vibration . 47 www.ck12.org MEDIA Click image to the left for more content. FIGURE 13.5 https://www.youtube.com/watch?v=tnS0SYF4pYE In Figure above , a set of pendulums are ﬁxed to a horizontal bar that can be easily jostled. Pendulums A and E have the same length. If one of them is set swinging, the horizontal bar will be forced into moving with a period equal to that of the pendulum, which, in turn, will cause the other pendulum of the same length to begin swinging. Any pendulum that is close in length to pendulums A and E , for example, pendulum D , will also begin to swing. Pendulum D will swing with smaller amplitude than pendulums A and E since its resonant frequency is not quite the same as pendulums A and E . Pendulums with lengths dramatically different from pendulums A and E will hardly move at all. You can try a similar demonstration out yourself with the following simulation: http://phet.colorado.edu/en/simulation/resonance Resonance is a very common phenomenon, especially with sound. The length of any instrument is related to what note it plays. If you blow into the top of a bottle, for example, the note will vary depending on the height of air in the bottle. This plays an important role in human voice generation. The length of the human vocal tube is between 17 cm and 18 cm. The typical frequencies of human speech are in the range of 100 Hz to 5000 Hz. FIGURE 13.6 48 www.ck12.org Concept 13. Resonance with Sound Waves By using the muscles in their throat, singers change the note they sing. A dramatic example of this is breaking glass with the human voice. By singing at exactly the resonant frequency of a delicate wine glass, the glass will resonate with the note and shatter. http://www.youtube.com/watch?v=7YmuOD5X4L8 The resonance of sound is also a mechanical analogue to how a radio set receives a signal. The Figure below shows one of the earliest radio designs, called a crystal radio because the element which detected the radio waves was a crystalline mineral such as galena. FIGURE 13.7 An old crystal radio set. In modern times, the air is ﬁlled with all manner of radio waves. In order to listen to your favorite radio station, you must tune your radio to resonate with only the frequency of the radio station. When you hear the tuning number of a radio station, such as “101.3 FM”, that is the measure in megahertz, 101:3 MHz = 1:013 108 Hz . The coiled wire (called an inductor) and the capacitor in Figure above act together to tune in a speciﬁc radio station . Effectively, the capacitor and inductor act analogously to a pendulum of a speciﬁc length that will only respond to vibrations of another pendulum of the same length. So when tuned, only a speciﬁc radio frequency will cause resonance in the radio antenna. Strings Fixed at Both Ends A case of natural frequency that you can observe plainly is when you pluck a string or stretched rubber band. Normally, the string will vibrate at a single widest point in the middle. This is called the fundamental or ﬁrst harmonic resonance of the string. This is the same as the natural frequency of a simple pendulum or mass on a spring. Because it vibrates all along its length, though, the string also lets us see further patterns of resonance. By vibrating the end of the string rather than just plucking it, we can force vibration at frequencies other than the ﬁrst harmonic. When the string is set into vibration, energy will travel down the string and reﬂect back toward the end where the waves are being generated. This steady pattern of vibration is called a standing wave . The points where the reﬂecting waves interfere destructively with the “generated’ waves are called nodes. The points where the reﬂecting waves interfere constructively with the generated waves are called anti-nodes. Figure below shows a string ﬁxed at both ends vibrating in its fundamental mode. There are two nodes shown and one antinode. The dashed segment represents the reﬂected wave. If you compare the wave shape of the ﬁrst harmonic to the wave shape of Figure above , it will be apparent that the ﬁrst harmonic contains one-half of a wavelength, l . Therefore L , the length of the unstretched string, is equal to one-half the wavelength, which is 1 2 l1 = L ! l1 = 2L: 49 www.ck12.org FIGURE 13.8 The second harmonic contains an entire wavelength 2 2 l2 = L ! l2 = L as shown in Figure below . FIGURE 13.9 And the third harmonic contains one and one-half wavelengths 3 2 l3 = L ! l3 = 2 3 L . FIGURE 13.10 If the pattern continues then the fourth harmonic will have a wavelength of 4 pressions for the length of the string in terms of the wavelength, a simple pattern emerges: L = 1 . We can express the condition for standing waves (and of resonance, as well) as L = n the length of string and n = 1; 2; 3 : : : . 2 l4 = L ! l4 = 1 2 L . Looking at the ex2 l2; 3 2 l3; 4 2 l4 : : : n L , where L is 2 l1; 2 2 ln or ln = 2 Check Your Understanding 1. How many nodes and anti-nodes are shown in Figure above .? Answer: There are three nodes and two anti-nodes. 2. If the length of the unstretched string is 20 cm, what is the wavelength for the 10th harmonic? Answer: ln = 2 5 (20 cm) = 4 cm n L ! l10 = 2 10 L = 1 Strings Fixed at One End and Opened at One End A string ﬁxed at one only end displays a different standing wave pattern. In this case, the unbounded end of a string of length L is an antinode. The fundamental mode (the ﬁrst harmonic) for the length L of string contains only one- 50 www.ck12.org Concept 13. Resonance with Sound Waves fourth of a wavelength as shown in Figure below below. Therefore L, the length of the unstretched string, is equal to one-quarter the wavelength, which is 1 4 l1 = L ! l1 = 4L . FIGURE 13.11 The second harmonic contains three-quarters of a wavelength 3 4 l2 = L ! l2 = 4 3 L as shown in Figure above FIGURE 13.12 The third harmonic contains ﬁve-fourths of a wavelength 5 4 l2 = L ! l2 = 4 5 L as shown in Figure above The third harmonic contains ﬁve-fourths wavelengths as shown in Figure below . If the pattern continues, then the fourth harmonic will have a wavelength of 7 expressions for the length of the string in terms of the wavelength, a simple pattern emerges 1 We can express the condition for resonance as L = n . 7 L . Looking at the 4 l3; 7 4 l4 : : : . n L , where L is the length of string and n = 1; 3; 5 : : : 4 l4 = L ! l4 = 4 4 ln or ln = 4 4 l2; 5 4 l1; 3 As long as the tension in the string remains ﬁxed, the velocity of the wave along the string remains constant. Does it seem reasonable that a sagging string will not support the same wave velocity as a taut string? Since v = l f product l f is constant as long as the wave velocity remains constant. Therefore, for a string vibrating in many different modes, we have v = l1 f1 = l2 f2 = l3 f3 : : : . 51 www.ck12.org FIGURE 13.13 Illustrative Example 1 1a. If the frequency of the ﬁrst har
monic for a string ﬁxed at both ends is f1 , determine the frequency for successive harmonics in terms of f1 . Answer: We know that ln = 2 v = l1 f1 = 2L f1 and substituted into fn = n fn = n n L and v = l f . Combining, we have v = 2 2L v . But v can be expressed n L f ! fn = n 2L v , giving 2L (2L f1) = n f1 ! fn = n f1 1b. If the ﬁrst harmonic has frequency of 261 Hz, what frequencies do the second and third harmonics have? Answer: Since fn = n f1 ! 2(261 Hz) = 522 Hz fn = n f1 ! 3(261 Hz) = 783 Hz All whole number multiples of the ﬁrst harmonic (the fundamental) are called harmonics. String instruments, as well as non-string instruments, can actually vibrate with many different frequencies simultaneously (called modes). For example, a string may vibrate with frequencies 261 Hz, 522 Hz and 783 Hz simultaneously. One of attributes of the “quality” or “timbre” of musical instruments depends upon the combination of the various overtones produced by the instrument. Check Your Understanding 1. A tuning fork has a frequency of 512 Hz stamped on it. When it is struck, a student claims she can hear higher frequencies from the tuning fork. Is this possible? Answer : Yes, it is. The tuning fork may be producing harmonics, in which case the student may be hearing frequencies in multiples of 512 Hz, such as 1,024 Hz and 1,536 Hz. 2. A string with a fundamental frequency of 220 Hz vibrates in its third harmonic with a wavelength of 60 cm. What is the wave velocity on the string? 52 www.ck12.org Concept 13. Resonance with Sound Waves Answer : v = l f but f = 3 f1 = 3(220 Hz) = 660 Hz , so l = 0:60 m v = (660 Hz)(0:60 m) = 396 m/s . Open and Closed Pipes and Tubes In our discussions of pipes, the length of the pipe will be assumed to be much greater than the diameter of the pipe. An open pipe, as the name implies, has both ends open. Though open pipes have antinodes at their ends, the resonant conditions for standing waves in an open pipe are the same as for a string ﬁxed at both ends. Thus for an open pipe we have: for n = 1; 2; 3:::; L = n 2 ln , or ln = 2 n L . There is a simple experiment your instructor may have you do in class that demonstrates resonance in an open tube. Roll two sheets of long paper into two separate tubes and use a small amount of tape to keep them rolled. Have the diameter of one tube just small enough to ﬁt inside the other tube so the inside tube can freely slide back and forth. Hold a struck tuning fork (your instructor will make sure the frequency is adequate) close to the end of the outer tube while the inside tube is moved slowly. When the total length of the tubes is the proper length to establish resonance, you’ll hear a noticeable increase in the volume of the sound. At this moment, there are standing waves present in the tubes. A closed pipe is closed at only one end. Closed pipes have the same standing wave patterns as a string ﬁxed at one end and unbound at the other end. They therefore have the same resonant conditions as a string ﬁxed at only one end, for n = 1; 3 ln or ln = 4 n L . A closed pipe supporting the ﬁrst harmonic (the fundamental frequency) will ﬁt one-fourth of the wavelength, the second harmonic will ﬁt three-fourths, and so on, as shown in Figure below . Compare these pictures to those in Figure above for a string ﬁxed at only one end FIGURE 13.14 A standard physics laboratory experiment is to ﬁnd the velocity of sound by using a tuning fork that vibrates over a closed pipe as shown in Figure Figure above . The water level in a pipe is slowly changed until the ﬁrst harmonic is heard. http://demonstrations.wolfram.com/ResonanceInOpenAndClosedPipes/ 53 www.ck12.org FIGURE 13.15 Illustrative Example 2 Resonance is established in a hollow tube similar to that shown in Figure above with a tuning fork of 512 Hz. The distance from the tube opening to the water level is 16.8 cm. a. What is the velocity of sound according to this experiment? Answer: The wave velocity equation is v = l f . One-fourth of the wave occupies the length of the tube for the ﬁrst harmonic. So the wavelength of the resonant wave must be four times the length of the hollow tube. That is, l1 = 4L = 4 16:8 cm = 67:2 cm = 0:672 m v = (0:672 m)(512 Hz) = 344 m/s b. The velocity of sound changes with temperature as given by the formula v = 330 + 0:6T , where T is the temperature in degrees centigrade. Using the result of part A, determine the temperature at the location the experiment was conducted. Answer: We simply set the result of part A equal to the given equation: 344 = 330 + 0:6T ! T = 23:3C (or about 74F ). References 1. Evan Long (Flickr: Clover_1). http://www.ﬂickr.com/photos/clover_1/4915240660/ . CC-BY-NC 2.0 54 www.ck12.org Concept 13. Resonance with Sound Waves 2. Barney Elliot, Prelinger Archives. struction.ogg . Public Domain http://commons.wikimedia.org/wiki/File:Tacoma_Narrows_Bridge_de 3. . http://commons.wikimedia.org/wiki/File:Tacoma-narrows-bridge-collapse.jpg . Public Domain 4. Russell James Smith (Flickr: russelljsmith). http://www.ﬂickr.com/photos/russelljsmith/2146210247/ . CC-BY 2.0 5. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou, using public domain image by Mariana Ruiz Villarreal (Wikimedia: . CC-BY- http://commons.wikimedia.org/wiki/File:Respiratory_system_complete_en.svg LadyofHats). NC-SA 3.0 7. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 8. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 9. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 10. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 11. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 12. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 13. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 14. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 15. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 55 CONCEPT 14 www.ck12.org Sound in a Tube Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube. Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube. Key Equations v = l f for a tube closed at one end f = nv=4L , where n is always odd for a tube open at both ends f = nv=2L , where n is an integer Guidance In the case of a tube that is open at one end, a node is forced at the closed end (no air molecules can vibrate up and down) and an antinode occurs at the open end (here, air molecules are free to move). A different spectrum of standing waves is produced. For instance, the fundamental standing sound wave produced in a tube closed at one end is shown below. In this case, the amplitude of the standing wave is referring to the magnitude of the air pressure variations. This standing wave is the ﬁrst harmonic and one can see that the wavelength is l = 4L . Since v = l f , the frequency of oscillation is f = v=4L . In general, the frequency of oscillation is f = nv=4L , where n is always odd. Example 1 Question The objects A, B, and C below represent graduated cylinders of length 50 cm which are ﬁlled with water to the depths of 10, 20 and 30 cm, respectively as shown. 56 www.ck12.org Concept 14. Sound in a Tube a) If you blow in each of these tubes, which (A,B,C) will produce the highest frequency sound? b) What is the wavelength of the 1st harmonic (i.e. fundamental) of tube B? c) The speed of sound at room temperature is about 343 m/s. What is the frequency of the 1st harmonic for tube B? Solution a) The water forms the bottom of the tube and thus where the node of the wave will be. Thus the air column is where the sound wave can exist. The larger the air column, the larger the wavelength. Frequency is inversely proportional to wavelength, thus the tube with the smallest air column will have the highest frequency. So the answer is tube C. b) The air column is 50 cm - 20 cm = 30 cm. The ﬁrst harmonic has a quarter wavelength in the tube. Thus l = 4 L . Thus, l = 4 30cm = 120cm c) Using the wave equation for the ﬁrst harmonic (thus, n = 1) of a tube open at one end we get f = v 286Hz 4L = 343m/s 1:2m = Watch this Explanation MEDIA Click image to the left for more content. 57 www.ck12.org MEDIA Click image to the left for more content. Time for Practice 1. 2. Aborigines, the native people of Australia, play an instrument called the Didgeridoo like the one shown above. The Didgeridoo produces a low pitch sound and is possibly the world’s oldest instrument. The one shown above is about 1.3 m long and open at both ends. a. Knowing that when a tube is open at both ends there must be an antinode at both ends, draw the ﬁrst 3 harmonics for this instrument. b. Calculate the frequency of the ﬁrst 3 harmonics assuming room temperature and thus a velocity of sound of 340 m/s. Then take a shot at deriving a generic formula for the frequency of the n th standing wave mode for the Didgeridoo, as was done for the string tied at both ends and for the tube open at one end. 3. Students are doing an experiment to determine the speed of sound in air. They hold a tuning fork above a large empty graduated cylinder and try to create resonance. The air column in the graduated cylinder can be adjusted by putting water in it. At a certain point for each tuning fork a clear resonance point is heard. The students adjust the water ﬁnely to get the peak resonance then carefully measure the air column from water to top of air column. (The assumption is that the tuning fork itself creates an anti-node and the water creates a node.) The following data were collected: TABLE 14.1: Wavelength (m) Speed of sound (m/s) Frequency fork (Hz) 184 328 384 512 1024 of tuning Length of air column (cm) 46 26 22 16 24 (a) Fill out the last two columns in the data table. (b) Explain major inconsistencies in the data or results. (c) The graduated cylinder is 50 cm high. Were there other resonance points that could have been heard? If so what would be the length of the wavelength? (d)
What are the inherent errors in this experiment? 3. Peter is playing tones by blowing across the top of a glass bottle partially ﬁlled with water. He notices that if he blows softly he hears a lower note, but if he blows harder he hears higher frequencies. (a) In the 120 cm long tubes below draw three diagrams showing the ﬁrst three harmonics produced in the tube. Please draw the waves as transverse even though we know sound waves are longitudinal (reason for this, obviously, is that it 58 www.ck12.org Concept 14. Sound in a Tube is much easier to draw transverse waves rather than longitudinal). Note that the tube is CLOSED at one end and OPEN at the other. (b) Calculate the frequencies of the ﬁrst three harmonics played in this tube, if the speed of sound in the tube is 340 m/s. (c) The speed of sound in carbon dioxide is lower than in air. If the bottle contained CO2 instead of air, would the frequencies found above be higher or lower? Knowing that the pitch of your voice gets higher when you inhale helium, what can we say about the speed of sound in He . Answers: 1. (b) 131 Hz, 262 Hz, 393 Hz; formula is same as closed at both ends 2. Discuss in class 3. (b) 70.8 Hz, 213 Hz, 354 Hz (c) voice gets lower pitch. Speed of sound in He must be faster by same logic. 59 Physics Unit 11: Electromagnetic Waves Patrick Marshall Jean Brainard, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 6 9 12 16 21 24 27 32 35 Contents 1 Electromagnetic Waves 2 Properties of Electromagnetic Waves 3 Electromagnetic Spectrum 4 Law of Reﬂection 5 Refraction of Light 6 Total Internal Reﬂection 7 Single Slit Interference 8 Double Slit Interference 9 Diffraction Gratings 10 Wave-Particle Theory iv www.ck12.org Concept 1. Electromagnetic Waves CONCEPT 1 Electromagnetic Waves • Deﬁne electromagnetic wave and electromagnetic radiation. • Describe the electric and magnetic ﬁelds of an electromagnetic wave. • Explain how electromagnetic waves begin and how they travel. • State how electromagnetic waves may interact with matter. • Identify sources of electromagnetic waves. Did you ever wonder how a microwave works? It directs invisible waves of radiation toward the food placed inside of it. The radiation transfers energy to the food, causing it to get warmer. The radiation is in the form of microwaves, which are a type of electromagnetic waves. What Are Electromagnetic Waves? Electromagnetic waves are waves that consist of vibrating electric and magnetic ﬁelds. Like other waves, electromagnetic waves transfer energy from one place to another. The transfer of energy by electromagnetic waves is called electromagnetic radiation . Electromagnetic waves can transfer energy through matter or across empty space. For an excellent video introduction to electromagnetic waves, go to this URL: http://www.youtube.com/watch?v=cfXz wh3KadE MEDIA Click image to the left for more content. Q: How do microwaves transfer energy inside a microwave oven? A: They transfer energy through the air inside the oven to the food. 1 May the Force Be with You A familiar example may help you understand the vibrating electric and magnetic ﬁelds that make up electromagnetic waves. Consider a bar magnet, like the one in the Figure 1.1 . The magnet exerts magnetic force over an area all around it. This area is called a magnetic ﬁeld. The ﬁeld lines in the diagram represent the direction and location of the magnetic force. Because of the ﬁeld surrounding a magnet, it can exert force on objects without touching them. They just have to be within its magnetic ﬁeld. www.ck12.org FIGURE 1.1 Q: How could you demonstrate that a magnet can exert force on objects without touching them? A: You could put small objects containing iron, such as paper clips, near a magnet and show that they move toward the magnet. An electric ﬁeld is similar to a magnetic ﬁeld. It is an area of electrical force surrounding a positively or negatively charged particle. You can see electric ﬁelds in the following Figure 1.2 . Like a magnetic ﬁeld, an electric ﬁeld can exert force on objects over a distance without actually touching them. How an Electromagnetic Wave Begins An electromagnetic wave begins when an electrically charged particle vibrates. The Figure 1.3 shows how this happens. A vibrating charged particle causes the electric ﬁeld surrounding it to vibrate as well. A vibrating electric ﬁeld, in turn, creates a vibrating magnetic ﬁeld. The two types of vibrating ﬁelds combine to create an electromagnetic wave. You can see animations of electromagnetic waves at these URLs: http://www.youtube.com/ http://www.phys.hawaii.edu/~teb/java/ntnujava/emWave/emWave.htm watch?v=Qju7QnbrOhM&feature=related l 2 www.ck12.org Concept 1. Electromagnetic Waves FIGURE 1.2 FIGURE 1.3 How an Electromagnetic Wave Travels As you can see in the diagram above, the electric and magnetic ﬁelds that make up an electromagnetic wave are perpendicular (at right angles) to each other. Both ﬁelds are also perpendicular to the direction that the wave travels. Therefore, an electromagnetic wave is a transverse wave. However, unlike a mechanical transverse wave, which can only travel through matter, an electromagnetic transverse wave can travel through empty space. When waves travel through matter, they lose some energy to the matter as they pass through it. But when waves travel through space, no energy is lost. Therefore, electromagnetic waves don’t get weaker as they travel. However, the energy is “diluted” as it travels farther from its source because it spreads out over an ever-larger area. 3 Electromagnetic Wave Interactions When electromagnetic waves strike matter, they may interact with it in the same ways that mechanical waves interact with matter. Electromagnetic waves may: www.ck12.org • reﬂect, or bounce back from a surface; • refract, or bend when entering a new medium; • diffract, or spread out around obstacles. Electromagnetic waves may also be absorbed by matter and converted to other forms of energy. Microwaves are a familiar example. When microwaves strike food in a microwave oven, they are absorbed and converted to thermal energy, which heats the food. Sources of Electromagnetic Waves The most important source of electromagnetic waves on Earth is the sun. Electromagnetic waves travel from the sun to Earth across space and provide virtually all the energy that supports life on our planet. Many other sources of electromagnetic waves depend on technology. Radio waves, microwaves, and X rays are examples. We use these electromagnetic waves for communications, cooking, medicine, and many other purposes. Summary • Electromagnetic waves are waves that consist of vibrating electric and magnetic ﬁelds. They transfer energy through matter or across space. The transfer of energy by electromagnetic waves is called electromagnetic radiation. • The electric and magnetic ﬁelds of an electromagnetic wave are areas of electric or magnetic force. The ﬁelds can exert force over objects at a distance. • An electromagnetic wave begins when an electrically charged particle vibrates. This causes a vibrating electric ﬁeld, which in turn creates a vibrating magnetic ﬁeld. The two vibrating ﬁelds together form an electromagnetic wave. • An electromagnetic wave is a transverse wave that can travel across space as well as through matter. When it travels through space, it doesn’t lose energy to a medium as a mechanical wave does. • When electromagnetic waves strike matter, they may be reﬂected, refracted, or diffracted. Or they may be absorbed by matter and converted to other forms of energy. • The most important source of electromagnetic waves on Earth is the sun. Many other sources of electromag- netic waves depend on technology. Vocabulary • electromagnetic radiation : Transfer of energy by electromagnetic waves across space or through matter. • electromagnetic wave : Transverse wave consisting of vibrating electric and magnetic ﬁelds that can travel across space. Practice Watch the electromagnetic wave animation at the following URL, and then answer the questions below. http://www. youtube.com/watch?v=4CtnUETLIFs 4 www.ck12.org Concept 1. Electromagnetic Waves MEDIA Click image to the left for more content. 1. Identify the vibrating electric and magnetic ﬁelds of the wave.
2. Describe the direction in which the wave is traveling. Review 1. What is an electromagnetic wave? 2. Deﬁne electromagnetic radiation. 3. Describe the electric and magnetic ﬁelds of an electromagnetic wave. 4. How does an electromagnetic wave begin? How does it travel? 5. Compare and contrast electromagnetic and mechanical transverse waves. 6. List three sources of electromagnetic waves on Earth. References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Christopher Auyeung. . CC BY-NC 3.0 3. Christopher Auyeung. . CC BY-NC 3.0 5 CONCEPT 2 Properties of Electromagnetic Waves www.ck12.org • State the speed of light. • Describe wavelengths and frequencies of electromagnetic waves. • Relate wave frequency to wave energy. • Show how to calculate wavelength or wave frequency if the other value is known. What do these two photos have in common? They both represent electromagnetic waves. These are waves that consist of vibrating electric and magnetic ﬁelds. They transmit energy through matter or across space. Some electromagnetic waves are generally harmless. The light we use to see is a good example. Other electromagnetic waves can be very harmful and care must be taken to avoid too much exposure to them. X rays are a familiar example. Why do electromagnetic waves vary in these ways? It depends on their properties. Like other waves, electromagnetic waves have properties of speed, wavelength, and frequency. Speed of Electromagnetic Waves All electromagnetic waves travel at the same speed through empty space. That speed, called the speed of light , is about 300 million meters per second (3.0 x 10 8 m/s). Nothing else in the universe is known to travel this fast. The sun is about 150 million kilometers (93 million miles) from Earth, but it takes electromagnetic radiation only 8 minutes to reach Earth from the sun. If you could move that fast, you would be able to travel around Earth 7.5 times in just 1 second! You can learn more about the speed of light at this URL: http://videos.howstuffworks.com/discove ry/29407-assignment-discovery-speed-of-light-video.htm MEDIA Click image to the left for more content. Wavelength and Frequency of Electromagnetic Waves Although all electromagnetic waves travel at the same speed across space, they may differ in their wavelengths, frequencies, and energy levels. 6 www.ck12.org Concept 2. Properties of Electromagnetic Waves • Wavelength is the distance between corresponding points of adjacent waves (see the Figure 2.1 ). Wavelengths of electromagnetic waves range from longer than a soccer ﬁeld to shorter than the diameter of an atom. • Wave frequency is the number of waves that pass a ﬁxed point in a given amount of time. Frequencies of electromagnetic waves range from thousands of waves per second to trillions of waves per second. • The energy of electromagnetic waves depends on their frequency. Low-frequency waves have little energy and are normally harmless. High-frequency waves have a lot of energy and are potentially very harmful. FIGURE 2.1 Q: Which electromagnetic waves do you think have higher frequencies: visible light or X rays? A: X rays are harmful but visible light is harmless, so you can infer that X rays have higher frequencies than visible light. Speed, Wavelength, and Frequency The speed of a wave is a product of its wavelength and frequency. Because all electromagnetic waves travel at the same speed through space, a wave with a shorter wavelength must have a higher frequency, and vice versa. This relationship is represented by the equation: Speed = Wavelength Frequency The equation for wave speed can be rewritten as: Frequency = Speed Wavelength or Wavelength = Speed Frequency Therefore, if either wavelength or frequency is known, the missing value can be calculated. Consider an electromagnetic wave that has a wavelength of 3 meters. Its speed, like the speed of all electromagnetic waves, is 3.0 10 8 meters per second. Its frequency can be found by substituting these values into the frequency equation: Frequency = 3:0108 m/s 3:0 m = 1:0 108 waves/s , or 1.0 10 8 Hz Q: What is the wavelength of an electromagnetic wave that has a frequency of 3.0 10 8 hertz? A: Use the wavelength equation: Wavelength = 3:0108 m/s 3:0108 waves/s = 1:0 m You can learn more about calculating the frequency and wavelength of electromagnetic waves at these URLs: htt p://www.youtube.com/watch?v=GwZvtfZRNKk and http://www.youtube.com/watch?v=wjPk108Ua8k 7 www.ck12.org Summary • All electromagnetic waves travel across space at the speed of light, which is about 300 million meters per second (3.0 x 10 8 m/s). • Electromagnetic waves vary in wavelength and frequency. Longer wavelength electromagnetic waves have lower frequencies, and shorter wavelength waves have higher frequencies. Higher frequency waves have more energy. • The speed of a wave is a product of its wavelength and frequency. Because the speed of electromagnetic waves through space is constant, the wavelength or frequency of an electromagnetic wave can be calculated if the other value is known. Vocabulary • speed of light : Speed at which all electromagnetic waves travel through space, which is 3.0 10 8 m/s. Practice Use the calculator at the following URL to ﬁnd the frequency and energy of electromagnetic waves with different wavelengths. Use at least eight values for wavelength. Record and make a table of the results. http://www.1728.org /freqwave.htm Review 1. What is the speed of light across space? 2. Describe the range of wavelengths and frequencies of electromagnetic waves. 3. How is the energy of an electromagnetic wave related to its frequency? 4. If the frequency of an electromagnetic wave is 6.0 10 8 Hz, what is its wavelength? References 1. Christopher Auyeung. . CC BY-NC 3.0 8 www.ck12.org Concept 3. Electromagnetic Spectrum CONCEPT 3 Electromagnetic Spectrum • Describe electromagnetic radiation and its properties. • Give an overview of the electromagnetic spectrum. It’s a warm sunny Saturday, and Michael and Lavar have a big day planned. They’re going to ride across town to meet their friends and then go to the zoo. The boys may not realize it, but they will be bombarded by electromagnetic radiation as they ride their bikes and walk around the zoo grounds. The only kinds of radiation they can detect are visible light, which allows them to see, and infrared light, which they feel as warmth on their skin. Q: Besides visible light and infrared light, what other kinds of electromagnetic radiation will the boys be exposed to in sunlight? A: Sunlight consists of all the different kinds of electromagnetic radiation, from harmless radio waves to deadly gamma rays. Fortunately, Earth’s atmosphere prevents most of the harmful radiation from reaching Earth’s surface. You can read about the different kinds of electromagnetic radiation in this article. Electromagnetic Radiation Electromagnetic radiation is energy that travels in waves across space as well as through matter. Most of the electromagnetic radiation on Earth comes from the sun. Like other waves, electromagnetic waves are characterized by certain wavelengths and wave frequencies. Wavelength is the distance between two corresponding points on adjacent waves. Wave frequency is the number of waves that pass a ﬁxed point in a given amount of time. Electromagnetic waves with shorter wavelengths have higher frequencies and more energy. A Spectrum of Electromagnetic Waves Visible light and infrared light are just a small part of the full range of electromagnetic radiation, which is called the electromagnetic spectrum . You can see the waves of the electromagnetic spectrum in the Figure 3.1 . At the top 9 of the diagram, the wavelengths of the waves are given. Also included are objects that are about the same size as the corresponding wavelengths. The frequencies and energy levels of the waves are shown at the bottom of the diagram. Some sources of the waves are also given. For a video introduction to the electromagnetic spectrum, go to this URL: http://www.youtube.com/watch?NR=1&feature=endscreen&v=cfXzwh3KadE www.ck12.org FIGURE 3.1 • On the left side of the electromagnetic spectrum diagram are radio waves and microwaves. Radio waves have the longest wavelengths and lowest frequencies of all electromagnetic waves. They also have the least amount of energy. • On the right side of the diagram are X rays and gamma rays. They have the shortest wavelengths and highest frequencies of all electromagnetic waves. They also have the most energy. • Between these two extremes are waves that are commonly called light. Light includes infrared light, visible light, and ultraviolet light. The wavelengths, frequencies, and energy levels of light fall in between those of radio waves on the left and X rays and gamma rays on the right. Q: Which type of light has the longest wavelengths? A: Infrared light has the longest wavelengths. Q: What sources of infrared light are shown in the diagram? A: The sources in the diagram are people and light bulbs, but all living things and most other objects give off infrared light. Summary • Electromagnetic radiation travels in waves through space or matter. Electromagnetic waves with shorter wavelengths have higher frequencies and more energy. • The full range of electromagnetic radiation is called the electromagnetic spectrum. From longest to shortest wavelengths, it includes radio waves, microwaves, infrared light, visible light, ultraviolet light, X rays, and gamma rays. Vocabulary • electromagnetic spectrum : Full range of wavelengths of electromagnetic waves, from radio waves to gamma rays. 10 www.ck12.org Practice Concept 3. Electromagnetic Spectrum At the ﬁrst URL below, read about electromagnetic waves with different frequencies. Then use the information to complete the table at the second URL. http://www.darvill.clara.net/emag/index.htm and http://www.darvill.clar a.net/emag/gcseemag.pdf Review 1. Describe the relationship between the wavelength and frequ
ency of electromagnetic waves. 2. What is the electromagnetic spectrum? 3. Which electromagnetic waves have the longest wavelengths? 4. Identify a source of microwaves. 5. Which type of light has the highest frequencies? 6. Explain why gamma rays are the most dangerous of all electromagnetic waves. References 1. NASA. . public domain 11 www.ck12.org Law of Reﬂection CONCEPT 4 • Deﬁne reﬂection and image. • Compare and contrast regular and diffuse reﬂection. • State the law of reﬂection. These dancers are practicing in front of a mirror so they can see how they look as they performs. They’re watching their image in the mirror as they dance. What is an image, and how does it get “inside” a mirror? In this article, you’ll ﬁnd out. Reﬂected Light and Images Reﬂection is one of several ways that light can interact with matter. Light reﬂects off surfaces such as mirrors that do not transmit or absorb light. When light is reﬂected from a smooth surface, it may form an image. An image is a copy of an object that is formed by reﬂected (or refracted) light. Q: Is an image an actual object? If not, what is it? A: No, an image isn’t an actual object. It is focused rays of light that make a copy of an object, like a picture projected on a screen. Regular and Diffuse Reﬂection If a surface is extremely smooth, as it is in a mirror, then the image formed by reﬂection is sharp and clear. This is called regular reﬂection (also called specular reﬂection). However, if the surface is even slightly rough or bumpy, an image may not form, or if there is an image, it is blurry or fuzzy. This is called diffuse reﬂection. Q: Look at the boats and their images in the Figure 4.1 . Which one represents regular reﬂection, and which one represents diffuse reﬂection? 12 www.ck12.org Concept 4. Law of Reﬂection FIGURE 4.1 A: Reﬂection of the boat on the left is regular reﬂection. The water is smooth and the image is sharp and clear. Reﬂection of the boat on the right is diffuse reﬂection. The water has ripples and the image is blurry and wavy. In the Figure 4.2 , you can see how both types of reﬂection occur. Waves of light are represented by arrows called rays. Rays that strike the surface are referred to as incident rays, and rays that reﬂect off the surface are known as reﬂected rays. In regular reﬂection, all the rays are reﬂected in the same direction. This explains why regular reﬂection forms a clear image. In diffuse reﬂection, the rays are reﬂected in many different directions. This is why diffuse reﬂection forms, at best, a blurry image. You can see animations of both types of reﬂection at this URL: htt p://toolboxes.ﬂexiblelearning.net.au/demosites/series5/508/Laboratory/StudyNotes/snReﬂectionMirrors.htm FIGURE 4.2 Law of Reﬂection One thing is true of both regular and diffuse reﬂection. The angle at which the reﬂected rays leave the surface is equal to the angle at which the incident rays strike the surface. This is known as the law of reﬂection . The law is illustrated in the Figure 4.3 and also in this animation: http://www.physicsclassroom.com/mmedia/optics/lr.cfm Summary • Reﬂection is one of several ways that light can interact with matter. When light is reﬂected from a smooth surface, it may form an image. An image is a copy of an object that is formed by reﬂected (or refracted) light. • Regular reﬂection occurs when light reﬂects off a very smooth surface and forms a clear image. Diffuse reﬂection occurs when light reﬂects off a rough surface and forms a blurry image or no image at all. • According to the law of reﬂection, the angle at which light rays reﬂect off a surface is equal to the angle at which the incident rays strike the surface. 13 www.ck12.org FIGURE 4.3 Vocabulary • image : Copy of an object that is formed by reﬂected or refracted light. • law of reﬂection : Law stating that the angle at which reﬂected rays of light bounce off a surface is equal to the angle at which the incident rays strike the surface. • reﬂection : Bouncing back of waves from a barrier they cannot pass through. Practice At the following URL, review the law of reﬂection and watch the animation. Then ﬁll in the blanks in the sentence below. http://www.physicsclassroom.com/mmedia/optics/lr.cfm 1. When a ray of light strikes a plane mirror, the light ray __________ off the mirror. 2. Reﬂection involves a change in __________ of a light ray. 3. The angle of incidence equals the angle between the incident ray and ________. 4. The angle of __________ equals the angle of incidence. 5. The normal line is __________ to the mirror. Review 1. What is an image? 2. Identify the object and the image in the Figure 4.4 . Which type of reﬂection formed the image: regular reﬂection or diffuse reﬂection? How do you know? FIGURE 4.4 3. What is the law of reﬂection? 4. Label the angle of incidence and the angle of reﬂection in the Figure 4.5 . 14 www.ck12.org Concept 4. Law of Reﬂection FIGURE 4.5 References 1. Left: Kenneth Baruch; Right: Damian Gadal. . CC BY 2.0 2. Joy Sheng. . CC BY-NC 3.0 3. Christopher Auyeung. . CC BY-NC 3.0 4. Mike Baird. . CC BY 2.0 5. Zachary Wilson. . CC-BY-NC-SA 3.0 15 www.ck12.org CONCEPT 5 Refraction of Light • Deﬁne refraction. • Given data about the optical density of the media, predict whether the light will bend toward the normal or away from the normal. • State Snell’s Law and solve refraction problems using it. • Solve problems using the relationship between the index of refraction and the velocity of light in the media. • Explain effects caused by the refraction of light. When a light ray passes at an angle through the boundary between optically different media, the light does not travel in a straight line. The pencil in the glass of liquid shown above is a normal straight pencil. The light that travels from the pencil through the liquid, through the glass, and into the air is bent differently than light from the portion of the pencil that is not in the liquid. Your eye assumes the light from both portions of the pencil moved in a straight line, but the two portions of the pencil do not appear to be lined up. Your eye thinks the pencil is broken. Refraction of Light The speed of light is different in different media. If the speed of light is slower in a particular medium, that medium is said to be more optically dense . When a wave front enters a new medium at an angle, it will change directions. If the light is entering a more optically dense medium, the light bends toward the normal line. If the light is entering a less optically dense medium, the light will bend away from the normal line. Remember that the normal line is the line perpendicular to the medium interface. 16 www.ck12.org Concept 5. Refraction of Light In the sketch below, light wave fronts are moving upward from the bottom of the page and encounter a boundary into a more optically dense medium. The light waves bend toward the normal line. Because the right end of the wave fronts enter the new medium ﬁrst, they slow down ﬁrst. When the right side of the wave front is moving more slowly that the left side, the wave front will change directions. When light is traveling from air into another medium, Snell’s Law states the relationship between the angle of incidence and angle of refraction is n = sin qi sin qr where qi is the angle of incidence, qr is the angle of refraction, and n is the ratio of the two sines and is called the index of refraction . Snell’s Law may be stated that a ray of light bends in such a way that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The index of refraction is also related to the relative speeds of light in a vacuum and in the medium. n = speed of light in a vacuum speed of light in the medium When a ray of light is traveling from medium into another medium, Snell’s Law can be written as ni sin qi = nr sin qr . Table of Indices of Refraction Medium Vacuum Air Water Ethanol Crown Glass Quartz Flint Glass Diamond TABLE 5.1: n 1.00 1.0003 1.36 1.36 1.52 1.54 1.61 2.42 17 www.ck12.org Example Problem: A ray of light traveling through air is incident upon a slab of Flint glass at an angle of 40:0 . What is the angle of refraction? Solution: n = sin qi sin qr so sin qr = sin qi 1:61 = 0:399 n = 0:643 The angle of refraction = sin1:399 = 23:5 Example Problem: What is the speed of light in a diamond? Solution: speed of light in diamond = speed of light in a vacuum n = 3:00108 m=s 2:42 speed of light in diamond = 1:24 108 m=s Effects of Refraction Bending the Sun’s Rays Because air is slightly more optically dense than a vacuum, when sunlight passes from the vacuum of space into our atmosphere, it bends slightly towards the normal. When the sun is below the horizon and thus not visible on a direct line, the light path will bend slightly and thus make the sun visible by refraction. Observers can see the sun before it actually comes up over the horizon, or after it sets. Mirages In the Figure below, the sun shines on the road, heating the air just above the road. The difference in density between the hot air over the road and the surrounding air causes the hot air to refract light that passes through it. When you look at the road, you see a mirage . What appears to be water on the road is actually light coming from the sky that has been refracted as it passes through the hot air above the road. This phenomenon is common on hot roads and in the desert. Summary • The speed of light is different in different media. • When a wave front enters a new medium at an angle, it will change directions. If the light is entering a more optically dense medium, the light bends toward the normal line. If the light is entering a less optically dense medium, the light will bend away from the normal line. 18 www.ck12.org Concept 5. Refraction of Light FIGURE 5.1 • When light is traveling from air into another medium, Snell’s Law states that n = sin qi sin qr • The index of refraction is also related to the r
elative speeds of light in a vacuum and in the medium. n = . speed of light in a vacuum speed of light in the medium • When a ray of light is traveling from one medium into another medium, Snell’s Law can be written as ni sin qi = nr sin qr . Practice Use the video on refraction to answer the questions below. http://video.mit.edu/watch/mit-physics-demo-refraction-a-total-internal-reflection-12044/ MEDIA Click image to the left for more content. 1. What happens to the path of a light beam when it enters a new medium at an angle? 2. Light moving from air into water is bent ___________ the normal. 3. Light moving from water into air is bent ___________ the normal. Review 1. Light moving through air is incident on a piece of crown glass at an angle of 45 . What is the angle of refraction? 2. A ray of light passes from air into water at an incident angle of 60:0 . Find the angle of refraction. 3. Light passes from water into a block of transparent plastic. The angle of incidence from the water is 31 and the angle of refraction in the block is 27 . What is the index of refraction for the plastic? 4. The index of refraction of water is 1.36. What is the speed of light in water? 19 www.ck12.org 5. If the speed of light in a piece of plastic is 2:00 108 m=s , what is the index of refraction for the plastic? • refraction : The change of direction of a ray of light or sound in passing obliquely from one medium into another in which its wave velocity is different. • optically dense: Refers to the ability of a material to slow the light waves. The greater the optical density of a material, the greater the slowing effect. • Snell’s Law: For a light ray incident on the interface of two media, the sine of the angle of incidence times the index of refraction of the ﬁrst medium is equal to the sine of the angle of refraction times the index of refraction of the second medium. • index of refraction: The ratio of the speed of light in a vacuum to the speed of light in a medium under consideration. • mirage: An optical phenomenon that creates the illusion of water, often with inverted reﬂections of distant objects, and results from the refraction of light by alternate layers of hot and cool air. References 1. Image copyright leonello calvetti, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 4. Michael Gil (Flickr: MSVG). http://www.ﬂickr.com/photos/msvg/5994891327/ . CC-BY 2.0 20 www.ck12.org Concept 6. Total Internal Reﬂection CONCEPT 6 Total Internal Reﬂection • Describe total internal reﬂection. • Use the critical angle to determine when total internal reﬂection will occur. Total internal reﬂection allows the light to travel down the optical ﬁber and not pass through the sides of the tube. The light continuously reﬂects from the inside of the tube and eventually comes out the end. Optical ﬁbers make interesting lamps but they are also used to transport telephone and television signals. Total Internal Reﬂection We already know that when light passes from one medium into a second medium where the index of refraction is smaller, the light refracts away from the normal. In the image below, the light rays are passing into an optically less dense medium; therefore, the rays bend away from the normal. As the angle of incidence increases, the light ray bends even further away from the normal. Eventually, the angle of incidence will become large enough that the angle of refraction equals 90 , meaning the light ray will not enter the new medium at all. 21 www.ck12.org Consider a ray of light passing from water into air. The index of refraction for air is 1.00 and for water is 1.36. Using Snell’s Law, ni sin qi = nr sin qr , and allowing the angle of refraction to be 90 , we can solve for the angle of incidence which would cause the light ray to stay in the old medium. ni sin qi = nr sin qr (1:36)(sin qi) = (1:00)(sin 90) sin qi = 0:735 and qi = 47 This result tells us that when light is passing from water into air, if the angle of incidence exceeds 47 , the light ray will not enter the new medium. The light ray will be completely reﬂected back into the original medium. This is called total internal reﬂection . The minimum angle of incidence for total internal reﬂection to occur is called the critical angle . Total internal reﬂection is the principle behind ﬁber optics . A bundle of ﬁbers made out of glass or plastic only a few micrometers in diameter is called a light pipe since light can be transmitted along it with almost no loss. Light passing down the ﬁbers makes glancing collisions with the walls so that total internal reﬂection occurs. Summary • When light passes from one medium into a second medium with a smaller index of refraction, the light refracts away from the normal. • If the angle of incidence becomes large enough that the angle of refraction equals 90 , the light ray will not enter the new medium with the smaller angle of refraction. • Total internal reﬂection means the light ray will not enter the new medium but will be completely reﬂected back into the original medium. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=hBQ8fh_Fp04 In this video, a hole is drilled in the side of a plastic bottle ﬁlled with liquid. An arc of liquid shoots out through the hole. A laser pointer is aimed through the opposite side of the bottle so that the light also exits through the hole. The stream of liquid acts like an optical ﬁber and the light undergoes total internal reﬂection as it follows the stream of liquid. As the amount of liquid in the bottle decreases, the arc of the stream of liquid changes and the direction of the light follows the stream of liquid. Toward the end, the light beam is shining almost 90 from the direction of the laser pointer. The following video discusses total internal reﬂection. Use this resource to answer the questions that follow. http://www.khanacademy.org/science/physics/waves-and-optics/v/total-internal-reﬂection# 22 www.ck12.org Concept 6. Total Internal Reﬂection MEDIA Click image to the left for more content. 1. What phenomenon occurs when the light does not enter the new medium and remains in the old medium? 2. When does this phenomenon occur? Review 1. Find the critical angle for light passing from diamond into air, given ndiamond = 2:42 2. When two swimmers are under water in a swimming pool, it is possible for the interface between the water and the air to act as a mirror, allowing the swimmers to see images of each other if they look up at the underside of the surface. Explain this phenomenon. 3. Robert shines a laser beam through a slab of plastic and onto the interface between the slab of plastic and the air on the other side. The index of refraction for the plastic is 1.62. If the angle of incidence in the plastic is 54 , will the laser beam pass out of the plastic into the air? • total internal reﬂection: When light is passing from a medium of higher index of refraction into a medium of lower index of refraction is completely reﬂected by the boundary between the two media. • critical angle: The smallest angle of incidence at which a light ray passing from one medium to another less refractive medium will be totally reﬂected from the boundary between the two. • ﬁber optics: The science or technology of light transmission through very ﬁne, ﬂexible glass or plastic ﬁbers without energy loss making use of the principle of total internal reﬂection. References 1. Roshan Nikam (Flickr: roshan1286). http://www.ﬂickr.com/photos/31916678@N07/4753800195/ . CC-BY 2.0 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 23 CONCEPT 7 Single Slit Interference • Explain how single slit diffraction patterns occur. • Use single slit diffraction patterns to calculate wavelength. www.ck12.org Though it looks like a double slit interference pattern, the pattern on the screen are actually the results of light diffracting through a single slit with the ensuing interference. Single Slit Interference Interference patterns are produced not only by double slits but also by single slits, otherwise known as single slit interference . In the case of a single slit, the particles of medium at both corners of the slit act as point sources, producing circular waves from both edges. These circular waves move across to the back wall and interfere in the same way that interference patterns were produced by double slits. In the sketch at below, the black lines intersect at the center of the pattern on the back wall. This center point is equidistanct from both edges of the slit. Therefore, the waves striking at this position will be in phase; that is, the waves will produce constructive interference. Also shown in the sketch, just above the central bright spot where the red lines intersect, is a position where destructive interference occurs. One of these red lines is one-half wavelength longer than the other, causing the two waves to hit the wall out of phase and undergo destructive interference. A dark bank appears at this position. 24 www.ck12.org Concept 7. Single Slit Interference Just as in double slit interference, a pair of similar triangles can be constructed in the interference pattern. The pertinent values from these triangles are the width of the slit, w , the wavelength, l, the distance from the central bright spot to the ﬁrst dark band, x , and the distance from the center of the slit to back wall, L . The relationship of these four values is L or l = wx w = x L . l Example Problem: Monochromatic light of wavelength 605 nm falls on a slit of width 0.095 mm. The slit is located 85 cm from a screen. How far is the center of the central bright band to the ﬁrst dark band? Solution: x = lL w = (6:05107 m)(0:85 m) (9:5105 m) = 0:0054 m Summary • Interference patterns can also be produced by single slits. • In the case of a single slit, the p
articles of medium at both corners of the slit act as point sources, and produce circular waves from both edges. • The wavelength can be determined by this equation: l w = x L or l = wx L . Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=_HO7LJDcqos Follow up questions. 1. The interference pattern appears as the slit becomes ___________ (wider, thinner). Review 1. The same set up is used for two different single slit diffraction experiments. In one of the experiments, yellow light is used, and in the other experiment, green light is used. Green light has a shorter wavelength than yellow light. Which of the following statements is true? 25 (a) The two experiments will have the same distance between the central bright band and the ﬁrst dark band. (b) The green light experiment will have a greater distance between the central bright band and the ﬁrst dark band. (c) The yellow light experiment will have a greater distance between the central bright band and the ﬁrst www.ck12.org dark band. 2. Why are the edges of shadows often fuzzy? (a) Interference occurs on the wall on which the shadow is falling. (b) Light diffracts around the edges of the object casting the shadow. (c) The edges of the object casting the shadow is fuzzy. (d) Light naturally spreads out. 3. Monochromatic, coherent light passing through a double slit will produce exactly the same interference pattern as when it passes through a single slit. (a) True (b) False 4. If monochromatic light passes through a 0.050 mm slit and is projected onto a screen 0.70 m away with a distance of 8.00 mm between the central bright band and the ﬁrst dark band, what is the wavelength of the light? 5. A krypton ion laser with a wavelength of 524.5 nm illuminates a 0.0450 mm wide slit. If the screen is 1.10 m away, what is the distance between the central bright band and the ﬁrst dark band? 6. Light from a He-Ne laser (l = 632:8 nm) falls on a slit of unknown width. In the pattern formed on a screen 1.15 m away, the ﬁrst dark band is 7.50 mm from the center of the central bright band. How wide is the slit? • single slit interference: When monochromatic, coherent light falls upon a small single slit it will produce a pattern of bright and dark fringes. These fringes are due to light from one side of the slit interacting (interfering) with light from the other side. References 1. Luiz Sauerbronn. http://commons.wikimedia.org/wiki/File:Fresnel_Diffraction_experiment_DSC04573.JPG . Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 26 www.ck12.org Concept 8. Double Slit Interference CONCEPT 8 Double Slit Interference • Deﬁne diffraction of light. • Deﬁne wave interference. • Describe double slit interference patterns. • Explain Thomas Young’s contibutions to physics. • Calculate the wavelength from a double slit interference pattern. FIGURE 8.1 When waves strike a small slit in a wall, they create circular wave patterns on the other side of the barrier. This is seen in the image above, where ocean waves create precise circular waves. The circular waves undergo constructive and destructive interference, which generates a regular interference pattern. Diffraction and Interference When a series of straight waves strike an impenetrable barrier, the waves stop at the barrier. However, the last particle of the medium at the back corner of the barrier will create circular waves from that point, called the point source. This can be seen in the image below. This phenomenon is called diffraction , and it occurs in liquid, sound, and light waves. While the waves become circular waves at the point source, they continue as straight waves where the barrier does not interfere with the waves. Any two waves in the same medium undergo wave interference as they pass each other. At the location where the two waves collide, the result is essentially a summation of the two waves. In some places, a wave crest from one source will overlap a wave crest from the other source. Since both waves are lifting the medium, the combined wave crest will be twice as high as the original crests. Nearby, a wave trough will overlap another wave trough and the new 27 trough will be twice as deep as the original. This is called constructive interference because the resultant wave is larger than the original waves. Within the interference pattern, the amplitude will be twice the original amplitude. Once the waves pass through each other and are alone again, their amplitudes return to their original values. In other parts of the wave pattern, crests from one wave will overlap troughs from another wave. When the two waves have the same amplitude, this interaction causes them to cancel each other out. Instead of a crest or a trough, there is nothing. When this cancellation occurs, it is called destructive interference . www.ck12.org FIGURE 8.2 It is easy to see how waves emanating from multiple sources, such as drops of rainwater in still water, create interference patterns. But a single source of waves can create interference patterns with itself as a result of diffraction. The Double Slit Experiment A similar situation to the raindrops above occurs when straight waves strike a barrier containing two slits. These waves are cut off everywhere except for where the waves that pass through the two slits. The medium in the slits again acts as a point source to produce circular waves on the far side of the barrier. As long as these two circular waves have the same wavelength, they interfere constructively and destructively in a speciﬁc pattern. This pattern is called the wave interference pattern and is characterized by light and dark bands. The light bands are a result of constructive interference, and the dark bands occur because of destructive interference. 28 www.ck12.org Concept 8. Double Slit Interference In the early 1800’s, light was assumed to be a particle. There was a signiﬁcant amount of evidence to point to that conclusion, and famous scientist Isaac Newton’s calculations all support the particle theory. In 1803, however, Thomas Young performed his famous Double Slit Experiment to prove that light was a wave. Young shined a light onto the side of a sealed box with two slits in it, creating an interference pattern on the inside of the box opposite the slits. As seen above, interference patterns are characterized by alternating bright and dark lines. The bright lines are a result of constructive interference, while the dark lines are a result of destructive interference. By creating this interference pattern, Young proved light is a wave and changed the course of physics. Calculating Wavelength from Double Slit Pattern Using the characteristics of the double slit interference pattern, it is possible to calculate the wavelength of light used to produce the interference. To complete this calculation, it is only necessary to measure a few distances. As can be seen below, ﬁve distances are measured. In the sketch, L is the distance from the two slits to the back wall where the interference pattern can be seen. d is the distance between the two slits. To understand x , look again at the interference pattern shown above. The middle line, which is the brightest, is called the central line . The remaining lines are called fringes . The lines on either side of the central line are called the ﬁrst order fringes, the next lines are called the second order fringes, and so on. x is the distance from the central line to the ﬁrst order fringe. r 1 and r 2 are the distances from the slits to the ﬁrst order fringe. We know that the fringes are a result of constructive interference, and that the fringe is a result of the crest of two waves interfering. If we assume that r 2 is a whole number of wavelengths (conﬁrm for yourself that this is a logical assumption), then r 1 must be one more wavelength. This is because r 1 and r 2 are the distances to the ﬁrst order fringe. Mathematically, we can let r2 = nl and r1 = nl + l , where l is the wavelength and n is a constant. Using this relationship, we determine that r1 r2 = l . Looking again at the diagram, the red and blue triangles are similar, which means that the ratios of corresponding sides are the same. The ratio of x to L in the red triangle is equal to the ratio of l to d in the blue triangle. For proof of this, visit http://www.physicsclassroom.com/class/light/u12l3c.cfm . From this, we can determine that the wavelength is dependent on x, d, and L: 29 l = xd L Example Problem: Monochromatic light falls on two narrow slits that are 0.0190 mm apart. A ﬁrst order fringe is 21.1 mm from the central line. The screen (back wall) is 0.600 m from the slits. What is the wavelength of the light? www.ck12.org Solution: l = xd L = (0:021 m)(0:000019 m) (0:600 m) = 6:68 107 m Summary • The last particle of medium at the back corner of an impenetrable barrier will act as a point source and produce circular waves. • Diffraction is the bending of waves around a corner. • Constructive interference occurs when two wave crests overlap, doubling the wave amplitude at that location. • Destructive interference occurs when a wave crest overlaps with a trough, causing them to cancel out. • Light is a wave, and creates an interference pattern in the double slit experiment. • An interference pattern consists of alternating bright and dark lines; the bright lines are called fringes. • In a double slit experiment, the wavelength can be calculated using this equation: l = xd L Practice http://www.youtube.com/watch?v=AMBcgVlamoU Follow up questions. 1. When the amplitude of waves add, it is called _________________ interference. 2. When the amplitude of waves subtract, it is called _________________ interference. 3. What do we call the phenomenon of light bending around a corner? Review 1. Destructive interference in waves occurs when (a) two troughs overlap. (b) crests and troughs align. (
c) two crests overlap. (d) a crest and a trough overlap. 2. Bright bands in interference patterns result from (a) destructive diffraction. (b) destructive interference. (c) constructive diffraction. (d) constructive interference. 3. In a double slit experiment with slits 1:00 105 m apart, light casts the ﬁrst bright band 3:00 102 m from the central bright spot. If the screen is 0.650 m away, what is the wavelength of this light? (a) 510 nm (b) 390 nm (c) 430 nm (d) 460 nm 4. Violet light falls on two slits separated by 1:90 105 m . A ﬁrst order bright line appears 13.2 mm from the central bright spot on a screen 0.600 m from the slits. What is the wavelength of the violet light? 5. Suppose in the previous problem, the light was changed to yellow light with a wavelength of 5:96 107 m while the slit separation and distance from screen to slits remained the same. What would be the distance from the central bright spot to the ﬁrst order line? 30 www.ck12.org Concept 8. Double Slit Interference 6. Light with a wavelength of 6:33 107 m is used in a double slit experiment. The screen is placed 1.00 m from the slits and the ﬁrst order line is found 65.5 mm from the central bright spot. What is the separation between the slits? • diffraction: Change in the directions and intensities of a group of waves after passing by an obstacle or through an aperture whose size is approximately the same as the wavelength of the waves. • monochromatic: Light having only one wavelength. • constructive interference: The interference of two or more waves of equal frequency and phase, resulting in their mutual reinforcement and producing a single amplitude equal to the sum of the amplitudes of the individual waves. • destructive interference: The interference of two waves of equal frequency and opposite phase, resulting in their cancellation where the negative displacement of one always coincides with the positive displacement of the other. References 1. . . CC BY-NC-SA 2. . Kathy Shield . CC BY-NC-SA 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 4. Pieter Kuiper. http://commons.wikimedia.org/wiki/File:SodiumD_two_double_slits.jpg . Public Domain 5. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 31 CONCEPT 9 www.ck12.org Diffraction Gratings • Understand the structure of a diffraction grating. • Explain the interference pattern formed by diffraction gratings. • Use diffraction grating interference patterns to calculate the wavelength of light. Suppose we had a light bulb that emitted exactly four frequencies of light; one frequency in each of the colors red, yellow, green, and blue. To our eye, this bulb would appear white because the combination of those four colors produces white light. If viewed through a diffraction grating, however, each color of light would be visible. The original white light bulb is visible in the center of the image, and interference causes the light bulb to appear in each color to the left and the right. Diffraction Gratings Diffraction gratings are composed of a multitude of slits lined up side by side, not unlike a series of double slits next to each other. They can be made by scratching very ﬁne lines with a diamond point on glass, or by pressing plastic ﬁlm on glass gratings so that the scratches are replicated. The clear places between the scratches behave as slits similar to the slits in a double slit experiment and the gratings form interference patterns in the same general way that double slits do. With more slits, there are more light waves out of phase with each other, causing more destructive interference. Compared to the interference pattern of a double slit, the diffraction grating interference pattern’s colors are spread out further and the dark regions are broader. This allows for more precise wavelength determination than with double slits. The image below shows the diffraction pattern emanating from a white light. 32 www.ck12.org Concept 9. Diffraction Gratings Also in this image is the measurement for q, which can be used to calculate the wavelength of the original light source. The equation from the double slit experiment can be adjusted slightly to work with diffraction gratings. Where l is the wavelength of light, d is the distance between the slits on the grating, and q is the angle between the incident (original) light and the refracted light, l = xd L = d sin q (Note that x Looking at the equation, x = lL d , it should be apparent that as the distance between the lines on the grating become smaller and smaller, the distance between the images on the screen will become larger and larger. Diffraction gratings are often identiﬁed by the number of lines per centimeter; gratings with more lines per centimeter are usually more useful because the greater the number of lines, the smaller the distance between the lines, and the greater the separation of images on the screen. L = sin q , using the small angle approximation theorem.) Example Problem: A good diffraction grating has 2500 lines/cm. What is the distance between two lines on the grating? Solution: d = 1 2500 cm1 = 0:00040 cm Example Problem: Using a diffraction grating with a spacing of 0.00040 cm, a red line appears 16.5 cm from the central line on the screen. The screen is 1.00 m from the grating. What is the wavelength of the light? Solution: l = xd L = (0:165 m)(4:0106 m) 1:00 m = 6:6 107 m Summary • Diffraction gratings can be made by blocking light from traveling through a translucent medium; the clear places behave as slits similar to the slits in a double slit experiment. • Diffraction gratings form interference patterns much like double slits, though brighter and with more space between the lines. • The equation used with double slit experiments to measure wavelength is adjusted slightly to work with diffraction gratings. l = xd L = d sin q 33 www.ck12.org Practice http://vimeo.com/39495562 Follow up questions. 1. How does a diffraction grating differ from single or double slit? 2. What happens when you increase the number of slits in a diffraction grating? Review 1. White light is directed toward a diffraction grating and that light passes through the grating, causing its monochromatic bands appear on the screen. Which color will be closest to the central white? 2. Three discrete spectral lines occur at angles of 10.1°, 13.7°, and 14.8° respectively in the ﬁrst order spectrum. If the grating has 3660 lines/cm, what are the wavelengths of these three colors of light? 3. A 20.0 mm section of diffraction grating has 6000 lines. At what angle will the maximum bright band appear if the wavelength is 589 nm? 4. Laser light is passed through a diffraction grating with 7000 lines/cm. The ﬁrst order maximum on the screen is 25° away from the central maximum. What is the wavelength of the light? • diffraction grating: A glass, plastic, or polished metal surface having a large number of very ﬁne parallel grooves or slits and used to produce optical spectra by diffraction. References 1. CK-12 Foundation - Samantha Bacic, using light bulb images copyright Ruslan Klimovich, 2013. http:// www.shutterstock.com . Used under license from Shutterstock.com 2. Candace (Flickr: cosmiccandace). http://www.ﬂickr.com/photos/candace/315205005/ . CC-BY 2.0 34 www.ck12.org Concept 10. Wave-Particle Theory CONCEPT 10 Wave-Particle Theory • State the wave-particle theory of electromagnetic radiation. • Describe a photon. • Identify evidence that electromagnetic radiation is both a particle and a wave. What a beautiful sunset! You probably know that sunlight travels in waves through space from the sun to Earth. But do you know what light really is? Is it just energy, or is it something else? In this article you’ll ﬁnd out that light may be more than it seems. The Question Electromagnetic radiation, commonly called light, is the transfer of energy by waves called electromagnetic waves. These waves consist of vibrating electric and magnetic ﬁelds. Where does electromagnetic energy come from? It is released when electrons return to lower energy levels in atoms. Electromagnetic radiation behaves like continuous waves of energy most of the time. Sometimes, however, electromagnetic radiation seems to behave like discrete, or separate, particles rather than waves. So does electromagnetic radiation consist of waves or particles? The Debate This question about the nature of electromagnetic radiation was debated by scientists for more than two centuries, starting in the 1600s. Some scientists argued that electromagnetic radiation consists of particles that shoot around like tiny bullets. Other scientists argued that electromagnetic radiation consists of waves, like sound waves or water waves. Until the early 1900s, most scientists thought that electromagnetic radiation is either one or the other and that scientists on the other side of the argument were simply wrong. Q: Do you think electromagnetic radiation is a wave or a particle? A: Here’s a hint: it may not be a question of either-or. Keep reading to learn more. 35 www.ck12.org A New Theory In 1905, the physicist Albert Einstein developed a new theory about electromagnetic radiation. The theory is often called the wave-particle theory . It explains how electromagnetic radiation can behave as both a wave and a particle. Einstein argued that when an electron returns to a lower energy level and gives off electromagnetic energy, the energy is released as a discrete “packet” of energy. We now call such a packet of energy a photon . According to Einstein, a photon resembles a particle but moves like a wave. You can see this in the Figure 10.1 . The theory posits that waves of photons traveling through space or matter make up electromagnetic radiation. FIGURE 10.1 Energy of a Photon A photon isn’t a ﬁxed amount of energy. Instead, the amount of energy in a photon depends on the frequency of the electromagnetic wave. The frequency of a wave is the number of waves that pass a ﬁxed point in a given amount
of time, such as the number of waves per second. In waves with higher frequencies, photons have more energy. Evidence for the Wave-Particle Theory After Einstein proposed his theory, evidence was discovered to support it. For example, scientists shone laser light through two slits in a barrier made of a material that blocked light. You can see the setup of this type of experiment in the sketch below. Using a special camera that was very sensitive to light, they took photos of the light that passed through the slits. The photos revealed tiny pinpoints of light passing through the double slits. This seemed to show that light consists of particles. However, if the camera was exposed to the light for a long time, the pinpoints accumulated in bands that resembled interfering waves. Therefore, the experiment showed that light seems to consist of particles that act like waves. 36 www.ck12.org Concept 10. Wave-Particle Theory FIGURE 10.2 Summary • Electromagnetic radiation behaves like waves of energy most of the time, but sometimes it behaves like particles. From the 1600s until the early 1900s, most scientists thought that electromagnetic radiation consists either of particles or of waves but not both. • In 1905, Albert Einstein proposed the wave-particle theory of electromagnetic radiation. This theory states that electromagnetic energy is released in discrete packets of energy—now called photons—that act like waves. • After Einstein presented his theory, scientists found evidence to support it. For example, double-slit experi- ments showed that light consists of tiny particles that create patterns of interference just as waves do. Vocabulary • photon : Tiny “packet” of electromagnetic radiation that is released when an electron returns to a lower. • wave-particle theory : Theory proposed by Albert Einstein that electromagnetic energy is released in discrete packets of energy (now called photons) that act like waves. Practice Watch the animation “Let There Be Light” at the following URL. Then create a timeline of ideas and discoveries about the nature of light. http://www.abc.net.au/science/explore/einstein/lightstory.htm Review 1. Why did scientists debate the nature of electromagnetic radiation for more than 200 years? 2. State Einstein’s wave-particle theory of electromagnetic radiation. 3. What is a photon? 4. After Einstein proposed his wave-particle theory, how did double-slit experiments provide evidence to support the theory? 37 References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Zachary Wilson. . CC-BY-NC-SA 3.0 www.ck12.org 38 Physics Unit 12: Static Electricity Patrick Marshall Jean Brainard, Ph.D. Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Static Electricity and Static Discharge 2 Electric Charge and Electric Force 3 Transfer of Electric Charge 4 Forces on Charged Objects 5 Coulomb’s Law 6 The Electric Field 7 Electric Potential 1 4 7 12 19 22 25 iv www.ck12.org Concept 1. Static Electricity and Static Discharge CONCEPT 1 Static Electricity and Static Discharge • Describe static electricity. • Explain static discharge. • Outline how lightning occurs. You’re a thoughtful visitor, so you wipe your feet on the welcome mat before you reach out to touch the brass knocker on the door. Ouch! A spark suddenly jumps between your hand and the metal, and you feel an electric shock. Q: Why do you think an electric shock occurs? A: An electric shock occurs when there is a sudden discharge of static electricity. What Is Static Electricity? Static electricity is a buildup of electric charges on objects. Charges build up when negative electrons are transferred from one object to another. The object that gives up electrons becomes positively charged, and the object that accepts the electrons becomes negatively charged. This can happen in several ways. One way electric charges can build up is through friction between materials that differ in their ability to give up or accept electrons. When you wipe your rubber-soled shoes on the wool mat, for example, electrons rub off the mat onto your shoes. As a result of this transfer of electrons, positive charges build up on the mat and negative charges build up on you. Once an object becomes electrically charged, it is likely to remain charged until it touches another object or at least comes very close to another object. That’s because electric charges cannot travel easily through air, especially if the air is dry. 1 www.ck12.org Q: You’re more likely to get a shock in the winter when the air is very dry. Can you explain why? A: When the air is very dry, electric charges are more likely to build up objects because they cannot travel easily through the dry air. This makes a shock more likely when you touch another object. Static Discharge What happens when you have become negatively charged and your hand approaches the metal doorknocker? Your negatively charged hand repels electrons in the metal, so the electrons move to the other side of the knocker. This makes the side of the knocker closest to your hand positively charged. As your negatively charged hand gets very close to the positively charged side of the metal, the air between your hand and the knocker also becomes electrically charged. This allows electrons to suddenly ﬂow from your hand to the knocker. The sudden ﬂow of electrons is static discharge . The discharge of electrons is the spark you see and the shock you feel. Watch the animation “John Travoltage” at the following URL to see an example of static electricity and static discharge. http://www.cabrillo.edu/~jmccullough/Physics/Electric_Forces_Fields.html How Lightning Occurs Another example of static discharge, but on a much larger scale, is lightning. You can see how it occurs in the following diagram and animation as you read about it below. http://micro.magnet.fsu.edu/electromag/java/lightning/index.html FIGURE 1.1 During a rainstorm, clouds develop regions of positive and negative charge due to the movement of air molecules, water drops, and ice particles. The negative charges are concentrated at the base of the clouds, and the positive charges are concentrated at the top. The negative charges repel electrons on the ground beneath them, so the ground below the clouds becomes positively charged. At ﬁrst, the atmosphere prevents electrons from ﬂowing away from areas of negative charge and toward areas of positive charge. As more charges build up, however, the air between the oppositely charged areas also becomes charged. When this happens, static electricity is discharged as bolts of lightning. At the URL below, you can watch an awesome slow-motion lightning strike. Be sure to wait for the real-time lightning strike at the end of the video. You’ll be amazed when you realize how much has occurred during that split-second discharge of static electricity. 2 www.ck12.org Concept 1. Static Electricity and Static Discharge http://www.youtube.com/watch?v=Y8oN0YFAXWQ&feature=related MEDIA Click image to the left for more content. Summary • Static electricity is a buildup of electric charges on objects. It occurs when electrons are transferred from one object to another. • A sudden ﬂow of electrons from one charged object to another is called static discharge. • Examples of static discharge include lightning and the shock you sometimes feel when you touch another object. Vocabulary • static discharge : Sudden ﬂow of electrons from an object that has a buildup of charges. • static electricity : Buildup of charges on an object that occurs through induction. Practice Watch the video at the following URL. Then answer the discussion questions. Read the background essay if you need help with any of the questions. http://www.teachersdomain.org/resource/phy03.sci.phys.mfe.zsnap/ Review 1. What is static electricity? 2. How does static discharge occur? 3. Explain why a bolt of lightning is like the spark you might see when you touch a metal object and get a shock. References 1. Zachary Wilson. . CC BY-NC 3.0 3 CONCEPT 2 Electric Charge and Electric Force www.ck12.org • Deﬁne electric charge. • Describe electric forces between charged particles. A lightning bolt is like the spark that gives you a shock when you touch a metal doorknob. Of course, the lightning bolt is on a much larger scale. But both
the lightning bolt and spark are a sudden transfer of electric charge. Introducing Electric Charge Electric charge is a physical property of particles or objects that causes them to attract or repel each other without touching. All electric charge is based on the protons and electrons in atoms. A proton has a positive electric charge, and an electron has a negative electric charge. In the Figure 2.1 , you can see that positively charged protons (+) are located in the nucleus of the atom, while negatively charged electrons (-) move around the nucleus. Electric Force When it comes to electric charges, opposites attract, so positive and negative particles attract each other. You can see this in the diagram below. This attraction explains why negative electrons keep moving around the positive nucleus of the atom. Like charges, on the other hand, repel each other, so two positive or two negative charges push apart. This is also shown in the diagram. The attraction or repulsion between charged particles is called electric force . The strength of electric force depends on the amount of electric charge on the particles and the distance between 4 www.ck12.org Concept 2. Electric Charge and Electric Force FIGURE 2.1 them. Larger charges or shorter distances result in greater force. You can experiment with electric force with the animation at the following URL. http://www.colorado.edu/physics/2000/waves_particles/wavpart2.html FIGURE 2.2 Q: How do positive protons stay close together inside the nucleus of the atom if like charges repel each other? A: Other, stronger forces in the nucleus hold the protons together. Summary • Electric charge is a physical property of particles or objects that causes them to attract or repel each other without touching. • Particles that have opposite charges attract each other. Particles that have like charges repel each other. The force of attraction or repulsion is called electric force. 5 www.ck12.org Vocabulary • electric charge : Physical property of particles or objects that causes them to attract or repel each other without touching; may be positive or negative. • electric force : Force of attraction or repulsion between charged particles. Practice Read the ﬁrst four boxes of text at the following URL. Then write a concise paragraph explaining why direction E is the correct answer to the quick quiz. http://www.physics.wisc.edu/undergrads/courses/208-f07/Lectures/lect6.pdf Review 1. What is electric charge? 2. Make a simple table summarizing electric forces between charged particles. References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Zachary Wilson. . CC BY-NC 3.0 6 www.ck12.org Concept 3. Transfer of Electric Charge CONCEPT 3 Transfer of Electric Charge • Describe how the transfer of electrons changes the charge of matter. • Relate the transfer of electrons to the law of conservation of charge. • Compare and contrast three ways that electric charge can be transferred. Why is this girl’s hair standing straight up? She is touching a device called a van de Graaff generator. The dome on top of the device has a negative electric charge. When the girl places her hand on the dome, she becomes negatively charged as well—right down to the tip of each hair! You can see a video demonstrating a van de Graff generator at this URL: http://www.youtube.com/watch?v=SREXQWAIDJk MEDIA Click image to the left for more content. Q: Why is the man’s hair standing on end? A: All of the hairs have all become negatively charged, and like charges repel each other. Therefore, the hairs are pushing away from each other, causing them to stand on end. 7 www.ck12.org Transferring Electrons The man pictured above became negatively charged because electrons ﬂowed from the van de Graaff generator to him . Whenever electrons are transferred between objects, neutral matter becomes charged. This occurs even with individual atoms. Atoms are neutral in electric charge because they have the same number of negative electrons as positive protons. However, if atoms lose or gain electrons, they become charged particles called ions. You can see how this happens in the Figure 3.1 . When an atom loses electrons, it becomes a positively charged ion, or cation. When an atom gains electrons, it becomes a negative charged ion, or anion. FIGURE 3.1 Conservation of Charge Like the formation of ions, the formation of charged matter in general depends on the transfer of electrons, either between two materials or within a material. Three ways this can occur are referred to as conduction, polarization, and friction. All three ways are described below. However, regardless of how electrons are transferred, the total charge always remains the same. Electrons move, but they aren’t destroyed. This is the law of conservation of charge . Conduction The transfer of electrons from the van de Graaff generator to the man is an example of conduction. Conduction occurs when there is direct contact between materials that differ in their ability to give up or accept electrons. A van de Graff generator produces a negative charge on its dome, so it tends to give up electrons. Human hands are positively charged, so they tend to accept electrons. Therefore, electrons ﬂow from the dome to the man’s hand when they are in contact. You don’t need a van de Graaff generator for conduction to take place. It may occur when you walk across a wool carpet in rubber-soled shoes. Wool tends to give up electrons and rubber tends to accept them. Therefore, the carpet transfers electrons to your shoes each time you put down your foot. The transfer of electrons results in you becoming negatively charged and the carpet becoming positively charged. 8 www.ck12.org Polarization Concept 3. Transfer of Electric Charge Assume that you have walked across a wool carpet in rubber-soled shoes and become negatively charged. If you then reach out to touch a metal doorknob, electrons in the neutral metal will be repelled and move away from your hand before you even touch the knob. In this way, one end of the doorknob becomes positively charged and the other end becomes negatively charged. This is called polarization. Polarization occurs whenever electrons within a neutral object move because of the electric ﬁeld of a nearby charged object. It occurs without direct contact between the two objects. The Figure 3.2 models how polarization occurs. FIGURE 3.2 Q: What happens when the negatively charged plastic rod in the diagram is placed close to the neutral metal plate? A: Electrons in the plate are repelled by the positive charges in the rod. The electrons move away from the rod, causing one side of the plate to become positively charged and the other side to become negatively charged. Friction Did you ever rub an inﬂated balloon against your hair? You can see what happens in the Figure 3.3 . Friction between the balloon and hair cause electrons from the hair to “rub off” on the balloon. That’s because a balloon attracts electrons more strongly than hair does. After the transfer of electrons, the balloon becomes negatively charged and the hair becomes positively charged. The individual hairs push away from each other and stand on end because like charges repel each other. The balloon and the hair attract each other because opposite charges attract. Electrons are transferred in this way whenever there is friction between materials that differ in their ability to give up or accept electrons. Watch the animation “Balloons and Static Electricity” at the following URL to see how electrons are transferred by friction between a sweater and a balloon. http://www.cabrillo.edu/~jmccullough/Physics/Electr ic_Forces_Fields.html Q: If you rub a balloon against a wall, it may stick to the wall. Explain why. A: Electrons are transferred from the wall to the balloon, making the balloon negatively charged and the wall positively charged. The balloon sticks to the wall because opposite charges attract. Summary • Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. 9 www.ck12.org FIGURE 3.3 • Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same. This is the law of conservation of charge. • Conduction occurs when there is direct contact between materials that differ in their ability to give up or accept electrons. • Polarization is the movement of electrons within a neutral object due to the electric ﬁeld of a nearby charged object. It occurs without direct contact between the two objects. • Electrons are transferred whenever there is friction between materials that differ in their ability to give up or accept electrons. Vocabulary • law of conservation of charge : Law stating that charges are not destroyed when they are transferred between two materials or within a material, so the total charge remains the same. Practice At the following URL, review how charges are transferred through friction. Watch the animation and read the list of more-positive to less-positive materials. Then answer the questions below. http://www.regentsprep.org/regents/physics/phys03/atribo/default.htm 1. If you rub glass with a piece of plastic wrap, will the glass become positively or negatively charged? 2. Assume that after you pet your dog with very dry hands, you touch a metal doorknob and get a shock. Is electric charge transferred from your hand to the doorknob or the other way around? Review 1. How is charge transferred by a van de Graaff generator? 2. Compare and contrast the formation of cations and anions. 3. State the law of conservation of charge. 4. Explain how conduction and polarization occur, using the example of walking across a wool carpet in rubber- soled shoes and then reaching out to touch a metal doorknob. 10 www.ck12.org Concept 3. Transfer of Electric Charge 5. Predict what will happen to the charges of a plastic
comb and a piece of tissue paper if you rub the tissue paper on the comb. ( Hint : Plastic tends to accept electrons and tissue paper tends to give up electrons.) References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Christopher Auyeung. . CC BY-NC 3.0 3. Flickr:olga.palma. . CC BY 2.0 11 CONCEPT 4 Forces on Charged Objects www.ck12.org • Describe the changes that occur in the sub-atomic arrangement in matter when charged. • Describe how to charge an object. • Deﬁne conductors and insulators. • Understand the difference between conduction and induction. • Summarize the forces between charged objects. Lightning is the discharge of static electricity that has built up on clouds. Every year, the earth experiences an average of 25 million lightning strikes. Lightning bolts travel at speeds up to 60,000 miles per second, and can reach temperatures of 50,000°F, which is ﬁve times the temperature of the surface of the sun. The energy contained in a single lightning strike could light a 100 Watt light bulb 24 hours per day for 90 days. Forces on Charged Objects Electric charges exist within the atom. At the turn of the 20th century, J. J. Thomson and Ernest Rutherford determined that atoms contain very light-weight negatively charged particles called electrons and more massive, positively charged particles called protons . The protons are lodged in the nucleus of the atoms, along with the neutrally charged particles called neutrons, while the electrons surround the nucleus. When the number of electrons in the electron cloud and the number of protons in the nucleus are equal, the object is said to be neutral . Changes to the nucleus of an atom require tremendous amounts of energy, so protons are not easily gained or lost by atoms. Electrons, on the other hand, are held fairly loosely and can often be removed quite easily. When an object loses some electrons, the remaining object is now positively charged because it has an excess of protons. The electrons may either remain free or may attach to another object. In that case, the extra electrons cause that object to become negatively charged. Atoms that have lost electrons and become positively charged are called positive ions, and atoms that have gained electrons and become negatively charged are called negative ions . Electrons can be removed from some objects using friction, simply by rubbing one substance against another substance. There are many examples of objects becoming charged by friction, including a rubber comb through 12 www.ck12.org Concept 4. Forces on Charged Objects hair, and a balloon on a sweater. In both these instances, the electrons move from the second object to the ﬁrst, causing the ﬁrst object to become negatively charged and the second one positively charged. Friction between the tires on a moving car and the road cause the tires to become charged, and wind causes friction between clouds and air which causes clouds to become charged and can result in tremendous bolts of lightning. A common method of producing charge in the lab is to rub cat or rabbit fur against stiff rubber, producing a negative charge on the rubber rod. If you hold a rubber rod on one end and rub only the tip of the other end with a fur, you will ﬁnd that only the tip becomes charged. The electrons you add to the tip of the rod remain where you put them instead of moving around on the rod. Rubber is an insulator. Insulators are substances that do not allow electrons to move through them. Glass, dry wood, most plastics, cloth, and dry air are common insulators. Materials that allow electrons to ﬂow freely are called conductors. Metals have at least one electron that can move around freely, and all metals are conductors. Forces are exerted on charged objects by other charged objects. You’ve probably heard the saying "opposites attract," which is true in regards to charged particles. Opposite charges attract each other, while like charges repulse each other. This can be seen in the image below. When two negatively charged objects are brought near each other, a repulsive force is produced. When two positively charged objects are brought near each other, a similar repulsive force is produced. When a negatively charged object is brought near a positively charged object, an attractive force is produced. Neutral objects have no inﬂuence on each other. A laboratory instrument used to analyze and test for static charge is called an electroscope. Seen below, an electroscope consists of a metal knob connected by a metal stem to two very lightweight pieces of metal called leaves, shown in yellow. The leaves are enclosed in a box to eliminate stray air currents. 13 www.ck12.org When a negatively charged object is brought near the knob of a neutral electroscope, the negative charge repels the electrons in the knob, and those electrons move down the stem into the leaves. Excess electrons ﬂow from the rod into the ball, and then downwards making both leaves negatively charged. Since both leaves are negatively charged, they repel each other. When the rod is removed, the electroscope will remain charged because of the extra electrons added to it. Conversely, if the rod is brought near the knob but doesn’t touch it, the electroscope will appear the same while the rod is near. That is, the negative charge in the rod repels the electrons in the ball, causing them to travel down to the leaves. The leaves will separate while the rod is nearby. No extra electrons were added to the electroscope, meaning that the electrons in the electroscope will redistribute when the negatively charged rod is taken away. The leaves return to neutral, and they stop repelling each other. If the rod touches the knob, the electroscope leaves are permanently charged but if the rod is brought near but does not touch the knob, the electroscope leaves are only temporarily charged. If the leaves are permanently charged and the rod removed, the electroscope can then be used to determine the type of unknown charge on an object. If the electroscope has been permanently negatively charged, and a negatively charge object is brought near the knob, the leaves will separate even further, showing the new object has the same charge as the leaves. If a positively charged object is brought near a negatively charged electroscope, it will attract some of the excess electrons up the stem and out of the leaves, causing the leaves to come slightly together. Similar to the results of a negatively charged rod, if a positively charged rod is brought near the knob of a neutral electroscope, it will attract some electrons up from the leaves onto the knob. That process causes both of the leaves to 14 www.ck12.org Concept 4. Forces on Charged Objects be positively charged (excess protons), and the leaves will diverge. If the positively charged rob is actually touched to the knob, the rob will remove some electrons and then when the rob is removed, the electroscope will remain positively charged. This is a permanent positive charge. Charging an object by touching it with another charged object is called charging by conduction. By bringing a charged object into contact with an uncharged object, some electrons will migrate to even out the charge on both objects. Charging by conduction gives the previously uncharged object a permanent charge. An uncharged object can also be charged using a method called charging by induction . This process allows a change in charge without actually touching the charged and uncharged objects to each other. Imagine a negatively charged rod held near the knob, but not touching. If we place a ﬁnger on the knob, some of the electrons will escape into our body, instead of down the stem and into the leaves. When both our ﬁnger and the negatively charged rod are removed, the previously uncharged electroscope now has a slight positive charge. It was charged by induction. Notice that charging by induction causes the newly charged object to have the opposite charge as the originally charged object, while charging by conduction gives them both the same charge. Summary • Electric charges exist with the atom. • Atoms contain light-weight, loosely held, negatively charged particles called electrons and heavier, tightly- held, positvely charged particles called protons. • When the number of electrons and the number of protons are equal, the object is neutral. • The loss of electrons gives an ion a positive charge, while the gain of electrons gives it a negative charge. • Materials that allow electrons to ﬂow freely are called conductors, while those that do not are called insulators. • Opposite charges attract, and like charges repel. • Charging an object by touching it with another charged object is called charging by conduction. Practice The following video shows a young woman placing her hands on a Van de Graf generator which then gives her a static charge. Use this resource to answer the two questions that follow. http://www.youtube.com/watch?v=87DqbdqBx8U 15 www.ck12.org MEDIA Click image to the left for more content. 1. What happens to her hair when she touches a ground? 2. What happens to her hair when she steps off the platform? This video shows the static charge from the Van de Graf generator. http://www.youtube.com/watch?v=prgu6AvauuI MEDIA Click image to the left for more content. This video demonstrates superconductivity that occurs at extremely low temperatures. http://www.youtube.com/watch?feature=player_embedded&v=nWTSzBWEsms MEDIA Click image to the left for more content. Additional Practice Questions: 1. When a glass rod is rubbed with a silk cloth and the rod becomes positively charged, (a) electrons are removed from the rod. (b) protons are added to the silk. (c) protons are removed from the silk. (d) the silk remains neutral. 2. Electric charge is (a) found only in a conductor. (b) found only in insulators. (c) conserved. (d) not conserved. 3. When two objects are rubbed together and they become oppositely charged, t
hey are said to be charge by (a) conduction. (b) induction. 16 www.ck12.org Concept 4. Forces on Charged Objects (c) friction. (d) grounding. 4. Two objects each carry a charge and they attract. What do you know about the charge of each object? (a) They are both charged positively. (b) They have opposite charged from each other. (c) They are both charged negatively. (d) Any of the above are possible. 5. A material that easily allows the ﬂow of electric charge through it is called a(n) (a) insulator. (b) conductor. (c) semiconductor. (d) heat sink. 6. What is the most common way of acquiring a positive static electrical charge? (a) by losing electrons (b) by gain protons (c) by losing protons (d) by gaining electrons (e) by switching positions of electrons and protons in the atom Review 1. How does friction generate static electricity? (a) Friction heats the materials, thus causing electricity. (b) Rubbing materials together displaces atoms, causing sparks to ﬂy. (c) Rubbing materials together can strip electrons off atoms, causing one material to become positive and the other to become negative. (d) Rubbing materials together causes neutrons and electrons to trade places. (e) None of the above. 2. What electrical charge does an electron have? (a) A negative charge. (b) A positive charge. (c) A neutral charge. (d) May be any of the above. (e) None of the above. 3. What happens when opposite charges get close to each other? (a) They repel each other. (b) They attract each other. (c) Nothing happens. (d) They attract surrounding objects. (e) They repel surrounding objects. 4. What is an electrical conductor? (a) A material that allows electrons to travel through it freely. (b) A material that doesn’t allow electrons to travel through it freely. (c) A material that melts at low temperature. (d) A material that creates free electrons. (e) None of the above. 5. Which of the following is a good insulator of electricity? 17 www.ck12.org (a) Copper (b) Iron (c) Rubber (d) Salt water (e) None of these. • electrons: A fundamental sub-atomic particle, meaning it cannot be broken into smaller particles. Electrons are found in the “electron cloud” surrounding an atomic nucleus, or they may break free and exist as a free electron. • protons: A stable, positively charged, sub-atomic particle, found in atomic nuclei in numbers equal to the atomic number of the element. • neutral: A neutral particle, object, or system is one that has a net electric charge of zero. • conductors: Materials through which electric charge can pass. • insulator: Substances that block or retard the ﬂow of electrical current or charge. • positive ions: An atom or a group of atoms that has acquired a net positive charge by losing one or more electrons. • negative ions: An atom or a group of atoms that has acquired a net negative charge by gaining one or more electrons. • ions: An atom or a group of atoms that has acquired a net electric charge by gaining or losing one or more electrons. • electroscope: An instrument used to detect the presence and sign of an electric charge by the mutual attraction or repulsion of metal foils. • charging by conduction: Involves the contact of a charged object to a neutral object. • charging by induction: A method used to charge an object without actually touching the object to any other charged object. References 1. Courtesy of NOAA Photo Library, NOAA Central Library; OAR/ERL/National Severe Storms Laboratory (NSSL). http://www.photolib.noaa.gov/htmls/nssl0016.htm . Public Domain 2. Sweater: Image copyright Sibiryanka, 2013; Balloon: Image copyright simpleman, 2013. http://www.shut terstock.com . Used under licenses from Shutterstock.com 3. CK-12 Foundation - Christopher Auyeung. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 5. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 6. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 18 www.ck12.org Concept 5. Coulomb’s Law CONCEPT 5 Coulomb’s Law • State Coulomb’s Law. • Describe how electric force varies with charge and separation of charge. • State the SI unit of charge. • Solve problems using Coulomb’s Law. Electric cars are becoming more popular. One large advantage for electric cars is the low cost of operation, which may become an ever bigger advantage as gas prices climb. Energy costs for electric cars average about one-third of the cost for gasoline engine cars, but they can only travel about 200 miles per charge at this point. These cars run using the science of electrical charges and forces. Coulomb’s Law The questions regarding the relationship between the electrical force, the size of the charge, and the separation between the charges were solved by Charles Coulomb in 1785. He determined that electrical force between two charges is directly related to the size of the charges and inversely proportional to the distance between the charges. This is known as Coulomb’s Law. Fe = Kq1q2 d2 In this equation, q 1 and q 2 are the two charges, d is the distance between the two charges, and K is a constant of proportionality. F e is the electric force , which occurs as a result of interactions between two charged particles. For the purpose of calculating electric forces, we assume all charge is a point charge , in which the entire charge of the particle is located in a massless point. The SI unit of charge is the coulomb, C , which is the charge of 6:25 1018 electrons. The charge on a single electron is 1:60 1019 C . The charge on a single electron is known as the elementary charge . The charge on a proton is the same magnitude but opposite in sign. When the charges are measured in coulombs, the distance in meters, and the force in Newtons, the constant K is 9:0 109 N m2=C2 . 19 www.ck12.org The electrical force, like all forces, is a vector quantity. If the two charges being considered are both positive or both negative, the sign of the electrical force is positive and this force is repulsive. If the two charges are opposite in sign, the force will have a negative sign and the force is attractive. Example Problem: Object A has a positive charge of 6:0 106 C . Object B has a positive charge of 3:0 106 C . If the distance between A and B is 0.030 m, what is the force on A? Solution: Fe = Kq1q2 d2 = (9:0109 Nm2=C2)(6:0106 C)(3:0106 C) = 180 N (0:030 m)2 The positive sign of the force indicates the force is repulsive. This makes sense, because both objects have a positive charge. Example Problem: 6:0 106 C . Calculate the total force on q 2 . In the sketch below, the charges are q1 = 10:0 106 C; q2 = 2:0 106 C , and q3 = Solution: Fe = Kq1q2 d2 = (9:0109 Nm2=C2)(10:0106 C)(2:0106 C) (2:0 m)2 = 0:045 N (towards q3) Fe = Kq2q3 d2 = (9:0109 Nm2=C2)(2:0106 C)(6:0106 C) (4:0 m)2 = 0:007 N (towards q3) Since the two forces act in the same direction, their absolute values can be added together; the total force on q 2 is 0.052 N towards q 3 . Summary • Coulomb determined that electrical force between two charges is directly related to the size of the charges and inversely proportional to the distance between the charges: Fe = Kq1q2 d2 • The SI unit of charge is the coulomb, C , which is the charge of 6:25 1018 electrons. • The charge on a single electron is 1:60 1019 C and is known as the elementary charge. • The electrical force is a vector quantity that is positive in repulsion and negative in attraction. Practice The following video covers Coulomb’s Law. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=rYjo774UpHI MEDIA Click image to the left for more content. 1. What happens when like charges are placed near each other? 2. What happens when opposite charged are placed near each other? 3. What happens to the force of attraction if the charges are placed closer together? 20 www.ck12.org Concept 5. Coulomb’s Law Practice problems on Coulomb’s Law. http://physics.info/coulomb/problems.shtml Review 1. Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m: (a) Will the charges attract or repel? (b) What is the magnitude of the force between them? (c) If the distance between them is doubled, what does the force become? 2. What is the electrical force between two balloons, each having 5.00 C of charge, that are 0.300 m apart? 3. Two spheres are charged with the same charge of -0.0025 C and are separated by a distance of 8.00 m. What is the electrical force between them? 4. A red foam ball and a blue foam ball are 4.00 m apart. The blue ball has a charge of 0.000337 C and is attracting the red ball with a force of 626 N. What is the charge on the red ball? • Coulomb’s Law: States the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them. References 1. Image copyright testing, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic. . CC BY-NC-SA 3.0 21 CONCEPT 6 • Deﬁne an electric ﬁeld. • Solve problems relating to ﬁeld, force, and charge. www.ck12.org The Electric Field A plasma globe, such as the one pictured above, is ﬁlled with a mixture of noble gases and has a high-voltage electrode at the center. The swirling lines are electric discharge lines that connect from the inner electrode to the outer glass insulator. When a hand is placed on the surface of the globe, all the electric discharge travels directly to that hand. The Electric Field Coulomb’s Law gives us the formula to calculate the force exerted on a charge by another charge. On some occasions, however, a test charge suffers an electrical force with no apparent cause. That is, as observers, we cannot see or detect the original charge creating the electrical force. Michael Faraday dealt with this problem by developing the concept of an electric ﬁeld. According to Faraday, a charge creates an electric ﬁeld a
bout it in all directions. If a second charge is placed at some point in the ﬁeld, the second charge interacts with the ﬁeld and experiences an electrical force. Thus, the interaction we observe is between the test charge and the ﬁeld and a second particle at some distance is no longer necessary. The strength of the electric ﬁeld is determined point by point and can only be identiﬁed by the presence of test charge. When a positive test charge, q t , is placed in an electric ﬁeld, the ﬁeld exerts a force on the charge. The ﬁeld strength can be measured by dividing the force by the charge of the test charge. Electric ﬁeld strength is given the symbol E and its unit is Newtons/coulomb. 22 E = Fonqt qt www.ck12.org Concept 6. The Electric Field The test charge can be moved from location to location within the electric ﬁeld until the entire electric ﬁeld has been mapped in terms of electric ﬁeld intensity. Example Problem: A positive test charge of 2:0 105 C is placed in an electric ﬁeld. The force on the test charge is 0.60 N. What is the electric ﬁeld intensity at the location of the test charge? Solution: E = F q = 0:60 N 2:0105 C = 3:0 104 N/C Summary • An electric ﬁeld surrounds every charge and acts on other charges in the vicinity. • The strength of the electric ﬁeld is given by the symbol E , and has the unit of Newtons/coulomb . • The equation for electric ﬁeld intensity is E = F q . Practice The following video covers electric ﬁelds. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=lpb94QF0_mM MEDIA Click image to the left for more content. 1. What does it mean when a force is called a non-contact force? 2. What symbol is used to represent electric ﬁeld strength? 3. What is the relationship between the direction of the electric ﬁeld and the direction of the electric force? Review 1. The weight of a proton is 1:64 1026 N . The charge on a proton is +1:60 1019 C . If a proton is placed in a uniform electric ﬁeld so that the electric force on the proton just balances its weight, what is the magnitude and direction of the ﬁeld? 2. A negative charge of 2:0 108 C experiences a force of 0.060 N to the right in an electric ﬁeld. What is the magnitude and direction of the ﬁeld? 3. A positive charge of 5:0 104 C is in an electric ﬁeld that exerts a force of 2:5 104 N on it. What is the magnitude of the electric ﬁeld at the location of the charge? 4. If you determined the electric ﬁeld intensity in a ﬁeld using a test charge of 1:0 106 C and then repeated the process with a test charge of 2:0 106 C , would the forces on the charges be the same? Would you ﬁnd the value for E ? 5. A 0.16 C charge and a 0.04 C charge are separated by a distance of 3.0 m. At what position between the two charges would a test charge experience an electric ﬁeld intensity of zero? • electric ﬁeld: A region of space characterized by the existence of a force that is generated by an electric charge. • electric ﬁeld intensity: Electric ﬁeld intensity or ﬁeld strength is described as the ratio of force to the amount of test charge. 23 References 1. User:Slimsdizz/Wikipedia. http://commons.wikimedia.org/wiki/File:Glass_plasma_globe.jpg . Public Domain www.ck12.org 24 www.ck12.org Concept 7. Electric Potential CONCEPT 7 Objectives The student will: Electric Potential • Understand how to solve problems using electric potential energy. • Understand how to solve problems using voltage differences. • Understand how to solve problems in a uniform electric ﬁeld. Vocabulary • electric potential: Energy per unit charge. • electric potential difference: The difference in electric potential between two points within an electric ﬁeld. • voltage: The amount of work done by the electric ﬁeld per unit charge in moving a charge between two points in the electric ﬁeld DV = W q , also known as a change in potential energy. Introduction In order to draw an analogy between gravitational potential energy and electrical potential energy , we liken the electric ﬁeld E to the gravitational acceleration g and the mass m of a particle to the charge q of a particle. Of course, g is assumed constant (uniform) when we remain close to the surface of the Earth. As of yet, we have not encountered an example of a uniform electric ﬁeld E . But that won’t stop us from making a prediction! Since the gravitational potential energy of a mass m in a uniform gravitational ﬁeld is PEgravity = (mg)h , we predict the electric potential energy (PEelectric) of a charge (q) in a constant electric ﬁeld is PEelectric = (qE)h . Furthermore, the electric potential (the energy per unit charge) can be deﬁned as Ve = PEelectric will be dropped from now on.) We will discuss electric potential later. (The subscript “e” q It must be understood that, just as with gravity, the electric potential energy and the electric potential are measured at the same point. If a point charge q has electric potential energy PEx1 at point x1 , the electric potential at x1 is Vx1 PEx1 = qVx1 ! Vx1 = PEx1 q q PEi q . Again, only differences in electric potential and electric potential energy are meaningful. That is, DPE or DV ! Vf Vi = PE f The unit of electric potential is called the volt and from the deﬁnition above we see that the volt is equivalent to Coulomb ! V = J Joules C . 25 Electric Potential Difference The electric potential difference is the difference in electric potential between two points within an electric ﬁeld. For example, a 1.5-volt battery has a potential difference of 1.5 volts (written 1.5 V ) between its positive and negative terminals. www.ck12.org Parallel Plate Conductors: A uniform Electric Field The equation E = k q r2 for the electric ﬁeld holds for point charges or for a charge distribution that effectively acts as a point charge. It turns out, however, that if opposite charges are placed on two parallel conducting plates, the electric ﬁeld between the plates is more or less uniform as long as the distance between the plates is much smaller than the dimensions of the plates. The plates can be charged by connecting them to the positive and negative terminals of a battery. A battery contains a substance (called an electrolyte) which causes two dissimilar metals to acquire opposite charges. The two dissimilar metals form the positive and negative terminals of the battery. If a metal plate is connected to the positive terminal of the battery, and another metal plate is connected to the negative terminal of the battery, and the two plates brought closely together, a parallel plate arrangement (parallel-plate conductors) can be constructed with a uniform electric ﬁeld between the plates (seen edge on) in Figure below . We will see later that parallel plate conductors are also referred to as capacitors . FIGURE 7.1 Parallel plates. Just as in the case of the battery, one of the plates of the parallel-plate conductor will be at a higher potential than the other plate. Think, for example, of a standard AA battery with a voltage rating of 1.5V, Figure below . See the link below to learn more about how a battery works. http://phet.colorado.edu/en/simulation/battery-voltage 26 www.ck12.org Concept 7. Electric Potential FIGURE 7.2 volt battery Electrical Potential Energy In our gravitational analogy, the energy that a charge possesses at the plate with the higher potential is analogous to the energy a mass possesses above the ground. Additionally, now that we have found a way to create a uniform electric ﬁeld, we have an analog to a uniform gravitational ﬁeld. If a positive charge +q is placed at the positive plate in Figure below , it will be repelled by the positive charges on the plate and move toward the negative plate. (Think of +q as the object m falling toward the ground.) FIGURE 7.3 A positive charge moving toward the negative plate What is the force acting on the +q charge? Recall that the Coulomb force on a charge placed in an electrostatic ﬁeld is F = qE . The work that the electric ﬁeld does on the charge is equal to the negative change in the potential energy of the charge, just as in the gravitational case. We can ﬁnd an expression for the electric potential energy by ﬁnding the work that is done on the charge. Recall that W = FDx . We write 27 www.ck12.org Wf ield = FDx = (qE)Dx = DPE ! qE(x f xi) = DPE ! qEx f qExi = DPE The expression for the electric potential energy is thus: PEelectrical = qEx . Recall that the equation for the gravitational potential energy is PEgravitational = mgh . We can compare the terms in the gravitational and electrical cases as follows Thus, we see that our prediction for the equation of electric potential energy stated in the introduction of the lesson, was correct! Check Your Understanding 1a. The electrical potential at the negative plate in Figure above is deﬁned as zero volts. What is the electrical potential energy of a charge +q = 15:0µC at the positive plate if the electric ﬁeld between the plates is 25:0 N The positive plate has position 6:00cm = 6:00 102 m according to Figure above . Answer : PEpositive plate = qEx = (15:0 106 C) 25:0 N C 1b. What is the change in the electrical potential energy DPE of the charge +q = 15:0µC if its potential changes from 1.5 V to 1.0 V ? (6:00 102 m) = 3:75 104 J C ? Answer : Just as in the case of a change in gravitational potential energy, the charge must lose potential energy, since it gains kinetic energy. The charge moves from the position xi = 6:00 102 m (1:5 V ) to the position xi = 4:00 102 m (1:0 V ) . DPE = qEx f qExi = qE(x f xi) = N C (15:0 106 C) 25:0 (4:00 102 m 6:00 102 m) = 7:50 106 J: 1c. What is the work done on the charge by the electric ﬁeld? Answer: Wf ield = DPE = (7:50 106) = 7:50 106 J Notice that the electric ﬁeld does positive work on the charge, since the electric force and the displacement of the charge have the same direction. We should recall a very important point: It is only the change in potential energy that is meaningful, whether we are discu
ssing the gravitational potential energy or the electrical potential energy. 28 www.ck12.org Concept 7. Electric Potential 2. An electron placed at the negative plate of a parallel-plate conductor will move toward the positive plate. The potential energy of the electron: A. Decreases B. Increases C. Remains the same. Answer : The correct answer is A. The electron is repelled by the negative charges of the conducting plate and therefore gains kinetic energy. Just as an object that is dropped gains kinetic energy and loses potential energy, so does the electron. Recall our discussion of the conservation of energy. As long as the total energy remains conserved, the sum of the initial kinetic and potential energies must equal the sum of the ﬁnal kinetic and potential energies: KEi + PEi = KE f + PE f ! DKE = DPE The gain in kinetic energy occurs due to the loss in potential energy. In order for the charges of the same sign to be brought together, as in the example above, positive work must be done by an external force against the electrostatic repulsion between the charges. The work increases the potential energy stored in the electric ﬁeld. When the charges are released, the potential energy of the ﬁeld is converted into the kinetic energies of the charges. The link below may be helpful in learning more about the work done upon charges in electric ﬁelds. http://www.youtube.com/watch?v=elJUghWSVh4 Electric Potential Difference in a Uniform Electric Field In working with the change in potential energy above, we wrote the equation DPE = qEx f qExi = qE(x f xi) ! Let us call this Equation A. Recall that the electric potential was deﬁned at a speciﬁc point Vx1 = We therefore see that PEx1 = qVx1 ! Let us call this Equation B. Comparing Equation A and Equation B, we see that the electric potential can be expressed as Vx1 = Ex1 . If the electric potential is deﬁned as V = 0 at x = 0 , then the potential at any point in the electric ﬁeld is V = Ex . (Assuming that vector E is directed along the x axis). . PEx1 q Note: It is common to write V = Ed , where V is understood to mean the voltage (or potential difference) between the plates of a parallel-plate conductor, and d is the distance between the plates. Check Your Understanding Verify that the potential difference between the plates in Figure above is 1.5 V . Recall that the electric ﬁeld is E = 25:0 N Answer : V = Ed = 25 N C (6:00 102 m 0:00m) = 1:5 V C . Work We state again: 1. The electric potential is deﬁned as the energy per unit charge ! Vx1 = PEx1 q . 29 www.ck12.org 2. The electric potential difference (the voltage) is Vf Vi = PE f 3. An arbitrary reference level must be established for zero potential (just as in the case of gravitational potential q PEi q energy). 4. The units of electric potential and electric potential difference are J C since Vx1 = PEx1 q . It is often useful to express the voltage in terms of the work done on a charge. From Vf Vi = PE f But the work done on a charge by the ﬁeld is Wf ield = DPE . q , we have PE f PEi = q(Vf Vi) ! DPE = q(Vf Vi) . q PEi Combining DPE = q(Vf Vi) and Wf ield = DPE gives Wf ield = q(Vf Vi) . An external force that does work on a charge in an electric ﬁeld exerts a force in the opposite direction to the ﬁeld (just as the external force acting on a spring acts opposite to the spring force). The work that an external force does is therefore Wexternal f orce = q(Vf Vi) . The voltage can be thought of as the amount of work done by the electric ﬁeld per unit charge in moving a charge between two points in the electric ﬁeld DV = W q . We often refer to a change in potential as simply “the voltage.” In computing the work, it is often easier to ignore the sign in the equation and simply see if the force and displacement on the charge are in the same direction (positive work) or opposite to one another (negative work). Recall that the force and displacement need not be in the same direction or oppositely directed. In general, work is expressed as W = Fx cos q http://www.youtube.com/watch?v=F1p3fgbDnkY Other Units for the Electric Field ! E = C , since ! F q . But the electric ﬁeld has been also deﬁned using the scalar equation x . So the units of the electric ﬁeld can be also expressed as The electric ﬁeld has units N V = Ex . Transposing terms, the electric ﬁeld is E = V volts per meter volts If we compare the units for the electric ﬁeld N m , we see that a (N m) is equivalent to a (C V ) . A Joule can therefore be expressed as a Coulomb-Volt. Recall that work, measured in Joules, is the product of charge and voltage W = qDV . meter ! V m . C and V Illustrative Example 16.2.1 All questions refer to Figure above . a. What is the potential at x = 2:0 cm in Figure below ? The electric ﬁeld is E = 25:0 N C . Answer : The potential V varies directly with the position x between the plates (V = Ex) . Thus, V = 25:0 N 0:50 V . C (2:0 102m) = b. Sketch a graph showing the relationship between the potential and the position. Answer: 30 www.ck12.org Concept 7. Electric Potential c. How much work is done by an external force F moving a 2:0 106 C charge from the positive plate to the negative plate? Answer: An external force must pull the charge away from the positive plate so the force will be in the same direction as the displacement. W = q(Vf Vi) = (2:0 106C)(0:00V 1:50V ) = 3:0 106 J d. What is the magnitude of the Coulomb force acting on the charge? Answer: W = F ! 3:0 106 J = F(6:00 102 m) F = 3:0 106 J 6:00 102 m = 5:0 105 N Illustrative Example 16.2.2 a. A particle of mass m of 2:00 105 kg is has a charge q of +3:00 103 C . If the particle is released from the positive plate of a parallel-plate conductor with an electric ﬁeld E of 1:30 105 N C , determine the acceleration of the particle, see Figure below . Answer : If we ignore gravity, the only force acting on the particle is the electric force F = qE . Using Newton’s Second Law, the net force on the particle is equal to F = ma ! qE = ma . The acceleration is a = Eq m = (1:30105 V m )(3:00103 C) 2:00105 kg = 1:95 107 V C kgm . b. Show that the units V C kgm are equivalent to the units m s2 . Answer : V C kg m = J kg m = N m kg m = N kg = kg m s2 kg = m s2 c. The plates have separation of 8.00 mm. Determine the velocity of the particle when it reaches the negative plate Answer : 31 www.ck12.org FIGURE 7.4 Illustrative Example 16.3.2 This is a kinematics problem, where the displacement and acceleration are known and the velocity is to be found. Recall the equation v2 f = v2 i + 2aDx . ! v2 1:95 107 m s2 f = 0 + 2 v = 558:6 ! 5:59 102 m s : (8:00 103 m) = 312; 000 m2 s2 d. What is the potential difference between the plates? Answer : V = Ex = 1:30 105 V m (8:00 103 m) = 1:04 103 V e. How much work has the ﬁeld done on the particle as it moved from one plate to the other? W = qDV = (3:00 103 C)(1:04 103 V ) = 3:12 J Illustrative Example 16.2.3 An electron is accelerated from rest through a potential difference of 30,000 V. The mass of the electron is 9:11 1031 kg and the charge of the electron is 1:60 1019 C . Find its velocity. Answer : Recall that the Work-Energy Principle states that W = DKE . 32 www.ck12.org Concept 7. Electric Potential W = DKE W = qDV DKE = qDV 1 2 i = qDV mv2 1 2 mv2 f v f = Illustrative Example 16.2.4 vi = 0 ! v2 s f = 2qDV m ! v f = r 2qDV m ! 2(1:60 1019 C)(3:00 104 V ) 9:11 1031 kg = 1:026 108 ! 1:03 108 m s What magnitude of an electric ﬁeld is required to balance the gravitational force acting on an electron in Figure below ? FIGURE 7.5 Illustrative Example 16.2.4-An electron suspended in an electric ﬁeld. Answer : Draw a Free-Body-Diagram (FBD) of the situation. The electrostatic force that acts on the electron points upward and the gravitational force that acts upon on the electron points downward. The electron is suspended motionless (or moves with a constant velocity) when the net force on the electron is zero. 33 The net force on the electron must be zero, thus www.ck12.org F = 0 ! eF = mg ! F = mg e mg e mg = eE ! E = e2 ! but F = eE ! (9:11 1031 kg) 9:81 m s2 (1:60 1019 C)2 E = = 3:49 108 V m http://www.youtube.com/watch?v=wT9AsY79f1k The Electron-Volt It is often convenient when dealing with small particles such as electrons, protons, and ions to express the energy of these particles with a smaller unit of measure. The electron-volt is deﬁned as the change in potential energy that an electron acquires when moving through a potential difference of 1 V, or equivalently, its change in kinetic energy after moving through a potential difference of 1 V . That is, PE = eV = (1:60 1019 C)(1:00 V ) = 1:60 1019 J . The energy 1:60 1019 J is deﬁned as one electron-volt. We write one-electron-volt as 1 eV = 1:60 1019 J . Check Your Understanding 1. What is the change in kinetic energy KE when an electron is released at the negative plate of a parallel plate conductor with a potential difference of 3,500 V? Express your answer in eV. Answer : The electron is repelled at the negative plate and therefore gains kinetic energy (and loses potential energy). The change in KE is positive and equal to = x eV 1 eV 1 V 3; 500 V DKE = 3; 500 eV ! x = 3; 500 eV It is simplest to think that for every one volt of potential difference the particle experiences, it gains (or loses) 1 eV . 2. An alpha-particle (the nucleus of a helium atom) is ﬁred toward the positive plate of a parallel plate conductor and passes through a potential difference of 1,500 V. What is the change in its kinetic energy? Express your answer in eV. Answer : Protons are the only charges inside the nucleus of an atom and so the alpha particle must be positively charged. A helium nucleus contains two protons (and two neutrons) with a total charge of 2(1:60 1019 C) . The alpha particle must slow down due to the electrostatic repulsion from the positive plate. It must, therefore, lose kinetic energy and gain potential energy. Each proton loses 1,500 eV of kinetic energy. DKE = 3; 000 eV . 3. An e
lectron and a proton both gain kinetic energy of 1 eV . True or False: Their speeds must be the same, since they both gained the same amount of energy. 34 www.ck12.org Concept 7. Electric Potential Answer : False. The mass of a proton is nearly 2000 times greater than the mass of an electron. Remember that kinetic energy depends on both the speed and mass of an object. Therefore, the ﬁnal speed of the electron will be much greater. Illustrative Example 16.2.5 a. An electron and a proton both gain kinetic energy of 1 eV. What is the ratio of the electron’s speed to the proton’s speed? Answer : As discussed above, though both particles gain the same kinetic energy, their speeds will not be the same, since they have different masses. The mass of the proton is nearly 2000 times as great as the electron’s so: = 1 2 mev2 e 1 2 mpv2 p = 1 ! mev2 e = mpv2 p ! v2 e v2 p = mp me = 2000me me = 2; 000 ! = p 2; 000 = 44:7 ! 45 KEe KEp ve vp The electron will move about 45 times faster than the proton. b. What is the speed of a proton which has a kinetic energy of 37 MeV ? The mass of a proton is 1:67 1027 kg . Answer : Because the electron-volts are a very small unit, they are typically expressed in KeV (1000 electron-volts) and MeV (one million electron-volts). The electron-volt is a convenient unit of measure but it is not an SI unit. In order to ﬁnd the velocity of a particle if its energy is given in units of eV , we must convert back into Joules. 37 MeV = (37 106)(1:60 1019 J) = 5:92 1014 J 1 2 1 2 (1:67 1027 kg)v2 = 5:92 1014 J ! mpv2 = 5:92 1014 J ! v = 8:4 106 m s References 1. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 2. User:Asim18/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:02_-_Single_Energizer_Ba ttery.jpg . CC-BY 3.0 3. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 5. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 6. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 7. CK-12 Foundation - Raymond Chou. . CC-BY-NC-SA 3.0 35 Physics Unit 13: Circuits Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org 1 4 7 10 12 15 19 22 25 28 32 35 Contents 1 Electric Circuits 2 Electric Current 3 Electric Resistance 4 Ohm’s Law 5 Resistance and Ohm’s Law 6 Energy Transfer in Electric Circuits 7 Ammeters and Voltmeters 8 Series Circuits 9 Resistors in Series 10 Parallel Circuits 11 Resistors in Parallel 12 Combined Series-Parallel Circuits iv www.ck12.org Concept 1. Electric Circuits CONCEPT 1 Electric Circuits • Deﬁne electric circuit. • Describe the parts of an electric circuit. • Show how to represent a simple electric circuit with a circuit diagram. Jose made this sketch of a battery and light bulb for science class. If this were a real set up, the light bulb wouldn’t work. The problem is the loose wire on the left. It must be connected to the positive terminal of the battery in order for the bulb to light up. Q: Why does the light bulb need to be connected to both battery terminals? A: Electric current can ﬂow through a wire only if it forms a closed loop. Charges must have an unbroken path to follow between the positively and negatively charged parts of the voltage source, in this case, the battery. Electric Circuit Basics A closed loop through which current can ﬂow is called an electric circuit . In homes in the U.S., most electric circuits have a voltage of 120 volts. The amount of current (amps) a circuit carries depends on the number and power of electrical devices connected to the circuit. Home circuits generally have a safe upper limit of about 20 or 30 amps. 1 Parts of an Electric Circuit All electric circuits have at least two parts: a voltage source and a conductor. They may have other parts as well, such as light bulbs and switches, as in the simple circuit seen in the Figure 1.1 . To see an animation of a circuit like this one, go to: http://www.rkm.com.au/animations/animation-electrical-circuit.html www.ck12.org FIGURE 1.1 • The voltage source of this simple circuit is a battery. In a home circuit, the source of voltage is an electric power plant, which may supply electric current to many homes and businesses in a community or even to many communities. • The conductor in most circuits consists of one or more wires. The conductor must form a closed loop from the source of voltage and back again. In the circuit above, the wires are connected to both terminals of the battery, so they form a closed loop. • Most circuits have devices such as light bulbs that convert electrical energy to other forms of energy. In the case of a light bulb, electrical energy is converted to light and thermal energy. • Many circuits have switches to control the ﬂow of current. When the switch is turned on, the circuit is closed and current can ﬂow through it. When the switch is turned off, the circuit is open and current cannot ﬂow through it. Circuit Diagrams When a contractor builds a new home, she uses a set of plans called blueprints that show her how to build the house. The blueprints include circuit diagrams. The diagrams show how the wiring and other electrical components are to be installed in order to supply current to appliances, lights, and other electric devices. You can see an example of a very simple circuit in the Figure 1.2 . Different parts of the circuit are represented by standard circuit symbols. An ammeter measures the ﬂow of current through the circuit, and a voltmeter measures the voltage. A resistor is any device that converts some of the electricity to other forms of energy. For example, a resistor might be a light bulb or doorbell. The circuit diagram on the right represents the circuit drawing on the left. Below are some of the standard symbols used in circuit diagrams. 2 www.ck12.org Concept 1. Electric Circuits FIGURE 1.2 Q: Only one of the circuit symbols above must be included in every circuit. Which symbol is it? A: The battery symbol (or a symbol for some other voltage source) must be included in every circuit. Without a source of voltage, there is no electric current. Summary • An electric circuit is a closed loop through which current can ﬂow. • All electric circuits must have a voltage source, such as a battery, and a conductor, which is usually wire. They may have one or more electric devices as well. • An electric circuit can be represented by a circuit diagram, which uses standard symbols to represent the parts of the circuit. Vocabulary • electric circuit : Closed loop through which current can ﬂow. Practice Take the electric circuit quiz at the following URL. Be sure to have your answers corrected. Try the quiz again if any of your answers are incorrect. http://www.myschoolhouse.com/courses/O/1/68.asp Review 1. What is an electric circuit? 2. Which two parts must all electric circuits contain? 3. Sketch a simple circuit that includes a battery, switch, and light bulb. Then make a circuit diagram to represent your circuit, using standard circuit symbols. References 1. Christopher Auyeung. . CC BY-NC 3.0 2. Christopher Auyeung. . CC BY-NC 3.0 3 www.ck12.org Electric Current CONCEPT 2 Objectives The student will: • Understand how electric current is deﬁned • Solve problems involving electric current Vocabulary • electric current: A ﬂow of charges under the inﬂuence of an electric ﬁeld, such as between the terminals of a battery. The rate I = DQ Dt at which charges ﬂow within a conducting wire past any point in the wire. Introduction The term electrical current is familiar to most people. Many electrical devices have electrical speciﬁcations printed on them. Figure below shows a typical AC adapter (“plug”) with its “specs.” Can you guess what the terms 5 VDC and 500 mA printed on the adapter mean? FIGURE 2.1 An electrical plug. Electric Current An electric current is a ﬂow of charges under the inﬂuence of an electric ﬁeld. A ﬂow of charges can be established, for instance, between the terminals of a battery, as in Figure below . The rate I = DQ Dt at which charges ﬂow within 4 www.ck12.org Concept 2. Electric Current a conducting wire past any point in the wire is deﬁned as the electric current. The unit of current is coulombs
Ampere, 1775-1836), Figure below . second which is called the ampere or amp (named for the French physicist Andre’-Marie The symbol A is used to represent the ampere. A rate of one coulomb per second is equivalent to one ampere: 1C 1s = 1A FIGURE 2.2 Figure above shows a ﬂow of electrons (e) from the positive terminal of a battery through a lightbulb to the negative terminal of a battery. FIGURE 2.3 Andre’-Marie Ampere One ampere is a very large current. The current of 1 A can easily kill a person. In fact, about 0.20 A can kill rather easily. Even relatively small voltage can produce these currents, which is why care must always be taken when dealing with all electrical appliances and any electrical device that is plugged into a wall outlet. A typical 12-V car battery can also be dangerous. Under the right circumstances, it does not take a huge voltage to cause deadly currents. It is common to express current in milliamperes 1 mA = 103 A , or microamperes 1 ¯A = 106 A . http://demonstrations.wolfram.com/ElectricCurrent/ 5 Illustrative Example 17.4.1 A total of 7:9 1012 electrons move past a point in a conducting wire every 1.45 s. What is the average current in the wire? Answer : www.ck12.org The total charge moving past the point is the product of the electric charge of an electron and the number of electrons moving past the point. The total charge is: ! Q = 1:6 1019 (7:9 1012 electrons) = 12:6 107 ! 13 107 C The current is I = DQ 1:45 s = 8:69 107 A ! 0:87 µA . Dt = 12:6107 C electron References 1. Ray Dehler (Flickr: raybdbomb). http://www.ﬂickr.com/photos/raybdbomb/2200741209/ 2. CK-12 Foundation - Ira Nirenberg. . CC-BY-NC-SA 3.0 3. . http://commons.wikimedia.org/wiki/File:Andre-marie-ampere2.jpg . public domain . CC-BY 2.0 6 www.ck12.org Concept 3. Electric Resistance CONCEPT 3 Electric Resistance • Deﬁne resistance and identify the SI unit for resistance. • List factors that affect resistance. • Explain why resistance can be a help or a hindrance. These athletes are playing rugby, a game that is similar to American football. The players in red and blue are trying to stop the player in orange and black from running across the ﬁeld with the ball. They are resisting his forward motion. This example of resistance in rugby is a little like resistance in physics. What Is Resistance? In physics, resistance is opposition to the ﬂow of electric charges in an electric current as it travels through matter. The SI unit for resistance is the ohm. Resistance occurs because moving electrons in current bump into atoms of matter. Resistance reduces the amount of electrical energy that is transferred through matter. That’s because some of the electrical energy is absorbed by the atoms and changed to other forms of energy, such as heat. Q: In the rugby analogy to resistance in physics, what do the players on each team represent? A: Factors that Affect Resistance How much resistance a material has depends on several factors: the type of material, its width, its length, and its temperature. • All materials have some resistance, but certain materials resist the ﬂow of electric current more or less than other materials do. Materials such as plastics have high resistance to electric current. They are called electric insulators. Materials such as metals have low resistance to electric current. They are called electric conductors. 7 www.ck12.org • A wide wire has less resistance than a narrow wire of the same material. Electricity ﬂowing through a wire is like water ﬂowing through a hose. More water can ﬂow through a wide hose than a narrow hose. In a similar way, more current can ﬂow through a wide wire than a narrow wire. • A longer wire has more resistance than a shorter wire. Current must travel farther through a longer wire, so there are more chances for it to collide with particles of matter. • A cooler wire has less resistance than a warmer wire. Cooler particles have less kinetic energy, so they move more slowly. Therefore, they are less likely to collide with moving electrons in current. Materials called superconductors have virtually no resistance when they are cooled to extremely low temperatures. Is Resistance Good or Bad? Resistance can be helpful or just a drain on electrical energy. If the aim is to transmit electric current through a wire from one place to another, then resistance is a drawback. It reduces the amount of electrical energy that is transmitted because some of the current is absorbed by particles of matter. On the other hand, if the aim is to use electricity to produce heat or light, then resistance is useful. When particles of matter absorb electrical energy, they change it to heat or light. For example, when electric current ﬂows through the tungsten wire inside an incandescent light bulb like the one in the Figure 3.1 , the tungsten resists the ﬂow of electric charge. It absorbs electrical energy and converts some of it to light and heat. FIGURE 3.1 What’s wrong with this picture? (Hint: How does current get to the light bulb?) Q: The tungsten wire inside a light bulb is extremely thin. How does this help it do its job? A: Summary • In physics, resistance is opposition to the ﬂow of electric charges that occurs as electric current travels through matter. The SI unit for resistance is the ohm. • All materials have resistance. How much resistance a material has depends on the type of material, its width, its length, and its temperature. • Resistance is a hindrance when a material is being used to transmit electric current. Resistance is helpful when a material is being used to produce heat or light. 8 www.ck12.org Vocabulary Concept 3. Electric Resistance • resistance : Opposition to the ﬂow of electric charges that occurs when electric current travels through matter. Review 1. What is resistance? Name the SI unit for resistance. 2. Explain what causes resistance. 3. Describe properties of a metal wire that would minimize its resistance to electric current. 4. Extend the rugby analogy to explain why a longer wire has greater resistance to electric current. 5. Copper wires have about one-third the resistance of tungsten wires. Why would copper be less suitable than tungsten as a ﬁlament in an incandescent light bulb? References 1. lenetstan. . Used under license from Shutterstock.com 9 CONCEPT 4 • Explain Ohm’s law. • Use Ohm’s law to calculate current from voltage and resistance. www.ck12.org Ohm’s Law Look at the water spraying out of this garden hose. You have to be careful using water around power tools and electric outlets because water can conduct an electric current. But in some ways, water ﬂowing through a hose is like electric current ﬂowing through a wire. Introducing Ohm’s Law For electric current to ﬂow through a wire, there must be a source of voltage. Voltage is a difference in electric potential energy. As you might have guessed, greater voltage results in more current. As electric current ﬂows through matter, particles of matter resist the moving charges. This is called resistance, and greater resistance results in less current. These relationships between electric current, voltage, and resistance were ﬁrst demonstrated in the early 1800s by a German scientist named Georg Ohm, so they are referred to as Ohm’s law. Ohm’s law can be represented by the following equation. Current(amps) = Voltage(volts) Resistance(ohms) Understanding Ohm’s Law Ohm’s law may be easier to understand with an analogy. Current ﬂowing through a wire is like water ﬂowing through a hose. Increasing voltage with a higher-volt battery increases the current. This is like opening the tap wider so more water ﬂows through the hose. Increasing resistance reduces the current. This is like stepping on the hose so less water can ﬂow through it. If you still aren’t sure about the relationships among current, voltage, and resistance, watch the video at this URL: http://www.youtube.com/watch?v=KvVTh3ak5dQ 10 www.ck12.org Concept 4. Ohm’s Law Using Ohm’s Law to Calculate Current You can use the equation for current (above) to calculate the amount of current ﬂowing through a circuit when the voltage and resistance are known. Consider an electric wire that is connected to a 12-volt battery. If the wire has a resistance of 2 ohms, how much current is ﬂowing through the wire? Current = 12 volts 2 ohms = 6 amps Q: If a 120-volt voltage source is connected to a wire with 10 ohms of resistance, how much current is ﬂowing through the wire? A: Substitute these values into the equation for current: Current = 120 volts 20 ohms = 12 amps Summary • According to Ohm’s law, greater voltage results in more current and greater resistance results in less current. • Ohm’s law can be represented by the equation:’ Current(amps) = Voltage(volts) Resistance(ohms) • This equation can be used to calculate current when voltage and resistance are known. Vocabulary • Ohm’s law : Law stating that current increases as voltage increases or resistance decreases. Practice Review Ohm’s law and how to calculate current at the following URL. Then try to solve the two problems at the bottom of the Web page. Be sure to check your answers against the correct solutions. http://www.grc.nasa.gov /WWW/k-12/Sample_Projects/Ohms_Law/ohmslaw.html Review 1. State Ohm’s law. 2. An electric appliance is connected by wires to a 240-volt source of voltage. If the combined resistance of the appliance and wires is 12 ohms, how much current is ﬂowing through the circuit? 11 www.ck12.org CONCEPT 5 Resistance and Ohm’s Law • Deﬁne resistance. • Understand the unit for resistance: ohms. • Use Ohm’s Law to solve problems involving current, potential difference, and resistance. The bands of color on a resistor are a code that indicates the magnitude of the resistance of the resistor. There are four color bands identiﬁed by letter: A, B, C, and D, with a gap between the C and D bands so that you know which end is A. This particular resistor has a red A band, blue B band, green C ban
d, and gold D band, but the bands can be different colors on different resistors. Based on the colors of the bands, it is possible to identify the type of resistor. the A and B bands represent signiﬁcant digits; red is 2 and blue is 6. The C band indicates the multiplier, and green indicates 10 5 . These three together indicate that this particular resistor is a 26,000 Ohm resistor. Finally, the D band indicates the tolerance, in this case 5%, as shown by the gold band. These terms will be explained over the course of this lesson. Resistance and Ohm’s Law When a potential difference is placed across a metal wire, a large current will ﬂow through the wire. If the same potential difference is placed across a glass rod, almost no current will ﬂow. The property that determines how much current will ﬂow is called the resistance. Resistance is measured by ﬁnding the ratio of potential difference, V , to current ﬂow, I . 12 R = V I www.ck12.org Concept 5. Resistance and Ohm’s Law When given in the form V = IR , this formula is known as Ohm’s Law, after the man that discovered the relationship. The units of resistance can be determined using the units of the other terms in the equation, namely that the potential difference is in volts (J/C) and current in amperes (C/s): R = volts amperes = joules/coulomb coulombs/second = joules seconds coulombs2 = ohms The units for resistance have been given the name ohms and the abbreviation is the Greek letter omega, W. 1.00 W is the resistance that will allow 1.00 ampere of current to ﬂow through the resistor when the potential difference is 1.00 volt. Most conductors have a constant resistance regardless of the potential difference; these are said to obey Ohm’s Law. There are two ways to control the current in a circuit. Since the current is directly proportional to the potential difference and inversely proportional to the resistance, you can increase the current in a circuit by increasing the potential or by decreasing the resistance. Example Problem: A 50.0 V battery maintains current through a 20.0 W resistor. What is the current through the resistor? Solution: I = V R = 50:0 V 20:0 W = 2:50 amps Summary • Resistance is the property that determines the amount of current ﬂow through a particular material. • V = IR is known as Ohm’s Law. • The unit for resistance is the ohm, and it has the abbreviation W. Practice The following video covers Ohm’s Law. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=uLU4LtG0_hc MEDIA Click image to the left for more content. 1. What happens to current ﬂow when voltage is increased? 2. What happens to current ﬂow when resistance is increased? This website contains instruction and guided practice for Ohm’s Law. http://www.wisc-online.com/Objects/ViewObject.aspx?ID=DCE11904 Review 1. If the potential stays the same and the resistance decreases, what happens to the current? 13 www.ck12.org (a) increase (b) decrease (c) stay the same 2. If the resistance stays the same and the potential increases, what happens to the current? (a) increase (b) decrease (c) stay the same 3. How much current can be pushed through a 30.0 W resistor by a 12.0 V battery? 4. What voltage is required to push 4.00 A of current through a 32.0 W resistor? 5. If a 6.00 volt battery will produce 0.300 A of current in a circuit, what is the resistance in the circuit? • resistance: Opposition of a circuit to the ﬂow of electric current. • Ohm’s Law: V = IR . • Ohms: A resistance between two points of a conductor when a constant potential difference of 1 volt, applied to these points, produces in the conductor a current of 1 ampere. References 1. Image copyright Robert Spriggs, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 14 www.ck12.org Concept 6. Energy Transfer in Electric Circuits CONCEPT 6 Energy Transfer in Electric Circuits • Explain how devices convert electrical energy to other forms. • Use P = I2R and E = I2Rt for calculations involves energy transfer in electrical circuits. • Describe the reason for the use of high voltage lines for transmitting electrical energy. • Deﬁne the kilowatt-hour. Part of the electrical grid, an electrical transmission sub-station receives extremely high current levels, then passes the electrical energy on to as many as 200,000 homes. Approximately 5000 megawatt-hours of energy passes through this particular substation each year. Energy Transfer in Electric Circuits Electric power is the energy per unit time converted by an electric circuit into another form of energy. We already know that power through a circuit is equal to the voltage multiplied by the current in a circuit: P = V I . It is possible to determine the power dissipated in a single resistor if we combine this expression with Ohm’s Law, V = IR . This becomes particularly useful in circuits with more than one resistor, to determine the power dissipated in each one. Combining these two equations, we get an expression for electric power that involves only the current and resistance in a circuit. P = I2R The power dissipated in a resistor is proportional to the square of the current that passes through it and to its resistance. 15 www.ck12.org Electrical energy itself can be expressed as the electrical power multiplied by time: E = Pt We can incorporate this equation to obtain an equation for electrical energy based on current, resistance, and time. The electrical energy across a resistor is determined to be the current squared multiplied by the resistance and the time. E = I2Rt This equation holds true in ideal situations. However, devices used to convert electrical energy into other forms of energy are never 100% efﬁcient. An electric motor is used to convert electrical energy into kinetic energy, but some of the electrical energy in this process is lost to thermal energy. When a lamp converts electrical energy into light energy, some electrical energy is lost to thermal energy. Example Problem: A heater has a resistance of 25.0 W and operates on 120.0 V. a. How much current is supplied to the resistance? b. How many joules of energy is provided by the heater in 10.0 s? Solution: a. I = V b. E = I2Rt = (4:8 A)2(25:0 W)(10:0 s) = 5760 joules 25:0 W = 4:8 A R = 120:0 V Think again about the power grid. When electricity is transmitted over long distances, some amount of energy is lost in overcoming the resistance in the transmission lines. We know the equation for the power dissipated is given by P = I2R . The energy loss can be minimized by choosing the material with the least resistance for power lines, but changing the current also has signiﬁcant effects. Consider a reduction of the current by a power of ten: How much power is dissipated when a current of 10.0 A passes through a power line whose resistance is 1.00 W? P = I2R = (10:0 A)2(1:00 W) = 100: Watts How much power is dissipated when a current of 1.00 A passes through a power line whose resistance is 1.00 W? P = I2R = (1:00 A)2(1:00 W) = 1:00 Watts The power loss is reduced tremendously by reducing the magnitude of the current through the resistance. Power companies must transmit the same amount of energy over the power lines but keep the power loss minimal. They do this by reducing the current. From the equation P = V I , we know that the voltage must be increased to keep the same power level. The Kilowatt-Hour Even though the companies that supply electrical energy are often called “power” companies, they are actually selling energy. Your electricity bill is based on energy, not power. The amount of energy provided by electric current can be calculated by multiplying the watts (J/s) by seconds to yield joules. The joule, however, is a very small unit of energy and using the joule to state the amount of energy used by a household would require a very large number. For that reason, electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh). A kilowatt hour is exactly as it sounds - the number of kilowatts (1,000 W) transferred per hour. 1:00 kilowatt hour = (1000 J=s)(3600 s) = 3:6 106 J Example Problem: A color television uses about 2.0 A when operated on 120 V. a. How much power does the set use? b. If the TV is operated for 8.00 hours per day, how much energy in kWh does it use per day? c. At $0.15 per kWh, what does it cost to run the TV for 30 days? 16 www.ck12.org Solution: Concept 6. Energy Transfer in Electric Circuits a. P = V I = (120 V )(2:0 A) = 240 W b. E = (240 J=s)(8 h)(3600 s=h) c. Cost = (1:92 kW h)(30)($0:15) = $8:64 = 1:92 kW h 3:6106 J=kW h Summary • Electric power is the energy per unit time converted by an electric circuit into another form of energy. • The formula for electric power is P = I2R . • The electric energy transferred to a resistor in a time period is equal to the electric power multiplied by time, E = Pt , and can also be calculated using E = I2Rt . • Electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh) which is equivalent to 3:6 106 J . Practice The following video is on electrical energy and power. Use this resource to answer the questions that follow. http://youtu.be/NWcYBvHOiWw MEDIA Click image to the left for more content. 1. What is this video about? 2. What is the deﬁnition of electrical power? 3. What happens to the electrical energy that is not converted into work? Instruction and practice problems related to the energy delivered by an electric circuit: http://www.physicsclassroom.com/Class/circuits/u9l2d.cfm Review 1. A 2-way light bulb for a 110. V lamp has ﬁlament that uses power at a rate of 50.0 W and another ﬁlament that uses power at a rate of 100. W. Find the resistance of these two ﬁlaments. 2. Find the power dissipation of a 1.5 A lamp operating on a 12 V battery. 3. A high voltage (4:0 105 V ) power transmission line delivers electrical energy from a generating station to a substation at a rate of 1:5 109 W
. What is the current in the lines? 4. A toaster oven indicates that it operates at 1500 W on a 110 V circuit. What is the resistance of the oven? • electrical energy: Energy is the ability to do work, so electrical energy is the work done by an electrical circuit. • kilowatt hour: An amount of energy equal to 3:6 106 Joules . 17 References 1. User:Cutajarc/Wikipedia. http://en.wikipedia.org/wiki/File:Melbourne_Terminal_Station.JPG . Public Do- main www.ck12.org 18 www.ck12.org Concept 7. Ammeters and Voltmeters CONCEPT 7 Ammeters and Voltmeters • Describe the primary difference between the construction of ammeters and voltmeters. • Describe whether ammeters should be placed in circuits in series or parallel and explain why. • Describe whether voltmeters should be placed in circuits in series or parallel and explain why. This photo is of the interior of the control room for a nuclear power plant. Many of the meters are reading information about the water temperature and the nuclear reaction that is occurring, but the majority of the meters are reading data about the electric energy being generated. Ammeters and Voltmeters Ammeters and voltmeters are cleverly designed for the way they are used. Ammeters measure the current of a circuit, and voltmeters measure the voltage drop across a resistor. It is important in the design and use of these meters that they don’t change the circuit in such a way as to inﬂuence the readings. While both types of meters are technically resistors, they are speciﬁcally designed to make their readings without changing the circuit itself. 19 www.ck12.org Ammeter An ammeter measures the current traveling through the circuit. They are designed to be connected to the circuit in series, and have an extremely low resistance. If an ammeter were connected in parallel, all of the current would go through the ammeter and very little through any other resistor. As such, it is necessary for the ammeter to be connected in series with the resistors. This allows the ammeter to accurately measure the current ﬂow without causing any disruptions. In the circuit sketched above, the ammeter is m 2 . Voltmeter In contrast, a voltmeter is designed to be connected to a circuit in parallel, and has a very high resistance. A voltmeter measures the voltage drop across a resistor, and does not need to have the current travel through it to do so. When a voltmeter is placed in parallel with a resistor, all the current continues to travel through the resistor, avoiding the very high resistance of the voltmeter. However, we know that the voltage drop across all resistors in parallel is the same, so connecting a voltmeter in parallel allows it to accurately measure the voltage drop. In the sketch, the voltmeter is m 1 . Summary • Ammeters measure the current through a resistor. • Ammeters have low resistances and are placed in the circuit in series. • Voltmeters measure the voltage drop across a resistor. • Voltmeters have high resistances and are placed in the circuit in parallel. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=liwan6-w-Pw In this video, a circuit is established with a power supply, which also has an attached voltmeter, and a lamp (resistor). After the circuit is established, a voltmeter and an ammeter are alternately placed in the circuit. Follow up questions: 1. What happens when the ammeter is connected in parallel with the lamp? 2. Why do the problems occur when the narrator in the video places the ammeter in parallel with the lamp? 20 www.ck12.org Review Concept 7. Ammeters and Voltmeters 1. In the sketch at above, there are four positions available for the placement of meters. Which position(s) would be appropriate for placement of an ammeter? a. 1 b. 3 c. 4 d. All of them. e. None of them. 2. Which position(s) would be appropriate for placement of a voltmeter? a. 1 b. 2 c. 3 d. All of them. e. None of them. 3. Which position could hold an ammeter that would read the total current through the circuit? a. 1 b. 2 c. 3 d. 4 e. None of them. 4. Which position could hold a voltmeter that would read the total voltage drop through the circuit? a. 1 b. 2 c. 3 or 4 d. All of them. e. None of them. • ammeter: A measuring instrument used to measure the electric current in a circuit. • voltmeter: An instrument used for measuring electrical potential difference between two points in an electric circuit. References 1. Image copyright rtem, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 21 CONCEPT 8 www.ck12.org Series Circuits • Describe a series circuit. • Understand current as it passes through a series circuit. • Understand voltage drops in a series circuit. • Understand resistance in a series circuit with multiple resistors. • Calculate current, voltage drops, and equivalent resistances for devices connected in a series circuit. Resistors, including electrical appliances, have a particular current at which they operate most effectively and safely, and excessive current can cause irreparable damage. Therefore, it is important to limit the amount of current that may pass through a particular electrical circuit. There are a number of safety devices used in electrical circuits to limit the current; fuses, circuit breakers, and surge suppressors. When fuses, such as those shown above, are placed in an electrical circuit, all the current must pass through the wire in the fuse. Series Circuits Electrical circuits are often modeled by using water in a river. The potential energy of the water is the highest at the source of the river and decreases as the water ﬂows down the river toward the end. When the water reaches the ocean, its potential energy has become zero. The circuit shown above has a similar situation. The current in this circuit is drawn in the direction of the electron ﬂow. It starts at the battery on the left, where electrons leave the 22 www.ck12.org Concept 8. Series Circuits negative terminal and travel around the circuit. Since all of the current travels across each resistor, these resistors are said to be in series . A series circuit is one in which all of the current must pass through every resistor in the circuit. Returning to the water analogy, there is only one riverbed from the top of the mountain to the ocean. Consider the series circuit sketched above. This circuit has a voltage drop for the entire circuit of 120 V and has three resistors connected in series. The current in this circuit is drawn in terms of electron ﬂow. The electrons leave the potential difference source at the negative terminal and ﬂow through the three resistors, starting with R 3 . Though they have a small amount of resistance, the resistance of the connecting wires is so small in relation to the resistors that we ignore it. Therefore, we say that there is no voltage drop when the current passes through the connecting wires. The voltage drops occur when the current passes through each of the resistors and the total voltage drop for the entire circuit is equal to the sum of the voltage drops through the three resistors. VT = V1 +V2 +V3 The current through each of the resistors must be exactly the same because the current in a series circuit is the same everywhere. The current is moving in the entire circuit at the same time. IT = I1 = I2 = I3 Since the current passes through each resistor, the total resistance in the circuit is equal to the sum of the resistors. In the circuit above, the total resistance is: RT = R1 + R2 + R3 = 30 W + 15 W + 15 W = 60 W Therefore, the total current and the current through each resistor is I = V R = 120 V 60 W = 2:0 A: The individual voltage drops can be calculated using the current through each resistor and each resistor’s individual resistance. V1 = I1R1 = (2:0 A)(30 W) = 60 V V2 = I2R2 = (2:0 A)(15 W) = 30 V V3 = I3R3 = (2:0 A)(15 W) = 30 V Example Problem: Four 15 W resistors are connected in series with a 45 V battery. What is the current in the circuit? Solution: RT = 15 W + 15 W + 15 W + 15 W = 60 W I = V R = 45 V 60 W = 0:75 A 23 www.ck12.org Summary • A series circuit is one in which all of the current must pass through every resistor in the circuit. • VT = V1 +V2 +V3 • IT = I1 = I2 = I3 • RT = R1 + R2 + R3 Practice The following video is on series circuits. Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=qO391knBRjE 1. How do the voltage drops across the two light bulbs in the video related to the total voltage drop for the entire circuit? 2. In the video, what was the assumed voltage drop for the connecting wires and the switch? 3. What was the current through the second light bulb as compared to the current through the ﬁrst light bulb. Review 1. There are three 20.0 W resistors connected in series across a 120 V generator. (a) What is the total resistance of the circuit? (b) What is the current in the circuit? (c) What is the voltage drop across one of the resistors? 2. A 5.00W, a 10.0W, and a 15.0W resistor are connected in a series across a 90.0 V battery. (a) What is the equivalent resistance of the circuit? (b) What is the current in the circuit? (c) What is the voltage drop across the 5.00W resistor? 3. A 5.00W and a 10.0W resistor are connected in series across an unknown voltage. The total current in the circuit is 3.00 A. (a) What is the equivalent resistance of the circuit? (b) What is the current through the 5.00W resistor? (c) What is the total voltage drop for the entire circuit? • series circuit: One in which all of the current must pass through every resistor in the circuit. References 1. Image copyright sevenke, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 24 www.ck12.org Concept 9. Resistors in Series CO
NCEPT 9 Resistors in Series Students will learn how to analyze and solve problems involving circuits with resistors in series. Students will learn how to analyze and solve problems involving circuits with resistors in series. Key Equations Guidance Rtotal = R1 + R2 + R3 + : : : Resistors in Series: All resistors are connected end to end. There is only one river, so they all receive the same current. But since there is a voltage drop across each resistor, they may all have different voltages across them. The more resistors in series the more rocks in the river, so the less current that ﬂows. Example 1 A circuit is wired up with two resistors in series. Both resistors are in the same ’river’, so both have the same current ﬂowing through them. Neither resistor has a direct connection to the power supply so neither has 20V across it. But the combined voltages across the individual resistors add up to 20V. Question: What is the total resistance of the circuit? Answer: The total resistance is Rtotal = R1 + R2 = 90 W + 10 W = 100 W Question: What is the total current coming out of the power supply? Answer: Use Ohm’s Law (V = IR) but solve for current (I = V =R) . Itotal = Vtotal Rtotal = 20V 100 W = 0:20 A Question: How much power does the power supply dissipate? Answer: P = IV , so the total power equals the total voltage multiplied by the total current. Thus, Ptotal = ItotalVtotal = (0:20 A)(20V ) = 4:0 W . So the Power Supply is outputting 4W (i.e. 4 Joules of energy per second). Question: How much power does each resistor dissipate? Answer: Each resistor has different voltage across it, but the same current. So, using Ohm’s law, convert the power formula into a form that does not depend on voltage. 25 www.ck12.org P = IV = I(IR) = I2R: P90 W = I2 P10 W = I2 90 WR90 W = (0:2 A)2(90 W) = 3:6W 10 WR10 W = (0:2 A)2(10 W) = 0:4W Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved. Question: How much voltage is there across each resistor? Answer: In order to calculate voltage across a resistor, use Ohm’s law. V90 W = I90 WR90 W = (0:2 A)(90 W) = 18V V10 W = I10 WR10 W = (0:2 A)(10 W) = 2V Note: If you add up the voltages across the individual resistors you will obtain the total voltage of the circuit, as you should. Further note that with the voltages we can use the original form of the Power equation (P = IV ) , and we should get the same results as above. P90 W = I90 WV90 W = (18V )(0:2 A) = 3:6W P10 W = I10 WV10 W = (2:0V )(0:2 A) = 0:4W MEDIA Click image to the left for more content. Watch this Explanation 26 www.ck12.org Simulation Concept 9. Resistors in Series • http://simulations.ck12.org/Resistor/ Time for Practice 1. Regarding the circuit below. a. If the ammeter reads 2 A , what is the voltage? b. How many watts is the power supply supplying? c. How many watts are dissipated in each resistor? 2. Five resistors are wired in series. Their values are 10W , 56W , 82W , 120W and 180W . a. If these resistors are connected to a 6 V battery, what is the current ﬂowing out of the battery? b. If these resistors are connected to a 120 V power suppluy, what is the current ﬂowing out of the battery? c. In order to increase current in your circuit, which two resistors would you remove? 3. Given the resistors above and a 12 V battery, how could you make a circuit that draws 0.0594 A? Answers to Selected Problems 1. a. 224 V b. 448 W c. 400 W by 100 W and 48 W by 12 W 2. a. 0.013 A b. 0.27 A c. 120W and 180W 3. need about 202W of total resistance. So if you wire up the 120W and the 82W in series, you’ll have it. 27 CONCEPT 10 www.ck12.org Parallel Circuits • Describe a parallel circuit. • Understand current as it passes through a parallel circuit. • Understand voltage drops in a parallel circuit. • Understand resistance in a parallel circuit with multiple resistors. • Calculate voltage drops, currents, and equivalent resistances when devices are connected in a parallel circuit. Electrical circuits are everywhere: skyscrapers, jumbo jets, arcade games, lights, heating, and security . . . very few complex things work without electrical circuits. Since the late 1970s, electrical circuits have primarily looked like this. The circuits are formed by a thing layer of conducting material deposited on the surface of an insulating board. Individual components are soldered to the interconnecting circuits. Circuit boards are vastly more complicated than the series circuits previously discussed, but operate on many similiar principles. Parallel Circuits Parallel circuits are circuits in which the charges leaving the potential source have different paths they can follow to get back to the source. In the sketch below, the current leaves the battery, passes through the orange switch, and then has three different paths available to complete the circuit. Each individual electron in this circuit passes through only one of the light bulbs. After the current passes through the switch, it divides into three pieces and each piece passes through one of the bulbs. The three pieces of current rejoin after the light bulbs and continue in the circuit to the potential source. 28 www.ck12.org Concept 10. Parallel Circuits In the design of this parallel circuit, each resistor (light bulb) is connected across the battery as if the other two resistors were not present. Remember that the current going through each resistor goes through only the one resistor. Therefore, the voltage drop across each resistor must be equal to the total voltage drop though the circuit. VT = V1 = V2 = V3 The total current passing through the circuit will be the sum of the individual currents passing through each resistor. IT = I1 + I2 + I3 If we return to the analogy of a river, a parallel circuit is the same as the river breaking into three streams, which later rejoin to one river again. The amount of water ﬂowing in the river is equal to the sum of the amounts of water ﬂowing in the individual streams. Ohm’s Law applies to resistors in parallel, just as it did to resistors in a series. The current ﬂowing through each resistor is equal to the total voltage drop divided by the resistance in that resistor. I1 = VT R1 and I2 = VT R2 and I3 = VT R3 Since IT = I1 + I2 + I3 , + VT then IT = VT R3 R1 + VT = VT and VT R3 R1 RT + VT R2 + VT R2 , . If we divide both sides of the ﬁnal equation by VT , we get the relationship between the total resistance of the circuit and the individual parallel resistances in the circuit. The total resistance is sometimes called the equivalent resistance. 1 RT = 1 R1 + 1 R2 + 1 R3 Consider the parallel circuit sketched below. The voltage drop for the entire circuit is 90. V. Therefore, the voltage drop in each of the resistors is also 90. V. 29 www.ck12.org The current through each resistor can be found using the voltage drop and the resistance of that resistor: I1 = VT R1 = 90: V 60: W = 1:5 A I2 = VT R2 = 90: V 30: W = 3:0 A I3 = VT R3 = 90: V 30: W = 3:0 A The total current through the circuit would be the sum of the three currents in the individual resistors. IT = I1 + I2 = I3 = 1:5 A + 3:0 A + 3:0 A = 7:5 A The equivalent resistance for this circuit is found using the equation above. 1 RT = 1 R1 + 1 R2 + 1 R3 = 1 60: W + 1 30: W + 1 30: W = 1 60: W + 2 60: W + 2 60: W = 5 60: W RT = 60: W 5 = 12 W The equivalent resistance for the circuit could also be found by using the total voltage drop and the total current. RT = VT IT 7:5 A = 12 W = 90: W Summary • Parallel electrical circuits have multiple paths the current may take. • VT = V1 = V2 = V3 . • IT = I1 + I2 + I3 . + 1 • R3 = 1 R1 = 1 R2 1 RT . Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=apHkG4T6QHM Follow up questions: 1. Why do the light bulbs glow less brightly when connected across a 120 V source in a series circuit than when connected across the same 120 V source in a parallel circuit? 2. Why do the other bulbs go dark when one bulb is removed in the series circuit but the other bulbs do not go dark when one bulb is removed in the parallel circuit? 30 www.ck12.org Review Concept 10. Parallel Circuits 1. Three 15.0W resistors are connected in parallel and placed across a 30.0 V potential difference. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through a single branch of the circuit? 2. A 12.0W and a 15.0W resistor are connected in parallel and placed across a 30.0 V potential. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? 3. A 120.0W resistor, a 60.0W resistor, and a 40.0W resistor are connected in parallel and placed across a potential difference of 12.0 V. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? • parallel circuit: A closed electrical circuit in which the current is divided into two or more paths and then returns via a common path to complete the circuit. • equivalent resistance: A single resistance that would cause the same power loss as the actual resistance values distributed throughout a circuit. References 1. Image copyright vilax, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 2. Light bulb: Image copyright Snez, 2013; composite created by CK-12 Foundation - Samantha Bacic. http:// www.shutterstock.com . Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 31 CONCEPT 11 Resistors in Parallel Students will learn how to analyze and solve problems involving circuits with resistors in parallel. Students will learn how to analyze and solve problems involving circuits with resistors in parallel. www.ck12.org Key Equations Guidance 1 Rtotal = 1 R
1 + 1 R2 + 1 R3 + : : : Resistors in Parallel: All resistors are connected together at both ends. There are many rivers (i.e. The main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage. Example 1 A circuit is wired up with 2 resistors in parallel. Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different ’rivers’ so they have different current ﬂowing through them. Lets go through the same questions and answers as with the circuit in series. Question: What is the total resistance of the circuit? = 1 Answer: The total resistance is Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit. 90W thus, Rtotal = 90W 90W = 10 10W = 1 90W + 9 90W + 1 = 1 R1 + 1 R2 10 = 9W 1 Rtotal Question: What is the total current coming out of the power supply? Answer: Use Ohm’s Law (V = IR) but solve for current (I = V =R) . Itotal = Vtotal Rtotal = 20V 9W = 2:2A Question: How much power does the power supply dissipate? Answer: P = IV , so the total power equals the total voltage multiplied by the total current. Thus, Ptotal = ItotalVtotal = (2:2A)(20V ) = 44:4W . So the Power Supply outputs 44W (i.e. 44 Joules of energy per second). 32 www.ck12.org Concept 11. Resistors in Parallel Question: How much power is each resistor dissipating? Answer: Each resistor has different current across it, but the same voltage. So, using Ohm’s law, convert the power formula into a form that does not depend on current. P = IV = V R Substituted I = V =R into the power R formula. P90W = V 2 10W = 40W 90W R90W Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved. 90W = 4:4W ; P10W = V 2 R10W = (20V )2 V = V 2 = (20V )2 10W Question: How much current is ﬂowing through each resistor? Answer: Use Ohm’s law to calculate the current for each resistor. I90W = V90W R90W = 20V 90W = 0:22A I10W = V10W R10W = 20V 10W = 2:0A Notice that the 10W resistor has the most current going through it. It has the least resistance to electricity so this makes sense. Note: If you add up the currents of the individual ’rivers’ you get the total current of the of the circuit, as you should. Watch this Explanation MEDIA Click image to the left for more content. Simulation • http://simulations.ck12.org/Resistor/ 33 Time for Practice 1. Three 82 W resistors and one 12 W resistor are wired in parallel with a 9 V battery. a. Draw the schematic diagram. b. What is the total resistance of the circuit? 2. What does the ammeter read and which resistor is dissipating the most power? www.ck12.org 3. Given three resistors, 200 W; 300 W and 600 W and a 120 V power source connect them in a way to heat a container of water as rapidly as possible. a. Show the circuit diagram b. How many joules of heat are developed after 5 minutes? Answers to Selected Problems 1. b. 8:3 W 2. 0:8A and the 50 W on the left 3. part 2 43200J . 34 www.ck12.org Concept 12. Combined Series-Parallel Circuits CONCEPT 12 Combined Series-Parallel Circuits • Solve problems of combined circuits. Electrical circuits can become immensely complicated. This circuit is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various x values. Combined Series-Parallel Circuits Most circuits are not just a series or parallel circuit; most have resistors in parallel and in series. These circuits are called combination circuits . When solving problems with such circuits, use this series of steps. 1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them. 2. For resistors in series, calculate the single equivalent resistance that can replace them. 3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The voltage drops and currents though individual resistors can then be calculated. Example Problem: the total current through the circuit, and ﬁnd the current through each individual resistor. In the combination circuit sketched below, ﬁnd the equivalent resistance for the circuit, ﬁnd 35 www.ck12.org Solution: We start by simplifying the parallel resistors R2 and R3 . = 1 180 W 1 R23 R23 = 99 W + 1 220 W = 1 99 W We then simplify R1 and R23 which are series resistors. RT = R1 + R23 = 110 W + 99 W = 209 W We can then ﬁnd the total current, IT = VT RT All the current must pass through R1 , so I1 = 0:11 A . = 24 V 209 W = 0:11 A The voltage drop through R1 is (110 W)(0:11 A) = 12:6 volts . Therefore, the voltage drop through R2 and R3 is 11.4 volts. I2 = V2 220 W = 0:052 A R2 180 W = 0:063 A and I3 = V3 R3 = 11:4 V = 11:4 V Summary • Combined circuit problems should be solved in steps. Practice Video teaching the process of simplifying a circuit that contains both series and parallel parts. MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=In3NF8f-mzg Follow up questions: 1. In a circuit that contains both series and parallel parts, which parts of the circuit are simpliﬁed ﬁrst? 2. In the circuit drawn below, which resistors should be simpliﬁed ﬁrst? 36 www.ck12.org Concept 12. Combined Series-Parallel Circuits Review 1. Two 60.0W resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0W resistor. The combination is placed across a 120. V potential difference. (a) Draw a diagram of the circuit. (b) What is the equivalent resistance of the parallel portion of the circuit? (c) What is the equivalent resistance for the entire circuit? (d) What is the total current in the circuit? (e) What is the voltage drop across the 30.0W resistor? (f) What is the voltage drop across the parallel portion of the circuit? (g) What is the current through each resistor? 2. Three 15.0 Ohm resistors are connected in parallel and the combination is then connected in series with a 10.0 Ohm resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit. • combined or combination circuits: A route for the ﬂow of electricity that has elements of both series and parallel circuits. References 1. User:Linkaddict/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Huge_circuit.JPG . Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 37 Physics Unit 14: Magnetism Patrick Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Properties of Magnets 2 Magnetic Fields 1 6 iv www.ck12.org Concept 1. Properties of Magnets CONCEPT 1 Properties of Magnets • Describe magnetic ﬁelds around permanent magnets. • Understand ferromagnetism and magnetic domains. • Describe some properties of magnets. Some countries are using powerful electromagnets to develop high-speed trains, called maglev, or magnetic levitation, trains. These trains use the repulsive force of magnets to ﬂoat over a guide way, removing the friction of steel wheels and train tracks. Reducing this friction allows the trains to travel at much higher speeds. Properties of Magnets Any magnet , regardless of its shape, has two ends called poles where the magnetic effect is strongest. If a magnet is suspended by a ﬁne thread, it is found that one pole of the magnet will always point toward the north. This fact has been made use of in navigation since the eleventh century. The pole of the magnet that seeks the north pole is called the north pole of the magnet, while the opposite side is the south pole. It is a familiar fact that when two magnets are brought near one another, the magnets exert a force on each other. The magnetic force can be either attractive or repulsive. If two north poles or two south poles are brought near each other, the force will be repulsive. If a north pole is brought near a south pole, the force will be attractive. 1 www.ck12.org The Earth’s geographic north pole (which is close to, but not
exactly at the magnetic pole) attracts the north poles of magnets. We know, therefore, that this pole is actually the Earth’s magnetic south pole. This can be seen in the image above; the geographic north and south poles are labeled with barber shop poles, and the Earth’s magnetic poles are indicated with the double-headed arrow. Only iron and few other materials such as cobalt, nickel, and gadolinium show strong magnetic effects. These materials are said to be ferromagnetic . Other materials show some slight magnetic effect but it is extremely small and can be detected only with delicate instruments. Ferromagnetic Domains Microscopic examination reveals that a magnet is actually made up of tiny regions known as magnetic domains , which are about one millimeter in length and width. Each domain acts like a tiny magnet with a north and south pole. 2 www.ck12.org Concept 1. Properties of Magnets When the ferrous material is not magnetized, the domains are randomly organized so that the north and south poles do not line up and often cancel each other. When the ferrous material is placed in a magnetic ﬁeld, the domains line up with the magnetic ﬁeld so that the north poles are all pointed in the same direction and the south poles are all pointed in the opposite direction. In this way, the ferrous material has become a magnet. In many cases, the domains will remain aligned only while the ferrous material is in a strong magnetic ﬁeld; when the material is removed from the ﬁeld, the domains return to their previous random organization and the ferrous material loses any magnetic properties. Magnets that have magnetic properties while in the ﬁeld of another magnet but lose the magnetic properties when removed from the ﬁeld are called temporary magnets . Under certain circumstances, however, the new alignment can be made permanent and the ferrous substance becomes a permanent magnet . That is, the ferrous object remains a magnet even when removed from the other magnetic ﬁeld. The formation of temporary magnets allows a magnet to attract a non-magnetized piece of iron. You have most likely seen a magnet pick up a paper clip. The presence of the magnet aligns the domains in the iron paper clip and it becomes a temporary magnet. Whichever pole of the magnet is brought near the paper clip will induce magnetic properties in the paper clip that remain as long as the magnet is near. Permanent magnets lose their magnetic properties when the domains are dislodged from their organized positions and returned to a random jumble. This can occur if the magnet is hammered on or if it is heated strongly. Magnetic Fields When we were dealing with electrical effects, it was very useful to speak of an electric ﬁeld that surrounded an electric charge. In the same way, we can imagine a magnetic ﬁeld surrounding a magnetic pole. The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic ﬁeld of the other magnet. Magnetic ﬁeld lines go from the north magnetic pole to the south magnetic pole. We deﬁne the magnetic ﬁeld at any point as a vector (represented by the letter B ) whose direction is from north to south magnetic poles. 3 Summary • Any magnet has two ends called poles where the magnetic effect is strongest. • The magnetic pole found at the north geographical pole of the earth is a south magnetic pole. • The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic ﬁeld of the other magnet. • Magnetic ﬁeld lines go from the north magnetic pole to the south magnetic pole. www.ck12.org Practice MEDIA Click image to the left for more content. http://www.darktube.org/watch/crealev-levitating-ﬂoating-flying-hovering-bouncing This video demonstrates magnetic levitation. 1. In the video, one object rests on top of the magnetic ﬁeld of another. Compare the friction between these two objects to the friction between a saucer and a table the saucer rests on. Review 1. The earth’s magnetic ﬁeld (a) has a north magnetic pole at exactly the same spot as the geographical north pole. (b) is what causes compasses to work. (c) is what causes electromagnets to work. (d) all of these are true. (e) none of these are true. 2. A material that can be permanently magnetized is generally said to be (a) magnetic. (b) electromagnetic. (c) ferromagnetic. (d) none of these are true. 3. The force between like magnetic poles will be (a) repulsive. (b) attractive. (c) could be repulsive or attractive. 4. Why is a magnet able to attract a non-magnetic piece of iron? 5. If you had two iron rods and noticed that they attract each other, how could you determine if both were magnets or only one was a magnet? • magnet: A body that can attract certain substances, such as iron or steel, as a result of a magnetic ﬁeld. 4 www.ck12.org Concept 1. Properties of Magnets • magnetic pole: Either of two regions of a magnet, designated north and south, where the magnetic ﬁeld is strongest. Electromagnetic interactions cause the north poles of magnets to be attracted to the south poles of other magnets, and conversely. The north pole of a magnet is the pole out of which magnetic lines of force point, while the south pole is the pole into which they point. • ferromagnetic: A body or substance having a high susceptibility to magnetization, the strength of which depends on that of the applied magnetizing ﬁeld, and which may persist after removal of the applied ﬁeld. This is the kind of magnetism displayed by iron, and is associated with parallel magnetic alignment of adjacent domains. • magnetic ﬁeld: A ﬁeld of force surrounding a permanent magnet or a moving charged particle. • magnetic domain: An atom or group of atoms within a material that have some kind of “net” magnetic ﬁeld. • temporary magnet: A piece of iron that is a magnet while in the presence of another magnetic ﬁeld but loses its magnetic characteristics when the other ﬁeld is removed. • permanent magnet: A piece of magnetic material that retains its magnetism after it is removed from a magnetic ﬁeld. References 1. User:JakeLM/Wikipedia. http://commons.wikimedia.org/wiki/File:Maglev_june2005.jpg . CC-BY 2.5 2. Globe: Image copyright pdesign, 2013; Poles: Image copyright lineartestpilot, 2013; Composite created by http://www.shutterstock.com . Used under licenses from Shutter- CK-12 Foundation - Samantha Bacic. stock.com 5 CONCEPT 2 www.ck12.org Magnetic Fields Students will learn the idea of magnetic ﬁeld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic ﬁeld at an arbitrary distance from the wire. Students will learn the idea of magnetic ﬁeld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic ﬁeld at an arbitrary distance from the wire. Key Equations µ0I 2pr Where µ0 = 4p 107 Tm/A Bwire = Magnetic ﬁeld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Guidance Permanent magnets (like refrigerator magnets) consist of atoms, such as iron, for which the magnetic moments (roughly electron spin) of the electrons are “lined up” all across the atom. This means that their magnetic ﬁelds add up, rather than canceling each other out. The net effect is noticeable because so many atoms have lined up. The magnetic ﬁeld of such a magnet always points from the north pole to the south. The magnetic ﬁeld of a bar magnet, for example, is illustrated below: If we were to cut the magnet above in half, it would still have north and south poles; the resulting magnetic ﬁeld would be qualitatively the same as the one above (but weaker). Charged particles in motion also generate magnetic ﬁelds. The most frequently used example is a current carrying wire, since current is literally moving charged particles. The magnitude of a ﬁeld generated by a wire depends on distance to the wire and strength of the current (I) (see ’Key Equations’ section) : Meanwhile, its direction can be found using the so called ﬁrst right hand rule : point your thumb in the direction of the current. Then, curl your ﬁngers around the wire. The direction your ﬁngers will point in the same direction as the ﬁeld. Be sure to use your right hand! 6 www.ck12.org Concept 2. Magnetic Fields Sometimes, it is necessary to represent such three dimensional ﬁelds on a two dimensional sheet of paper. The following example illustrates how this is done. In the example above, a current is running along a wire towards the top of your page. The magnetic ﬁeld is circling the current carrying wire in loops which are perpendicular to the page. Where these loops intersect this piece of paper, we use the symbol J to represent where the magnetic ﬁeld is coming out of the page and the symbol N to . This convention can be used for all vector quantities: represent where the magnetic ﬁeld is going into the page ﬁelds, forces, velocities, etc. Example 1 You are standing right next to a current carrying wire and decide to throw your magnetic ﬁeld sensor some distance perpendicular to the wire. When you go to retrieve your sensor, it shows the magnetic ﬁeld where it landed to be 4 105 T . If you know the wire was carrying 300A, how far did you throw the sensor? Solution To solve this problem, we will just use the equation given above and solve for the radius. 7 www.ck12.org B = r = r = µoI 2pr µoI 2pB 4p 107 Tm/A 300 A 2p 4 105 T r = 1:5 m MEDIA Click image to the left for more content. Watch this Explanation Simulation Magnet and Compass (PhET Simulation) Time for Practice 1. Sketch the magnetic ﬁeld lines for the horseshoe magnet shown here. Then, show the direction in which the two compasses (shown as circles) should point considering their positions. In other words, draw an arrow in the compass that represents North in relation to the compass magnet. 8 www.ck12.org Concept 2. Magnetic Fields 2.
Find the magnetic ﬁeld a distance of 20 cm from a wire that is carrying 3 A of electrical current. 3. In order to measure the current from big power lines the worker simply clamps a device around the wire. This provides safety when dealing with such high currents. The worker simply measures the magnetic ﬁeld and deduces the current using the laws of physics. Let’s say a worker uses such a clamp and the device registers a magnetic ﬁeld of 0.02 T. The clamp is 0.05 m from the wire. What is the electrical current in the wire? Answers to Selected Problems 1. Both pointing away from north 2. 3 106T 3. 5000 A 9 Physics Unit 15: Electromagnetism Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: April 6, 2014 iii Contents www.ck12.org 1 4 8 13 18 22 26 Contents 1 Electromagnetic Induction 2 Electromagnets 3 Current and Magnetism 4 Electric Motors 5 Electromotive Force 6 Electric Generators 7 Transformers iv www.ck12.org Concept 1. Electromagnetic Induction CONCEPT 1 Electromagnetic Induction • Deﬁne electromagnetic induction. • Explain how electromagnetic induction occurs. • Describe the current produced by electromagnetic induction. • Identify ways that electromagnetic induction is used. The girl on the left in this photo is riding a stationary bike. She’s getting exercise, but that’s not the real reason she’s riding the bike. She’s using her muscle power to generate electricity through a process called electromagnetic induction. What Is Electromagnetic Induction? Electromagnetic induction is the process of generating electric current with a magnetic ﬁeld. It occurs whenever a magnetic ﬁeld and an electric conductor, such as a coil of wire, move relative to one another. As long as the conductor is part of a closed circuit, current will ﬂow through it whenever it crosses lines of force in the magnetic ﬁeld. One way this can happen is illustrated in the Figure 1.1 . The sketch shows a magnet moving through a wire coil. You can watch an animated version of the illustration at this URL: http://jsticca.wordpress.com/2009/09/01 /the-magnet-car/ Q: What is another way that a coil of wire and magnet can move relative to one another and generate an electric current? A: The coil of wire could be moved back and forth over the magnet. The Current Produced by a Magnet The device with the pointer in the circuit above is an ammeter. It measures the current that ﬂows through the wire. The faster the magnet or coil moves, the greater the amount of current that is produced. If more turns were added to the coil or a stronger magnet were used, this would produce more current as well. 1 www.ck12.org FIGURE 1.1 The Figure 1.2 shows the direction of the current that is generated by a moving magnet. If the magnet is moved back and forth repeatedly, the current keeps changing direction. In other words, alternating current (AC) is produced. Alternating current is electric current that keeps reversing direction. FIGURE 1.2 How Electromagnetic Induction Is Used Two important devices depend on electromagnetic induction: electric generators and electric transformers. Both devices play critical roles in producing and regulating the electric current we depend on in our daily lives. Electric generators use electromagnetic induction to change kinetic energy to electrical energy. They produce electricity in power plants. Electric transformers use electromagnetic induction to change the voltage of electric current. Some transformers increase voltage and other decrease voltage. Q: How do you think the girl on the exercise bike in the opening photo is using electromagnetic induction? A: As she pedals the bike, the kinetic energy of the turning pedals is used to move a conductor through a magnetic ﬁeld. This generates electric current by electromagnetic induction. 2 www.ck12.org Summary Concept 1. Electromagnetic Induction • Electromagnetic induction is the process of generating electric current with a magnetic ﬁeld. It occurs whenever a magnetic ﬁeld and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic ﬁeld. • The current produced by electromagnetic induction is greater when the magnet or coil moves faster, the coil has more turns, or the magnet is stronger. If the magnet or coil is moved back and forth repeatedly, alternating current is produced. • Electric generators and electric transformers use electromagnetic induction to generate electricity or change the voltage of electric current. Vocabulary • electromagnetic induction : Process of generating electric current with a changing magnetic ﬁeld. Practice Simulate electromagnetic induction at the following URL. Then answer the questions below. http://micro.magnet .fsu.edu/electromag/java/faraday2/ 1. How is electric current created in the simulation? What type of current is it? 2. How is electric current measured in the simulation? 3. What happens when you stop moving the magnet? Review 1. What is electromagnetic induction? When does it occur? 2. How could you increase the amount of current produced by electromagnetic induction? 3. Explain how a moving magnet and a coil of wire can be used to produce alternating current. 4. List two devices that use electromagnetic induction. References 1. Christopher Auyeung. . CC-BY-NC-SA 3.0 2. Christopher Auyeung. . CC-BY-NC-SA 3.0 3 CONCEPT 2 www.ck12.org Electromagnets • Deﬁne and describe a solenoid. • Deﬁne and describe an electromagnet. • Determine the direction of the magnetic ﬁeld inside a solenoid given the direction of current ﬂow in the coil wire. • Understand why an electromagnet has a stronger magnetic ﬁeld than a solenoid. One of the most famous electric car companies is Tesla, named after Nikola Tesla. These electric cars, and all others, require an electromagnet to run the engine. Electromagnets A long coil of wire consisting of many loops of wire and making a complete circuit is called a solenoid . The magnetic ﬁeld within a solenoid can be quite large since it is the sum of the ﬁelds due to the current in each individual loop. 4 www.ck12.org Concept 2. Electromagnets The magnetic ﬁeld around the wire is determined by a hand rule. Since this description doesn’t mention electron ﬂow, we must assume that the current indicated by I is conventional current (positive). Therefore, we would use a right hand rule. We grasp a section of wire with our right hand pointing the thumb in the direction of the current ﬂow and our ﬁngers will curl around the wire in the direction of the magnetic ﬁeld. Therefore, the ﬁeld points down the cavity in these loops from right to left as shown in the sketch. If a piece of iron is placed inside the coil of wire, the magnetic ﬁeld is greatly increased because the domains of the iron are aligned by the magnetic ﬁeld of the current. The resulting magnetic ﬁeld is hundreds of time stronger than the ﬁeld from the current alone. This arrangement is called an electromagnet . The picture below shows an electromagnet with an iron bar inside a coil. Our knowledge of electromagnets developed from a series of observations. In 1820, Hans Oersted discovered that a current-carrying wire produced a magnetic ﬁeld. Later in the same year, André-Marie Ampere discovered that a coil of wire acted like a permanent magnet and François Arago found that an iron bar could be magnetized by putting it inside coil of current-carrying wire. Finally, William Sturgeon found that leaving the iron bar inside the coil greatly increased the magnetic ﬁeld. Two major advantages of electromagnets are that they are extremely strong magnetic ﬁelds, and that the magnetic ﬁeld can be turned on and off. When the current ﬂows through the coil, it is a powerful magnet, but when the current is turned off, the magnetic ﬁeld essentially disappears. Electromagnets ﬁnd use in many practical applications. Electromagnets are used to lift large masses of magnetic materials such as scrap iron, rolls of steel, and auto parts. 5 www.ck12.org The overhead portion of this machine (painted yellow) is a lifting electromagnet. It is lowered to the deck where steel pipe is stored and it picks up a length of pipe and moves it to another machine where it is set upright and lowered in
to an oil well drill hole. Electromagnets are essential to the design of the electric generator and electric motor and are also employed in doorbells, circuit breakers, television receivers, loudspeakers, electric dead bolts, car starters, clothes washers, atomic particle accelerators, and electromagnetic brakes and clutches. Electromagnets are commonly used as switches in electrical machines. A recent use for industrial electromagnets is to create magnetic levitation systems for bullet trains. Summary • A solenoid is a long coil of wire consisting of many loops of wire that makes a complete circuit. • An electormagnet is a piece of iron inside a solenoid. • While the magnetic ﬁeld of a solenoid may be quite large, an electromagnet has a signiﬁcantly larger magnetic ﬁeld. • Electromagnets’ magnetic ﬁelds can be easily turned off by just halting the current. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=S6oop6RXg9w Follow up questions. 1. What components are needed to make a homemade electromagnet? 2. What objects were attracted by the electromagnet in the video? Review 1. Magnetism is always present when electric charges ___________. 2. What happens to the strength of an electromagnet if the number of loops of wire is increased? 3. What happens to the strength of an electromagnet if the current in the wire is increased? 4. Which direction does the magnetic ﬁeld point in the solenoid sketched here? 6 www.ck12.org Concept 2. Electromagnets • solenoid: A current-carrying coil of wire that acts like a magnet when a current passes through it. • electromagnet: A temporary magnet consisting of an iron or steel core wound with a coil of wire, through which a current is passed. • magnetic levitation: The suspension of an object above a second object by means of magnetic repulsion. References 1. Image copyright Dongliu, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 2. Coil: Image copyright Giant Stock, 2013; modiﬁed by CK-12 Foundation - Samantha Bacic. http://www. shutterstock.com . Used under license from Shutterstock.com 3. Image copyright Zigzag Mountain Art, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com 4. Image copyright Ingvar Tjostheim, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 5. . . CC BY-NC-SA 6. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 7 CONCEPT 3 Current and Magnetism Students will learn to analyze and solve problems involving current carrying wires in magnetic ﬁelds. Students will learn to analyze and solve problems involving current carrying wires in magnetic ﬁelds. www.ck12.org Key Equations Fwire = LIB sin(q) Force on a Current Carrying Wire In this equation, L refers to the length of the wire, I to the electric current, B the magnitude of the magnetic ﬁeld and q is the angle between the direction of the current and the direction of the magnetic ﬁeld. µ0I 2pr Where µ0 = 4p 107 Tm/A Bwire = Magnetic ﬁeld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Force on a Wire Since a wire is nothing but a collection of moving charges, the force it will experience in a magnetic ﬁeld will simply be the vector sum of the forces on the individual charges. If the wire is straight — that is, all the charges are moving in the same direction — these forces will all point in the same direction, and so will their sum. Then, the direction of the force can be found using the second right hand rule, while its magnitude will depend on the length of the wire (denoted L ), the strength of the current, the strength of the ﬁeld, and the angle between their directions: Two current-carrying wires next to each other each generate magnetic ﬁelds and therefore exert forces on each other: 8 Concept 3. Current and Magnetism MEDIA Click image to the left for more content. www.ck12.org Example 1 Example 2 A wire loop and an inﬁnitely long current carrying cable are placed a distance r apart. The inﬁnitely long wire is carrying a current I1 to the left and the loop is carrying a current I2 CCW. The dimensions of the wire loop are shown in the diagram illustrating the situation below. What is the magnitude and direction of the net force on the loop (the mass of the wires are negligible)? Solution In this problem, it is best to start by determining the direction of the force on each segment of the loop. Based on the ﬁrst right hand rule, the magnetic ﬁeld from the inﬁnite cable points into the page where the loop is. This means that the force on the top segment of the loop will be down toward the bottom of the page, the force on the left segment will be to right, the force on the bottom segment will be toward the top of the page, and the force on the right segment will be to the left. The forces on the left and right segments will balance out because both segments are the same distance from the cable. The forces from the top and bottom section will not balance out because the wires are different distances from the cable. The force on the bottom segment will be stronger than the one on the top segment because the magnetic ﬁeld is stronger closer to the cable, so the net force on the loop will be up, toward the top of the page. Now we will begin to calculate the force’s magnitude by ﬁrst determining the strength of the magnetic ﬁeld at the bottom and top segments. All we really have to do is plug in the distances to each segment into the equation we already know for the magnetic ﬁeld due to a current carrying wire. 9 www.ck12.org B = Bbottom = Btop = µoI 2pr µoI1 2pR µoI1 2p2R Now we will calculate the net force on the loop using the equation given above. We’ll consider up the positive direction. SF = I2L( SF = Fbottom Ftop SF = I2LBbottom I2LBtop SF = I2L(Bbottom Btop) µoI1 2pR µoI1I2L 2pR µoI1I2L 4pR µoI1 2p2R 1 2 SF = SF = (1 ) ) start by summing the forces on the loop substitute in the values for each of the force terms factor the equation substitute in the values for the magnetic ﬁeld factor the equation again simplify to get the answer Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A vertical wire, with a current of 6:0 A going towards the ground, is immersed in a magnetic ﬁeld of 5:0 T pointing to the right. What is the value and direction of the force on the wire? The length of the wire is 2:0 m . 10 www.ck12.org Concept 3. Current and Magnetism 2. 3. A futuristic magneto-car uses the interaction between current ﬂowing across the magneto car and magnetic ﬁelds to propel itself forward. The device consists of two ﬁxed metal tracks and a freely moving metal car (see illustration above). A magnetic ﬁeld is pointing downward with respect to the car, and has the strength of 5:00 T . The car is 4:70 m wide and has 800 A of current ﬂowing through it. The arrows indicate the direction of the current ﬂow. a. Find the direction and magnitude of the force on the car. b. If the car has a mass of 2050 kg , what is its velocity after 10 s , assuming it starts at rest? c. If you want double the force for the same magnetic ﬁeld, how should the current change? 4. A horizontal wire carries a current of 48 A towards the east. A second wire with mass 0:05 kg runs parallel to the ﬁrst, but lies 15 cm below it. This second wire is held in suspension by the magnetic ﬁeld of the ﬁrst wire above it. If each wire has a length of half a meter, what is the magnitude and direction of the current in the lower wire? 5. Show that the formula for the force between two current carrying wires is F = µoLi1i2 2pd , where d is the distance between the two wires, i1 is the current of ﬁrst wire and L is the segment of length of the second wire carrying a current i2 . (Hint: ﬁnd magnetic ﬁeld emanating from ﬁrst wire and then use the formula for a wire immersed in that magnetic ﬁeld in order to ﬁnd the force on the second wire.) 6. Two long thin wires are on the same plane but perpendicular to each other. The wire on the y axis carries a current of 6:0 A in the y direction. The wire on the x axis carries a current of 2:0 A in the +x direction. Point, P has the co-ordinates of (2:0; 2; 0) in meters. A charged particle moves in a direction of 45o away from the origin at point, P , with a velocity of 1:0 107 m=s: a. Find the magnitude and direction of the magnetic ﬁeld at point, P . b. If there is a magnetic force of 1:0 106 N on the particle determine its charge. c. Determine the magnitude of an electric ﬁeld that will cancel the magnetic force on the particle. 7. A long straight wire is on the x axis and has a current of 12 A in the x direction. A point P , is located 2:0 m above the wire on the y axis. a. What is the magnitude and direction of the magnetic ﬁeld at P . b. If an electron moves through P in the x direction at a speed of 8:0 107 m=s what is the magnitude and direction of the force on the electron? Answers to Selected Problems 1. Down the page; 60 N 2. a. To the right, 1:88 104 N b. 91:7 m=s c. It should be doubled 3. East 1:5 104 A 4. . 5. a. 8 107 T b. 1:3 106 C 11 6. a. 1:2 106 T; +z b. 1:5 1017 N; y www.ck12.org 12 www.ck12.org Concept 4. Electric Motors CONCEPT 4 Electric Motors • Explain the design and operation of an electric motor. As gas prices continue to rise, electric cars and hybrids are becoming increasingly popular. These cars are certainly a part of our future. On the left in the image above is an all-electric vehicle, and on the right is a hybrid vehicle that uses gas part time and electricity part time. Electric Motors In an earlier concept, we described and calculated the force that a magnetic ﬁeld exerts on a current carrying wire. Since you are familiar with Newton’s third law of motion, you know that if the magnetic ﬁeld exerts a force on the current carrying wire, then the current carrying wire also exerts a force of equal magnitude and opposite direction on the magnetic ﬁeld. In the sketch abo
ve, a circuit is connected to a battery, with one part of the circuit placed inside a magnetic ﬁeld. When current runs through the circuit, a force will be exerted on the wire by the magnetic ﬁeld, causing the wire to 13 www.ck12.org move. If we choose to consider electron current in this case, the electrons ﬂow from the back of the sketch to the front while the magnetic ﬁeld is directed upward. Using the left hand rule for this, we ﬁnd that the force on the wire is to the right of the page. Had we chosen to consider the current to be conventional current, then the current would be ﬂowing from the front of the sketch to the back and we would use the right hand rule. The force on the wire would, once again, be toward the right. This movement is harnessed in electrical motors . Electrical motors change electrical energy into mechanical energy. The motor consists of an electrical circuit with part of the wires inside a magnetic ﬁeld. This can be seen below. Positive charges move through the circuit in the direction of the light purple arrows. When the charges move up through the part of the coil that is right next to the north pole, the right hand rule tells us that the wire suffers the force, F , pushing the wire in the direction of the blue arrow, toward the back of the sketch. On the other side of the coil, where the charges are moving down through the ﬁeld, the right hand rule shows the force would push this side of the coil toward the front. These two forces are working together, rotating the coil in the direction of the circular red arrow. Where the rotating coil (in grey) meets the wires attached to the power source (black), we ﬁnd a split ring commutator. The coil turns, but the commutator and power source do not. As the coil turns, it moves off of the blue box connector and as it continues to turn, it connects to the other blue box connector. As the coil turns, it reverses its connections to the external circuit. Therefore, inside the coil, the current is always ﬂowing in the same direction because the left side of the coil is always attached to the left side of the external circuit. This allows the coil, or armature, to continue to spin the same direction all the time. In electrical motors, these coils often consist of not just one, but many wires, as can be seen here: 14 www.ck12.org Concept 4. Electric Motors Summary • When current runs through the circuit, a force will be exerted on the wire. • An electrical motor changes electrical energy into mechanical energy. • A split ring commutator keeps the current in the coil ﬂowing in the same direction even though the coil changes sides every half turn. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=elFUJNodXps In the video, a simple electrical motor is constructed. Follow up questions. 1. Who ﬁrst built an electric motor? 2. What size battery was used in the video motor? 3. The rover Curiosity is on the surface of what body? Review 1. Which way will the wire be pushed when current passes through the wire? a. up b. down 15 c. left d. right e. None of these. www.ck12.org 2. Which way will the coil spin when current passes through the wire? a. clockwise b. counterclockwise • electric motor: An electricmotor converts electricity into mechanical motion. • commutator: A split - ring commutator (sometimes just called a commutator) is a simple and clever device for reversing the current direction through an armature every half turn. • armature: A revolving structure in an electric motor or generator, wound with the coils that carry the current. References 1. Myrtle Beach TheDigitel and User:Mariordo/Wikimedia Commons. http://commons.wikimedia.org/wiki/F ile:Nissan_Leaf_%26_Chevy_Volt_charging_trimmed.jpg . CC-BY 2.0 16 www.ck12.org Concept 4. Electric Motors 2. . . CC BY-NC-SA 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic. . CC BY-NC-SA 3.0 5. Image copyright Sim Kay Seng, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 6. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 7. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 17 CONCEPT 5 www.ck12.org Electromotive Force • Deﬁne EMF. • Calculate EMF for a wire moving in a magnetic ﬁeld. Electrical generators convert mechanical energy into electrical energy. Every electrical generator needs some method for spinning the coil inside the magnetic ﬁeld. Hydroelectric generators use water pressure to spin the coil while windmills, of course, use the wind to spin the coil. The image here is a combination of steam turbine and generator. The steam can be produced by burning coal or diesel fuel or by a nuclear reaction and the steam then turns the coil and generates electricity. Electromotive Force When an individual charge ﬂies through a magnetic ﬁeld, a force is exerted on the charge and the path of the charge bends. In the case shown in the sketch below, the charge is positive and the right hand rule shows us the force will be upward, perpendicular to both the ﬁeld and the path of the charge. 18 www.ck12.org Concept 5. Electromotive Force If a wire that is part of a complete circuit is moved through a magnetic ﬁeld, the force on the individual electrons in the wire occurs in exactly the same manner. Since the electrons in the wire are negatively charged, the force would be in the opposite direction but otherwise the situation is the same. When the wire is pulled downward through the magnetic ﬁeld, the force on the electrons cause them to move within the wire. Since the charges are negative, the left hand rule shows that the electrons would move as diagrammed in the sketch. (Point ﬁngers in the direction of the magnetic ﬁeld, point thumb in the direction of wire movement, and palm shows direction of electron ﬂow.) No current will ﬂow, of course, unless the section of wire is part of a complete circuit. This process allows us to convert mechanical energy (the motion of the wire) into electrical energy (the current). This is the opposite of what happens in an electric motor where electrical energy is converted into mechanical energy. In order to maintain a constant current ﬂow, it is necessary to have a potential difference or voltage in the circuit. The voltage or potential difference is also frequently referred to electromotive force . The term electromotive force, 19 like many historical terms, is a misnomer. Electromotive force is NOT a force, it is a potential difference or potential energy per unit charge and is measured in volts. The potential difference in the case of moving a wire through a magnetic ﬁeld is produced by the work done on the charges by whatever is pushing the wire through the ﬁeld. The EMF (or voltage) depends on the magnetic ﬁeld strength, B , the length of the wire in the magnetic ﬁeld, l , and the velocity of the wire in the ﬁeld. www.ck12.org EMF = Blv This calculation is based on the wire moving perpendicularly through the ﬁeld. If the wire moves an angle to the ﬁeld, then only the component of the wire perpendicular to the ﬁeld will generate EMF . Example Problem: A 0.20 m piece of wire that is part of a complete circuit moves perpendicularly through a magnetic ﬁeld whose magnetic induction is 0.0800 T. If the speed of the wire is 7.0 m/s, what EMF is induced in the wire? Solution: EMF = Blv = (0:0800 N=A m)(0:20 m)(7:0 m=s) = 0:11 N m=C = 0:11 J=C = 0:11 V Summary • If a wire that is part of a complete circuit is moved through a magnetic ﬁeld, the magnetic ﬁeld exerts a force on the individual electrons in the wire, which causes a current to ﬂow. • The potential difference in the case of moving a wire through a magnetic ﬁeld is produced by the work done on the charges by whatever is pushing the wire through the ﬁeld. • The EMF (or voltage) depends on the magnetic ﬁeld strength, B , the length of the wire in the magnetic ﬁeld, l , and the velocity of the wire in the ﬁeld, EMF = Blv . Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=0OHmMVBLXTI Follow up questions. 1. We have been discussing the process of generating electricity by moving a wire through a magnetic ﬁeld. What happens if the wire is held steady and the magnetic ﬁeld moves instead? 2. When a loop of wire is turned circularly in a magnetic ﬁeld, what type of current is produced? Review 1. Which of the following units are equivalent to those of EMF produced in a generator? (a) T m=s (b) V m2=s 20 www.ck12.org (c) J=s (d) A W (e) T m Concept 5. Electromotive Force 2. A straight wire 0.500 m long is moved straight up through a 0.400 T magnetic ﬁeld pointed in the horizontal direction. The speed of the wire is 20.0 m/s. (a) What EMF is induced in the wire? (b) If the wire is part of a circuit with a total resistance of 6:00 W , what is the current in the circuit? 3. A straight wire, 25.0 m long, is mounted on an airplane ﬂying at 125 m/s. The wire moves perpendicularly through earth’s magnetic ﬁeld (B = 5:00 105 T ) . What is the EMF induced in the wire? 4. A straight wire, 30.0 m long, moves at 2.00 m/s perpendicularly through a 1.00 T magnetic ﬁeld. (a) What is the induced EMF ? (b) If the total resistance of the circuit is 15:0 W , what is the current in the circuit? • electromotive force: The potential energy per unit charge that produces a ﬂow of electricity in a circuit, expressed in volts. • induced current: The electric current generated in a loop of conducting material by movement of the loop across a magnetic ﬁeld. References 1. Courtesy of the NRC. http://commons.wikimedia.org/wiki/File:Modern_Steam_Turbine_Generator.jpg . Public Domain 2. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 21 CONCEPT 6 Electric Generators www.ck12.org • Explain how an electric generator works. • Explain the difference between an electric generator and an electric motor. • Explain the difference between peak and effective vol
tage and current from a generator. These large machines are electric generators. This particular row of generators is installed in a hydroelectric power station. The insides of these generators are coils of wire spinning in a magnetic ﬁeld. The relative motion between the wire and the magnetic ﬁeld is what generates electric current. In all generators, some mechanical energy is used to spin the coil of wire in the generator. In the case of hydroelectric power, the coil of wire is spun by water falling from higher PE to lower PE . Windmills and steam turbines are used in other types of power generators to spin the coil. Electric Generators Electric generators convert mechanical energy to electric energy. The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic ﬁeld. The loops of wire and the iron core are called the armature . The armature is mounted so that it can rotate freely inside the magnetic ﬁeld. Mechanical energy is used to spin the armature in the ﬁeld so that the wire loops cut across the ﬁeld and produce electric current. The EMF of this current is calculated by EMF = Blv . 22 www.ck12.org Concept 6. Electric Generators Consider the coil and magnetic ﬁeld sketched above. When the right hand side of the coil moves up through the ﬁeld, the left hand rule indicates that the electron ﬂow will be from the front to the back in that side of the coil. The current generated will have the greatest EMF as the wire is cutting perpendicularly across the ﬁeld. When the wire reaches the top of its arc, it is moving parallel to the ﬁeld and therefore, not cutting across the ﬁeld at all. The EMF at this point will be zero. As that same wire then cuts down through the ﬁeld as it continues to spin, the left hand rule indicates that the electron ﬂow will be from the back to the front in that side of the coil. In this second half of the arc, the direction of the electron ﬂow has reversed. The magnitude of the EMF will reach maximum again as the wire cuts perpendicularly down through the ﬁeld and the EMF will become zero again as the wire passes through the bottom of the arc. The current produced as the armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current . By having more and more loops of wire on the armature, the crests and troughs overlap and ﬁll in until a constant current is produced. A direct current is one that always ﬂows in the same direction rather than alternating back and forth. Batteries produce direct currents. A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. Generators and motors are almost identical in construction but convert energy in opposite directions. Generators convert mechanical energy to electrical energy and motors convert electrical energy to mechanical. Because of the alternating direction in alternating current, the average value is less than the power supplied by a direct current. In fact, the average power of an AC current is one-half its maximum power and one-half the power of an equivalent DC current. The effective current of an AC generator is 0.707 times its maximum current. The same is true for the effective voltage of an AC generator. 23 Ieff = 0:707 Imax Veff = 0:707 Vmax Example Problem: An AC generator develops a maximum voltage of 34.0 V and delivers a maximum current of 0.170 A. www.ck12.org a. What is the effective voltage of the generator? b. What is the effective current delivered by the generator? c. What is the resistance in the circuit? Solution: a. Veff = 0:707 Vmax = (0:707)(34:0 V ) = 24:0 V b. Ieff = 0:707 Imax = (0:707)(0:17 A) = 0:120 A c. R = V 0:120 A = 200: W I = 24:0 V Summary • Electric generators convert mechanical energy to electric energy. • The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic ﬁeld. • The loops of wire and the iron core are called the armature. • The armature is mounted so that it can rotate freely inside the magnetic ﬁeld. • Mechanical energy is used to spin the armature in the ﬁeld so that the wire loops cut across the ﬁeld and produce electric current. • The current produced as the armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current. • A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. • The effective current of an AC generator is 0.707 times its maximum current. • The effective voltage of an AC generator is 0.707 times its maximum voltage. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=RFOMpOM1WHQ Follow up questions. 1. Which of the two generators in the video (an AC generator and a DC generator) involves a magnetic ﬁeld? 2. Which of the two generators in the video involves a wire-wrapped armature? 3. What is the difference between the DC generator and the AC generator? 24 Concept 6. Electric Generators www.ck12.org Review 1. What three things are necessary to produce EMF mechanically? (a) magnet, force lines, and magnetic ﬁeld (b) EMF , conductor, and magnetic ﬁeld (c) conducting wire, magnetic ﬁeld, and relative motion (d) conducting wire, electrical ﬁeld, and relative motion (e) none of these will produce EMF mechanically 2. Increasing which of the following will increase the output of a generator? (a) EMF (b) strength of the magnetic ﬁeld (c) resistance of the conductor (d) load on the meter (e) none of these 3. The current in the rotating coil of all generators is (a) AC (b) DC (c) pulsating AC (d) pulsating DC 4. A generator in a power plant develops a maximum voltage of 170. V. (a) What is the effective voltage? (b) A 60.0 W light bulb is placed across the generator. A maximum current of 0.70 A ﬂows through the bulb. What effective current ﬂows through the bulb? (c) What is the resistance of the light bulb when it is working? 5. The effective voltage of a particular AC household outlet is 117 V. (a) What is the maximum voltage across a lamp connected to the outlet? (b) The effective current through the lamp is 5.50 A. What is the maximum current in the lamp? • direct current: An electric current ﬂowing in one direction only. • alternating current: An electric current that reverses direction in a circuit at regular intervals. • electric generator: An electric generator is a device that converts mechanical energy to electrical energy. • armature: The rotating part of a generator, consisting essentially of copper wire wound around an iron core. References 1. Image copyright James L. Davidson, 2013. http://www.shutterstock.com . Public Domain 2. Galvanometer: Image copyright scropy, 2013; composite created by CK-12 Foundation. http://www.shutters tock.com . Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0 25 CONCEPT 7 • Describe the function of a transformer. • Explain the relationship between turns and voltage ratio. • Solve mathematical problems involving transformers. www.ck12.org Transformers Power loss in long transmission lines is related to the magnitude of the current. Speciﬁcally, the power loss can be decreased by decreasing the magnitude of the current. The amount of power passed through transmission lines can be calculated by multiplying voltage by current. The same power can be transmitted using a very high voltage and a very low current as with a low voltage and high current. Since power companies do not wish to waste power as it is transmitted to homes and businesses, they deliberately ’step up’ the voltage and reduce the current before transmitting the power over extended distances. That type of power transmits well without great loss of energy but it cannot be used in household appliances. It becomes necessary to convert it back (’step down’) to low voltage and high current for household use. That is the job of electrical transformers – those big gray barrels you see on power poles. Transformers When we move a wire through a magnetic ﬁeld, a force is exerted on the charges in the wire and a current is induced. Essentially the same thing happens if we hold the wire steady and move the magnetic ﬁeld by moving the magnet. Yet a third way of causing relative motion between the charges in a wire and a magnetic ﬁeld is to expand or contract the ﬁeld through the wire. When a current begins to ﬂow in a wire, a circular magnetic ﬁeld forms around the wire. Within the ﬁrst fractional second when the current begins to ﬂow, the magnetic ﬁeld expands outward from the wire. If a second wire is placed nearby, the expanding ﬁeld will pass through the second wire and induce a brief current in the wire. 26 www.ck12.org Concept 7. Transformers Consider the sketch above. When the knife switch is closed, current begins to ﬂow in the ﬁrst circuit and therefore, a magnetic ﬁeld expands outward around the wire. When the magnetic ﬁeld expands outward from the wire on the right side, it will pass through the wire in the second circuit. This relative motion between wire and ﬁeld induces a current in the second circuit. The magnetic ﬁeld expands outward for only a very short period of time and therefore, only a short jolt of cu
rrent is induced in the second circuit. You can leave the knife switch closed and the current will continue to ﬂow in the ﬁrst circuit but no current is induced in the second circuit because the ﬁeld is constant and therefore there is no relative motion between the ﬁeld and the wire in the second circuit. When the knife switch is opened, the current in the ﬁrst circuit ceases to ﬂow and the magnetic ﬁeld collapses back through the wire to zero. As the magnetic ﬁeld collapses, it passes through the wire and once again we have relative motion between the wire in the second circuit and the magnetic ﬁeld. Therefore, we once again have a short jolt of current induced in the second circuit. This second jolt of induced current will be ﬂowing in the opposite direction of the ﬁrst induced current. We can produce an alternating current in the second circuit simply by closing and opening the knife switch continuously in the ﬁrst circuit. Obviously, a transformer would have little use in the case of DC current because current is only induced in the second circuit when the ﬁrst circuit is started or stopped. With AC current, however, since the current changes direction 60 times per second, the magnetic ﬁeld would constantly be expanding and contracting through the second wire. A transformer is a device used to increase or decrease alternating current voltages. They do this with essentially no loss of energy. A transformer has two coils, electrically insulated from each other as shown in the sketch. One coil is called the primary coil and the other is called the secondary coil . When the primary coil is connected to a source of AC voltage, the changing current creates a varying magnetic ﬁeld. The varying magnetic ﬁeld induces a varying EMF in the secondary coil. The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. 27 www.ck12.org secondary voltage primary voltage VS VP = = number of turns on secondary number of turns on primary NS NP If the secondary voltage is larger than the primary voltage, the transformer is called a step-up transformer . If the voltage out of the transformer is smaller than the voltage in, then the transformer is called a step-down transformer . In an ideal transformer, the electric power put into the primary equals the electric power delivered by the secondary. The current that ﬂows in the primary depends on how much current is required by the secondary circuit. VPIP = VSIS IS IP = VP VS = NP NS Example Problem: A particular step-up transformer has 200 turns on the primary coil and 3000 turns on the secondary coil. a. If the voltage on the primary coil is 90.0 V, what is the voltage on the secondary coil? b. If the current in the secondary circuit is 2.00 A, what is the current in the primary coil? c. What is the power in the primary circuit? d. What is the power in the secondary circuit? Solution: = NP NS = NP NS VS = (90:0 V )(3000) a. VP VS IS IP = 30:0 A b. IP c. PP = VPIP = (90:0 V )(30:0 A) = 2700 W d. PS = VSIS = (1350 V )(2:00 A) = 2700 W = 200 3000 = 200 3000 90:0 V VS 2:00 A IP (200) = 1350 V 28 www.ck12.org Summary Concept 7. Transformers • When a current begins to ﬂow in a wire, a circular magnetic ﬁeld forms around the wire. • Within the ﬁrst fractional second when the current begins to ﬂow, the magnetic ﬁeld expands outward from the wire. • If a second wire is placed nearby, the expanding ﬁeld will pass through the second wire and induce a brief current in the wire. • A transformer is a device used to increase or decrease alternating current voltages. • A transformer has two coils, electrically insulated from each other. One coil is called the primary coil and the other is called the secondary coil. • The varying magnetic ﬁeld induces a varying EMF in the secondary coil. • The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. secondary voltage primary voltage = number of turns on secondary number of turns on primary Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=-v8MYAFl7Mw Follow up questions. 1. What type of transformer is used at the power station where the electric power is generated? 2. What type of transformer is used at power sub-stations? 3. What type of transformer is used inside cell phone chargers? Review 1. A step-down transformer has 7500 turns on its primary and 125 turns on its secondary. The voltage across the primary is 7200 V. (a) What is the voltage across the secondary? (b) The current in the secondary is 36 A. What current ﬂows in the primary? 2. The secondary of a step-down transformer has 500 turns. The primary has 15,000 turns. (a) The EMF of the primary is 3600 V. What is the EMF of the secondary? (b) The current in the primary is 3.0 A. What is the current in the secondary? 3. An ideal step-up transformer’s primary coil has 500 turns and its secondary coil has 15,000 turns. The primary EMF is 120 V. (a) Calculate the EMF of the secondary. (b) If the secondary current is 3.0 A, what is the primary current? (c) What power is drawn by the primary? 29 www.ck12.org • transformer: An electric device consisting essentially of two or more windings wound on the same core, which by electromagnetic induction transforms electric energy from one circuit to another circuit such that the frequency of the energy remains unchanged while the voltage and current usually change. • primary coil: A coil to which the input voltage is applied in a transformer. • secondary coil: The coil in a transformer where the current is induced. • step-up transformer: A step-up transformer is one that increases voltage. • step-down transformer: A step-down transformer is one that decreases voltage. References 1. Image copyright Sylvie Bouchard, 2013. http://www.shutterstock.com . Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic, using image copyright Sergii Korolko, 2013. http://www.shutterstock .com . Used under license from Shutterstock.com 3. CK-12 Foundation - Christopher Auyeung. . CC-BY-NC-SA 3.0 30
rors are also used in supermarkets, underground parking areas etc. to cover a wider field of view to be observed. Fig. 1.35: Parabolic dish aerial. (vii) Shaving mirror A concave mirror is used as a shaving or a make-up mirror. The mirror is placed at a distance less than its focal length, from the face. A virtual upright, magnified image of the face is seen in the mirror (Fig. 1.36). B face F Fig. 1.36: Shaving mirror. C (b) Convex mirrors A convex mirror is used as a rear view driving mirror. Any object in front of the mirror, i.e, lying behind the car, forms an upright diminished virtual image between F and P (Fig. 1.37). For example, another car behind is seen upright and small in size in the mirror. The driver of a car has a wider view of all the objects lying behind. 24 image C F M I P B O object (car) v u Fig. 1.37: Rear view driving mirror. A convex mirror covers a wider field of view so that the light rays from a wide angle can be observed. A plane mirror of the same size as the convex mirror helps us to see the objects lying in a limited area (smaller). 1.8 Project work Construction of a solar concetrator Suggested materials: Mild steel wires of about 3 mm in thickness, aluminium foil, superglue Assembly 1. Cut the mild steel wire so as to make three circles of diameter 1.0 m each. 2. Make the three circles using the wire. 3. Weld the wires to form a spherical framework. 4. Cover the framework using the aluminium foil. 5. Cut the framework to produce two equal ½ sphere. 25 6. Mount one of the ½ sphere on a stand such that it can be rotated in all directions. 7. Using a thicker mild steel wire, make a framework that will be able to support a small sufuria and connect it to the framework above. Make sure that the sauce pan is at ½ centre of the sphere i.e. at focal point. 8. Direct the ½ sphere and the sauce pan towards the sun. 9. Fill the sauce pan with water at 15 minutes interval. 10. Measure the temperature at 15 minutes interval. Such an arrangement is called a solar concentrator. 26 Topic summary • The reflecting surface of a plane mirror is a straight plane, concave mirror curves in, convex mirror curves outwards and parabolic mirror is a section of a paraboloid. • The centre of the reflecting surface is the pole, P, of a curved mirror. • The centre of curvature, C, of a curved mirror is the centre of the sphere of which the mirror forms a part. • The principal axis of a curved mirror is a line passing through the pole, P, and the centre of curvature, C. • The radius of curvature, r, of a curved mirror is the radius of the sphere of which the mirror forms a part. • The principal focus, F, is a fixed point on the principal axis where a set of incident rays parallel and close to the principal axis converge in a concave mirror and appear to diverge on in a convex mirror. • The principal focus is real for a concave mirror and virtual for a convex mirror. • The focal length, f, of a curved mirror is the distance from its pole to the principal focus. • Focal length of a concave mirror is half the radius of curvature of the mirror. • A parabolic concave reflector has a single point focus and the caustic curve is not formed. • Curved mirrors obey the laws of reflection just like plane mirrors. • The focal length, f, of a curved mirror is half the radius of curvature, r. • Magnification (m) is defined as the ratio of the height of the image to the height of the object. Magnification = height of image (IM) height of object (OB) = image distance (v) object distance (u) m = 1 f IM OB 1 = u = v u + 1 v • The nature, size and the position of the image formed in a concave mirror depends on the position of the object in front of the mirror. See the Table 1.1 below. 27 Table 1.1 Object Image Nature of image Size of image compared to object F At infinity (far away) Beyond C At C Between C and F Beyond C At F Between F and C At C At infinity (far away) Real and inverted Diminished Real and inverted Diminished Real and inverted Same size Real and inverted Magnified Real and inverted Magnified Between F and P Behind the mirror Virtual and upright Magnified • In a convex mirror, for all the positions of the object at a measurable distance from the mirror, an upright, diminished, virtual image is always formed between F and P. • Cinema projectors, search lights, car head lights, astronomical telescopes, solar concentrators, dish aerials are a few applications of curved reflecting surfaces. 28 Topic Test 1 1. The focal length of a curved mirror is the distance between the principal focus and the centre of curvature. the pole of the mirror and the centre of curvature. the pole of the mirror and the principal focus. the object and the image. A. B. C. D. 2. In a curved mirror, how is the focal length (f) related to the radius of curvature (r) of the mirror? A. f = r 2 B. f = 2r C. f = r D. None of the above. 3. In a concave mirror, when the object is placed between the principal focus and the pole, the image formed is A. real and diminished C. virtual and upright B. real and magnified D. virtual and inverted 4. In a concave mirror, when the object is placed at 12 cm from the pole a real image is formed at 36 cm from the pole of the mirror. The magnification produced by the mirror is A. 0.25 B. 4.00 C. 0.33 D. 3.00 5. The effect of formation of a caustic curve in a concave mirror can be minimized by A. cutting off the marginal rays. B. cutting off the axial rays. C. cutting off all the rays parallel to the principal axis. D. cutting off the rays passing through the centre of curvature of the mirror. 6. In a convex mirror, the image formed is always A. C. real and upright. real and inverted. B. virtual and upright. D. virtual and inverted. 7. In a concave mirror, when a real object is placed at 2f (Fig. 1.38), the image formed will be A. at infinity. B. between p and f . C. between P and 2f. D. at 2f. 8. Draw a sketch of a concave mirror of radius of curvature 20 cm. Mark on the diagram the pole, the centre of curvature, the principal axis, the principal focus and the focal length. 29 PF2FobjectFig. 1.38 9. Describe a parabolic mirror. What is the advantage of a parabolic mirror over a spherical concave mirror. 10. Define the following terms: pole, principal axis, principal focus, focal length, focal plane in relation to a concave mirror. 11. Copy and complete the following diagrams (Fig. 1.39 (a) to (d)) to show the path of the reflected rays. Fig. 1.39 12. A concave mirror has a focal length of 6 cm and an object 2 cm tall is placed 10 cm from the pole of the mirror. By means of an accurate ray diagram, find the position and size of the image formed. Is the image real or virtual? Explain your answer. 13. Describe an experiment to determine the magnification of an image in a concave mirror. 14. Fig. 1.40 shows the graph of real image distance v against the object distance u for a concave mirror. Fig. 1.40 Explain why the coordinate P is for a magnified image. 15. (a) Define magnification. 30 PFC(a)PFC(b)PFC(c)PFC (d)0uPv (b) An object of height 3 cm is placed at 20 cm in front of a concave mirror. The real image formed is 10 cm from the pole of the mirror. Calculate the height of the image formed. 16. A concave mirror has a focal length of 6 cm and a real object 3 cm tall is placed 15 cm from the pole. Calculate the distance of the image from the pole if the size of the image is 2 cm. 17. Explain, with the aid of a neat ray diagram how a concave mirror can be used as a ‘make-up’ mirror. (The object may be represented by an arrow-head). 18. Motorists use a convex mirror, rather than a plane mirror, as a rear-view mirror. State one advantage and one disadvantage of using a convex mirror. 19. Name the type of mirror used in the following: (a) car head lights. (b) solar concentrators. (c) underground car parking area. 20. For the following Figure 1.41 (a) and (b), copy and complete the ray diagram to locate the position of the point image I for the point object placed at O. (a) Fig. 1.41 (b) 21. The graph (Fig. 1.42) shows how the image distance v and the object distance u vary for a concave mirror. (a) When the object is placed at a distance of 60 cm from the mirror Fig. 1.42 31 PCOCOv(cm)0u(cm)10203040506070802030405060708010 (i) what is the distance of the image from the mirror? (ii) Is the image magnified, diminished or the same size as the object? (iii) what is the magnification produced by the mirror? (b) The object is now moved until it is 20 cm from the mirror. State and explain what happens to (i) the image distance. (ii) the magnification produced by the mirror. 22. A man uses a showing mirror with a focal length of 72 cm to view the images of his face. If his face is 18 cm from the mirror, determine the image distance and the magnification of his face. 23. The real image produced by a concave mirror is observed to be six times longer than the object when the object is 34.2 cm in front of a mirror. Determine the radius of curvature of this mirror. 24. An object and a concave mirror are used to produce a sharp image of the object on the screen. The table below shows the magnification and image distance of the object. Magnification m Image distance v (cm) 0.25 20 1.5 40 2.5 56 3.5 72 Table 1.2 (a) Plot a graph of m against v (b) Use the graph to find: i. The image distance when M=1.0 ii. The object distance iii. The focal length of the mirror 32 TOPIC 2 Refraction of light in thin lenses Unit Outline • Definition of a lens • Types of lenses • Terms used in thin lenses • Image formation by converging and diverging lenses • The lens formula • Sign convetion • Magnification • Power of lenses • Defects of lenses • Simple telescope Introduction In secondary 1, we learnt about refraction of light through a medium. In this topic, we are going to specifically focus refraction in thin lenses. How the thin lenses are applied in optical instruments such as microscope, glasses, cameras and others. 2.1 Definition of a lens Activi
ty 2.1 What is a lens? (Work in pairs or in groups) Materials • Water • Round bottomed flask • Plain paper • Retort stand • Sun • Magnifying lens Steps 1. What is a lens? 2. Fill a volumetric flask with clear water. 3. Cork the flask. 4. Tilt the flask such that the neck of the cork is horizontal. 5. Place a source of light (sun, bulbs) above the flask. 33 6. Place a white paper under the flask preferably on the ground. 7. Move the flask to or away the white paper. 8. What happens on the plain paper? 9. Discuss the results in your groups. 10. Replace the round bottomed flask with a magnifying lens. What happens to the paper? Explain. Sun Clamp Round bottomed flask Stand Cork Water Piece of paper Fig. 2.1 A lens is a transparent medium bounded by two spherical surface or a planed curved surface. 2.2 Types of lenses To identify and describe types and shapes of lenses Activity 2.2 (Work in groups) Materials • Charts showing converging and diverging beam through lens. • Convex lenses Steps • Plane lenses • Spherical lenses 1. Place some lenses available in your school on a labeled white plane paper. Trace their outlines. What is the shape of the lenses? 2. Feel the lenses in Fig. 2.2 below with your fingers. What do you feel? What are their shapes? Why do you think they are made in such shapes? 34 3. Name the above lenses. Fig. 2.2 4. Identify the lenses in Fig. 2.3 below. What are their shapes? Why do you think they are made in such shapes? Name them. (a) (b) (c) Fig. 2.3: Types of concave lenses There are two main groups of lenses. A type that is thick in the middle and thin at the edges, causing rays of light to converge. This is called converging or convex lenses. The other type is thin in middle and thick at the edges causing the rays of light to diverge. This lens is called diverging or concave lens. Concave lenses are of different shapes as shown in Fig. 2.4. A bi-convex or double convex lens has both its surfaces ‘curving out’. (Fig. 2.4). (a) (b) (c) Bi-convex or double convex Plano - convex Concavo - convex Concave lenses are also of different shapes (See Fig. 2.5) Fig. 2.4: Types of convex lenses (a) (b) Bi-concave plano-concave Fig. 2.5: Types of lenses 35 A bi-concave or double concave lens has both its surfaces ‘curving in’. Other concave lenses are plano-concave and convexo-concave or diverging meniscus (Fig. 2.5). When a beam of light is incident on the lenses, rays tend to converge at a point when they pass through a convex lens and diverge through a concave lens. (Fig. 2.6) (a) Incident rays Refracted rays Principal axis P F C 2F (b) C 2F Incident rays Refracted rays Lens F P Principal axis f f Fig. 2.6: Refraction of light through lenses 2.3 Terms used in thin lenses Activity 2.3 To find out the meaning of the terms used in thin lenses (Work in pairs or in groups) Materials • Magnifying lens • Double convex lenses • Sunshine Steps • Piece of paper 1. This activity is done outside the classroom during daylight. 2. Place a dry tissue paper (or dry leaves) on a flat open ground. 3. Place the lens on the tissue paper. 4. Slowly lift the lens upwards away from the paper until a spot of light is formed on the paper. Fig. 2.7: Burning a paper using a lens 36 5. Hold the lens at this position for some time. 6. Observe what happens to the paper. Explain why the paper burns. 7. Repeat the activity with a concave lens. What do you observe? What happens to the beam of light? The paper starts to burn. This shows that a convex lens brings to a focus point light energy from the sun and since light is in form of energy, a lot of it is concentrated at a point. This point where the rays are brought together after passing through the convex lens is called principal focus. This point is real. When the activity is repeated with a concave lens, nothing happens. That means the principal focus of a concave lens is virtual. The following are other common terms used in thin lenses: (a) The centre of curvature (C) The centre of curvature of the surface of a lens is the centre of the sphere of which the lens forms a part (Fig. 2.8 (a) and (b)). For each spherical lens there are two centres of curvature (C1, C2) due to the two curved surface. (b) The radius of curvature (r) The radius of curvature of the surface of a lens is the radius of the sphere of which the surface forms a part (Fig. 2.8 (a) and (b)). Each surface has its own radius of curvature (r1 or r2). (c) Principal axis The principle axis of a lens is a line passing through the two centres of curvature (c1 and c2) as shown in Fig. 2.8. r1 c1 c2 r2 r1 c1 c2 r2 (a) Convex lens (b) Concave lens Fig. 2.8: Principal axis (d) The principal focus A prism always deviates the light passing through it towards its base. A convex lens may be regarded as being made up of large portions of triangular prisms as 37 shown below. The emergent beam, therefore, becomes convergent in a convex lens (Fig. 2.9 (a)). The reverse is the effect in a concave lens (Fig. 2.9(b)). (a) (b) Fig. 2.9: Action of lenses compared with prisms (i) Principal focus of a convex lens Consider a set of incident rays parallel and close to the principal axis of a convex lens (Fig. 2.10). These rays, after refraction through the lens, pass through point F on the principal axis. Since all the rays converge at this point, it is called principal focus. Since this point can be projected on a screen, it is said to be a real principal focus. Incident rays Refracted rays Principal axis P F C 2F f Fig. 2.10: Principal focus on a convex lens (ii) Principal focus of a concave lens For a set of incident rays parallel and close to the principal axis of a concave lens, the refracted rays appear to diverge from a fixed point on the principal axis. This point is called the principal focus F, of a concave lens (Fig. 2.11). This principal focus is virtual since it cannot be projected on a screen. Incident rays Refracted rays Lens C 2F F P Principal axis f Fig. 2.11: Principal focus on a concave lens 38 (e) The focal plane When a set of parallel rays are incident on a convex lens at an angle to the principal axis, as shown in Fig. 2.12, the refracted rays converge to a point, on a line passing through F and perpendicular to the principal axis. The plane passing through F is the focal plane. Lens axis Focal plane 90º F Fig. 2.12: Focal plane of a convex lens 3 1 F F (f) The optical centre (P) F 2 P The optical centre of a lens is a point which lies exactly in the middle of the lens (PA = PB) as shown in Fig. 2.13(a) and 2.13(b). Light rays going through this point go straight through without any deviation or displacement. (a) (b) A P B P Fig. 2.13: Optical centre of a convex and concave lenses (g) The focal length of a lens (f) This is the distance from the optical centre to the principal focus of the lens (see Fig. 2.10(a) and 2.11(b). Biconvex and biconcave lenses have a focal length on each side of the lens. The concept of centres of curvature of the surfaces is required only in drawing the principal axis. Otherwise these points are referred to as 2F, as they are situated at a distance twice the focal length from the centre of the lens (PC = 2PF). 39 Exercise 2.1 1. Distinguish between converging and diverging lenses. 2. Define the following: (a) Principal axis (b) Optical centre 3. Differentiate between the principal focus of the concave and convex lens. 4. How many principal foci does a biconcave lens have? 2.4 Image formation by converging lenses Activity 2.4 (Work in groups) To find out and describe the image formed by converging lenses Materials • Convex lenses • Tree, Screen (white wall can act as screen) Steps 1. Place a convex lens between a screen and a far away object e.g. a candle. (See fig 2.14). 2. Adjust the distance between the lens and the screen until the image of the candle is observed on the screen. 2F F v F 2F u Fig. 2.14: Image formation by a far object 3. What are the characteristics of the image formed? 4. In groups, discuss the formation of the image using ray diagrams. Ray diagrams are used to illustrate how and where the image is formed. The following are the important incident rays and their corresponding refracted rays used in the construction of ray diagrams. 40 Ray 1: A ray of light parallel and close to the principal axis, passes through the principal focus F (Fig. 2.15). P F F P Ray 2: A ray of light through the principal focus F emerges parallel to the principal axis after refraction (Fig. 2.16). Fig. 2.15: Ray 1 F P P F Fig. 2.16: Ray 2 Ray 3: A ray through the optical centre, P is undeviated after refraction through the lens (Fig. 2.17). P P Fig. 2.17: Ray 3 2.5 Locating images by simple ray diagrams and describing their characteristics To locate the image of an object, we need a minimum of two incident rays from the object. From the three standard rays discussed above, any two incident rays and their corresponding refracted rays can be drawn to locate the image. If the refracted rays converge, a real image is obtained. If the refracted rays diverge, then a virtual image is obtained. 41 2.5.1 Convex lens Activity 2.5 (Work in groups) Materials To design and describe the characteristics of images formed by convex lenses when the object is at infinity • White screen • Lens • An object at infinity (landscape or a tree) • Metre rule Instructions 1. In this activity, you will design and carry out an investigation to describe the characteristics of images formed by convex lenses when the object is at infinity. 2. Modify the set-up, we used in Activity 1.6 with materials provided to you. Sketch the new set-up. 3. Write a brief procedure for your investigation. Conduct the investigations. Draw the image formed. 4. Describe its characteristics. 5. Suggest some possible sources of errors in your investigation and explain how they can be minimised. Write a report and present it in a class discussion. Note: The distance from the centre of the lens to the screen is nearly equal to the focal length, f,
of the lens. (a) Object far away from the lens (at infinity) Since the object is at infinity, all the rays from the object, incident on the lens are almost parallel. The refracted rays converge at a point on the focal plane, as shown in Fig. 2.18. Image characteristics A diminished, real, inverted image is formed at F. F P F I M Fig. 2.18: Object OB at infinity 42 (b) Object OB just beyond C (2F) Activity 2.6 (Work in groups) Materials • Screen Steps To describe images formed by convex lens when the object is beyond 2F and at 2F • Lens • Candle 1. Mark the positions of the principal focus F and 2F on both the sides of the lens with a piece of chalk. 2. Place a lit candle on the table along the principal axis of the lens, slightly away from 2F. 3. Place a white screen, on the other side of the lens, perpendicular to the principal axis of the lens and adjust its position to and fro to the screen and observe what happens. What are the characteristics of the image formed? Real image Lens Object (candle) Screen 2F F v P F u 2F Fig. 2.19: Object beyond 2F 4. Repeat step 3 by placing the candle at 2F and observe what happens. What are the characteristics of the images formed? Fig 2.20 shows the ray diagram to locate the images when the object is beyond C. B 0 2F F P F 2F I M Fig. 2.20: Object OB just beyond 2F 43 Image characteristics A diminished, real, inverted image is formed between F and 2F. (c) Object OB at 2F The ray diagram when the object (candle) was placed at 2F is as shown in Fig. 2.22 below. B 0 2F F P F 2F I M Fig. 2.21: Object OB at 2F Image characteristics A real, inverted image of the same size as the object is formed at 2F. (d) Object OB between 2F and F To design and describe the images formed by convex lens when the object is between F and 2F and at F Activity 2.7 (Work in groups) Materials • Candle • Lens • Screen Instructions 1. Modify the set-up as used in Activity 2.6 by placing the candle between 2F and F. 2. Draw the set-up. 3. Write a brief procedure for your investigation. Carry out the investigation and describe the characteristics of the images formed. 4. Suggest some possible sources of errors in your investigation and explain how they can be minimised. 5. Write a report and present it to a class. 44 The simple ray diagram when the object is between F and 2F is as shown in Fig. 2.22. B 0 2F F P F 2F I M Fig. 2.22: Object OB between 2F and F Image characteristics A real, inverted and magnified image is formed beyond 2F. (e) Object OB at F When the object was at F, the refracted rays are nearly parallel and converge at infinity as shown in Fig. 2.23 below. B O F P F parallel rays Fig. 2.23: Object OB at F Image characteristics A real, inverted, magnified image is formed far away from the lens i.e. at infinity. (cannot be described) (f) Object OB between F and P Activity 2.8 (Work in groups) Materials To describe the image formed by convex lens when the object is between F and P • Candle • Lens • Screen Steps 1. Repeat Activity 2.7 keeping the candle close to the lens, between F and P. Can you get an image on the screen? Describe its characteristics. 45 2. Is the image real or virtual? 3. Where is the image formed? 4. Explain your observations. Lens Eye F Object F Upright, virtual and enlarged image Fig. 2.24: Object between F and P The image formed is virtual and cannot be projected on the screen. An enlarged, upright image can be seen through the lens on the same side with the object (Fig. 2.24) above. A simple ray diagram to locate the image when the object is placed between F and P is as shown in Fig. 2.25 below. M B I F O P F Fig. 2.25: Object OB between F and the lens. Image characteristics A magnified, upright and virtual image is formed on the same side as the object. 2.5.2 Concave lens When the object is at infinity, an upright, diminished and virtual image is formed at principal focus F. For all other positions of the object OB, an upright, diminished, virtual image is always formed between F and P (Fig. 2.26). 46 B M I P O F Fig. 2.26: Image formation by a concave lens. Example 2.1 A convex lens has a focal length of 2 cm and a real object 6 cm tall is placed 18 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and nature of the image formed. Solution Using rays 1 and 3 of the image construction, two incident rays are drawn from B and the corresponding refracted rays through the lens. The refracted rays converge at M where the image of B is formed. Scale chosen for object and image values: 1 cm = 6 cm. BBBBBB OOOOOO FFFFFF 2F2F2F2F PPPP FFFF 2F2FF2F2FF2F2FF III uuuu vvvv MMMMMM Fig. 2.27: Graphical construction of images formed by convex lens The image of O is magnified, inverted and formed at I. IM is the real image formed at 6 cm from the lens. The height of the image is 2 cm. Since the scale is 1 cm represents 6 cm, the image is 36 cm from the lens and the height of the image is 12 cm (Fig. 2.27) above. Example 2.2 A concave lens has a focal length of 2 cm and real object 1.0 cm tall is placed at 3 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and the nature of the image formed. 47 Solution Scale chosen: 1 cm to represent 1 cm Similar to Example 2.1, draw minimum two incident rays from B and the corresponding refracted rays. Since the refracted rays diverge, a virtual image is formed. The image is 1.2 cm from the lens and the height of the image is 0.4 cm (Fig. 2.28). It is diminished, upright and virtual. Fig. 2.28 Exercise 2.2 1. Name two features of the image formed by a convex lens when: (a) The object is between F and optical centre (b) (c) The object is at infinity. The object is at F. 2. Sketch a ray diagram to show image formation for an object placed between 2 F and F of a converging lens. State four characteristics of the image. 3. (a) If a convex lens picks up rays from a very distant object, where is the image formed? (b) If the object is moved towards the lens, what happens to the position and size of the image? 4. An object 2 cm high is placed 2 cm away from a convex lens of focal length 6 cm. By using an accurate drawing on graph paper, find the position, height and type of the image. 2.6 The lens formula Activity 2.9 Lens formula (Work in groups) 1. Using reference materials or internet research about the relationship between focal length, f, object distance, u, and image distance, v. 2. What is the lens formular? Derive it. 48 BuvOFFPIM 3. How is it important to learning of convex and concave mirrors or curved reflecting surfaces? The lens formula is a formula relating the focal length, image and object distance. Consider a convex lens of focal length, f, which forms a real image IM of an object OB as shown in Fig. 2.29. B 2F O D F P u I 2F M F v f Fig. 2.29: Lens formula Triangles OBP and IMP are similar (3 angles are equal) ∴ OB IM = OP IP ……………………………………… (1) Draw a line DP perpendicular to the principal axis where DP = BO. Triangles PDF and IMF are similar (3 angles are equal) ∴ DP IM = PF IF …………………………………… (2) Since DP = OB, from equations (1) and (2), = OP IP u v = PF IF f v – f Cross multiplying, uv – uf = vf Dividing both sides by uvf uv uvf 1 – = ⇒ – = u uf uvf vf uvf 1 f 1 v . Hence 1 f 1 = + u 1 v . This is the Lens formula, where u stands for the distance of the object from the optical centre. v stands for the distance of the image from the optical centre. f stands for the focal length of the lens. 49 2.7 Sign Convention (Real is positive) We can adopt a method or a convention to describe the upward motion and downward motion. For example let the distances up be negative and down positive or vice versa. ∴ 3 m up = -3 m 3 m down = +3 m There are several sign conventions used when the distances of the object and the image are measured from the lens. In this book, we shall adopt the real is positive in which: 1. All the distances are measured from the optical centre. 2. The distances of the real objects and the real images measured from the optical centre are taken as positive, while those of virtual objects and virtual images are taken as negative. From this convention, the focal length of a convex lens is positive and that of a concave lens is negative. See Fig. 2.30 (a) and (b). P F F +f (a) P – f (b) Fig. 2.30: Real and virtual focal lengths of lenses Example 2.3 An object is placed 24 cm from the centre of a convex lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From 20 6 – 5 120 – = 1 u 1 24 = 1 120 The image distance (v) = 120 cm 50 Example 2.4 An object is placed 2 cm from the centre of a concave lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From lens formula 20 1 – 10 20 – = = –9 20 v = –20 9 = –2.2 cm, v = 2.2 cm v is negative because the image is virtual. 2.8 Magnification formula of the lens The term magnification refers to how many times an image is bigger than the object. Linear magnification (m) is defined as the ratio of the height of the image to the height of the object. To derive magnification formula Activity 2.10 (Work in groups) Material • Graph papers Steps 1. Draw three vertical lines on a graph paper. 18 cm 6 cm 2 cm A B C Fig. 2.31 2. How many times is line B bigger than A. 3. How many times is line B bigger than C. 4. What are the units of these comparisons? 51 Earlier in this unit we have done activities where we saw that the size of images formed by lenses are either bigger or smaller than the object. The increase or decrease in size of an object is called magnification. That is Linear magnification (m) = height of the image height of the object = = image distance (v) object distance (u) = h1 h0 v u Note: Since magnification is a ratio, it does not have units Sometimes it becomes difficult to measure the height of the image or the object accurately. In such cases, magnification can be calculated in terms of distances u and v. For example, consider a
convex lens where a magnified image is formed (Fig. 2.32). B 2F O F P F 2F I R u v M Fig. 2.32: Magnification Since triangles OBP and IMP are similar (3 angles are equal), the ratios of corresponding sides are equal i.e, IM IP IM OB = OB OP = IP OP = v u ∴ Hence magnification, m = IM OB = v u Magnification (m) = image distance (v) object distance (u) or m = v u 52 v Therefore m = = u also equal to the ratio of image to object distances v – u h1 h0 , therefore, the ratio of image to object sizes hi –– ho is measured from the optical centre. Example 2.5 An object of height 2 cm is placed 20 cm infront of a convex lens. A real image is formed 80 cm from the lens. Calculate the height of the image. Solution hi h0 m = = ∴ hi = hi 2 = 80 20 = 8 cm ⇒ v u 8 1 80 × 2 20 10 1 Example 2.6 An object placed 30 cm from a convex lens produces an image of magnification 1. What is the focal length of the lens? Solution Magnification, m = OB IM = OP IP = 1. (Fig. 2.33) Since m = 1; then v = u This occurs when object is at 2f. Hence 2f = 30 ∴ f = 15 cm B O P 30 cm I M u v Fig. 2.33: Image formed by convex lens 53 Example 2.7 An object of height 1.2 cm is placed 2 cm from a convex lens and real image is formed at 36 cm from the lens. Calculate (a) the focal length of the lens (b) magnification produced by the lens (c) the size of the image. Solution (a) From lens formula 36 = 1 f 1 f 18 + 1 36 = 19 36 = 1 f 36 19 = 1 f ⇒ f = 1.89 cm Focal length of the lens = 9 cm (b) m = v u = 36 2 = 3 (c) m = hi h0 ∴ hi = 3 × 1.2 = 3.6 Size of the image = 3.6 cm Example 2.8 An object of height 2 cm is placed 8 cm from a convex lens and a virtual image is formed on the same side as the object at 24 cm from the lens. Calculate (a) the focal length of the lens (b) the height of the image formed. Solution (a) From lens formula24 = 3 – 1 24 = 1 f 1 f ∴ focal length, f = 12 cm (v = –24 cm because the image is virtual) ⇒ 1 12 = 1 f 54 (b) Magnification m = v u = h1 h0 ⇒ –24 8 = h1 2 (negative sign indicate image is virtual) ∴ hi = 24 × 2 8 = 6 cm Example 2.9 A convex lens produces a real image of an object and the image is 3 times the size of the object. The distance between the object and the image is 80 cm. Calculate the focal length of the lens. Solution Magnification m = v u = 3 ∴ v = 3u ………… (1) u + v = 80 ………… (2) Solving equations (1) and (2) u + 3u = 80 ⇒ 4u = 80 ∴ u = 20 cm Hence v = 3u = 60 cm From lens formula 1 f 1 = + u 3 + 1 60 = ∴ focal length, f = 15 cm 2.9 Power of a lens 1 v = + = = 1 20 4 60 1 60 1 15 To explain what is the power of a lens Activity 2.11 (Work in groups) Material • A lens Steps 1. Discuss with your classmates what the power of the lens is. 2. Is it possible to increase the power of a lens? Discuss. 3. Share your findings with other classmates. 55 The ability to collect rays of light and focus them at a point in the case of a converging, or to diverge them so that they appear to come from a point in the case of diverging lens is called the power of a lens. It is calculated from its focal length using the formula Power = 1 f The unit for power is the dioptre represented by the symbol D. The f must be in S.I units of length. Example 2.10 A lens has a focal length of 25 cm. Find the power of the lens. Solution f = 25 cm = 0.25 m. The focal length of convex lens = +ve (It forms real image) 1 ∴ Power = NB: For a concave lens f = -ve (because a concave lens forms a virtual image) +0.25 = +4 m–1 ∴ Power = 1 -0.25 = -4 m–1 Exercise 2.3 1. Define the terms: principal axis, optical centre and focal length of a convex lens. 2. With the help of a diagram, show the action of a convex lens as a converging lens. 3. The focal length of a diverging lens is 15 cm. With the help of a diagram explain the meaning of this statement. 4. Fig. 2.34 below shows a convex lens of focal length 15 cm and two rays of light parallel to the principal axis. Copy and complete the diagram to show the path of these rays as they pass through the lens. Label the position of the principal focus as F. P axis Fig. 2.34 5. Draw ray diagrams showing how a convex lens could be used to produce (a) a real and diminished image (b) a virtual and magnified image of a real object. 56 6. Fig. 2.35 is drawn to scale. One incident ray from the object is parallel to the principal axis and other ray passes through the principal focus of a convex lens. Copy and complete the diagram to show the path of the ray through the lens. Hence determine (i) position of the image (ii) the magnification produced by the lens. Convex lens Object Fig. 2.35 7. Copy the table below and put a tick () in three of the boxes to describe the image formed by a diverging lens. Table 2.1 Magnified Diminished Upright Inverted Virtual Real 8. Draw a diagram to show how a convex lens produce a virtual image. 9. Fig. 2.36 shows two rays of light approaching a thin diverging lens. Copy and complete the diagram and show the path of the rays as they pass through and emerge out of the lens. Label the position of the principal focus F. Fig. 2.36 10. Fig. 2.37 is drawn to scale. An object OB placed in front of a convex lens of focal length 5.0 cm. Copy and complete the diagram and (a) show the position of the image (b) find the size of the image 57 Convex lens B O F F Fig. 2.37 11. A convex lens is used to form an upright, magnified image of an object placed 6 cm from the lens. Calculate the focal length of the lens, if the magnification produced is 4. 12. An object 3 cm high is placed 20 cm from a lens of focal length –25 cm. Find the position, size and the nature of the image formed. 13. At what distance must an object be placed from a convex lens of focal length 20 cm so as to get real image 4 times the size of the object? 14. An object 3 cm high is placed 30 cm from a convex lens of focal length 20 cm (a) Find the position, size and the nature of the image formed (b) If the same object is now moved by 20 cm towards the lens, calculate the magnification produced by the lens. 15. A convex lens forms a focused image on a screen when the distance between an illuminated object and the screen is 1 m. The image is 0.25 times the size of the object. Calculate (a) the object distance from the lens (b) the focal length of the lens used. 16. An object 3 cm high is placed 150 cm from a screen. Calculate the focal length of the lens that has to be placed between the object and the screen, so as to produce a real image 6 cm high on the screen. 17. An object 6 cm high is placed 30 cm from a diverging lens of focal length 15 cm. With the help of a scale diagram determine (a) the position of the image. (b) the magnification produced by the lens. 18. A real object placed 8 cm in front of a converging lens produces an image at a distance of 2 cm from the lens and on the same side as the object. Calculate the focal length of the lens. 19. A diverging lens of focal length 24 cm forms an image at 18 cm from the lens. Calculate the distance of the object from the lens. 20. In an experiment to determine the focal length of a converging lens, a student obtains the results shown in Table 2.3. 58 Table 2.2 u (cm) v (cm) 21.0 50.0 24.0 40.0 33.0 22.5 36.0 25.0 45.0 22.0 60.0 20.0 (i) Plot a graph of u (x-axis) against v (y-axis) (ii) Using the graph, determine the focal length of the lens. 2.10 Applications of thin lenses 1. The human eye The human eye consists of a nearly spherical ball of about 2.5 cm diameter except for a slight bulge at the front. Fig. 2.38 shows the cross-section of the human eye with the optic nerve leading to the brain. Retina Optic nerve Suspensory ligaments Iris Lens Pupil Cornea ciliary muscles Fig. 2.38: A cross-section of a human eye. The front portion of the eye is known as the cornea and is slightly bulged outwards and is transparent in nature. Behind the cornea, there is a diaphragm called the iris, with a hole in the middle known as the pupil. Behind the iris is a crystalline lens. This is a biconvex converging lens made of a large number of jelly-like layers which are flexible and transparent in nature. The lens is suspended inside the eye by the help of suspensory ligaments which fasten it to the ciliary muscles. These muscles control the shape of the lens. The lens forms a real, diminished and inverted image on the retina. Light falling on the retina produces a sensation in the cells which then send the electrical signals to the brain by the nerve known as the optic nerve. The amount of light reaching the retina is regulated by the size of the pupil. When a bright object is viewed, the iris reduces the size of the pupil so as to admit less light, whereas in dark light the iris contracts so as to admit as much light as possible. An image formed by the eye lens leaves an impression on the retina for about 0.1 second. This persistence of the vision enables us to “see” cinema or television pictures which appear to change smoothly from one image to the next without 59 any interruption. In a cinema theatre or television screen about 20 pictures are projected per second. During the time interval between the pictures, the eye “remembers” the previous picture. Image formation in the eye When one looks at far objects, such as a tree, the eye lens becomes thinner and the focal length of the lens increases. The ciliary muscle is relaxed and the lens has the longest focal length. It is able to focus rays from distant objects onto the retina (Fig. 2.39 (a)). To view the objects close to the eye, the lens becomes thicker and the focal length of the lens decreases. The contraction of the ciliary muscle reduces tension in the lens and the lens becomes more curved with short focal length and is more powerful. The lens now focuses images of near objects onto the retina (Fig. 2.39 (b)). For both far view and close view the image formed is real, diminished and inverted. This process by which the lens of the eye changes its focal length and produces focused images of both distant and near objects on the retina is ca
lled accommodation. (a) Far view – thin lens (b) Near view – thick lens Fig. 2.39: Accommodation of the eye In the eye, the distance between the eye lens and the retina remains the same, whereas the lens automatically changes its focal length according to the distance of the object. This effect is brought about by changing the shape of the ciliary muscle attached to the lens. Defects of vision A normal human eye can accommodate the range of distances from far off objects to objects close to the eye. There is, however, a limit to the power of accommodation of the eye. As a person grows older, the power of accommodation gradually decreases. Also, despite the ability of the eye to adjust its focal length by changing the lens shape, some eyes cannot produce clear images over the normal range of vision. This type of defect may arise due to the eyeball being slightly too long or too short compared to the normal spherical ball or due to the curvature of the cornea being defective. 60 The most common defects of vision are short-sightedness (myopia) and longsightedness (hypermetropia) (see Fig. 2.40, (a) (b) ). (a) Short-sightedness (b) Long-sightedness Fig. 2.40: Short-sightedness and long-sightedness (a) Short-sightedness or Myopia A person suffering from short-sightedness can only see nearby objects. The image of a distant object is formed in front of the retina as shown in Fig. 2.33 (a). This defect arises due to the eyeball being too long or more refraction takes place at the cornea and hence the focal length of the eye lens becomes short. In order to correct this defect, a concave lens of appropriate focal length should be used. This lens diverges the rays from a distant objects so that they appear to come from a virtual image formed at a point closer to the lens. The eye can focus on this virtual image, as shown in Fig. 2.41 (b). Far off objects at infinity convex lens Virtual image (a) Myopic eye (b) Corrective measure Fig. 2.41: Short-sightedness and corrective measure. (b) Long-sightedness or Hypermetropia A person suffering from long-sightedness can see distant objects clearly, but cannot see distinctly objects lying closer than a certain distance. The image of a nearby object is formed behind the retina as shown in Fig. 2.42 (a). The defect arises due to the eyeball being too short or due to the curvature of the cornea being defective and the focal length of the eye lens becoming longer. In order to correct this defect, a convex lens of appropriate focal length should be used. This lens converges the ray from a near object so that they appear to come from a virtual image formed at a point far off from the lens. The eye focuses on this virtual image as shown in Fig. 2.42 (b). 61 Near object Virtual image (a) Hypermetropic eye (b) Corrective measure Fig. 2.42: Long-sightedness and the corrective measure. 2. The lens camera A camera is a device used to take photographs. A human eye, though in principle, is similar to a camera, is far superior than the finest camera ever made by man. Fig. 2.43 (a) shows some parts of a lens camera. Fig. 2.43 (b) shows a commercial camera. Film Lens Diaphragm Shutter (a) (b) Fig. 2.43: A lens camera A camera consists of a converging lens and a light sensitive film or plate enclosed in a light-tight box, blackened from inside. The lens focuses light from an object to form a real, diminished and inverted image on the film. Focusing of objects is done, by adjusting the distance between the lens and the films. The amount of light entering the camera through the lens is regulated with the help of a diaphragm, with an adjustable opening in the middle. Light is admitted by the shutter, which opens for different required intervals of time and then closes automatically. During this interval of time, the film is exposed to light from the object. The film contains light sensitive chemicals that change on exposure to light. The film is developed to get what is called a negative. From the negative a photograph (positive) may be printed. Comparison of a lens camera with the human eye Similarities 1. Both use converging lenses 2. Both produce real, inverted, diminished images. 62 3. Both can control the amount of light entering the device. 4. Both are black inside. Differences Camera Eye 1. Focal length of the lens is constant. 1. Focal length of the lens changes with the thickness of the lens. 2. Distance between the lens and the 2. Distance between the lens and the retina film can be altered. is a constant. 3. Focuses objects between a few centimetres from the lens to infinity. 3. Focuses objects between 25cm from the lens to infinity. 4. Form permanent images at the film. 4. Form temporary images at the retina. Example 2.11 A lens camera is used to take photograph of a distant building. A well focused image is formed on the film. The lens of the camera is 6 cm from the film. (a) What is the focal length of the lens? Give reasons for your answer. (b) If the camera is then used to take a photograph of a person 2.0 m away from the lens, without moving the camera, in which direction should the lens be moved in order to produce the best possible image? Solution (a) Focal length of the lens = 6 cm. Since the object is a distant building,the light rays incident on the lens are almost parallel and are brought to focus at the principal focus of the lens. (b) As the object distance is only 2.0 m i.e. object distance, u, has decreased as compared to the distance in part (a). Hence the image distance, v, must be increased . To achieve this, the lens has to be moved forward towards the person. 3. Simple microscope A magnifying glass also known as a simple microscope is an instrument used to view the details of very small objects. It consists of a single converging lens of short focal length. When an object is placed within the focal length of such a lens, a magnified image which is virtual and upright is formed on the same side of the object. This image can be viewed by placing the eye close to the lens. The distance of the object from the lens is adjusted till an enlarged image is formed at a distance D, which is about 25 cm from the unaided eye. The distance D is referred to as 63 the least distance of distinct vision. Fig. 2.44 below illustrates the operations of a simple microscope. M I B O F D Fig. 2.44: A simple microscope. F Eye The action of a simple microscope The object OB, when viewed by an unaided eye, cannot be brought closer to the eye, than the distance D (Fig. 2.45 (a)). Otherwise the image as seen by the eye will not be clearly visible. When the same object is viewed through the magnifying glass, it moves nearer to the eye so that a magnified image is formed at the same distance D as before (Fig. 2.45 (b)). Therefore a simple microscope enables us to bring an object very close to the eye making it appear magnified and yet clearly visible. B O M I D v (a) B O (b) Eye F u Eye Fig. 2.45: Working of a simple microscope 64 Magnifying power of a simple microscope Linear magnification m = IM OB = v u Linear magnification of a lens is also called magnifying power of the instrument. In a simple microscope, the image distance v is negative, as the image is virtual. 1 f Hence, from the lens formula = – 1 v 1 u Multiplying throughout by v and simplifying From the above expression the shorter the focal length of the lens, the greater is the magnifying power of the instrument. Hence a simple microscope uses a converging lens of short focal length. Example 2.12 Calculate the magnification produced by a lens of focal length 5.0 cm used in a simple microscope, the least distance of distinct vision being 25 cm. Solution In this example, the image distance v = D = 25 cm. Magnification, m = 1 + v f = 1 + ( 25 5 ) = 1 + 5 = 6 Hence the magnification produced by the lens = 6. 4. Compound microscope In a simple microscope, the magnifying power cannot be increased beyond a certain limit, by decreasing the focal length of the lens. This is due to the mechanical difficulties of using a lens of very short focal length. A compound microscope uses two separate converging lenses, placed coaxially within two sliding tubes, to obtain a higher magnifying power. The lens O, nearer the object is called the objective lens and the lens E closer to the eye, is called the eyepiece lens. Though both these lenses are of short focal lengths, the eyepiece has a comparatively larger focal length than the objective lens. The final image formed is magnified, virtual and inverted as shown in Fig. 2.46. 65 Objective (O) Eye Eyepiece (E) 2F B O F I F 2F B' M Fig. 2.46: A compound microscope Action of a compound microscope The object OB is placed between F and 2F of the objective lens. A real, inverted, magnified image O′ B′ is formed beyond 2F of the objective lens. The position of the eyepiece lens is adjusted so that this image O′B′ falls within its focal length. The eyepiece then acts as a magnifying glass and produces a final magnified, virtual and inverted image IM at a distance of distinct vision D from the eye, placed very close to the eyepiece. If m1 is the magnification produced by the objective lens and m2 is the magnification produced by the eyepiece lens, then the magnification produced by the system of lenses m is given by, m = m1 × m2 If the first image O′B′ formed by the objective lens is exactly at the principal focus Fe of the eyepiece lens, then the final image IM will be formed at infinity. The image will be inverted and well enlarged. At this position, the compound microscope is said to be in normal adjustment. A good compound microscope produces a very high magnification. High magnification microscopes are usually used in research work in science (see Fig. 2.47). 66 Fig. 2.47: High magnification microscope for research work Example 2.13 In a compound microscope, the focal length of the objective lens is 2.0 cm and that of the eyepiece is 2.2 cm and they are placed at a distance of 8.0 cm. A real object of size 1.
0 mm is placed 3.0 cm from the objective lens. (a) Use the lens formula in turn for each lens to find the position of the final image formed. (b) Calculate (i) the magnification produced by the arrangement of these lenses and (ii) the size of the final image viewed by the eye? Solution (a) For the objective lens Solving this equation gives v = 6 cm As shown in Fig. 2.48, the real image I1M1 is formed at 6 cm from the objective lens. I1M1 acts as an object for the eyepiece (u = 2 cm). 1 For the eyepiece lens f 1 2.2 1 u 1 2 Solving this equation gives v = –22 cm 1 v 1 v = + = – 67 The negative sign shows that the image formed by the eyepiece is virtual and is formed on the same side as the object I1M1. The final image I2M2 is at a distance of 22 cm from the eyepiece (see Fig. 2.48). O 3 cm 6 cm 2 cm E Eye I2 M2 B O 22 cm I1 M1 8 cm Fig. 2.48: Arrangement of lenses in a compound microscope (b) (i) The magnification produced by the system of lenses m = m1 × m2 for the eyepiece. for the objective lens and m2 = where m1 = v u v u m = 6 3 × 22 2 = 2 × 11 = 22 (ii) The size of the final image = size of the object × m = 1 × 22 = 22 mm Topic summary • A lens is a transparent medium bound between two surfaces of definite geometrical shape. • Thin lenses may either be converging or diverging. • A convex lens is thicker at its centre than its edges and converges the light incident on it. • A concave lens is thicker at its edges than at the centre and diverges the light incident on it. • The following are some of the important terms used in spherical lenses: principal axis, optical centre, principal focus, focal length. • The focal length of a convex lens is positive, while that of a concave lens is negative. • The characteristics of the image formed by a converging lens depends on the position of the object (see Table 2.3). 68 Position of object At infinity (far away) Beyond 2F At 2F Between 2F and F At F Between F and P Table 2.3 Position of image F Nature of image formed real and inverted diminished Size of image formed compared to object Between F and 2F At 2F Beyond 2F At infinity (far away) Same side as object real and inverted diminished real and inverted Same size real and inverted Magnified real and inverted Magnified Virtual and upright Magnified • A diverging lens always forms a virtual, upright, diminished image between F and P (except when the object is at infinity). • Magnification (m) is defined as the ratio of the height of the image to the height of the object magnification = height of image height of object = • Lens formula is given by 1 u + 1 v = image distance object distance 1 f where u is the object distance, v the image distance and f the focal length of the lens. • Short-signtedness and long-sightedness are two most common defects of a human eye • A lens camera, simple microscope, compound microscope are some examples of optical instruments. 69 Topic Test 2 1. Describe an experiment to illustrate that white light is composite in nature. 2. Fig. 2.49, drawn to scale, shows two rays starting from the top of an object OB incident on a converging lens of focal length 2 cm. B O Fig. 2.49: Equilateral glass prism (a) Copy and complete the diagram to determine where the image is formed. (b) Add one more incident ray from B through the principal focus and draw the corresponding refracted ray through the lens. (c) Calculate the magnification produced by the lens. 3. Fig. 2.50 shows an object placed at right angles to the principal axis of a thin converging lens. Fig. 2.50: Equilateral glass prism (a) Calculate the position of the image formed. (b) Give an application of this arrangement of a lens. (c) Describe the nature of the image formed. 4. Describe with the aid of a ray diagram, how an image is formed in a (i) simple microscope (ii) lens camera. 5. A converging lens is used to form an upright image, magnified 5 times of an object placed 6 cm from the lens. Determine the focal length of the lens 70 FO6 cm8 cmLensBF 6. Fig. 2.51 shows two converging lenses L1 and L2 placed 8 cm from each other. The focal length of the lens L1 is 2 cm and that of L2 is 2.8 cm. An object 1.0 cm high is placed 3 cm from lens L1. Fig. 2.51: Equilateral glass prism (a) Construct a ray diagram to scale, on a graph paper to show the position of the final image as seen by the eye of a person. (b) Determine the magnification obtained by this arrangement. 71 ObjectEye3 cm8 cmL1L2 72 UNIT 2 Forces and Turning Effects Topics in the unit Topic 3: Moment of a Force Topic 4: Centre of Gravity and Equilibrium Learning outcomes Knowledge and Understanding • The effects of forces and centre of gravity. Skills • Design tests to locate the centre of gravity of regular objects by method of balancing and locate the centre of gravity of irregular shaped objects by means of a plumb-line. • Observe carefully. • Predict what might happen. • Use appropriate measures. • Draw a simple diagram to show moment of a force Interpret results accurately. • Calculate problems related to moments of forces. • Report findings appropriately. Attitudes • Appreciate the applications of moment of forces. Key inquiry questions • Why the pivot is important in taking moment of force? • What do you understand by couple forces? • Why cars are designed to have a wide base? • Why an object cannot be in equilibrium if it is in motion? • Why an overloaded vehicle is prone to overturn? 73 TOPIC 3 Moment of a Force Unit Outline • Moment of force • Principle of moments • Couple • Determination of centre of mass of a regular object • Applications of moment of a force Introduction One of the effects of a force we learnt about in Secondary 1 is that it produces a turning effect on body. But how can we quantify the turning effect? What are some of the applications of this effect in our daily lives? In this topic, we will seek answers to these questions. 3.1 Moment of a force A number of simple machines like levers, pliers, spanners and so on do work when a force produces a turning effect on some of their parts. It is important to know where the force should be applied for the machine to be more efficient in doing work. The following activity will help us to investigate the turning effect of a force. To investigate the turning effect of a force Activity 3.1 (Work in groups) Materials • A ruler. Steps 1. Balance a ruler on a finger. At what point did the ruler balance? Why do you think it balances at that point? 2. Press the ruler at one end. Observe what happens to the ruler. 3. Repeat the experiment by pressing the ruler on the other end. Why do you think the ruler behaves in such a manner? 74 Moment of a Force Fig. 3.1: Balancing a ruler on a finger In both cases, the ruler turns about the finger. When the force F1 is applied at one end, the ruler turns in anticlockwise direction about the finger. When the force F2 is applied at the other end, the ruler turns in the clockwise direction. Both F1 and F2 produce a turning effect on the ruler about the finger. The turning effect of a force about a point is called the moment of the force about that point. This moment depends on the force applied and its distance from the point. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. The moment of a force about a point is either clockwise or anticlockwise about the point. In Fig. 3.1, the anticlockwise moment about the finger is F1 × d1. The clockwise moment about the finger is F2 × d2. Moment of a force about a point = Force × Perpendicular distance from the point to the line of action of the force = F × d SI unit of moment of a force Moment of a force = Force (N) × Perpendicular distance (d). Therefore, the SI unit of moment is newton metre (Nm). Moment of a force is a vector quantity since it has both magnitude and direction. Example 3.1 A student applies a force of 10 N to the handle of a door, which is 0.8 m from the hinges of the door (Fig. 3.2). Calculate the moment of the force. 75 Solution Moment of a force about a point hinges = Force × perpendicular distance from the point to the force. = (10 × 0.8) Nm = 8 Nm in the clockwise direction. door 0.8 m F Fig. 3.2: Moment in opening the door Example 3.2 Calculate the moment of the force about the fulcrum when a pet dog of mass 10 kg is at a distance of 1.2 m from the fulcrum of the seesaw as shown in Fig. 3.3. mass = 10 kg fulcrum Solution F = Weight of dog = mg 1.2 m Fig. 3.3: Moment of force = 10 kg × 10 N/kg = 100 N Moment of the force about the fulcrum = Force × perpendicular distance from the fulcrum = 100 N × 1.2 m = 120 Nm in the clockwise direction. Exercise 3.1 1. Define ‘moment of a force’ and state its SI unit. 2. A force of 20 N is applied to open the gate of a fence as shown in Fig. 3.4. Calculate the moment of the force about the hinges if the force is applied at the edge of the gate. 76 hinges 1.2 m 0.3 m F 3. Give the scientific reasons for the following: Fig. 3.4: Moment in opening the gate (a) The handle of a door is fixed far from the hinges. (b) A pair of garden shears has small blades and long handles. (c) A lighter boy is able to produce same moment as that of a heavier girl on the seesaw as shown in Fig. 3.5 below. girl boy Fig. 3.5: Moment on a seesaw 4. A person applies a force of 500 N and produces a moment of force of 300 Nm about the wheels of a wheel cart (Fig. 3.6). Calculate the perpendicular distance, d, from the line of action of the force to the wheels. l o a d F = 500N weight d Fig. 3.6: Moments in a wheelcart 77 3.2 The principle of moments The principle of moment gives the relationship between two moments that are at the same turning point (fulcrum). Activity 3.2 (Work in groups) Materials To investigate the principle of moments • A metre rule • Three 100 g mass • string • support e.g. clamp Steps 1. Suspend a uniform metre rule from a firm support e.g. clamp, at the 50 cm mark, i.e. at its mid point G as
shown in Fig. 3.7 (a) using a string. 2. Suspend a 100 g mass at a point A as shown in Fig. 3.7 (b). Why do you think the ruler balances as shown? (a) O (b) 50 cm 100 cm 0 P P A 100 cm 100 g mass F = 1 N Fig. 3.7: Principle of moments. 3. Now suspend a 200 g mass at a point B near the 0 cm mark (Fig. 3.8 (a)). What has happened? The system turns in the anticlockwise direction. Now adjust the position of B till the system balances horizontally as shown in Fig. 3.8 (b). Explain the observations. (a) B O 200 g mass f = 2N P A 100 cm (b) 0 B P A F = 1N 2N F = 1N Fig. 3.8: Principle of moments. The metre rule turns in the clockwise direction. There is a moment of force in the clockwise direction due to the force acting vertically downwards at point A. Moment of the force in the clockwise direction = Force × perpendicular distance = 1.0 × PΑ 78 Moment due to the 2 N force about P is 2.0 × PB in the anticlockwise direction. Measure the distance PA and PB. Compare the values of 1.0 × PA and 2.0 × PB. What can you say about these values? We note that the two moments are equal in magnitude and opposite in direction. The clockwise moment of the 1 N force about point P is equal to the anticlockwise moment of the 2 N force about point P. Activity 3.3 (Work in groups) Materials To design and investigate the principle of moments with more than two forces • Four mass • A metre rule • string • Firm support e.g. clamp Instructions 1. In this activity, you will design and carry out an investigation to investigate the principle of moments with more than two forces. 2. By modifying the set-up, we used in Activity 3.2, with the materials provided, i.e using its masses, conduct an investigation. Sketch the new set-up. 3. Write a brief procedure and carefully execute the procedure to determine principle of moments with four masses. 4. By applying the relevant formula and relationships. Calculate the moments 0 about point P. 5. Compare your values with other groups. 6. What are some possible sources of errors? How can they be minimised in your investigation? Write a report and present it in a class discussion. 0 D F2 P 50 cm C F1 B A 100 cm F4 F3 Fig. 3.9: Balanced metre rule under action of forces. The sum of the clockwise moments = F3 × PA + F4 × PB. The sum of the anticlockwise moments = F1 × PC + F2 × PD. What can you say about F1 × PC + F2 × PD and F3 × PA + F4 × PB? 79 A B 100 cm C F1 P 50 cm D F2 Fig. 3.9: Balanced metre rule under action of forces. F3 F4 From Activity 3.2 and 3.3 we can conclude that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the metre rule is balanced. In Activity 3.3, we saw how a body can be balanced by a number of forces. When a body is balanced under the action of a number of forces, it is said to be in equilibrium. The results of Activities 3.2 and 3.3 are summarised in what is known as the principle of moments. It states that, when a body is in equilibrium under the action of forces, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. Activity 3.4 (Work in groups) Materials • A metre rule Steps To verify the principle of moments • Seven, 50 g masses 1. Consider a uniform metre rule suspended at its mid point P, which is the 50 cm mark. Suspend 3 masses; 200 g, 100 g and 50 g and adjust their positions A, B and C till the system is in equilibrium as shown in Fig. 3.10. 2. Calculate the distances PA, PB and PC in metres and enter the values in a table. 0 A F1 2N P B C 100 cm 50 cm F2 1N F3 0.5 N Fig. 3.9: Verifying the principle of moments. 3. Repeat the experiment by changing the positions of A or B or C or all the three so that the metre rule balances horizontally in each case. Record the results in Table 3.1. Table 3.1 F1 × PA (Nm) F2 × PB + F3 × PC (Nm) PA (m) PB (m) PC (m) 1 2 3 4 4. Complete the table and explain the results. 80 It is seen that the last two columns are equal for each set of results proving that the sum of the clockwise moments is equal to the sum of the anticlockwise moments about point P. F2 × PB + F3 × PC = F1 × PA Example 3.3 A uniform metre rule is pivoted at its centre P, and 3 masses are placed at A, B and C as shown in Fig. 3.10. Find the value for the weight W of the mass M placed at C so that the metre rule is balanced horizontally. 0 cm 50 cm A 2N 10 cm P B 1N 30 cm 40 cm Fig. 3.10: Principle of moment Mass M 100 cm C W Solution Taking moments about P, when the metre rule is in equilibrium. Sum of the clockwise moments = Sum of the anticlockwise moments W × 0.4 = (2.0 × 0.3) + (1.0 × 0.1) (2.0 × 0.3) + (1.0 × 0.1) 0.4 W = = 1.75 N Example 3.4 John, Joyce and Janet sat on a seesaw as shown in Fig. 3.11 below. Where is John, whose mass is 30 kg seated so that the seesaw is balanced horizontally if the masses of Joyce and Janet are 50 kg and 20 kg respectively? Joyce Janet P John 1m pivot 2 m d Fig. 3.11: See saw at balance Solution John’s weight = 600 N, Joyce’s weight = 500 N, Janet’s weight = 200 N Taking moments about the pivot, Sum of clockwise moments = Sum of the anticlockwise moments about the pivot about the pivot 81 600 × d = 500 × 2 + 200 × 1 600 d = 1 000 + 200 d = 1 200 600 = 2 m John should sit at a distance of 2 m from the pivot. Example 3.5 The uniform plank of wood in Fig. 3.12 is balanced at its center by the forces shown. Determine the value of W in kg. 2.6 N 24 cm 6 cm 2 N W Fig. 3.12 Solution Note that the 2.3 N produces an anticlockwise moment. Sum of clockwise moments = Sum of anticlockwise moments 0.24 × (2 + W) = 2.6 × 0.30 0.48 + 0.24W = 0.78 ⇒ W = 0.78 – 0.48 = 1.25 N 0.24 = 0.125 kg To determine the mass of an object using the principle of moments Activity 3.5 (Work in groups) Materials • A metre rule • A known mass Steps • An unknown mass • Support 1. Suspend a uniform metre rule at its mid point P. Suspend the object of mass m, using a string, from a point A. Suspend a known mass M on the other side of the metre rule and adjust the position of the mass M till the metre rule is horizontal as shown in Fig. 3.13. 82 0 cm A object of mass, m P 50 cm B 100 cm mass M mg Mg Fig. 3.13: Finding unknown mass m. 2. Record the distances PA and PB. Repeat the experiment by changing the position of the object or the mass M. Enter the readings of M, PA and PB in a tabular form as shown in Table 3.2. Table 3.2 Mass M(g) PA(cm) PB(cm) M.PB m = (g) PA 1 2 3 3. Calculate the mean value for the mass of the object from the last column. Mean Taking moments about P, (mg) × PA = (Mg) × PB, m × PA = M × PB, (g cancels out) M × PB PA ∴ m = Exercise 3.2 1. State the principle of moments. 2. A boy of mass 20 kg sits on one side of a log of wood and 10 m away from the pivot. A girl of mass 30 kg sits on the opposite side of the log. How far is the girl from the pivot. 3. State two conditions for a system to be in a state of equilibrium. 4. Weights of 25 N, 28 N and 8 N were suspended on a uniform plank of wood pivoted at its center on a knife-edge. Fig. 3.14 shows the plank immediately after it was placed on a knife-edge. 83 Fig. 3.14 (a) Work out: (i) Sum of Clockwise moments (ii) Sum of Anticlockwise moments (b) Is the bar in a state of equilibrium? Give a reason for your answer. 5. A uniform metre rule is pivoted at the center. It is balanced by weights of 8 N, F and 24 N suspended at 34 cm, 43 cm and 30 cm marks respectively (Fig. 3.15). Calculate the value of F. 50 cm 60 cm 46 cm 34 cm 0 cm 100 cm 8 N F 24 N Fig. 3.15 6. Fig. 3.16 below shows a modern bar balanced by forces of 210 N, 100 N and 25 N. Calculate the distance d. 210 N 6 m 5 m d m 25 N 100 N Fig. 3.16 3.3 Couple Couple refers to two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. Its effects is that it creates rotation without translation as shown in Fig. 3.17. 84 Fig. 3.17: A couple A couple produces a turning effect on a body. Moment of a couple Fig. 3.18 shows a couple acting on bar AB. F A d C B F Fig. 3.18: Moment of a couple Since the pivot is at C, the moment of the force F acting at point A = F × AC, in the clockwise direction. Similarly, the moment of the force, F acting at point B = F × BC, also in the clockwise direction. ∴ The moment of the couple = (F × AC) + (F × BC) = F (AC + BC), but AB = BC + BC = F × AB = F × Arm of the couple. The moment of the couple, called the torque which is defined as the total rotating effect of a couple and is given by the product one of the forces and the perpendicular distance between the forces. Hence, in Fig. 3.18, Torque = F × perpendicular distance AB. SI unit of torque is the newton-metre (Nm) 85 Example 3.6 In Fig. 3.19, each force is 4 N and the arm of the couple is 20 cm. Calculate the moment of the couple. F = 4 N 20 cm Fig. 3.19: Moment of a couple F = 4 N Solution The moment of the couple = F × perpendicular distance = 4 N × 0.20 m = 0.80 Nm. Some common real life examples of a couple are observed when: • Forces are applied by hands to turn a steering wheel of a motor car (Fig. 3.20 (a)) or the handle bars of a bicycle. • A water tap is opened or closed (Fig. 3.20 (b)). • A corkscrew is twisted into a cork in the mouth of a bottle. (Fig. 3.20 (c)). Fig. 3.20: Moment of couple 86 Exercise 3.3 1. Fig. 3.21 is a water tap in use. If the diameter of a circular path made by the tap (knob) when is open and closed is 20 cm. Calculate the moment of the couple. 2. Calculate the torque in Fig. 3.22 below. Fig. 3.21: Moment of couple of a tap 14 cm 14 cm 8N Fig. 3.22: Torque 3. A steering wheel of a truck has a diameter of 30 cm. If the driver is holding the wheel with both hands, while negotiating a corner, calculate the force applied by the right hand if the left hand is pulling the wheel by a force equal to 200 N. 3.4 Centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This forc
e of gravity is always directed towards the earth’s centre and is called the weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following experiment. 3.4.1 Determining centre of mass of regular objects Activity 3.6 (Work in groups) Materials To determine the weight of a beam (uniform metre rule) • Uniform metre rule • A mass, m • A knife edge (fulcrum) 87 Steps 1. Balance a uniform metre rule of mass m on a fulcrum and adjust its position until the metre rule is horizontal. Note the position P, where it is pivoted (Fig. 3.23(a)). 2. Move the fulcrum to a point A, say to the right of P. Observe what happens. Why do you think the ruler changes its state of equilibrium. 3. Place a mass M between A and 100 cm mark and adjust its position B until the metre rule is horizontal (Fig. 3.23(b)). (a) (b) Fig. 3.23: Determination of mass of a uniform metre rule 4. Find the lengths PA and BA. Repeat the experiment by changing the position of A or the mass M. 5. Record the mass of M, length PA and length BA in a table (Table 3.3). Table 3.3 Mass M(g) BA(cm) PA(cm) m = (g) M.BA PA 1 2 3 6. Take moments and determine the value of m (the mass of the metre rule). Mean m = Taking moments about A mg × PA = Mg × BA m = M × BA PA (g cancels out) Calculate the mean value for the mass of the metre rule from the last column of Table. 3.4. The weight of the metre rule = mg. 88 Example 3.7 A uniform metre rule pivoted at the 30 cm mark is kept horizontal by placing a 50 g mass on the 80 cm mark. Calculate the mass of the metre rule (Fig. 3.24). Fig. 3.24: Determining the mass of the metre rule Solution Let the mass of the metre rule be m Force due to m = m × g where g = 10 N/kg = 10 m newtons Force due to 50 g mass = PA = 10 cm = 0.1 m AB = 20 cm = 0.2 m 50 1 000 × g = 0.05 × g = 0.50 N. By the principle of moments, taking moments about point A, 10 m × 0.1 = 0.50 × 0.2 m = 0.50 × 0.2 10 × 0.1 = 0.10 1 = 0.1 kg or 100 g Example 3.8 A coffee table of mass 22 kg and length 1.3 m long is to be lifted off the floor on one of its shorter sides to slip a carpet underneath. Calculate the maximum force needed to lift the table. 89 Solution Fig. 3.25 shows the forces acting on the table where F is the lifting force and 220 N is the weight of the table acting at the center. F A 0.8 m 0.8 m 220 N Fig. 3.25 Taking the movement about point A Sum of Clockwise moments = Sum of Anticlockwise moments 220 N × 0.8 m = F × 1.6 m ⇒ F = 220 × 0.8 1.6 = 110 N ∴ The minimum force required = 110 N Exercise 3.4 1. A uniform metre rule of uniform width 2.5 cm and thickness 0.5 cm is suspended at the 78 cm mark and kept balanced by hanging a mass of 150 g at the 100 cm mark (Fig. 3.26). Calculate, (a) the mass of the metre rule, (b) the density of the material of the metre rule, (c) the tension T in the string. Fig. 3.26: A system at balance 2. A uniform metre rule is balanced at the 20 cm mark by a mass of 240 g placed at one end. (a) Draw a diagram to show the state of balance of the metre rule (b) Determine the weight and mass of the metre rule. 3. A non-uniform plank AB shown in Fig. 3.27 is balanced when a force of 200 N is applied at the end B. The centre of gravity, G, is shown. 90 Fig. 3.27: Centre of mass of a plank Calculate the: (a) weight and (b) mass of the plank. 3.5 Applications of moment of a force The following are some of the common examples which illustrate the turning effect of a force i.e moment of a force: 1. Opening or closing a door. 2. Opening a bottle using a bottle opener (Fig. 3.28(a)). 3. A pair of scissors or garden shears in use (Fig. 3.28(b)). 4. Children playing on a “see-saw”. 5. A wheelbarrow being used to carry some load (Fig. 3.28(c)). 6. A wheel cart being used to lift heavy loads. 7. A screwdriver being used to tighten/loosen a screw. 8. A crowbar being used to move large object (Fig. 3.28(d)). Load Pivot P Effort Effort P Load Pivot (a) Bottle opener (b) Scissors 91 Load Effort Pivot Load Pivot Effort (c) Wheelbarrow (d) Crowbar Fig. 3.28: Applications of moment of a force Note In all these tools the length of the handle determines the amount of effort to be used. Activity 3.1 will help us to understand this. Activity 3.7 (Work in groups) Materials To investigate the effect of length to the effort needed in using a tool Exercise books, 3 cm and 15 cm rulers, doors. Steps 1. Lift a book using a 15 cm ruler. Repeat the same using a 30 cm ruler. Why is it easier to lift a book using a longer ruler than the shorter one? 2. Open the classroom door with your hand near the door hinge. Now, open the same door with your hand far away from the door hinge. Of the activities above, when do you use less effort? We use less effort when using 30 cm ruler to lift a book. Several experiments have been done on several tools and machines and proved that: (a) The longer the handle, the lesser the effort used when using the machines. (b) The shorter the handle, the more the effort used while working with the tools and machines. Therefore, the manufacturers always design tools and machines such as bottle openers, see-saw, water taps, spanners and wheelbarrows with longer handles so that very little effort be used in working with them. 92 Exercise 3.5 Sketch and locate the effort, pivot, load in each of the following tools and machines. (a) (d) (g) (b) Water tap (e) (h) (c) See-saw (f) Hammer (i) Bottle opener Spanner Spade pliers Broom (j) Arm Spoon Topic summary • The moment of a force about a point is the turning effect of the force about the point. • The turning effect of the force depends upon the magnitude of the force applied and the perpendicular distance from the pivot. • The moment of a force is the product of the force applied and the perpendicular distance from the pivot to the line of action of the force. • The moment of a force is a vector quantity and its SI unit is Newton metre. • Principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the system is in equilibrium. Topic Test 3 1. (a) State the principle of moments. (b) Describe an experiment to prove this principle using two known masses and a uniform metre rule. 2. A uniform metre rule is suspended at the 50 cm mark and a stone at the 0 cm mark. The metre rule is balanced horizontally when a mass of 100 g is suspended at the 30 cm mark. Calculate the weight of the stone. 3. In the diagram shown (Fig. 3.29), calculate the value of the unknown mass M, when the plank is balanced horizontally. 0 M 1 m 10 kg 5 m 4 m 10 m 20 kg 8 m Fig. 3.29: A plank at balance 4. Jane and James are seated at 3 m and 2 m respectively from the centre of a seesaw on one side and Jack at 4 m from the centre on the other side. The seesaw is balanced horizontally. Find the weight of Jack, if the masses of Jane and James are 40 kg and 30 kg respectively. 93 5. Fig. 3.30 shows a uniform plank pivoted at its centre. Where should a 5 kg mass be attached if the plank is to be perfectly horizontal? 4 m 5 m 2 kg 1 kg Fig. 3.30: A uniform plank at balcon 6. A uniform metre rule is balanced horizontally at its centre. When a mass of 5 g is suspended at the 4 cm mark, the rule balances horizontally if a mass M is suspended at 30 cm mark. Calculate M. 7. Fig. 3.32 shows a uniform metallic metre rule balanced when pivoted at the 30 cm mark under the conditions shown on the diagram. Fig. 3.31: Moments in a uniform rule (a) Redraw the diagram showing all the forces acting on the metre. (b) Calculate the weight W of the metre rule. 8. Name four importance of moments in our daily life. 9. Fig. 3.32 represents a type of safety valve that can be fitted to the boiler of a model steam engine. 60 cm 20 cm Pivot F Steam Water Heat Fig. 3.32 (a) Briefly explain how the system works.. (b) What minimum pressure should the steam have in order to escape through the plate which has a cross sectional area 10.3 cm2? (Take atmospheric pressure = 105 Pa) 94 TOPIC 4 Centre of Gravity and Equilibrium Unit outline • Definition of centre of gravity and centre of mass. • Determination of centre of gravity of regular and irregular objects. • Effects of the position of centre of gravity on stability of objects. Introduction In our day to day experiences, we may have come across statements such as ‘‘that object is not stable on the table” or “that overloaded bus is not stable on the road”. Have you ever asked yourself what factors control the stability of an object? In this topic, we will study the factors that affect stability of objects and centre of gravity. Activity 4.1 To locate the centre of gravity of a book (Work individually or in groups) Materials: Exercise book Steps 1. Take your exercise book and try to balance it horizontally on your finger as shown in Fig 4.1 below. Fig. 4.1: Balancing a book (a) What do you observe? (b) Why do you think the book balances at only one point? (c) What do you think is special about the point where the book balances? 2. Discuss with your group members your observations and thoughts in 1(a),(b) and (c). 95 By going through the following discussions, you will be able to answer questions 1(a) to (c) in Activity 4.1. 4.1 Centre of gravity and centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This force of gravity is always directed towards the earth’s centre and is called the weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following activity. 4.1.1 To investigate where the weight of a body acts Activity 4.2 To find the centre of gravity of a regular body (Work in pairs or in groups) Materials: A table, thin rectangular card Steps 1. Place a thin rectangular cardboard near the edge of the bench top. 2. Pull the card slowly away from the bench until it is just about to topple over the
n released as shown in Fig. 4.2 (a). 3. Using a ruler, mark and draw the line AB along which the card balances. 4. Repeat the activity with the other side of the card, mark and draw the line CD along which the card balances. The lines AB and CD intersect at a point M (Fig. 4.2(b)). table thin rectangular card pull B M D C (a) A (b) Fig. 4.2: Location of a point where the weight of the body acts 4. Now, try to balance the card with the point M placed at the tip of your fore finger. What do you notice about the state of equilibrium of the card? Suggest a reason for this observation. From Activity 4.2 you should have observed that the cardboard balances horizontally at point M only. This shows that although the mass of the cardboard is distributed over the whole body, there is a particular point, M, where the 96 whole weight of the cardboard appears to be concentrated. When pivoted at this point the cardboard balances horizontally. This point, M, is called centre of gravity of the cardboard. The centre of gravity of a body is the point from which the whole weight of the body appears to act. The centre of gravity of an object is constant i.e. at the one location when a body in a place with uniform gravitational field strength. However, the centre of gravity of a body moves to a different location when the body is placed in a region with non-uniform gravitational field strength. Centre of mass of an object on the other hand is the point where all the mass of the object is concentrated. Since the mass of an object is constant and is not affected by pull of gravity, the location of the centre of mass of an object is constant i.e. does not change. In places like on earth where the gravitational field strength is uniform, the centre of mass and the centre of gravity coincide i.e. are at the same point. However, the two centres are at different locations for the same object if the object is placed in a place with non-uniform gravitational field strength. 4.2 Centre of gravity (c.g) of regular lamina Activity 4.3 To locate the centre of gravity of a regular lamina (Work in groups) Materials: manila paper, ruler, pencil Instructions 1. In this activity you will conduct an investigation to locate centre of gravity of a regular lamina. 2. Using your geometrical instruments, and the other materials provided, come up with a method of determining the centre of gravity of a regular lamina. Write a brief procedure and execute it. 3. Balance them at their centres of gravity on the tip of your pencil. 4. Practically locate their centres of gravities by drawing. In the previous topic, you did Activity 3.6 on determining the weight of a beam. (uniform metre ruler). You were able to realise that the weight of the uniform metre rule tends to act at central point of the ruler that is at 50 cm mark. A thin cardboard like the one used in Activity 4.2 is a lamina. The cover of a book is a lamina. The 97 set square or protractor in your mathematical set are all examples of laminae. Experiments have shown that bodies with uniform cross-section area and density have their centres of gravity located at their geometrical centres. For example, a metre rule of uniform cross-sectional area and density has its centre of gravity located at the 50 cm mark. Fig. 4.3(a) - (d) shows the centre of gravity (c.g) of rectangular, square, triangular, and circular laminae. c.g c.g c.g c.g (a) rectangle (b) square (c) isosceles triangle (d) circle Fig. 4.3: c.g. of regular lamina 4.3 Centre of gravity (c.g) of irregular lamina Activity 4.4 (Work in groups) To determine the centre of mass of an irregular lamina using a plumbline Materials: An irregular lamina, plumbline, a drawing pin Steps 1. Guess and mark the centre of gravity of an irregular lamina. 2. Make three holes P, Q and R on an irregularly shaped lamina as close as possible to the edges and far away from each other. The holes should be large enough to allow the lamina turn freely when supported through a drawing pin. 3. Suspend the lamina on the clamp using the drawing pin through each hole at a time. 4. Suspend a plumbline (a thin thread with a small weight at one end) from the point of support, P as shown in Fig. 4.4(a), and draw the line of the plumbline on the lamina by marking two points A and B far apart and joining them. 5. Repeat the steps with the support Q and mark the point M where the two lines intersect. 6. Check the accuracy of your method by suspending the lamina at R. What do you observe? Explain. 7. Comapre the c.g determined experimentally and that you guessed earlier. 98 irregular lamina P R A B Q smooth support e.g drawing pin hole string (plumbline) weight P A R M Q B (a) (b) Fig. 4.4: Locating centre of mass of a lamina using a plumbline 8. The plumbline pass through M (Fig. 4.4 (b)). Check the results again by balancing the lamina about point M. What do you observe? The lamina balances horizontally at point M. Point M is the centre of gravity of the lamina. This activity proves that when a body is freely suspended it rests with its centre of gravity vertically below the point of suspension. Activity 4.5 (Work in groups) To demonstrate how to locate the centre of gravity of an irregular object using a straight edge Materials: An irregular lamina, a prism Steps 1. Balance a lamina on the edge PQ of a prism as shown in Fig. 4.5(a). Mark the points A and B on the lamina and join them. 2. Repeat the activity for another position and note the points C and D on the lamina. Join C and D. knife edge P A lamina B Q A M C D B (a) (b) Fig. 4.5: Locating the centre of mass of a lamina using a straight edge 3. Label the point of intersection of lines AB and CD as point M. 4. Try balancing the lamina at point M on a sharp pointed support. What do you observe? Explain. 99 From Activiy 4.5, you should have observed that the lines AB and CD intersect at M; the centre of gravity of the lamina. 4.4 Effect of position of centre of gravity on the state of equilibrium of a body Activity 4.6 (Work in groups) To define and describe three states of equilibrium Materials: Internet, reference books Steps 1. Conduct a research from books and the internet on the meaning of the term equilibrium in regard to forces acting on an object. 2. State the three states of equilibrium. 3. Describe each of the three states of equilibrium. 4. Compare and discuss your findings with other groups in class. In your discussion, you should have noted that, the state of balance of a body is referred to as the stability of the body. Some bodies are in a more stable (balanced) state than others. The state of balance of a body is also called its state of equilibrium. Activities 4.7, will help us distinguish between the different states of equilibrium. 4.4.1 States of equilibrium Activity 4.7 To investigate the three states of equilibrium (Work in groups) Materials: plastic thistle funnel, bench Steps 1. Place the funnel upright with the wider mouth resting on the bench (Fig 4.6). 2. Displace the funnel slightly upwards as shown in Fig 4.6 (b) and then release it. What do you observe? 100 force F C W G (a) A C A force F G W (b) Fig 4.6: To show stable state 3. Explain the behaviour of the funnel in terms of the changes in the position of centre of gravity. 4. Place the funnel upright with the narrower mouth resting on the bench as shown in Fig 4.7 (a). 5. Displace the funnel slightly with your finger. What about change of state of the funnel? G C force F G W A (a) funnel topples down force F C G W (b) Fig 4.7: To show unstable state 6. Explain the behaviour of the funnel in terms of the change in position of its centre of gravity in this activity. 7. Place the funnel horizontally as shown in Fig 4.8 (a). 8. Displace the funnel gently by tapping it with a finger. What do you observe? table G W F G W F (a) (b) Fig. 4.8: To show neutral state 101 9. Explain the behaviour of the funnel in terms of the change in position of the centre of gravity when displaced slightly. When a body is resting with its centre of gravity at the lowest point, it is very stable. When displaced slightly; its centre of gravity is raised and when it is released, the object falls back to its original position to keep its centre of gravity as low as possible. This type of equilibrium is known as stable equilibrium. Thus, the funnel in Activity 4.7 Fig. 4.6 (a) was in stable equilibrium. Our finances and keeping the environment clean!! Note that we have used a plastic thistle funnel instead of a glass one. The latter has high chances of breaking. Any time we break a laboratory apparatus, we think of its effects on the school finances as it has to be replaced. Sometimes we may be required to pay ourselves hence affecting the finances of our parents. In case you use glass funnels, be careful when using them. As they may break and cause injuries to you or group members. If it breaks accidentally, collect the broken pieces and dispose them to keep the environment clean. When an object is resting with its centre of gravity at a very high position from the base support, it is unstable. When displaced slightly, it continues to fall up to the lowest possible position in order to lower its centre of gravity. This state of stability is known as unstable equilibrium. The funnel in Fig 4. 7(a) was in unstable equilibrium. When an object is resting such that the position of its centre of gravity remains at the same vertical position even when the object is displaces, it is said to be in neutral equilibrium. The funnel in Fig 4.8 (a) was in neutral equilibrium. 4.4.2 Relationship between position of centre of gravity and stability Activity 4.8 To investigate the relationship between position of centre of gravity of a body and its stability (Work in pairs or groups) Material: reference material Discuss with your group members why: 1. Buses sometimes carry heavy luggage at their roof tops. 2. Buses have their luggage bonets located underneath. 3. A person carrying two bucket
s full of water is more stable that one carrying one bucket. 102 4. Discuss and compare your explanations with your partner and report to the whole class. A body is more stable when its heavy part is as low as possible since it lowers the position of the centre of gravity. If the heavy part of the body is at high position or if the light part of the body in high position is made heavier than the lower position, the body becomes unstable and thus likely to topple over and can cause accidents like in the case of a vehicle carrying heavy luggage at its roof top. Exercise 4.1 1. Draw the figures below in your notebook and identify the centre of gravity of each. (a) (d) (b) (e) (g) (h) (c) (f) (i) Fig 4.9: Various types of figures 2. Fig 4.10 shows a Bunsen burner at different states of equilibrium. (i) (ii) (iii) Fig 4.10: A bunsen burner at different states of equilibrium 103 (a) Name the states in which the Bunsen burner is at in (i), (ii) and (iii). (b) Describe each state named in (a) above. 3. With aid of a diagram, describe how you can determine the centre of gravity of an irregular plane sheet of metal. 4. State and explain the states of equilibrium in Fig 4.11. Fig 4.11: A sphere 4.5 Factors affecting the stability of a body Activity 4.9 (Work in groups) To design and investigate factors that affect the stability of a body Materials: Plastic thistle funnels, Benches Instructions 1. By modifying the set-up, we used in Activity 4.7, using the materials provided, conduct an investigation on factors affecting stability of a body. 2. Sketch the step-up and write a brief procedure to investigate the factors. 3. Execute the procedure and answer the following questions (a) What happens to the funnel when the vertical line through the centre of gravity falls outside the base of the funnel? Deduce the factors that affect stability of the funnel. 4. Write a report and present it in a class discussions. 5. What are some of the sources of errors in the experiment and how can they be minimised? The funnel is more stable when its c.g is at a very low position and vice varsa. In addition, the activities show that the wider the base the more stable a body is. Activity 4.9 further shows that the funnel becomes unstable when the vertical line drawn through the centre of gravity falls outside the base that supports the body. In summary, a body is more stable if: 1. the centre of gravity is as low as possible. 2. the area of the base is as large as possible, and 3. the vertical line drawn from the centre of gravity falls within its base. 104 4.6 Applications of the position of centre of gravity Activity 4.10 (Work in groups) To describe the applications of the position centre of gravity Materials: reference books, Internet Steps 1. Conduct a research from Internet and reference books on the applications of position of centre of gravity. 2. In your research also find out: (a) Why a bird toy balances on its beak? (b) Why it is not advisable to stand on a small boat on the surface of the water? (c) Why one leans to the opposite direction when carrying a load? (d) Why the bus chassis is made heavier than the other parts of the bus? (e) How a tight-rope walker balances himself/herself? 3. Discuss your findings with other groups in class. Do you have the same explanations? 4. Have a class presentation on your findings from your research. In your research and discussion, you should have learnt the following: 1. The balancing bird is a toy that has its centre of gravity located at the tip of the beak. The bird balances with its beak resting on one finger or any other support placed underneath the beak, and the rest of the body in the air. This is because it is designed with its centre of gravity at that point (Fig 4.12). Fig. 4.12: Balancing bird toy 105 2. People in a small boat are advised neither to stand up nor lean over the sides while in the boat. This is because when they stand, they raise the position of the centre of gravity making the boat unstable and more likely to tip over (See Fig 4.13). Fig. 4.13: People in a boat 3. A person normally leans to the opposite direction when carrying heavy loads with one hand e.g. a bucket full water. This helps to maintain the position of the c.g to within the base of the person in order to maintain stability (See Fig 4.14) Fig. 4.14: Leaning while carrying heavy load 4. Most buses have their cargo in compartment in the basement instead of the roof rack in order to keep the centre of gravity of the buses as low as possible (Fig. 4.15). 106 Fig. 4.15: Buses carry their cargo below passengers’ level. 5. A tight-rope walker carries a pole to maintain stability. By swaying from side to side, he/she ensures that the vertical line drawn from his/her centre of gravity falls within the feet on the rope in order to maintain stability. (Fig. 4.16). tight-rope Fig. 4.16: A tight-rope walker carries a pole to maintain balance. Topic summary • The centre of gravity, c. g, of a body is the point where the whole weight of the body appears to act from. • Centre of mass of an objects is the point where all the mass of the object is concentrated. • The centre of gravity of a regular lamina or object is at its geometric centre. • The centre of gravity of a lamina can be found using a plumbline or by balancing it on a knife edge. • A body is said to be in stable equilibrium if it returns to its original position after being displaced slightly. • A body is in unstable equilibrium if on being slightly displaced, it does not return to its original position. • A body is said to be in neutral equilibrium it moves to a new position • but maintains the position of the c.o.g above its base support. Bodies can be made more stable if their centres of gravity are made as low as possible and the bases are made as broad as possible. 107 Topic Test 4 1. Define the term centre of gravity?. 2. Differentiate between centre of mass and centre of gravity. 3. Redraw the figures shown in Fig 4.17 below and indicate their centres of gravity. 4. Describe how you can determine the centre of gravity of the lamina shown in Fig 4.18. Fig. 4.17: Solids Fig. 4.18: Irregular shape 5. Fig. 4.19 shows a marble in three types of equilibrium. State and explain the type of equilibrium in each case. Fig. 4.19: Marble in three state of equilibrium 6. What is stability? 7. One vehicle which was travelling from Juba to Gulu was seen carrying heavy goods on its roof top and some of its passengers in the vehicle were standing. Discuss why the vehicle is likely to topple if it negotiates a corner at high speed. 108 8. Explain why a three-legged stool design is less stable than a four legged one. 9. Explain the following: (a) The passengers of a double-decker bus are not allowed to stand on the upper deck. (b) A racing car is made of a heavy chassis in its lower parts. (c) When one is alighting from a moving vehicle, it is advisable to spread out his/her legs. My safety Do not stand in a moving vehicle. Let us observe traffic rules. 109 110 UNIT 3 Work, Energy and Power Topics in the unit Topic 5: Work, Energy and Power Learning outcomes Knowledge and Understanding • Understand the concepts of work, energy and power. Skills • Design tests to relate work done to the magnitude of a force and the distance moved, power to work done and time taken, using appropriate examples. • Observe carefully. • Predict what might happen. • Use appropriate measures. • Collect and present results appropriate in writing or drawing. • Interpret results accurately and derive kinetic and potential energy formula. • Report findings appropriately and relate work, energy and power. Attitudes • Appreciate that food eaten is energy. Key inquiry questions • How does the world get its energy? • Why is energy not destroyed? • Why work done is energy? 111 TOPIC 5 Work, Energy and Power Unit outline • Forms of energy • Transformation of kinetic energy to potential energy and vice verse. • • Different ways to conserve energy • Law of conservation of mechanical energy Sources of energy Introduction Everyday, we do many types of work. We work in the offices, in the farms, in the factories etc. To make our work easier, we use machines ranging from simple tools to sophisticated machinery. Different machines or people do work at different rates (known as power). The ability and the rate of doing certain amount of work depends on how much energy is used. In this topic, we will seek to understand these three terms i.e work, energy and power from the science point of view. 5.1 Work Activity 5.1 To distinguish cases when work as defined in science is done or not (Work individually or in groups) Materials: a chart showing people carrying out different activities, pieces of chalk, pen, chair, desk. Steps 1. Conduct research from books on the scientific definition of work. 2. Walk from your chair to the chalkboard and write the word ‘work’ on the chalkboard. 3. Collect any litter in your classroom. Be responsible Always keep where you live clean. It is good for your health. 112 4. Carry your chair to the front of you classroom and sit on it. 5. Push against a rigid wall of your classroom. 6. Discuss with your colleagues whether scientifically speaking work, is done in steps 1, 2, 3 and 4. What do think is the meaning of ‘work’ ? 7. Now, look at the activities being performed by the people in Fig. 5.1 below. (a) (b) (c) (d) Fig. 5.1: People performing different tasks 8. According to the scientific definition of work, in which of the diagrams above is the person doing work? Explain. 9. Give other examples of doing work. Work is only said to have been done when an applied force moves the object through some distance in the direction of force. Therefore in Activity 5.1, work was done in steps 2, 3 and part of 4 (when carrying the desk). However, no work was done when you sat on your chair without moving in step 4 and pushing the wall without moving it in step 5. 113 Similary, in Fig 5.1, work is bein
g done in (a) and (d) only. When the girl applies a force to a wall in (b) and even becomes exhausted, she is not doing any work because the wall is not displaced. When the woman carries the basket on the head, she is not doing any work. This is because she exerts an upward force on the basket which is balanced by the weight hence there is no motion of the basket in the direction of the applied force. Definition of work Work is defined as the product of force and distance moved in the direction of the force. i.e Work = force × distance moved in the direction of the force W = F × d The SI unit of work is joule (J). A joule is the work done when a force of 1 newton moves a body through a distance of 1 metre. 1 joule =1 newton × 1 metre Bigger units used are kilojoules (1 kJ) = 1 000 J Megajoule (1 MJ) = 1 000 000 J Note: Whenever work is done, energy is transferred. Example 5.1 Find the work done in lifting a mass of 2 kg vertically upwards through 10 m. (g = 10 m/s2) Solution To lift the mass upwards against gravity, a force equal to its own weight is exerted. Applied force = weight = mg = 2kg × 10N/kg = 20 N Work done = F × d = 20 N × 10 m = 200 Nm = 200 J 5.1.1 Work done in pulling an object along a horizontal surface Activity 5.2 To design an investigation to determine the work done in pulling an object along a horizontal surface (Work in groups) Materials: a block, a weighing scale, and a tape measure/metre ruler, string. 114 Instructions 1. In this experiment you will design and carry out an investigation to determine work done in pulling an object on a horizontal surface. 2. Using the materials provided think of a set-up to do the investigation. Set-up the apparatus and sketch the set-up. 3. Write a brief procedure to execute in doing the activity. 4. Correctly execute the procedure and answer the following questions. 5. Using relevant formula, calculate the work done in pulling the block. What assumption did you make? Explain. 6. Compare and discuss your findings with other groups in class. 7. Explain how tractors pull large loads. Since the block was on a smooth surface, we assume that friction force is negligible hence the force applied is constant along the distance of motion, d. Work done in moving the block is given by: Work = force × distance W = F × d = Fd Example 5.2 A horizontal pulling force of 60 N is applied through a spring to a block on a frictionless table, causing the block to move by a distance of 3 m in the direction of the force. Find the work done by the force. Solution The work done = F × d = 60 N × 3 m = 180 Nm = 180 J Example 5.3 A horizontal force of 75 N is applied on a body on a frictionless surface. The body moves a horizontal distance of 9.6 m. Calculate the work done on the body. Solution Work = force × distance = 75 N × 9.6 m = 75 × 9.6 Nm = 720 J 115 Example 5.4 A towing truck was used to tow a broken car through a distance of 30 m. The tension in the towing chain was 2 000 N. If the total friction is 150 N, determine. (a) Work done by the pulling force. (b) Work done against friction. (c) Useful work done. Solution Fig. 5.2 shows the forces acting on the two cars. Fr = 150 N 2 000 N Fig. 5.2: Diagram of cars (a) Work done by the pulling force (b) Work done against friction W = F × d W = Fr × d (Fr is the frictional force) = 2 000 N × 30 m = 150 N × 30 m = 60 000 J = 4 500 J (c) Useful work done Useful work done = Fd – Frd = (60 000 – 4 500) J = 55 500 J Exercise 5.1 1. Explain why in trying to push a rigid wall, a person is said to be doing no work. 2. Define the term work and state its SI unit. 3. How much work is required to lift a 2 kilogram mass to a height of 10 metres (Take g=10 m/s2). 4. A garden tractor drags a plough with a force of 500 N at a distance of 2 metres in 20 seconds. How much work is done? 116 5.1.2 Work done against the force of gravity Activity 5.3 (Work in pairs) To determine the work done against the force of gravity Materials: Masses, meter rule (tape measure), a single fixed pulley, string, retort stand, newtonmeter. Steps 1. Hang a mass from a newtonmeter and record its weight. 2. Lift a mass from the ground or bench vertically upwards at a constant speed up to a certain point. For a pulley, tie the string on a mass and pass it around the pulley / clamped on a retort stand pull the other end of the thread. Explain why you must use some work in lifting a mass from the ground. 3. Have a friend to measure the height through which the mass has been raised using the meter ruler or tape measure. 4. The force needed to lift the mass is equal to its weight mg. 5. The work done on the mass is then w = Fd = mgh. 6. Repeat the steps with different masses and calculate the work. 7. When an object is thrown upwards, it raises upto a certain point and then drops. Explain why it raises and drops back. You can do this practically by throwing some objects (of different masses). Approximate amount of work done. The gravitational force (weight) acting on a body of mass m is equal to the product of mass and acceleration due to gravity, g, i.e. w = mg. Thus, to lift a body, work has to be done against the force of gravity (Fig. 5.3). – – – – – – – – – • – – – – – – – – mg – – – – – x = h • mass, m mg Ground Fig 5.3: Work done against gravity Work done against gravity to lift a body through height is given by: Work = Force × vertical height = mg × h = mgh 117 Example 5.5 Calculate the work done by a weight lifter in raising a weight of 400 N through a vertical distance of 1.4 m. Solution Work done against gravity = Force × displacement = mg × h = 400 N × 1.4 m = 560 J Example 5.6 A force of 200 N was applied to move a log of wood through a distance of 10 m. Calculate the work done on the log. Solution W = F × d = 200 N × 10 m = 2 000 J Exercise 5.2 1. Define work done and give its SI unit. 2. Calculate the work done by a force of 12 N when it moves a body through a distance of 15 m in the direction of that force. 3. Determine the work done by a person pulling a bucket of mass 10 kg steadily from the well through a distance of 15 m. 4 A car moves with uniform speed through a distance of 40 m and the net resistive force acting on the car is 3 000 N. (a) What is the forward driving force acting on the car? Explain your answer. (b) Calculate the work done by the driving force. (c) State the useful work done. 5. A student of mass 50 kg climbs a staircase of vertical height 6 m. Calculate the work done by the student. 6. A block was pushed by a force of 20 N through 9 m. Calculate the work done. 118 5.1.3 Work done along an inclined plane Activity 5.4 (Work in groups) Materials To determine the work done along an inclined plane Spring balance, one piece of wood of about 10 cm, a wedge, ruler, trolley/ piece of wood/mass hanger/stone. Steps 1. Make an inclined plane by putting a piece of wood on a wedge. 2. Attach the mass hanger/stone/trolley to a spring balance (calibrated in Newtons). What happens to the spring immediately when the mass is hanged on it? 3. Measure the length of the incline and record it down l= ..... cm. 4. Pull the spring balance with its object on from the bottom of the incline and note down the force used in pulling. ............N 5. Change the length in cm to m and find the work done using the formula, work = Fd = (J) 6. Using the above skills, approximate the amount of work you do when climbing a slope of 100 m long. Consider an inclined plane as shown in Fig. 5.4 below. A body of mass m moved up by a force f through a distance d li e d p F ( a θ Fig. 5.4: Work done along inclined plane. Work done by the applied force is given by Work done = F × d The work done against the gravitional force is given by: Work done = weight of the object × vertical height Work = mgh In case the inclined plane is frictionless force: Work done by the applied force = work done against gravity 119 In case there is some frictional force opposing the sliding of the object along the plane: Work done by the applied force > Work done against gravity Work done against friction = Work done by applied force – work done against gravity Example 5.7 A box of mass 100 kg is pushed by a force of 920 N up an inclined plane of length 10 m. The box is raised through a vertical distance of 6 m (Fig. 5.5). m = 100 kg Fig 5.5: Inclined plane (a) Calculate: (i) the work done by the applied force, (ii) the work done against the gravitational force. (iii) the difference in work done. (b) Why do the answers to (i) and (ii) in part (a) differ? Solution (a) (i) Work done by the applied force = F × d = 920 N × 10 m = 9 200 J (ii) Work done against gravity = F × d = mg × h = 100 kg × 10 N/kg × 6 m = 6 000 J (iii) The difference in work done = 9 200 J – 6 000 J = 3 200 J (b) This work done is used to overcome the friction between the box and surface of the incline plane. The useful work done is 6 000 N. 120 Exercise 5.3 1. A box of mass 50 kg is pushed with a uniform speed by a force of 200 N up an inclined plane of length 20 m to a vertical height for 8 m (Fig. 5.6). 0 N 0 F = 2 8 m Fig. 5.6: Work done along an inclined plane Calculate the: (a) Work done to move the box up the inclined plane. (b) Work done if the box was lifted vertically upwards. 2. A body of mass 85 kg is raised through a vertical height 6 m through an inclined plane as shown in Fig. 5.7. Calculate the: F = 150 N (a) Slant distance. (b) Work done by the force 150 N. 6 m (c) Work done, if the body was lifted vertically upwards. (d) Work done against friction. 85 kg 8 m Fig. 5.7: Work done on an inclined (e) Frictional force between the body and the track. 3. A block of mass 60 kg was raised through a vertical height of 7 m. If the slant height of a frictionless track is 21 m, and the force used to push the block up the plane is 800 N, calculate the work done in pushing the block. 4. A car engine offers a thrust of 2 500 N to ascend a sloppy road for 1.1 km. At the top of the slope, the driver realized that the attitude change was
200 m. If the mass of the car is 1.2 tonnes, calculate the; (a) Work done by the car engine. (b) Work done against resistance. 121 5.2 Power Activity 5.5 (Work in groups) To compare the time taken to do a piece of work by a person and a machine Materials: writing material, stopwatch and scientific calculator. Steps 1. By timing yourself, start solving the following problems without using calculator: (a) 37 6998 J × 276 J (b) 35 264 J × 469 J 2. Repeat step 1, but now with a scientific calculator and compare the time taken. Which one takes longer or shorter time to complete the task? Explain your answer. 3. Now, think of a man ploughing a square piece of land that measure 100 m by 100 m (a) by hand, (b) using a tractor. Which task do you think takes longer or shorter time to complete the activity? Suggest a reason. 4. What is power? In your groups discuss the meaning of power. In your discussion, you might have noted that sometimes work is done very quickly and sometimes very slowly. For instance, it takes a longer time to multiply the problems without a calculator in step 1 than with a calculator in step 2. Similarly, in step 3, a tractor will take few hours ploughing a piece of land while a man will take more hours ploughing the same piece of land. The person and the tractor are doing the same work but the tractor is doing it at a faster rate than the person does. This is because they have different power ratings. Different machines and engines have different power ratings. Engines with bigger power ratings are said to be powerful and operate very fast. Definition of power Power is the rate of doing work. i.e. Power = work done time taken = force × distance time 122 SI units of power are Watts. 1watt = 1 joule second Large units used are kilowatt and megawatt. 1 kilowatt = 1 000 W 1 megawatt = 1 000 000 W Example 5.8 What is the power of a boy lifting a 300 N block through 10 m in 10 s ? Solution Force = 300 N, Distance = 10 m, Time = 10 s Work done by the boy = F ×d = 300 ×10 Power = = 3 000 J work time = 3 000 J 10 s = 300 W 5.2.1 Estimating the power of an individual climbing a flight of stairs Activity 5.6 To estimate the power of an individual climbing a flight of stairs (Work in groups) Materials: stopwatch, weighing machine, tape measure Steps 1. Find a set of stairs that you can safely walk and run up. If there are no stairs in your school, create some at garden. 2. Count their number, measure the vertical height of each stair and then find the total height of the stairs in metres. 3. Let one member weigh himself/herself on a weighing machine and record the weight down. 4. Let him/her walk then run up the stairs. Using a stopwatch, record the time taken in seconds to walk and running up the stairs (Fig 5.8). 123 Fig. 5.8: Measuring one’s own power output 5. Calculate the work done in walking and running up the stairs. Let each group member do the activity. Is the work done by different members in walking and running up the stairs same? Explain the disparity of work done by various group members. 6. Calculate the power developed by each individual in walking and running up the stairs. Which one required more power, walking or running up the flight of stairs? Why? Note: (i) The disabled should be the ones to time others. Care must be taken on the stairs not to injure yourselves. (ii) Incase of lack of stairs, learners can perform other activities like lifting measured weights. From your discussion, you should have established that: Height moved up (h) = Number of steps (n) × height of one step (x) h = n × x = n x Time taken to move height (h) = t P = = P = Work done against gravity time W × n × x t Wnx t = mgh t = W × h t where, P = power, W = weight, x = height of first step, n = number of steps If x is in metres, W in newtons and t in seconds then power is in watts. 124 Example 5.9 A girl whose mass is 60 kg can run up a flight of 35 steps each of 10 cm high in 4 seconds. Find the power of the girl. (Take g = 10 m/s2). Solution Force overcome (weight) = mg = 60 kg × 10 N/kg = 600 N Total distance = 10 × 35 = 350 cm = 3.5 m Work done by the girl = F × d = 600 × 3.5 Power = = 2 100 J work time = 2 100 J 4 s The power of the girl is 525 W My health Do you know that regular exercises are good for your health. Do exercises regularly to keep your body healthy Exercise 5.4 1. Define power and give its SI unit. 2. To lift a baby from a crib, 100 J of work is done. How much power is needed 3. if the baby is lifted in 0.5 seconds. If a runner’s power is 250 W, how much work is done by the runner in 30 minutes? 4. The power produced by an electric motor is 700 W. How long will it take the motor to do 10,000 J of work? 5. Find the force a person exerts in pulling a wagon 20 m if 1 500 J of work is done. 6. A car’s engine produces 100 kw of power. How much work does the engine do in 5 seconds? 7. A color TV uses 120 W of power. How much energy does the TV use in 1 hour? 8. A machine is able to do 30 joules of work in 6.0 seconds. What is the power developed by the machine? 125 9. Rebecca is 42 kg. She takes 10 seconds to run up two flights of stairs to a landing, a total of 5.0 metres vertically above her starting point. What power does the girl develop during her run? 10. Student A lifts a 50 newton box from the floor to a height of 0.40 metres in 2.0 seconds. Student B lifts a 40 newton box from the floor to a height of 0.50 metres in 1.0 second. Which student has more power than the other? 11. Four machines do the amounts of work listed in Table 5.1 shown below. The time they take to do the work is also listed. Which machine develops the most power? Machine A B C D Work 1 000 joules 1 000 joules 2 000 joules 2 000 joules Table 5.1 Time 5 sec 10 sec 5 sec 10 sec 5.3 Energy Activity 5.7 (Work in pairs) To brainstorm about energy Materials: Reference materials, a pen, desk or slopy platform Steps 1. What enables your body to perform various functions besides keeping warm. Discuss this with your group member. 2. Define the term energy. 3. Push a pen on your desk such that it rolls for a small distance. What makes the pen roll? 4. State some examples of objects that posses energy in our environment. 5. Discuss the importance of energy in our lives. 6. With a heavy bag on your back, climb the stairs to the top most floor or the steep ground to the top most point. How do you feel? Have you done work? Did you use energy to do so? 7. What happens to a car that has been moving if it runs out of fuel (petrol or diesel)? Explain. 8. Discuss the observations and findings from the above activities in a class discussion. 126 Energy is one of the most fundamental requirement of our universe. It moves motorcycles, cars along roads, airplanes through air, and boats over water. It warms and lights our homes, makes our bodies grow and allows our minds to think. A person is able to push a wheelbarrow, a stretched catapult when released is able to make a stone in it move, wind mills are turned by a strong wind and cooking using electricity in a cooker. All these are possible because of energy. Therefore, for any work to be done, energy must be provided. But what is energy? Definition of energy Energy is the ability to do work. Work done = energy transferred SI unit of energy is joules (J). Relationship between energy and work In your discussion, you should have noted that you got exhausted because you did a lot of work against gravity to carry your body and the heavy bag to the top of the building. The work you did led to the loss of energy (chemical energy from the food) from your body. 5.4 Forms of energy Energy is not visible, it occupies no space and has neither mass nor any other physical property that can describe it. However, it exists in many forms, some of these forms include: 5.4.1 Solar energy Activity 5.8 (Work in groups) To investigate the effect of solar energy Materials: plastic basin, water, convex lens, thin piece of paper Steps 1. On a bright sunny day, fill a plastic basin with cold water and place it in an open place with no shade. Dip your hand into the water after 2 hours. What is the temperature difference of the water initially and after 2 hours? 2. Get a convex lens on the same day and put it horizontally with one surface facing the sun and another surface facing down. Place a thin paper below the lens. What do you observe after 5 minutes or more? 3. Discuss your observations in steps 1 and 2. 127 The water becomes hot in case 1 and in case 2, the paper burns because of the heat from the sun. These are some of the effects of solar energy. This energy from the sun is in form of radiant heat and light. In some countries where the sun shines throughout, large concave mirrors have been set to collect energy from the sun by focusing its rays on special boilers which provide power for running electric generators. 5.4.2 Sound energy Activity 5.9 (Work in pairs) To design and investigate the production of sound energy Materials: Two pens, a stone Instructions 1. In this activity, design and carry out an investigation on production of sound energy using the materials provided. 2. Write a brief procedure and execute it to produce sound energy. Then write a report about your investigation and then discuss it in a class presentation. In what form is the energy released by the pen and the stone? Discuss with your class partner. 3. From your discussion, you should have heard sound in steps 1 and 2. In each case, kinetic energy has been converted to sound and heat energy. Sound energy is the energy associated with the vibration or disturbance of bodies or matter. 5.4.3 Heat energy Activity 5.10 (Work in groups) To demonstrate heat energy Materials: Bunsen burner/candle, matchbox, a retort stand, a nail/metallic rod Steps 1. Light a Bunsen burner or a candle using a lighter (matchbox). 2 Clamp a nail (metallic rod) on a retort stand and bring it near the flame. 3. Carefully touch the other end of the nail after sometime.
What do you feel? Explain. 128 The other end of the nail is felt to be hot after sometime. The hotness is due to heat energy that has been transferred from the hot part to the cold part of the nail. Therefore, heat energy only travels from a hot object to a cooler one. Heat energy is a form of energy that is transferred from one body to another due to the difference in temperature. 5.4.4 Electrical energy Activity 5.11 To demonstrate production of light by electrical energy (Work in groups) Materials: bulb, electric wire, cells (battery), switch, bulb holder and cell holder Steps 1. Fix the battery/cells in their holders and the bulb too. 2. Connect one wire from one end of the cell holder to the bulb holder. Then connect the same wire from the bulb holder to the switch holder and then connect another wire from the other part/side of cell holder to the switch. Make sure the switch is open and the cells are fixed into their holder. 3. What do you observe after the connection? Explain. 4. Complete the circuit then close the switch and observe what happens. The bulb lights when the circuit is complete. Electrical energy is the energy produced by the flow of electric charges (electrons). Work is done when electrons move from one point to another in an electric circuit with electrical appliances such as bulbs. 5.4.5 Nuclear energy Activity 5.12 (Work in groups) To find out what is nuclear energy 1. Conduct a research from internet and reference books on the meaning of 2. nuclear energy. In your research, also find out advantages and disadvantages of nuclear energy. 3. Compare and discuss your findings with those of other groups in your class. You may consult your teacher for more guidance on your discussion. 129 In your discussion, you should have noted that nuclear energy is the energy that results from nuclear reactions in the nucleus of an atom. It is released when the nuclei are combined or split. 5.4.6 Chemical energy Activity 5.13 To investigate and demonstrate chemical energy (Work in groups) Materials: glass beaker, a small bowl, steel wool, white vinegar, thermometer Steps 1. Place the steel wool in the bowl and soak it in white vinegar for a couple of minutes. 2. Squeeze out excess vinegar and wrap the steel wool around the thermometer in a way that you are still able to read the temperature. 3. Put the steel wool in the beaker, then place a cover with a paper or small book on the top. 4. Record the temperature immediately, then again in a minute or so, and again every minute for about five minutes. What did you observe? 5. Discuss your observation with other groups in a class discussion. The thermometer records a higher temperature reading. The chemical reaction of vinegar and steel wool generates energy in form of heat. This causes temperature to rise as shown by the thermometer. Chemical energy is a type of energy stored in the bonds of the atoms and molecules that make up a substance. Once chemical energy is released by a substance, it is transferred into a new substance. Food and fuels like coal, oil, and gas are stores of chemical energy. Fuels release their chemical energy when they are burnt in the engine (e.g in a car engine). 5.4.7 Mechanical energy Activity 5.14 (Work in groups) To describe mechanical energy Materials: pen, a piece of chalk Steps 1. Raise a piece of chalk or pen from the ground to a position above your head and release it to fall to the ground. What do you observe with the change in height and the speed of the piece of chalk as it falls. 130 2. Throw two full pieces of chalk on the wall one at a time using different forces or at different speeds (one should move faster and another one slowly). Note the sound created by the piece of chalk after colliding with the wall. Which one makes more noise after collision? Mechanical energy is the energy possessed by a body due to its motion or due to its position. It can either be kinetic energy or potential energy of both. When an object is falling down through the air, it posses both potential energy (PE) due to its position above the ground, and kinetic energy (KE) due to its speed as it falls. The sum of its PE and KE is its mechanical energy. Mechanical energy = kinetic energy + potential energy. (a) Potential energy Activity 5.15 (Work in pairs) To demonstrate the forms of potential energy Materials: a catapult or a spring , a small stone. Steps 1. Raise a small stone from the ground or any other resting position upwards to a particular height above its resting surface. What kind of energy do you think it attains? 2. Now, release the stone and observe what happens. Explain your observations. 3. Compress a spring to a particular size. What kind of energy do you think it attains? Explain. 4. Release the spring and observe what happens to the spring. 5. Why does the spring move in such a manner? Discuss your observations in steps 1, 2, 3 and 4 with your colleagues and identify two types of potential energy from the activities. You should have observed that when the stone was released it moved down to the ground. This implies that the stone had stored energy due to its position that makes it to start moving down after it has been released. The energy possessed by a body (e.g. a stone) due to its position above the ground is called gravitational potential energy. In other words potential energy is energy by virtue of position. Similarly, when the compressed spring was released, it relaxed to a bigger size. This implies that the spring had stored energy due to compression. The energy possessed by a body due to compression (e.g. a spring) or stretch (e.g catapult) is called elastic potential energy. 131 Therefore, potential energy is in two forms; gravitational potential energy and elastic potential energy. (i) Gravitational potential energy Bodies which are at a given height above the ground posses gravitational potential energy. This energy depends on the position of objects above the ground. The following activity will help us to understand how to calculate potential energy of a body at a particular position above the ground. Activity 5.16 (Work in pairs) To determine gravitational potential energy of a raised object Materials: three bricks, meter ruler, beam balance, soft board bridge Steps 1. Conduct research from the Internet and books on the mathematical expression of potential energy. 2. Support the soft board on two bricks. 3. Measure the mass of the third brick by using a beam balance then place it on the soft board. 4. Now lift the third brick to a height h1. Let your partner measure the height h1, in metres. 5. Allow the brick to drop gently onto the soft board. Describe the energy possessed by the brick as it drops. 6 . Calculate the potential energy gained by the stone using the expression of potential energy you got from the research. 7 . Repeat the activity with the other two different heights h2 and h3. 8. Compare and discuss your observations and values of PE in the three cases and deduce a general conclusion from your discussion. If a stone is lifted upwards through a height h; and placed on a table (Fig 5.9), work is done against gravity. h F mg Fig. 5.9: Potential energy depends on height, h. 132 The work done to overcome gravity is equal to the gravitational potential energy gained by the stone. But work done = F × h ; F = mg ∴ work done = mg × h But, potential energy = work done. Therefore: P.E = mgh Example 5.10 A crane is used to lift a body of mass 30 kg through a vertical distance of 6.0 m. (a) How much work is done on the body? (b) What is the P.E stored in the body? (c) Comment on the two answers. Solution (a) Work done = F × d = mg × d = 30 × 10 × 6 = 300 × 6 = 1 800 J (b) P.E = mgh = 300 × 6 = 1 800 J (c) The work done against gravity is stored as P.E in the body. Caution A stone dropped from the roof of a building will cause more pain if it falls on someone’s foot than when the same stone falls from a table. This is because the one on the roof has more gravitation potential energy due to its greater height (position) above the ground. (ii) Elastic potential energy In Activity 5.15, we saw that a stretched catapult or compressed (Fig. 5.10(a)) has energy stored inform of elastic potential energy. When the stretched spring catapult is released it releases the energy that can be used to do work e.g. to throw a stone. (a) (b) Fig. 5.10: A compressed and a stretched spring 133 pull – – – – – – – – – – – – – – – – – – force = Work done in stretching the spring = Elastic P.E gained by the spring Fig.5.11: Elastic potential energy = average force × extension ) × e = ( 0 + F 2 1 2 Fe Work done is stored as elastic potential energy. = Note: Since the force is not uniform (F increases from 0 to F) we should use the average force in calculating the work done. Example 5.11 Calculate the elastic gravitational p.e stored in a spring when stretched through 4 cm by a force of 2 N. Solution Elastic P.E = 1 We will learn more about elastic potential energy and how to determine it later. 2 × 2 × (0.04) = 0.04 J 2 Fe = 1 (b) Kinetic energy Activity 5.17 (Work in pairs) To demonstrate kinetic energy Material: trolley, table Steps 1. Place a trolley on the table and give it a slight push. Observe what happens to it. Explain your observations. 2. Now, observe any moving objects or things around you. Which energy do you think they possess when they are in motion? Explain your answer. 134 From your discussion in activity above, you should have observed that the trolley starts to move once given a slight push. It possess energy as it moves. The energy which is possessed by a moving object due to its speed is called kinetic energy (KE). In other words we can define kinetic energy as energy by virtue of motion. Examples of objects that posses KE include moving air, rotating windmills, falling water, rotating turbines and a moving stone. In general, any moving body possesses energy called kinetic energy. The kinetic energy of a moving b
ody is given by: Kinetic energy = 1 2 mv2 , where m and v are the mass and velocity of the body respectively. Example 5.12 A body of mass 400 g falls freely from a tower and reaches the ground after 4 s. Calculate the kinetic energy of the mass as it hits the ground. (Take g = 10 m/s2) Solution The final velocity of the body as it hits the ground given by the equation of motion, v = u + at = u + gt = 0 + 10 × 4 = 40 m/s K.E = 1 2 mv2 (as the body hits the ground) 2 × 0.400 × 402 = 320 J = 1 Example 5.13 A car of mass 1 000 kg travelling at 36 km/h is brought to rest by applying brakes. Calculate the distance travelled by the car before coming to rest, if the frictional force between the wheels and the road is 2 000 N. Solution V = 36 km/h to m/s = 36 × 1 000 m/s 60 × 60 = 10 m/s K.E = work done against friction 135 1 2 mv2 = F × d 2 × 1000 × 102 = 2 000 × d ⇒ 1 ⇒ 50 000 = 2 000 d ∴ d = 50 000 2 000 = 25 m The stopping distance is 25 m. 5.5 Work and energy relationship When work is done a transfer of energy always occurs. For example carrying a box up the stairs/lifting something heavy from the ground, you transfer energy to the box which is stored as gravitational energy. Therefore its energy increases. Doing work is a way of transferring energy from one object to another. Just as power is the rate at which work is done when energy is transferred the power involved can be calculated by dividing the energy transferred by time needed for the transfer to occur. Power(in watts) = Energy transferred (in joules) time (in watts) P = E t or Energy = pt For example, when the light bulb is connected to an electric circuit, energy is transferred from the circuit to the light bulb filament. The filament converts the electrical energy supplied to the light bulb into heat and light. The power used by the bulb is the amount of electrical energy transferred to the light bulb each second. Exercise 5.5 1. Define the term energy. 2. State and explain briefly six forms of energy. 3. Differentiate between: (a) Potential energy and kinetic energy. (b) Gravitational potential energy and elastic potential energy. 4. A brick of mass 0.5 kg is lifted through a distance of 100 m to the top of a building. Calculate the potentials energy attained by the brick. 136 5. Explain what is meant by gravitational potential energy. 6. A force of 8.5 N stretches a certain spring by 6 cm. How much work is done in strecthing the spring by 10 cm. 7. A body is acted on by a varying force, F over a distance of 7 cm as shown in Fig. 5.121 –2 –3 –4 –5 –6 1 2 3 4 5 6 7 Distance (m) Calculate the total work done by the force. Fig.5.12 8. An object of mass 3.5 kg is released from a height of 7.0 m above the ground. (a) Calculate the gravitational potential energy of the object release. (b) Calculate the velocity of the object just before it strikes the ground. What assumption have you made in your calculation? 5.6 Sources of energy Activity 5.18 To find out sources of energy (Working in groups) Materials: Internet, reference books, a stream of water or a water tap Steps 1. Tell your group members the meaning of the terms ‘source’ and ‘energy source’. 137 2. Now, think of plants, animals, vehicles and so on. Where do you think their energy comes from? What of electricity used in your school and at home, where does it come from? 3. Compare and discuss your findings in step 1 and 2 with other groups in your class. 4. What is the meaning of primary sources of energy? 5. What is the meaning of secondary sources of energy? 6. Conduct a research from the Internet and reference books on primary and secondary sources of energy. 7. In your research find out: (a) The types of primary and secondary sources of energy. (b) The generation of energy from each source. 8. Let your secretary write down a summary of your discussion and present it to the whole class. 9. State and explain various types of primary and secondary sources of energy? The word ‘source’ means the beginning of something e.g. the stream begins from the mountain or hills around your school. Hey!! Are you aware that cutting down trees will lead to the loss of our forests in our country and consequently the loss of water sources? Let us protect our water sources by planting more trees. You should have also established that the energy source is a system which produces energy in a certain way. For instance, a hydroelectric station uses the motion of the water of the river to turn the turbines and thus producing electricity. There are two kinds of energy sources; 1. Primary sources 2. Secondary sources 138 5.6.1 Primary sources of energy From your research and discussion in activity 5.18, you should have established the following that Primary Sources are from sources which can be used directly as they occur in the natural environment. They include. 1. Flowing water 2. Nuclear 3. Sun 4. Wind 5. Geothermal ( interior of the earth) 6. Fuels 7. Minerals 8. Biomass (living things and their waste materials 1. Water (a) Hydropower - the flowing water from dams rotate turbines at the bottom of the dam which turn the generator resulting in generation of electricity. This water is kept behind a dam (reservoir) and released at a controlled rate downwards where it meets the turbines and turns them. An example is the falling water in the Fulla rapids on the River Nile in Nimule in South Sudan. (b) Waves - energy from water waves (generated by winds) is also used in generating electricity using sea wave converters. An example is pelamis wave energy converter, a technology that uses the motion of ocean surface waves to create electricity. 2. Nuclear energy Nuclear energy is created through reactions that involve the splitting or merging of the atoms of nuclei together. The process of splitting of large atoms such as those of uranium into smaller atoms is called fission. Fusion on the other hand, is the combining of two smaller atoms such as hydrogen or helium to produce a heavier atom. All these reactions release heat which is turned into electricity in nuclear power plants (Fig 5.13). An atomic bomb derives its energy from these kinds of reactions. 139 3. The sun Fig 5.13: Nuclear plant The sun is the biggest source of energy and has played an important role in shaping our life on earth since the dawn of time. The sun gives off radiant energy in form of electromagnetic waves. The light energy (visible spectrum) part of the spectrum can be converted directly into electricity in a single process using the photovoltaic (PV) cell otherwise known as the solar cell. The solar thermal energy is used for heating swimming pools, heating water for domestic use (solar heater) and heating of building. Solar thermal electricity generation is where the sun’s rays are used to heat a fluid for the production of high pressure and high temperature steam. The steam is in turn converted into mechanical energy in a turbine to generate electricity. (a) Solar panel (b) Solar heater Fig. 5.14: Solar panel and solar heater 140 4. Wind Wind is caused by the sun heating the earth unevenly. The air is heated differently causing hotter air to expand rise, and the colder one to condense and sink. This results to the movement of air and hence formation of wind. Modern wind turbines placed on the top of steel tubular towers harness the natural wind in our atmosphere and convert it into mechanical energy and then to electricity. Wind mills (Fig. 5.15) are also be used to pump water from the underground and do some other work. 5. Geothermal energy Fig 5.15: Wind mills Geothermal gradient is the difference in temperature between the core (interior) of the earth (planet) and its surface brings about conduction of heat from the core to the surface. The earth’s internal heat is generated from radioactive decay and continual heat loss from the earth’s formation. From hot springs, geothermal energy has been used for bathing to heal some diseases as in some cultures. Geothermal energy is also used to generate electricity at geothermal power stations where heat is used to heat water to get steam which in turn is used to turn the turbines to generate electricity. 141 Fig. 5.16 Geothermal power station 6. Thermal energy Thermal energy is the internal energy in a system by virtue of its temperature. It is defined as the average translational kinetic energy possessed by free particles in a system of free particles in a thermodynamic equilibrium. It can also include the potential energy of a system’s particle which may be an electron or an atom. Thermal (heat) energy is transferred of heat across the system boundaries. Thermal energy is important in our daily life, for example in warming the house, cooking, heating the water and drying the washed clothes. 7. Biogas Activity 5.19 (Work in groups) To conduct research on how to produce biogas Materials: bio-digester, reference materials (including this book), internet 1. Using reference materials (including this book) or internet, research about how biogas is produced. Make notes and present your finding in class. 2. Make a trip to a farm with biogas plant and take turn to ask about working of biodigester. 3. Write a report explaining how a biodigester works? 4. Hold a group discussion and discuss about biogass as a source of energy. Biomass is the total mass of organic matter in plant or animal. It is used to generate energy e.g. through burning to give heat energy. When bacteria acts on biomass, a gas called biogas is produced which is flammable hence is used as fuel to produce heat. It is a mixture of 65% methane and 35% carbondioxide. 142 A biogas plant or digester collects and directs the gas through pipes to the kitchen for cooking in a house or to a generator where electricity is produced. Fig. 5.17 shows a biogas plant. Biogas pipe Biogas Storage Tank Biogas inlet Biogas outlet Biogas Digester Fig. 5.17: Biogas plant 8. Fuels Fuels are substances which produce heat when burnt in the presence of oxygen. They include
kerosene, diesel, biogas, are sources of energy in homes, industries. In the process of combustion, the chemical energy in the fuel is converted into heat energy that is converted to other forms as desired. 9. Chemical energy Chemical energy is stored in the chemical bonds of atoms and molecules. It can only be seen when it is released in a chemical reaction. When chemical energy is released, from a substance, the substance is entirely changed into an entirely different substance. Some substances that store and release chemical energy are; (i) Electrolytes – the chemical reactions in an electrolyte in the batteries produce electricity. (ii) Petroleum – petroleum is made of molecules containing carbon and hydrogen. In vapour form, its natural gas and in liquid form, it is crude oil. Energy from petroleum is used to drive vehicles and to produce electricity. Examples include jet fuel, gasoline and others. (iii) Wood – dry wood acts as a store of chemical energy. This chemical energy is released when wood burns and it’s converted into heat and light energy. 143 (iv) Food – the chemical energy in food is released while the food is being digested. As the bonds between the atoms of the food break, new substances are created and chemical energy is given out. Warning Subjecting a battery to misuse or conditions for which it was not made for can result into battery failure or uncontrolled dangerous conditions which include explosion, fire and the emission of toxic fumes. Keep batteries well out of reach of children. 10. Light energy The potential of light to perform work is called light energy. It is formed through chemical radiation and mechanical means. It is a form of energy produced by hot bodies and travels in a straight line. It’s the only form of energy that we can see directly (visible light). It can be converted like sunlight energy is used during photosynthesis by plants to create chemical energy. UV lights are often used by forensic scientists to see details that are not seen by unaided eyes. 5.6.2 Secondary sources of energy Secondary sources are energy sources that are generated from primary sources. For instance, electricity is a secondary source because it is generated for example from solar energy using solar panels or from flowing water using the turbines to generate hydroelectricity. Other secondary sources of energy include; petroleum products, manufactured solid fuels, gases, heat and bio fuel. 5.7 Renewable and non renewable sources of energy Activity 5.20 (Work in pairs) To distinguish between renewable and nonrenewable sources of energy Materials: matchbox, reference books Steps 1. Take one matchstick from the matchbox and light it. 2. Leave it to burn for a few seconds and then put it off. 3. Use the same matchstick and try to lit it again. Observe and explain what happens. 4. From the knowledge of sources of energy, what do you think renewable and non-renewable sources of energy are? 144 5. Discuss your observations in steps 2 and 3 with your deskmate. 6. Now, conduct a research from internet and reference book on renewable and non-renewable energy source. 7. Compare and discuss your findings with other groups in your class. Hey!!! Be safe Always be careful with fire. It can cause massive damage which can result to loss of properties and lives. There are energy sources that cannot be used again once used to generate energy. They are called non renewable sources while those that can be used again without exhausting them are called renewable resources. (a) Renewable energy sources A renewable energy source is an energy source which can’t be depleted/exhausted. They exist infinitely i.e. never run out. They are renewed by natural processes. Examples include; (i) Sun However, some like trees they can also be depleted, like trees and animals if used too much more than the natural process can renew them. So it’s advisable to take precaution while using them, that is, they should be conserved. (iii) Geothermal (iv) Biomass (ii) Wind By doing so it will lead to access to affordable and reliable energy while increasing the share if renewable energy in our country. Hence contributing to affordable, reliable sustainable and modern energy for all. Achieving sustainable development goals (SDGs) by 2030 is a drive that countries across the world are working towards together. Gaining more of our energy from renewable sources is an important part of the strategy. (b) Non-renewable energy sources These are sources which can be depleted because they exists in fixed quantities. So they will run out one day. Examples are coal, crude oil, natural gas, and uranium. Fossil fuels like coal, crude oil, natural gas are mainly made up of carbon. They are usually found in one location because they are made through the same process and material. Millions of years ago dead sea organisms, plants, and animals settled on the ocean floor and in porous rocks. With time, sand, sediments and impermeable rock settle on the dead organic matter, as the matter continue to decay forming coal, oil and natural gas. Earth movements and rock shifts creates spaces that force these energy sources to collect at well-defined areas. With the help of technology, engineers are able to drill down into the sea bed to mine these sources and harness the energy stored in them. 145 5.8 Environmental effects of the use of energy sources Activity 5.21 (Work in groups) To investigate the environmental effects of the use of energy sources Materials: reference books, Internet Steps 1. Conduct research from the Internet and reference books on environmental 2. effects of the use of energy sources. In your research, identify the effects and suggest the measure to be taken to ensure safe use of those resources. 3. Record down your findings from your discussion and report them in a class discussion. In Activity 5.24, you should have learnt that, there is no such thing as a completely “clean” energy source. All energy sources have atleast an effect to the environment. Some energy sources have a greater impact than others. Energy is mostly lost into the environment in form of heat and sound. The following are some of the effects of use of the energy sources to the environment: • air and water pollution • climate change and global warming. • deforestation • land degradation (a) Air and water pollution Fossil fuels e.g petroleum, diesel are used in factories. Very harmful by-products may be released to the atmosphere or water bodies. Carbon monoxides, sulphur dioxide and carbon dioxide may be released to the atmosphere causing air pollution that may harm living things that depend on air. When human beings inhale some of the polluted air, they can develop respiratory diseases. The wastes disposed to the water bodies can cause death of living things in the water. It also make the water unsafe for human consumption. Factories and industries operators are encouraged to use bio-fuels which are less harmful to the environment. Most factories are trying to clean up the waste so as to reduce the environmental pollution. (b) Climate change and global warming Most energy sources e.g fossil fuels, coal etc, when used as sources of energy, produce wastes such as carbon dioxide, sulphur dioxide and mercury which are 146 the greenhouse gases. The accumulation of these gases in the atmosphere make the temperatures to be higher than the normal. This is referred to as global warming. Sometimes, these gases interfere with the climate causing very high temperature in the atmosphere, acidified rains, frequent droughts, floods etc. This results to climate change. The greenhouse gases e.g. excess carbon dioxide, sulphur dioxide etc destroy the ozone layer exposing living things to dangerous emissions from the sun e.g. UV rays. Release of these harmful gases into the atmosphere is a global problem and very many environmental agencies are encouraging on the proper disposal of these wastes. United Nations Conference on Environment and Development (UNICED) lead Nations to sign a joint treaties to pursue of economic development in ways that would protect the earth’s environment and non renewable resources but it is still a problem up to now. (c) Deforestation Using firewood and charcoal in most African countries lead to loss of biodiversity and erosion due to loss of forest cover. These may lead to deforestation i.e. the reduction of forest cover. Of great concern is that Africa is losing forest twice as fast as the rest of the world. Human beings are encouraged to use green energy that is renewable and have less effect to the environment. With your help we can support projects that help to train and educate forest communities so that they can use forests in a sustinable manner and protect their livelihoods for years. (d) Land degradation Land degradation is the process in which the value of bio-physical environment is affected by human – induced activity on the land. It is caused by over-cutting of vegetation e.g. forest, and woodland, for firewood and disposing factory wastes to the soil that may contaminate the soil. Use of non-biodegradable sources of energy is encouraged. Saving our energy Let’s adopt the use of biogas in cooking, energy saving stoves and reduce the use of firewood to the possible level and amount of smoke generated reducing the impact of indoor air pollution. This will reduce environmental impacts. 147 Exercise 5.6 1. Differentiate between energy and power. 2. In groups of two, identify any three primary sources of energy and hold a discussion on their: (a) definition and origin. (b) importance to us and our country. 3. Choose any renewable energy resource. Brainstorm on two to three jobs opportunities available in that field. 4. Distinguish between the terms renewable resource and non-renewable resources. 5. Give one example of a body with potential energy due to its state. 5.9 Energy transformations Activity 5.20 (Work in groups) To investigate energy transformation
Materials: an electric heater, radio, water in a basin. Steps 1. Place the electric heater in the basin with water and connect it to the socket. 2. Put on the switch. Observe and explain what happens after a couple of minutes. Suggest the types of energy involved in this case. 3. Now, disconnect the heater and connect the radio to the socket. 4. Turn the radio on and suggest the types of energy involved. 5. Repeat the activity by connecting wires, battery, switch and bulb. Observe and explain what happens when you make simple circuit and the switch is closed. 6. What is the meaning of energy transformation? Give five examples of energy transformation? 7. What is the name given to devices such as the radio, heater, battery, bulb etc. that converts energy from one form to another? 8. Discuss with your group members other forms of energy transformation and show with diagrams how energy is transformed from one form to another on the chalkboard. 148 Hey!!! Be safe Don’t touch water while an electrical heater is on, you may get an electrical shock. From your discussion, you should have observed that the water in the basin boils. Electrical energy has been converted to heat energy which boils the water. When the radio was connect to the socket and turned on, electrical energy is converted to sound energy. In step 5, when the wires are connected, the bulb is seen to give off light when you close the switch. This is because chemical energy in the battery has been converted to electrical energy which is then changed to light energy in the bulb and some part to heat energy. Therefore, energy in many of its forms may be used in its natural process or to provide some services to society such as heating, refrigeration, or performing mechanical work to operate machines. This is possible because energy can be changed from one form to another. This process of changing of energy from one form to another is called energy transformation. A device that converts energy from one form to another is known as a transducer. Fig 5.18 is a chart that shows some examples of energy transformation in our day to day activities. 149 Chemical n actio ar re ucle Electrical c tri c E l e e ll o l a r c S Nuclear Nuclear reactor Heat o c o u ple c tric h E le a t e T h er m Sound ti o r V i b g s s Mechanical Generator Electric motor W i n d m i l l Light Let us consider a few examples of energy transformation: Fig. 5.18: A flow chart of energy transformation 1. Hammering a nail Chemical energy in our bodies Potential energy of hammer Fig. 5.19: Energy transportation Heat K.E Sound 150 2. Lighting a bulb using a battery Chemical energy in the battery Electrical energy Radiant heat Light energy 3. Hydroelectric power Fig. 5.20: Energy transformation Potential energy of water in the water reservour Kinetic energy of falling water K.E of rotation of turbines Electrical energy Heat and sound Fig. 5.20: Potential energy and its transformation Other examples of energy transfomers. Wind turbines use wind energy to transform it into electricity. Energy from food (chemical energy) can be transformed in energy to play and run. A solar cell/ panel convert radiant energy of sunlight to electrical energy that can be used to give off lightning a bulb or to power a computer The sun gives the grass thermal energy which helps it to grow by transforming the energy into chemical energy using photosynthesis. Animals eat grass and help them to grow and have power to run. A microphone changes electric energy to sound energy and so on. One other example of energy transformations occurs when lightning strike. If it hits a tree, it’s electrical energy will be changed to heat and thermal energy. The tree will become hot and can even burn as a result of electric discharge, it can split and the leaves dry. 151 Exercise 5.7 1. Table 5.2 shows how energy is converted from form A to form B and the devices concerned. Complete the table. Form A Form B Electrical Sound Electrical Kinetic _ Sound _ _ Electrical Electrical Electrical _ Device Loudspeakers _ Photocell _ Thermocouple Heater Table 5.2: Forms of energy 2. Describe the energy changes that occurs in the following processes. (a) When you lift a brick to a certain height. (b) When you lift a brick and let it slide down a rough slope until it reaches the surface of the slope. 3. Describe the forms of energy shown in Fig 5.21. Fig. 5.21: Forms of energy 4. Name the changes in energy that take place when a torch is switched on. 5. Name the energy changes that take place when lighting a match box. 152 5.10 Law of conservation of energy To demonstrate the law of conservation of energy Activity 5.23 (Work in pairs) Materials: a ball Steps 1. Hold a football at a height of 1 m above the ground. What type of energy does the ball posses at that position? 2. Release the ball to start falling freely to the ground. What type of energies does the ball posses while falling? 3. What type of energy does the ball posses while just about to touch the ground. 4. Ignoring air resistance, compare the amount of energy possesed by the ball in step 1 and 3. What can you conclude? When the ball was stationary at a point 1 m above the ground in Activity 5.26, it possed P.E only. When released the P.E started being converted to K.E hence the ball dropped. When it was just about to touch the ground, all the P.E had been converted to K.E hence by ignoring air resistance, Height point of swing Height point of swing Maximum kinetic energy Fig. 5.22: Initial P.E = final K.E We say that energy has been conserved. This is summarised in the law of conservation of energy. The law of conservation of energy states that energy cannot be created or destroyed but is simply converted from one form into another. Or in other words we can state it that in a closed system the total amount of energy is conserved. Energy can be inter-converted among many forms, mechanical, chemical, nuclear, electric, and others but the total amount of it remains constant. For instance, in boiling water using a kettle, electrical energy drawn from the power source flows into the heating element of the kettle. As the current flows 153 through the element, the element rapidly heats up, so the electrical energy is converted to heat energy that is passed to the cold water surrounding it. After a couple of minutes, the water boils and (if the power source remains in the water) it starts to turn into steam. Most of the electrical energy supplied into the kettle is converted to heat energy in the water though some is used to provide latent heat of evaporation (the heat needed to turn a liquid into a gas without a change in temperature). If you add up the total energy supplied by the power source and the total energy gained by the water, you should find they are almost the same. The minor difference would be due to energy loss in other forms. Why aren’t they exactly equal? It’s simply because we don’t have a closed system. Some of the energy from the power source is converted to sound and wasted (kettles can be quit noisy). The kettles also give off some heat to their surrounding so that’s also wasted energy. Another example is a flying ball, that hits a window plane in a house, shattering the glass. The energy from the ball was transferred to the glass making it shatter into pieces and fly in various directions. 5.11 The law of conservation of mechanical energy Activity 5.24 (Work in pairs) To verify the law of conservation of mechanical energy Materials: A smooth metallic hemispherical bowl, a ball bearing Steps 1. Place the hemispherical bowl on the bench in a stable position. Mark at point A on the inside surface of the bowl at point A on the inside surface of the bowl 2. Place the ball bearing at point A and release it to slide downwards freely along the inside surface of the bowl as shown in Fig. 5.22. A h B C E D Fig. 5.22: A ball bearing sliding oscillating in a bowl 3. Mark point E where the ball rises to on the opposite side in the bowl. 154 4. Compare the vertical height of points A and E. What do you notice aboue the heights? 5. Repeat the activity with point A at a lower vertical height. 6. What type of energy does the ball bearings possess at points A, B, C, D and E. 7. Compare and comment on the total amount of energy possessed by the ball bearing at points A, B, C, D and E. 8. Make a conclusion based on your observation in step 7. You should have learnt that the law of conservation of mechanical energy states that The total mechanical energy (sum of potential energy and kinetic energy) in a closed system will remain constant/same. A closed system is one where there are no external dissipative forces (like friction, air resistance) which would bring about loss of energy. The sum of potential energy and kinetic energy anywhere during the motion must be equal to the sum of potential energy and kinetic energy anywhere else in the motion. To demonstrate the law of conservation of mechanical energy (a) A swinging pendulum Activity 5.28 (Work in pairs) To demonstrate the law of conservation of M.E using a swinging pendulum Materials: a bob, string Steps 1. Tie a string to the bob and fix it to a rigid object.( See Fig. 6.16). 2. Pull the bob to the right or left side at an angle and then release it. Observe the movement of the bob. 3. Draw a diagram for the motion of the pendulum and discuss with your class the energy changes at various points e.g. A, B, C, D and E shown in Fig. 5.24. A E B D Fig. 5.24: A swinging pendulum C 4. Explain the energy changes at points A, B, C, D 155 From the above activity, you should have noticed that the bob will attain a maximum potential energy due to its height above the ground at point A she have minimum kinetic energy because it is at rest. When it swings after letting it go, it will start loosing potential energy as it gain kinetic energy at point B because of its motion. As it passes through the lowest point point C, its potential energy is minim
um kinetic energy will be maximum. Because of its kinetic energy, it swings up to the other side and now its kinetic energy starts decreases as, potential energy increases at point D until when it reaches the maximum point E where it stops moving momentarily. At that point, it has maximum potential energy but minimum kinetic energy. At all positions, the total mechanical energy is constant (conserved). That is kinetic energy + potential energy = constant. Therefore, mechanical energy has been conserved. (b) A body thrown upwards Activity 5.26 To demonstrate the law of conservation of M.E using a ball thrown upwards (Work individually) Materials: tennis ball Steps 1. Throw a tennis ball upwards. Observe and describe the movement of the ball up to the maximum (highest) point. 2. Now, drop the ball from a high point e.g from top of the building or a cliff (see Fig 5.25). 3. Sketch its motion on a paper at three different intervals, starting from the lowest when thrown upwards or from highest when dropped from a cliff. 4. Explain why the ball falls back to the ground after thrown upwards. 5. Indicate the forms of energy at each stage. 6. Discuss your observations and drawing with your colleagues in class. In your discussion, you should have learnt that when a body (e.g. a ball) is thrown up vertically, it has maximum speed, (maximum kinetic energy) at the starting point. The ball moves up with a reducing speed because of the force of gravity acting on it downwards until it reaches the maximum point/ height where it stops moving momentarily and it falls back. 156 At maximum height, it has a maximum potential energy and minimum kinetic energy because the body is not moving. So the kinetic energy at the bottom is all turned into potential energy at the maximum point (Fig 5.25). P.Emax K.E = 0 cliff P.E = K.E object P.E = 0 K.Emax ground Fig. 5.25: Sketch of a ball thrown upwards The ball is under free fall because it is only being acted upon by the force of gravity. Initially the ball has maximum potential energy and no kinetic energy. As it falls down, its potential energy keeps on reducing as its position above the ground reduces but its kinetic energy is increasing because it speeds up as it falls downwards. The kinetic energy at the ground level is equal to the potential energy at the top of the wall. Hence mechanical energy is conserved. Exercise 5.8 1. A pendulum bob swings as shown in the diagram. Fig 5.26 Pendulum bob Start Fig 5.25: A pendulum swinging At which position (s) is: the kinetic energy of the pendulum bob least. (a) (b) the potential energy of the pendulum bob most. 157 (c) (d) the kinetic energy of the pendulum bob the most. the potential energy of the pendulum bob the least. 2. State the following laws: (a) (b) law of conservation of energy law of conservation of mechanical energy. 3. Describe how mechanical energy is conserved. 5.12 Ways of conserving energy Activity 5.27 (Work in groups) To do research about conservation of energy and identify ways of conserving energy Materials: internet, reference book (including this book) 1. What is the meaning of the word conservation of energy? 2. Conduct a research from Internet and reference books on ways of conserving 3. energy. In your research, identify different ways of conserving energy and find out what energy efficiency is. 4. Discuss your finding with other pairs in class and give a report of your findings to the whole class. From your research and discussion you should have established that energy conservation is the act of saving energy by reducing the length of use. In other words, to conserve energy, you need to cut back on your usage. For example, driving your car fewer miles per week, turning your thermostat down a degree or two in the winter time and unplugging your computer or home appliance when they are not in use. All these ways reduce the amount of energy you use by doing without or using less fuel or electricity. It can help reduce the monthly heating and electricity bills and save money at the gas pump. You also reduce the demand of fuels like coal, oil, and natural gas. Less burning of fuels means lower emissions of carbon dioxide, the primary contributor of to global warming and other pollutants. Other examples include: (i) Clean or replace air filters of cars as recommended. Energy is lost when air conditioners and hot air furnaces have to work harder to draw air through dirty filters. So save money by replacing old air filters with new (standard) ones which will take less electricity. 158 (ii) Select the most energy efficient models when you replace the old appliances. Look for the energy star label because the product saves energy and prevents pollution. (iii) Turn your refrigerator down. Refrigerators accounts for about 20% of the house hold electricity costs. (iv) Buy energy-efficient compact fluorescent bulbs for the lights you use most. Although they cost more, they save money in the long run because they only use a quarter the energy used by ordinary incandescent lamps and lasting 8-12 times longer. (v) Reduce the amount of waste you produce by buying minimally packaged goods, choosing reusable products over disposable ones, and recycling. Use 30% to 50% less paper products, 33% less glass and 90% less aluminum. (vi) People who live in colder areas should super insulate your walls and ceiling. It can save your the electricity of heating or fire wood costs. (vii) Plant shady trees and paint your house a light colour if you live in a hot place or paint them a dark colour if you live in cold conditions. If we do not conserve energy, it will be exhausted and we will have nothing to use. Energy conservation is also important when in managing climate change. Currently erratic climates and climatic changes are the greatest threats that we are facing today. Hence it is important to conserve energy. 5.13 Energy efficiency Energy efficiency is the act of saving energy but keeping the same level of service. For example, if you turn off the lights when you are leaving a room, that’s energy conservation, if you replace an efficient incandescent light bulb with a more efficient compact fluorescent bulb, you are practicing energy efficiency. Energy efficiency uses advances in sciences and technology to provide services and products that require the use of less energy. Exercise 5.9 1. (a) Demonstrate how mechanical energy is conserved. (b) What is energy efficiency? 2. By identifying practical activities in our daily lives, discuss how you can conserve energy. 159 Project work 1 Energy saving charcoal burner In most developing countries, wood is the most important source of energy mainly for cooking. The amount of wood consumed depends on the climate, culture and availability. Most people use open, three stone fireplaces for cooking. The fireplaces are often dirty, dangerous and inefficient. The smoke and soot settles on utensils, walls, ceiling and people. The smoke produced in fireplaces irritates people posing danger to health. The fireplaces have been found to be about 10-15% efficient. In view of the above, energy saving stoves have been designed. Most of these stoves use charcoal. Charcoal is preferred to wood in urban areas because of its portability, convenience and cleanliness. In designing energy saving stoves, one should try to minimise energy losses to the surrounding. One of the many advantages of a charcoal stove, is that the rate of charcoal burning can be controlled. Materials Metal sheets and clay Construction Cut the metal sheet into a circular sheet (Fig. 5.27(a)). The radius AO will depend on the size of the stove required. Mark arc AB which represents the circumference of the mouth of the charcoal burner. Draw AO and OB. Draw arc CD. The radius OD will depend on the area of the base on which the charcoal is to rest. Cut the section ACDB. Assembly Fold the plate ACDB in a shape of a cone as shown in Fig. 5.27(b). Rivet the sides AC and BD together. Repeat the procedure to construct the lower compartment. But this time make AC and DB shorter. A C B D O A, B C, D (a) Circular metal sheet (b) Upper compartment Fig. 5.27: Making the upper compartment of an energy saving charcoal burner Bring the two compartment together and join them by riveting Fig. 5.28(a). Cut off a small section of the lower compartment and construct a gate. Mould clay in such a shape that it fits into the upper compartment. Make the air holes while the clay is still wet. 160 Allow the clay to dry. Construct the stands for holding the cooking pot. A complete stove should look like the one shown in Fig. 5.28(b). Base with air inlet holes Gate Clay Metal Gate (a) Upper and lower compartment joined (b) Complete charcoal burner Fig. 5.29(a): Upper and lower compartments joined to make a complete charcoal burner Larger stove can be made by cutting the sheet as shown in Fig. 5.29. A C D B or O C D A B Fig.5.29: Larger jikos Project work 2 Solar heater Solar energy can be trapped with the help of solar heater and utilized to heat water. The most common type of solar water heater incorporates a flat-plate solar collector and a storage tank. The tank is positioned above the collector. Water from the tank is circulated through the collector and back to the tank by means of convectional currents caused by the heated water. Construction of a solar heater Suggested materials A 20 litre jerry can container, plastic pipes, cellophane paper, half open 20 litre jerry can, black paint or smoke soot and a wire mesh. Assembly Heat collector Paint the plastic pipes black. Use a wire mesh and curve the plastic pipes as shown in Fig. 5.30. The size of the wire mesh should be able to fit into an open 20 litre jerry can container. 161 1 2 Pipes Strings to tie the pipes onto the wire mesh Frame work Half open jerry can Fig. 5.30: Heat collector Heat exchanger Use another 20 litre jerry can (Fig. 5.31) and open at the top to allow the pipes to enter and then seal it using the
same material and a hot object. The hot object will make the materials to fuse together. Make provisions for water to enter and leave the heat exchanger when required. 1 Hot water Water gains energy Pipes Cold water 2 Fig. 5.31: Heat exchanger Join pipe 1 of the heater collector to pipe 1 of the heat exchanger. Do the same with pipe 2. Make sure the collector is inclined at a certain angle to allow water from the heat exchanger to flow freely. (Fig. 5.32). Cover the heat collector with a cellophane paper. 162 Hot water Heat exchanger Cold water Pump Cold water Stand Heat collector Hot water rises Transparent plastic paper θ Fig 5.33: Solar heater How to use Fill the pipes of the heat collector with water and expose them to the sun. Allow water from a reservoir to fill the heat exchanger. Topic summary • Work is the product of force and distance moved in the direction of the • force. A joule is the work done when a force of one newton acts on a body and makes it to move a distance of one metre in the direction of the force. • When work is done on an object, energy is transferred. Work is said to be done if a force acts on a body and makes it move (get displaced) in the direction of the force • Energy is the ability to do work. • Moving objects have kinetic energy that depends on the mass of the body • and the velocity. Potential energy is the energy possessed by a body due to its position. It depends on the objects height above the ground. • The total amount of kinetic energy and potential energy in a system is the mechanical energy of the system. Mechanical energy = KE + PE. • Falling, swinging, and projectile motion all involve transformations between kinetic and potential energy. 163 • • According to the law of conservation of energy, energy cannot be created or destroyed but can only be converted from one form to another. Energy is converted changes from one form to another by transducers such as light bulbs, hair driers. For example, a hair drier converts electrical energy into thermal energy, kinetic energy and sound energy. • Fuel is a substance which when burnt produces heat. Topic Test 5 1. Define the term power and give its SI unit. 2. A motor raised a block of mass 72 kg through a vertical height of 2.5 m in 28 s. Calculate the: (a) work done on the block. (b) useful power supplied by the motor. 3. A person of mass 40 kg runs up a flight of 50 stairs each of height 20 cm in 5 s. Calculate: (a) the work done. (b) the average power of the person. (c) explain why the energy the person uses to climb up is greater than the calculated work done. 4. A runner of mass 65 kg runs up a steep slope rising through a vertical height of 40 m in 65 s. Find the power that his muscles must develop in order to do so. 5. A fork-lift truck raises a 400 kg box through a height of 2.3 m. The case is then moved horizontally by the truck at 3.0 m/s onto the loading platform of a lorry. (a) What minimum upward force should the truck exert on the box? (b) How much P.E. is gained by the box? (c) Calculate the K.E of the box while being moved horizontally. (d) What happens to the K.E once the truck stops? 6. A stone falls vertically through a distance of 20 m. If the mass of the stone is 3.0 kg, (a) Sketch a graph of work done by the gravity against distance. (b) Find the power of the gravitational pull. 164 7. Mugisha climps 16 m rope in 20 s. If his mass is 60 kg, find the average power he developed. 8. A car is doing work at a rate of 8.0 × 104 W. Calculate the thrust of the wheels on the ground if the car moves with a constant velocity of 30 m/s. 9. Uwimbabazi took 55.0 s to climb a staircase to a height of 14.0 m. If her mass is 40 kg, find: (a) How much force did she exert in getting to the that level? (b) Her power? 10. In Fig. 5.33 three positions of a monkey swinging from a branch of a tree are shown. A B C Fig. 5.33: A monkey swinging (a) What kind of energy does the monkey have at each position? (b) What happens to the energy when the monkey is midway between A and C? (c) In which positions does the monkey have the least energy? What name is given to this type of energy? (d) What type of energy would the moneky have if it stopped swinging but still hanging? 165 11. A device which converts one form of energy to another is called a transducer. Name one transducer in each of the cases energy transformation given below. (a) Heat to kinetic energy (b) Electrical to light (c) Sound to electrical (e) Chemical to electrical (d) Potential energy to kinetic energy 12. Discuss the energy transformations in Fig. 5.34. Fig. 5.34: A boy jumping 13. (a) State the law of conservation of energy. (b) Differentiate between renewable and non-renewable sources of energy. Give two examples of each. (c) Explain the energy transformation in a hydroelectric power station. 166 UNIT 4 Machines Topics in the unit Topic 6: Machines Learning outcomes Knowledge and Understanding • Define machines and explain the dynamics of objects Skills • Design and carry out tests on pulleys and simple pulleys of different velocity ratios may be assembled Predict what might happen. • Observing carefully. • Use appropriate measures. • Collect and present results appropriate in writing or drawing. • Derive and Calculate mechanical advantage, velocity ratio and efficiency of a given machine. • Draw a labeled diagram to explain a lever. Attitudes • Appreciate use of machine to ease work. Key inquiry questions • Why do vehicles use low gears in steep places? • Why are pulleys important in load lifting? • Why does a cyclist get tired when cycling up-hill? • How can we design machines to enable humans to move masses greater than the human mass? 167 168 TOPIC 6 Machines Unit outline • Definition of simple machines. • Examples of simple machine (lever, pulley, wedge and axle, inclined plane, screw). • Working principal of simple machines. • Machine work out and friction in the machine. • Mechanical advantage and velocity ratio of a machine. • Determination of output of simple machine (efficiency). • Experiment to determine efficiency of simple machines. Introduction In everyday life, people perform various tasks in order to improve their standards of life, environment, quality of health, and understanding of natural phenomena in order to exploit and be in terms with them. Some of the tasks people do include; drawing of water from a well using a windless, construction of houses using timber, nail and harmer, loading and unloading of good into the ships for export, joining of timber and metal using screws, splitting of firewood using a wedge, digging a garden in preparation for planting, lifting heavy objects into tracks. The devices that help us to perform work easily are called machines. Machines can either be simple or compound. In topic 3, we learnt about some of the applications of moment of a force. Most simple machines apply the same principle in making work easier. In this topic we are going to learn, understand and apply the principles behind simple machines. 6.1 Definition of simple machines Activity 6.1 (Work in groups) To find out the definition and importance of a machine Materials: closed soda bottle, a bottle opener Steps 1. Open a soda bottle with your hand. Is it easy to open the bottle using your hand? 2. Now try opening the same bottle using an opener. Is it easier to open the bottle using an opener? Which of the two tasks is easier and why? 3. Based on your observation in steps 1 and 2, define a simple machine. 169 In topic 3, we learnt about moments and how it is applied in machines. In the above experiment the bottle opener applies moments to open a soda. It is a simple machine. A machine is a mechanical device or a system of devices that is used to make work easier. For example in loading an oil drum onto a truck, it is easier to roll it up an inclined plane (Fig. 6.1(a)) than lifting it up onto the truck (Fig. 6.1(b)). (a) Rolling up a drum into a truck (b) Lifting up a drum into a truck Fig. 6.1: Machines make our work easier A machine may be defined as any mechanical device that facilitates a force applied at one point to overcome another force at a different point in the system. Examples of simple machines include lever, pulley, wedge, wheel and axle, inclined plane, and screws. A simple machine is a machine that is made up of only one type of machine. Examples of simple machines are the screw, lever, inclined plane, pulley, wheel and axel and gears. A compound machine is made up of more than one simple machines working together to perform a particular task with ease. An example of a compound machine is the car engine. The car engine consist of pulley, belts, gears, wheel and axel, pistons and other simple machines working together to bring about the movement of the car. In mechanical machines, the force that is applied is called the effort (E) and the force the machine must overcome is called the load (L). Note that both the load and effort are forces which act on the machine. 6.2 Mechanical advantage, velocity ratio and efficiency of machines Activity 6.2 (Work in groups) To investigate and determine mechanical advantage, velocity ratio and efficiency of machines Materials: Internet, reference books, inclined plane Instructions 1. 2. In pairs conduct research from the Internet or reference books on the terms used to describe the ability of doing work easily by use of machines. In your research, find out what is mechanical advantage? What is velocity ratio? What is efficiency of machine? 3. Modify the set-up using locally available materials such as stones and timbers to make an inclined plane. Draw the set-up. 170 4. Write a brief procedure on how to determine M.A, V.R, and efficiency of the inclined plane. 5. Write a report about your investigation and explain how the experiment can be improved. 6. Present your findings in a class discussion. Mechanical advantage (M.A) of machines Machines overcome large loads by applying a small effort i.e.
the machines magnify the force applied. Mechanical advantage is the ratio of load to the effort. It describes how the applied force compares with the load to be moved. A machine with a mechanical advantage (M.A) of 1 does not change the force applied on it. A machine with a M.A of 2 can double your force, so you have to apply only half the force needed. Mechanical advantage = force applied by the machine to do the work (Load) force applied to the machine by the operator (Effort) ∴ Mechanical advantage (M.A) = load (N) effort (N) Since mechanical advantage is a ratio, it has no units. Velocity ratio (V.R) of a machine Velocity ratio of a machine is the ratio of the velocity of the effort to the velocity of the load. Velocity ratio (V.R) = velocity of the effort velocity of the load = displacement of effort time displacement of load time Since the effort and the load move at the same time, ∴ Velocity ratio (V.R) = displacement of effort displacement of load or (V.R) = effort distance load distance Velocity ratio has no units Efficiency of machines For a perfect machine, the work done on the machine by the effort is equal to the work done by the machine on the load. However, there is no such a machine because some energy is wasted in overcoming friction and in moving the moveable 171 parts of the machine. Hence, more energy is put into the machine than what is output by it. Thus, Work input = Useful work done + Wasted work done To describe the actual performance of a machine we use the term efficiency. Efficiency tells us what percentage of the work put into a machine is returned as useful work. The efficiency of a machine is defined as the ratio of its energy output to its energy input. Efficiency = useful energy output energy input × 100% or efficiency = useful work output work input × 100% = load × distance moved by load effort × distance moved by effort × 100% = load effort × distance load is moved distance moved by effort × 100% = M.A × 1 V.R × 100% ∴ Efficiency = M.A V.R × 100% Example 6.1 A machine whose velocity ratio is 8 is used to lift a load of 300 N. The effort required is 60 N. (a) What is the mechanical advantage of the machine? (b) Calculate the efficiency of the machine Solution (a) Mechanical advantage = load effort = 300 N 60 N = 5 (b) Efficiency = M.A V.R = 5 8 × 100% = 62.5% 172 Example 6.2 An effort of 250 N raises a load of 900 N through 5 m in a machine. If the effort moves through 25 m, find (a) the useful work done in raising the load (b) the work done by the effort (c) the efficiency of the machine Solution (a) Useful work done in raising the load = load × distance moved by load = (900 × 5) = 4 500 J (b) Work done by the effort = effort × distance moved by effort = 250 × 25 = 6 250 J (c) Efficiency = = work output work input × 100% 4 500 J 6 250 J × 100% = 72% Example 6.3 Calculate the efficiency of a machine if 8 000 J of work is done on the machine to lift a mass of 120 kg through a vertical height of 5 m. Solution Work done in lifting the load = 1 200 × 5 = 6 000 J Work input = 8 000 J Efficiency = work output work input × 100% = 6 000 J 8 000 J = 75% × 100% 173 6.3 Types of simple machines Activity 6.3 To identify types of simple machines (Work in pairs) Steps 1. Now, access the internet and reference books and conduct research on classification of simple machines. 2. Classify simple machines 3. Discuss your findings with other groups in your class. Simple machines are classified into two groups i.e. force multiplier and distance or speed multipliers. Force multipliers are those that allow a small effort to move a large load e.g. levers. Distance or speed multipliers are those that allow a small movement of the effort to produce a large movement of the load e.g. fishing rod, bicycle gear etc. Let us consider some simple machines and show how they operate. 6.3.1 Levers Activity 6.4 (Work in groups) To demonstrate the working of levers Materials: a nail, claw hammer, piece of cloth, a pair of scissors, groundnut, pliers. Part 1 Steps 1. Drive a long iron nail into a piece of timber. 2. Try to remove the nail from the timber using your fingers. Is it easy to remove the nail using your finger? 3. Use a claw hammer instead of your fingers. Explain? 4. When using fingers and a claw hammer, which task did you apply more effort? Explain why. Part 2 Steps 1. Cut piece of cloth into two pieces using your hands. 2. Use a pair of scissors instead of your hands. Between using your hands and using a pair of scissors, which task did you apply more effort? 174 Part 3 Steps 1. Crash a groundnut using your fingers. Are you able to crash it? 2. Now crush it using a nut cracker. Why is it easier and faster to crash a groundnut using a nut cracker than your hand. Using the simple machine, the work becomes easier. These types of machines used in the above activity are called levers. Levers are simple machines that apply the principle of moments. A lever consists of a rigid bar capable of rotating about a fixed point called the pivot. The effort arm is the perpendicular distance from pivot to the line of action of effort (See Fig. 6.2). There are 3 classes of levers. The difference between these types depends on the position of the pivot (fulcrum) with respect to the load and the effort. 1. First class. The pivot is between the load and the effort. Examples (Fig. 6.2). Load Pivot Pivot Effort Effort Load (a) Crowbar (b) Scissors Load Effort Pivot Pivot Load Effort (c) Claw hammer (d) Pliers Fig. 6.2: Pivot between the load and the effort 2. Second Class: The load is between the effort and the pivot. Examples (Fig. 6.3). Load Effort Pivot Load Pivot Effort (a) Wheelbarrow (b) Bottle opener Fig. 6.3: Load between efforts and pivots 175 3. Third class: The effort is between the load and the pivot. Examples (Fig. 6.4). Effort Pivot Load (a) Fishing rod Effort Pivot Load Effort (b) Tweezers Fig. 6.4: Efforts between load and pivots Mechanical advantage of levers Consider a lever with the pivot between the load and the effort (Fig. 6.5). load arm Pivot x L effort arm y E Fig. 6.5: Mechanical advantage for levers. Taking moment about the pivot load × load arm = effort × effort arm load effort = effort arm load arm , But load effort = mechanical advantage Mechanical Advantage, M.A = effort arm load arm = y x This also applies to the other types of levers. Since effort arm is usually greater than load arms, levers have mechanical advantage greater than 1. Velocity ratio (V.R) levers Consider three types of levers in which the load and the effort have moved a distance d L and d E respectively (Fig. 6.6). x x C B dL A L y y D dE E F C x y A dL B L C y D dE F Fig. 6.6: Determination of velocity ratio of levers x D dE F E A dL B 176 Triangles ABC and DFC are similar triangles. V.R = dE dL = y x In Fig. 6.6(a) and (b), y is greater than x. The velocity ratio is therefore greater than 1. However in (c), y is less than x, and therefore the velocity ratio is less than 1. Cases (a) and (b) are examples of force multipliers. All force multipliers have M.A and V.R greater than 1. Case (c) is an example of distance multiplier in which both the velocity ratio and mechanical advantage are less than 1. Example 6.4 A lever has a velocity ratio of 4. When an effort of 150 N is applied, a force of 450 N is lifted. Find: (a) mechanical advantage (b) efficiency of the lever. Solution (a) Mechanical advantage = (b) Efficiency = M.A V.R = 3 4 = 75% Example 6.5 load effort = 450 N 150 N = 3.0 × 100% A worker uses a crow bar 2.0 m long to lift a rock weighing 650 N (Fig. 6.7). 650 N x ( 2 – x ) m 250 N Fig. 6.7: Crow bar (a) Calculate the position of the pivot in order to apply an effort of 250 N. (b) Find the: (i) velocity ratio (ii) mechanical advantage and (iii) efficiency of the lever. (c) What assumptions have you made? Solution (a) Applying the principle of moments (b) (i) velocity ratio = effort distance load distance = 1.44 0.56 = 2.57 650x = 250(2 – x) 650x = 500 – 250x 900x = 500 x = 0.56 m from the end with 650 N. 177 (ii) mechanical advantage = 650 250 = 2.6 (iii) efficiency = M.A V.R × 100% = 2.6 2.6 × 100% = 100% (c) We have assumed that there is no friction and that the crowbar has weight. Exercise 6.1 1. A machine requires 6 000 J of energy to lift a mass of 55 kg through a vertical distance of 8 m. Calculate its efficiency. 2. A machine of efficiency 65% lifts a mass of 90 kg through a vertical distance of 3 m. Find the work required to operate the machine. 3. A machine used to lift a load to the top of a building under construction has a velocity ratio of 6. Calculate its efficiency if an effort of 1 200 N is required to raise a load of 6 000 N. Find the energy wasted when a load of 600 N is lifted through a distance of 3 m. 4. Define the following terms as applied to levers: (a) mechanical advantage (b) velocity ratio 5. Find the velocity ratio of the levers shown in Fig. 6.8. 5 c m 1 Load 85 cm Fig. 6.8: Levers 6.3.2 Inclined plane Activity 6.5 (Work in groups) To determine the work done when pulling an object on a flat surface and on an inclined plane Materials: piece of wood, a spring balance, tape measure, a trolley, a cardboard, reference books, Internet. Steps 1. Attach a spring balance on the trolley. Place a piece of wood in the trolley. 2. Pull the piece of wooden from the ground vertically upwards using the spring balance. Record the force reading on the spring balance. 178 h (a) A piece of wood moved vertically upwards through (h) s (b) Moving the load along the slope through s Fig 6.9: Determination of work done 3. Using a tape measure, measure the height (h) of a table. Calculate the amount of work done when the load is lifted from the floor to the top of table (Fig. 6.9(a)). Incline a wooden plank against the edge of the table. 4. 5. Measure the force needed to pull the load (piece of wood in a trolley) up along inclined plane at a constant speed up to the top of the table (Fig. 6.9(c)). 6. Measure the
distance (s) moved by the trolley along the inclined plane. 7. Determine the work done on the trolley when it is pulled up the inclined plane. 8. Discuss in your group, which of the three ways it was easier to lift the trolley. 9. Analyse what force balanced the force applied as the block was being pulled across the table. 10. Give examples where we use an iclined plane to lift loads. An inclined plane also known as a ramp is a flat supporting surface tilted at one angle, with one end higher than the other used for raising or lowering loads. Fig. 6.10 below is an example of an inclined plane. 179 ) E E ff o r t ( d θ A Fig. 6.10: Inclined plane C h B It is easier to lift a load from A to C by rolling or moving it along the plank than lifting it upwards from B to C. Velocity ratio of an inclined plane Velocity ratio (V.R) = distance moved by effort (d) distance moved by load (h) Mechanical advantage (M.A) of an inclined plane If the inclined plane is perfectly smooth (no friction), then the work done on load is equal to the work done by effort Work on load = Work done by effort load × h = effort × d load effort = d h Hence mechanical advantage = d h d h The ratio for an inclined plane is always greater than 1, hence its mechanical advantage is greater than 1. In practice, mechanical advantage is usually less than the calculated values due to frictional force. The effect of length of an inclined plane on its mechanical advantage Activity 6.6 To investigate how the length of an inclined plane affects its mechanical advantage Materials: A trolley, inclined plane, masses Steps 1. Measure the mass of a trolley. Place it on an inclined plane of length l, (Fig. 6.11). Add slotted masses until the trolley just begins to move up the plane. 180 2. Record the values of the load, effort and the length l of the inclined plane. 3. Repeat the activity with inclined planes of different lengths. Make sure the height, h, and the load are kept constant. Pulley Wire Trolley (load) h l Friction compensated slope of length l Slotted mass Effort = mg Fig. 6.11: How the length of inclined plane affects the mechanical advantage. 4. Record your results in Table 6.1. Table 6.1: Effort, length and MA values Effort E (N) Length, l Mechanical advantage = L E 5. What happens to the applied effort when the length of the inclined plane is increased? Work done on the load = load × distance moved by the load = L × h Work done on the effort = effort × distance moved by the effort = E × l ∴ E = But the work done on the load is equal to the work done by the effort i.e. El = L h Lh l mgh l But mgh is a constant: ∴ E α 1 l Therefore a small effort travels a long distance to overcome a large load. since L = mg = . 181 6.3.3 Screws and bolts Activity 6.7 (Work in pairs) To investigate the working of screws and bolts Materials: a screw, bolt, soft wood, a screw driver Steps 1. Take a taping screw and count the number of threads it has. 2. Use a screw driver to drive the screw into a soft wood. Once it reaches the end, remove it from the wood. 3. Feel the threads with your fingers. 4. Measure the depth of the hole made by the screw. 5. Measure the total length of all the threads. 6. Compare the length of the threads with the depth of the hole. 6. Count the number of threads. 8. Determine the distance between two consecutive threads (Suggest the name given to this distance). 9. How many revolutions does the screw head makes when the threads disappears completely into the wood? 10. Repeat the above steps using a bolt and a nut (Fig 6.12). 11. Discuss your findings with other groups in a class discussion. Fig 6.12(a) shows a screw, bolt and nut. Nut Bolt Top view Pitch Screw Bolt Fig. 6.12 (a): Screws, bolts and nut The distance between the two successive threads is called pitch. When the screw is turned through one revolution by a force applied at the screw head, the lower end moves up or down through a distance equal to its pitch. The working of screws and bolts is based on the principle of an inclined plane. 182 Velocity ratio of a bolt As the bolt is turned through one revolution, the screw moves one pitch up or down. The effort turns through a circle of radius R as the load is raised or lowered through a distance equivalent to one pitch (Fig. 6.13). Velocity ratio = distance moved by effort distance moved by load = circumference of a circle, C pitch (p) = 2πR p V.R = 2πR p Effort R pitch (p) Fig. 6.13: Velocity ratio for a bolt The effect of friction on mechanical advantage, velocity ratio and efficiency From activity 6.7 you noticed that the threads felt warm after being driven into the wood. This means some of the work done was wasted as heat due to friction. The mechanical advantage of a machine depends on the frictional forces present, since part of the effort has to be used to overcome friction. However, the velocity ratio does not depend on friction but rather on the geometry of the moving parts of the machine. Consequently a reduction of mechanical advantage by friction reduces the efficiency of a machine. 6.3.4 The wheel and axle Activity 6.8 (Work in groups) To demonstrate action of wheel and axle Materials: cylindrical rod, y-shaped branches, a stone, a string. Steps 1. Construct a wheel and axle using locally available materials as shown in Fig. 6.14 below. Cylindrical rod Tree branch A B eye C Fig 6.14: A wheel and axle 183 2. Turn the cylindrical rod at A to raise the stone. Is it easy to raise the stone by turning the road at this point? 3. Repeat turning the cylindrical rod but this time by turning at C. What do you realise when raising the stone at this point of turn as compared to the previous point? Explain. 4. Compare the energy needed to turn the cylindrical rod in the two cases. 5. Which feature of the set-up represent the wheel and axle? 6. Observe the setup from B and draw the wheel, axle, load and effort. 7. Using various loads, find the force which in each case will just raise the load. Record your results in tabular form as shown in Table 6.2 below. Load Effort M.A Table 6.2: Values of load, efforts and N.A 8. Draw a graph of M.A against load. Fig 6.15 shows simplified examples of wheel and axle. r • R effort load (b) wheel axle (a) load effort effort load (c) Fig 6.15: Simple wheel and axle 184 The wheel has a large diameter while the axle has a small diameter. The wheel and axle are firmly joined together and turn together on same axis. The effort is applied to the handle in the wheel. When the effort is applied, the axle turns, winding the load rope on the axle and consequently raising the load. Velocity ratio = distance moved by effort distance moved by load = 2π × radius of wheel(R) 2π × radius of axle (r) M.A may be obtained by taking moment Load × radius of axle = effort × radius of wheel M.A = Load Effort = radius of wheel radius of axle = R r Exercise 6.2 1. Give an example of a lever with a mechanical advantage less than 1. What is the real advantage of using such a machine? 2. Describe an experiment to determine the velocity ratio of a lever whose pivot is between the load and the effort. 3. An effort of 50 N is applied to drive a screw whose handle moves through a circle of radius 14 cm. The pitch of the screw thread is 2 mm. Calculate the: (a) velocity ratio of the screw. (b) load raised if the efficiency is 30%. 6.3.5 Pulleys Activity 6.9 (Work in groups) To demonstrate the action of a pulley Materials: Reference books, flag, a flag post Steps 1. Raise a flag up the flag post. 2. Explain how the flag post raises the flag. 3. We use a pulley to raise water from a well. Does it work the same way as the flag post? 4. Compare and discuss your findings with other groups in your class. 5. Let one of the group members present your findings to the whole class. 185 A pulley is usually a grooved wheel or rim. Pulleys are used to change the direction of a force and make work easy. There are three types of pulleys i.e. single fixed pulley, single movable pulley and block and tackle. (a) Single fixed pulley Fig. 6.16 shows a single fixed pulley being used to lift a load. This type of pulley has a fixed support which does not move with either the load or the effort. The tension in the rope is the same throughout. Therefore the load is equal to the effort if there is no loss of energy. The mechanical advantage is therefore 1. The only advantage we get using such a machine is convenience and ease in raising the load. Bucket Load (water) Effort Fig. 6.16: Single fixed pulley Since some energy is wasted due to friction and in lifting the weight of the rope, the mechanical advantage is slightly less than 1. The load moves the same distance as the effort and therefore the velocity ratio of a single fixed pulley is 1. Examples of real life applications of a single fixed pulley are as shown in Fig. 6.17. (b) Raising bricks (a) Raising a flag (c) Raising water from a well Fig. 6.17: Examples of single fixed pulley 186 My health Ensure you have covered the well/borehole in our homes after use. Its water may be polluted or even cause death due to accidents. The single movable pulley Fig.6.18 shows a single movable pulley. A movable pulley is a pulley-wheel which hangs in a loop of a rope. A simple movable pulley may be used alone or combined with a single fixed pulley. The total force supporting the load is given by the tension, T, plus effort, E, but since the pulley is moving up, the tension is equal to the effort. Therefore, the upwards force is equal to twice the effort (2E). Hence the load is equal to twice the effort (2E). Mechanical advantage = load effort = 2E E = 2 However, since we also have to lift the pulley, the mechanical advantage will be slightly less than 2. Experiments show that the effort moves twice the distance moved by the load. Therefore, velocity ratio = Distance moved by effort Distance moved by load = 2 Tension T Effort E Load L Fig. 6. 18: A single movable pulley A block and tackle A block and tackle consists
of two pulley sets. One set is fixed and the other is allowed to move. The pulleys are usually assembled side by side in a block or frame on the same axle as shown in Fig. 6.19 (a). The pulleys and the ropes are called the tackle. To be able to see clearly how the ropes are wound, the pulleys are usually drawn below each other as shown in Fig. 6.19 (b). E Block and tackle side by side E Upper fixed pulley block Lower moving pulley block (a) Pulley put side by side (b) Pulley drawn below each other Fig. 6.19: Block and tackle systems. 187 Velocity ratio of a block and tackle Activity 6.10 (Work in groups) To determine velocity ratio of a block and tackle Materials: A block and tackle pulley system, a load, a metre rule Steps 1. Set up a block and tackle system with two pulleys in the lower block and two pulleys in the upper block as shown in Fig. 6.19 (b). 2. Count the number of sections of string supporting the lower block. Raise the load by any given length, l, by pulling the effort downwards. Measure the distance, e, moved by the effort. Record the result in a table. (Table 6.3). 3. Repeat the activity by increasing the distance moved by the effort. How does change of length affect the effort? 4. Plot a graph of e, against, l (Fig. 6.20). Determine the gradient of the graph. Table 6.3: Distance by effort and distane by land Distance moved by effort (e) in cm Distance moved by load l cm 10 20 30 40 5. Compare the value of the gradient obtained with the number of sections of supporting strings. What do you notice? Explain. From Activity 6.10, you should have observed that the distance moved by the effort is distance moved by the load. 1 4 The graph of effort against the load is as shown in Fig 6.20 below. e ∆e ∆l l (cm) Fig. 6.20: Graph of the effort against the load. 188 The gradient ∆e which is the velocity ratio is found to be 4. When the value of ∆l the gradient is compared with the number of sections of string supporting the lower block, we note that they are the same i.e also 4. Tip: The velocity ratio of a pulley system is equal to the number of strings sections supporting the load. Precaution: The weight of the block in the lower section of the system has to be considered as this increases the load to be lifted. Mechanical advantage of a block and tackle Activity 6.11 (Work in groups) To determine the mechanical advantage of a block and tackle Materials: A block and tackle pulley, a load Steps 1. Assemble the apparatus as in Fig. 6.19 shown in Activity 6.10 and connect a spring balance on the effort string. For a given load, pull the string on the effort string until the load just begins to rise steadily. 2. Repeat the activity with other values of load. 3. Record the values of the effort in a table (Table 6.4). Table 6.4: Values of load (L), effort (E) and L E L E L E 4. For each set of load and effort, calculate the mechanical advantage. Plot a graph of mechanical advantage against the load (Fig. 6.21). Describe the shape of the graph. 189 Fig 6.21 shows a graph of mechanical advantage against the load. M.A Fig. 6.21: Graph of mechanical advantage against the load Load (N) As the load increases, the mechanical advantage also increases. When the load is less than the weight of the lower pulley block, most of the effort is used to overcome the frictional forces at the axle and the weight of the lower pulley block. That is, the effort does useless work. However, when the load is larger than the weight of the lower block, the effort is used to lift the load. This shows that the machine is more efficient when lifting a load that is greater than the weight of the lower block. Using the value of the velocity ratio obtained in Activity 6.11, calculate the efficiency of the pulley system. Plot a graph of efficiency against load (Fig. 6.22). Efficiency (%) Fig. 6.22: The graph of efficiency against load Load (N) The efficiency of the system improves with larger loads. Example 6.6 For each of the pulley systems shown in Fig. 6.23, calculate: (i) velocity ratio (ii) mechanical advantage (iii) efficiency 190 Solution (a) (i) velocity ratio = 2 (number of sections of string supporting the lower pulley) (ii) mechanical advantage 200 N 150 N = = 4 3 = 1.33 60 N 150 N (iii) efficiency = 4 3 × 1 2 × 100% 200 N (a) 210 N Fig. 6.23: Pulley system (b) = 66.7% (b) (i) velocity ratio = 5 (ii) mechanical advantage (iii) efficiency = load effort = 210 N 60 N = 3.5 = 3.5 5 × 100% = 70% Example 6.7 Draw a diagram of a single string block and tackle system with a velocity ratio of 6. Calculate its efficiency if an effort of 1 500 N is required to raise a load of 5 000 N. Solution See Fig. 6.24 Effort 1 500 N velocity ratio = 6 mechanical advantage = 5 000 N 1 500 N = 10 3 efficiency = M.A V.R × 100 = 10 3 × 1 6 × 100 Load 5 000 N = 55.6% Fig. 6.24: Block Tackle pulley system 191 Example 6.8 A block and tackle pulley system has a velocity ratio of 4. If its efficiency is 65%. Find the (a) mechanical advantage. (b) load that can be lifted with an effort of 500 N. (c) work done if the load is lifted through a vertical distance of 4.0 m. (d) average rate of working if the work is done in 2 minutes. Solution (a) efficiency = 65 = M.A V.R M.A 4 × 100% (b) MA = × 100% 2.6 = load effort load 500 mechanical advantage = 2.6 Load = 1 300 N (c) work = force × distance in the (d) Rate of doing work = Power direction of force = 1 300 × 4 = 5 200 J Power = work done time = 5 200 120 = 43.3 W Exercise 6.3 1. (a) Draw a system of pulleys with two pulleys in the lower and upper block. (b) Describe how you would find experimentally its mechanical advantage. 2. Fig. 6.25 shows a pulley system. Find; (a) (b) (c) (d) the velocity ratio of the pulley system. the mechanical advantage, if the system is 80% efficient. the effort. the work done by the effort in lifting the load through a distance of 0.6 m. (e) how much energy is wasted? 192 Effort 180 N Load Fig. 6.25: Pulleys system 3. A pulley system has a velocity ratio of 3. Calculate the effort required to lift a load of 600 N, if the system is 65% efficient. 4. A pulley system has a velocity ratio of 4. In this system, an effort of 68 N would just raise a load of 216 N. Find the efficiency of this system. Topic summary • A machine is a device that makes work easier. • Mechanical advantage (M.A) is the ratio of load to effort. • Mechanical advantage = load effort . The mechanical advantage of a machine depends on loss of energy of the moving parts of a machine. Mechanical advantage has no units. • Velocity ratio (V.R) is the ratio of distance the effort moves to that moved by the load. Velocity ratio = Distance moved by the load . Distance moved by the effort Velocity ratio is a ratio of similar quantities hence it has no units. • Theoretical value of velocity ratio may be obtained from the dimensions of the machine e.g. in pulleys–number of the sections of string supporting the load. Table 6.5: Expressions for velocity ratio of various machinery Machine Inclined plane Screw VR 1 sin θ 2πr pitch, P, Wheel and axle Radius of wheel, R Radius of axle, r = R r • Efficiency = work output work input × 100% = mechanical advantage velocity ratio × 100% 193 Topic Test 6 1. Define the following terms: (a) Power of a machine (c) Mechanical advantage (M.A) (b) Efficiency (d) Velocity ratio (V.R) 2. A farmer draws water from a well using the machine shown in Fig. 6.26 below. The weight of the bucket and water is 150 N. The force, F exerted by the farmer is 160 N. The bucket and its content is raised through a height of 15 m. (a) What is the name given to such a machine? (b) Why is the force, F, larger than the weight of the bucket and water? (c) What distance does the farmer pull the rope? (d) How much work is done on the bucket and water? Effort 150N (e) What kind of energy is gained by the Fig. 6.26: A simple pulley system bucket? (f) How much work is done by the farmer? (g) Where does the energy used by the farmer come from? (h) Show with a flow diagram the energy conversion in lifting the water from the well. 3. A factory worker lifts up a bag of cement of mass 50 kg, carries it horizontally then up a ramp of length 6.0 m onto a pick-up and finally drops the bag of cement on the pick-up (Fig. 6.27). Fig. 6.27: Worker lifting cement on the pick-up (a) Explain the energy changes in the various stages of the movement of the worker. (b) During which stages is the worker doing work on the bag of cement. 194 (c) If the worker has a mass of 60 kg and the ramp is 1.5 m high, find the (i) velocity ratio. (ii) efficiency of the inclined plane if the mechanical advantage is 3. 4. Fig. 6.28 shows the cross-section of a wheel and axle of radius 6.5 cm and 1.5 cm respectively used to lift a load. Calculate the efficiency of the machine. Effort 50 N Load 150 N Fig. 6.28: Wheel and axle 5. A student wanted to put 10 boxes of salt at the top of the platform using an inclined plane (Fig. 6.29). plat form If the resistance due to friction is 10 N, calculate (a) the work done in moving the box 10 boxes. (b) the efficiency of this arrangement. A W = 40.0 N 6 . 5 m B 3.0 m C ground (c) the effort required to raise one box to the platform. Fig. 6.29: A crane 6. A crane just lifts 9 940 N when an effort of 116 N is applied. The efficiency of the crane is 65%. Find its: (a) mechanical advantage (b) velocity ratio 7. Fig. 6.30 shows a pulley system. An effort of 113 N is required to lift a load of 180 N. (a) What distance does the effort move when the load moves 1 m? (b) Find the work done by the effort. (c) Find the work done on the load. (d) Calculate the efficiency of the system. 195 113 N 180 N Fig 6.30 A pulley system 8. The Fig. 6.31 shows a single fixed pulley. Calculate its: (a) V.R (b) Efficiency 9. In the system shown in Fig. 6.32, the winding machine exerts a force of 2.0 × 104 N in order to lift a load of 3.2 × 104 N. (a) What is the velocity ratio? (b) Calculate the M.A. (c) Find the efficiency. 15 000 N 20 000 N Fig 6.
31: Single fixed pulley 2.0 x 104 N Winding machine 3.2 × 104 N Fig. 6.32: A winding crane 10. Fig. 6.33 shows a pulley system. (a) What is the velocity ratio of the system? (b) Calculate the efficiency of the system. (c) Show the direction of the force on the string. 11. A block and tackle pulley system has five pulleys. It is used to raise a load through a height of 20 m with an effort of 100 N. It is 80% efficient. Effort 150 N Load 400 N Fig. 6.33: A block and (a) Is the end of the string attached to the upper or lower block of pulleys if the upper block has three pulleys? Show it in a diagram. (b) State the velocity ratio of the system. (c) Calculate the load raised. (d) Find the work done by the effort. (e) Find the energy wasted. tackle pulley 12. A man pulls a hand cart with a force of 1 000 N through a distance of 100 m in 100 s. Determine the power developed. 196 UNIT 5 The Properties of Waves Topics in the unit Topic 7: Introduction of waves Topic 8: Sound waves Learning outcomes Knowledge and Understanding • Understand and explain the motion, types and properties of waves Skills • Design tests to investigate waves using strings and ripple tanks. • Observe carefully • Predict cause and effect • Use appropriate measures • Collect and present results including representing waves in displacement- position and displacement-time graphs Interpret results accurately • • Report findings appropriately Attitudes • Appreciate the wave motion and that there are certain features common to all waves. • Appreciate use of ultrasound in medical diagnosis and radar. Key inquiry questions • What constitute a wave? • What parameters characterize a wave? • What evidence is there that wave exists? • How can we apply our knowledge of waves? 197 TOPIC 7 Introduction to Waves Unit outline • Oscillations • Characteristics of oscillations • Factors affecting oscillations • Types of waves • Characteristics of wave motion Introduction When a stone is dropped in a pool of still water, ripples spread out in a circular form. This constitutes what is called water waves. There are many different types of waves that we make use of such as light and sound waves, microwaves, infrared and radio waves used to transmit radio and television signals. In this unit, we shall study the production of waves and some common terms and properties used in describing wave motion. By understanding more about waves, more uses are made of them. The study of waves begin with the concept of oscillations. 7.1 Oscillations Movements form a major part in our lives. Movements can be regular or irregular. Some movements follow a fixed path and keep repeating. These kinds of movements are important in our lives as shown in the following examples: • A pendulum clock repeats movement to keep time. • Wheels of bicycles and vehicles keep repeating their movements round in a circular path and this makes them to move faster and easily to other places. • Heartbeats are also rhythmic movements that help us remain alive. • Swings in children’s playgrounds. All these and many others are repetitive to-and-fro movements called oscillations. Therefore, oscillations are repeated, regular movements that happen at a constant rate. 198 7.1.1 Characteristics of an oscillation • The displacement, d, of a vibrating body is the distance of that body from the mean/fixed position. • The amplitude, a, of a vibration is the maximum displacement from the fixed/ mean position in either direction. • Periodic time, T, is the time taken to complete one oscillation or cycle. • The frequency, f, is the number of complete oscillations (or cycles) made in one second.SI unit for frequency is the hertz, Hz. One hertz is defined as one oscillation per second or one cycle per second. Consider the following cases of oscillations: (a) A simple pendulum One oscillation is the movement Amplitude is the distance BA or BC. B Simpole Pendulum Fig. 7.1: A simple pendulum B Simpole Pendulum (b) A vibrating spring A Vibrationg String B A C t1 a t2 C Vibrating springs Vibrationg String B A One oscillation is the movement A C A t1 a A t2 a T B C t3 B a A A The amplitude is the distance BA or BC. C t4 Time B Displacementtime graph Fig. 7.2: A vibrating spring T !"#$%&'"()""+ Simpole Pendulum (c) A clamped metre rule a t3 t4 Time Displacementtime graph B A C C B A One oscillation is the movement A. C B B A Vibrationg String The amplitude is the distance BA or BC. B !"#$%&'"()""+ Fig. 7.3: A clamped rule A T a a 199 t3 Time t4 t1 a t2 Displacementtime graph !"#$%&'"()"*"+ .1.2 Factors affecting oscillations Activity 7.1 (Work in groups) Materials: Bob, stand, string To investigate how length affects the rate of oscillation Steps 1. Set up apparatus as shown in Fig. 7.4 showing a simple pendulum consisting of a bob attached at the end of a light cord. The other end of the string is clamped rigidly in position. 2. Displace the bob slightly to one side then displace slightly and release the bob Fig. 7.4: Simple pendulum 3. release it and observe what happens. Increase the length of the cord the change in vibration time. Why do you think there is change in vibration time? 4. Repeat this several times. We observe that the longer the cord of the pendulum, the slower it oscillates. Activity 7.2 (Work in groups) Materials: Stand, mass, spring and clamp To investigate how mass affects the rate of oscillation Steps 1. Attach a mass to one end of a spiral spring whose other end is rigidly clamped in position Fig. 7.5. Pull the mass slightly downwards then release it and observe what happens. 2. Repeat this activity three times each time using a bigger mass. Fig. 7.5: Hanged spiral spring 3. How does the change in mass affect the rate of oscillations? Explain. The bigger the mass, the slower the vibration of the spring. Activity 7.3 To investigate how frequency affects the rate of oscillation (Work in groups) Materials: G-clamp, mass, Steps 1. Fix a mass at the end of a metre rule and clamp the other end as shown in Fig. 7.6. 200 2. Displace the free end of the rule then release and observe what happens. Repeat this activity by attaching another mass two more times. -3. Repeat this activity using half of the metre rule. 4. How does frequency affect the rate of Oscillation? Explain. G clamp Fig. 7.6: Oscillations of a loaded metre rule. From the Activities (a) the bigger the mass, the slower the ruler swings. (b) the longer the ruler, the slower it swings. All the above activities indicate that the frequency of a vibrating system is affected by: (a) Length – the longer or larger the body the lower the frequency e.g a shortened guitar wire produces higher pitch. (b) Mass – the bigger the mass /thickness the lower the frequency (longer periodic time). This is usually the case in guitar wires, where thinner ones give higher pitch. Note that a pendulum is not affected by mass of the bob attached. (c) It can be shown that increase in tension increases the frequency /pitch of a vibrating body for example a string /wire. 7.2 The concept of a wave There are many cases in real life where energy produced at one place is consumed at a different place. In such cases, the energy need to be transferred from the place of production to the place of consumption. This can take place in a number of different ways including: •. Physically moving the matter carrying the energy and delivering it to the place where the energy is to be consumed. • Vibration of the particles of a medium leading to transfer energy from one particle to next. In this topic, will learn more about this mode of energy transfer. If a stone is thrown into a still swimming pool or pond, circular water ripples are seen moving from the point where the stone hit the water outwards to the banks. This implies that the energy from the stone is transferred from the hitting point to other regions through the ripples. Such ripples are examples of waves. 201 What is a wave? A wave can be defined as “a periodic disturbance (movement) that transfers energy from one point to another with no net movement of the medium particles.” Examples of waves include sound waves, water waves, light waves, radio waves, X-rays, gamma rays, seismic waves and microwaves. 7.2.1 Formation of waves and pulses Formation of wave motion As learnt earlier, waves transfer energy but not matter. This energy is transferred through pulses and waves. A pulse is a sudden short-lived disturbance in matter. Wave or wave train is a continuous disturbance of the medium which arises due to regular pulses being produced. The following experiments demonstrate wave motions. Formation of pulses A pulse is a single wave disturbance that moves through a medium from one point to the next point. Let us now demonstrate the formation of pulse in Activity 7.4. Activity 7.4 (Work in groups) To demonstrate the formation of pulses using a rope Materials: Rope, pins, nails, helical springs and table Steps 1. Fix one end of a rope to a wall. Hold the free end of the rope so that the rope is fully stretched. 2. Quickly move your hand upwards and then return to the original position as shown in Fig. 7.7(a). Observe what happens to the rope. 3. Now move your hand suddenly downwards and return to the original position as in Fig. 7.7(b). Observe what happens to the rope. (a) (b) Fig. 7.7: Production of a pulse using a rope 4. Tie one end of the rope to the fixed pole as shown in Figure 7.8. 202 wave motionwave motionwave pulsestringwave pulse fixed end Fig. 7.8: A rope fixed at one end 5. Hold the free end of the rope and shake it in an up and down motion. Observe how the rope behaves and explain the motion. 6. Place the helical spring to lie on a table and hold it firmly to the table on one end. 7. Gently pull the free end then push it repeatedly while keenly observing what happens. Explain your observation. What do you think can behave the same way as the spring when compressed? In Activity 7.4, we notice that pulses that move from one
end of the rope to the other are produced. If the disturbance is continuous waves or wave trains are formed. When pulses are produced regularly and give rise to a continuous wave motion. Waves or a wave train is a continuous disturbance of the medium which arises due to the regular pulses being produced. In Activity 7.4, when the hand (source) is moved continuously up and down or forward and backward, the particles of the rope or spring (medium) also move up and down or forward and backward. When the source is moved at regular intervals, the disturbance is also produced at regular intervals (Fig. 7.9). rope wave motion (a) wave motion (b) slinky spring fixed end fixed end Fig. 7.9: Production of continuous pulses in a string and a slinky 203 Continuous disturbance of a medium at a point produce continuous waves or wave trains. The waves or wave trains produced are of two types: transverse waves and longitudinal waves. Types of waves There are two types of waves namely; Mechanical waves and Electromagnetic waves. Electromagnetic waves These are waves that do not require a medium to travel from one point to another. They can travel through empty space (vacuum). Examples of electromagnetic waves are X-rays, gamma rays, visible light etc. They are produced by electric and magnetic fields. Mechanical waves These are waves which require a medium to travel from one point to another. They are produced by vibrating objects. They are transmitted by the vibration of the medium particles. Such waves can be seen or felt. Example of mechanical waves include waves on a rope or spring, water waves, sound waves in air, waves on a spring, seismic wave etc. A mechanical wave can be a progressive or stationary wave. A progressive (travelling) wave is a disturbance which carries energy from one place to another without transferring matter. There are two type of progressive mechanical waves:Transverse and Longitudinal waves. (a) Transverse waves Transverse waves are mechanical waves in which the particles of the medium move in a direction perpendicular to the direction of travel of the wave. Therefore, in a transverse wave, the direction of disturbance is at right angles to the direction of travel of the wave. From Activity 7.4, we notice that when the rope is shook up and down, it is seen to make rises and falls which move through the fixed end (Fig. 7.10). 204 Fig. 7.10: A rope in motion The rope particles are displaced up and down as they move towards the fixed end. These up and down disturbance are perpendicular to the direction of motion of the wave. The rises are known as crests while the falls are known as troughs (Fig. 7.11). Crest trough rope wave motion Fig. 7.11: Transersal waves fixed end (b) Longitudinal waves Longitudinal waves are mechanical waves in which particles of the medium move in direction parallel to the direction of the wave motion. The particles of the transmitting medium vibrates to and fro along the same line as that in which the wave is travelling. From Activity 7.4, we notice that when the spring is compressed gently, the coils are observed to move towards the fixed end. In some regions, the coils are close together while in other regions the coils are far apart as shown in Fig. 7.12. The region where the coil are close together are known as a compressions while the regions where they are far apart are known as rarefactions. (See Fig 7.12 (a) and (b)) (a) Rarefactions compressions 205 wave motionfixed endslinky spring (b) Key: C - Compressions R - Rarefaction Fig. 7.12: Longitudinal wave Thus, a longitudinal wave consists of compressions and rarefactions. Compressions is a region on a longitudinal wave with a high concentration of vibrating particles. On the other hand rarefaction is a region of the longitudinal wave with low concentration of vibrating particles. Example of longitudinal wave is sound waves. Figure 7.13 shows a longitudinal plane waves. Loud speaker Compression Rarefaction Compression Rarefaction Compression Wave λ λ Fig. 7.13: shows a longitudinal plane waves Differences between Transverse and Longitudinal waves Table 7.1: Difference between transverse and longitudinal waves Transverse waves Particles of the medium are displaced perpendicular to the direction of motion of the wave. Form crests and troughs. Example include: Electromagnetic waves, water waves, waves made by a rope when its moved up and down. Longitudinal waves Particles of the medium are displaced parallel to the direction of motion of the wave. Form compressions and rarefactions. Example include: sound waves, waves made by a spring when pushed. Exercise 7.1 1. What is an oscillation? 2. Distinguish between a pulse and wave train. 3. What factors affect the frequency of an oscillating: (b) mass – spring system (a) pendulum 206 RCCRCλλFixed pointwavelengthwavelength 4. Define the term ‘wave’. 5. Differentiate between transverse and longitudinal waves giving an example for each. 6. Name the type of wave found in the following activities: (a) Children playing rope jumping. (b) A spring being displaced forward and backward. (c) Waves due to dropping a stone into water on a basin. (d) A car moving on a bump. 7. Distinguish between compression and rarefaction 8. Briefly explain how a pulse in formed. 9. Name two factors that affect oscillations of an object. 7.3 Characteristics of wave motion Wavelength of transverse waves Consider a long rope with one of its ends rigidly tied to a peg while the other end is free. Produce a pulse by moving the hand upwards and notice the distance travelled by the disturbance. If the hand is moved up and down once through a complete cycle, the time taken by the hand is the periodic time (T). Fig. 7.14 shows a graph of displacement of particles against time. We see that the particles of the rope just vibrate up and down about their mean or rest position, but do not move with the wave. The disturbance is transferred from particle to particle. The distance travelled by the disturbance (wave energy) during each periodic time T is called the wavelength, λ, of the wave 10 11 12 13 7 T 2T 3T λ λ λ Fig. 7.14: Wavelength of a transverse wave. 15 particles time distance From the graph (Fig. 7.14), particles 2, 6, 10, 7, etc. are at similar positions and, move in the same direction. Such positions are called the crests of a wave. Similarly, particles 4, 8, 12 etc, are at similar positions and are moving in the same direction. Such positions are called the troughs of a wave. 207 Particles that are at similar positions and are moving in the same direction are said to be in phase. A crest is the position of maximum positive displacement, and a trough is the position of maximum negative displacement as shown in Fig. 7.15. The distance between two successive particles in phase such as two successive crests or troughs is equal to the wavelength of the wave. λ crest crest crest crest time + trough trough trough λ Fig. 7.15: Crest and troughs in a transverse wave Wavelength of a longitudinal wave Fig. 7.16 shows the energy propagation in a slinky spring Fig. 7.16: Compressions and rarefactions in a longitudinal wave. Just like the production of crests and troughs in a transverse wave, we have the regions of compressions (C) and rarefactions (R) in a longitudinal wave. A compression is a region where the particles of the medium are closely packed. In this region, the pressure of the particles of the medium is high, hence the density is high. A rarefaction is the region where the particles of the medium are spread out. In this region the pressure of the particles of the medium is low, hence the density is low. The wavelength of a longitudinal wave can be described as the distance between two successive compressions or rarefactions. Fig. 7.17 is a displacement -time graph for a wave. We will use it to describe other characteristics of waves. 208 Fig. 7.17: Displacement – Time graph Periodic time, T The time taken for one vibration /oscillation. It is also the time taken to cover a distance of one wave length. Thus, the value of T in Fig. 7.17 is the periodic time. By definition, periodic time is the duration for one complete oscillation. Amplitude, (A) As a body or particles vibrate, they change position from the mean rest position. The position of a point from the resting position at any given time is called its displacement. The maximum value of displacement is called amplitude (A) as shown on the Fig. 7.17. Frequency, (f) Frequency (f) is the number of cycles made per unit time. We can write this mathematically as, number of vibrations (n) Frequency (f) = –––––––––––––––––––– time taken (t) n In symbols, f = – t If n = 1 (i.e 1 oscillation), then t = T (periodic time) 1 1 Hence f, = – and T = – T f For example, if a newborn baby’s heart beats at a frequency of 120 times a minute, its frequency is f = ––– = 2 Hz and T = – = – = 0.5 s 120 I I 60 f 2 209 A C B B A C Vibrationg String B Simpole Pendulum Speed of wave This is the distance covered by a wave per unit time. It is measured in metres per second, (m/s). The speed of wave is given by: T A Wave speed = a distance travelled by a wavetrain time taken a Phase of a wave – is the fraction of wave cycle which has elapsed relative to the origin. Time t4 t3 t2 t1 The wave equation T Consider a source that produces n waves of wavelength (λ) in period time (T) seconds (Fig. 7.18) Displacementtime graph nλ = d !"#$%&'"()"*"+ Fig. 7.18: Displacement – distance graph The distance travelled by a wave train in one period time is the wavelength of a wave. Wave speed, v = distance travelled time taken Thus, the velocity of the wave is given by: nλ T .........................................(i) velocity (v) = but n T = f Substituting for T in (i), we get v = Thus, v = fλ = λf λ 1 f The speed of a wave is given by: Frequency × wavelength The equation v = fλ is called the wave equation. This formula holds for all waves. 210 Example 7.1 A slinky is made to vibrate in a transve
rse mode with a frequency of 4 Hz. If the distance between successive crests of the wave train is 0.7 m calculate the speed of the waves along the slinky. Solution λ = 0.7m, ƒ = 4Hz Wave speed = frequency × wavelength = 4Hz × 0.7m = 2.8 m/s Example 7.2 Calculate the frequency of a wave if its speed is 30 cm/s and the wavelength is 6 cm. Solution Wave speed = frequency × wavelength v = ƒ × λ ƒ = – = –– = 5 Hz v 30 λ 6 Example 7.3 A source of frequency 256 Hz is set into vibrations. Calculate the wavelength of the waves produced, if the speed of sound is 332 m/s in air. Solution v = ƒ × λ v 332 ƒ 256 λ = – = ––– = 1.30 m. Example 7.4 The speed of a certain wave in air is 3 × 108 m/s. The wavelength of that wave is 5 × 10–7m. Calculate the frequency of that wave. Solution v = f λ ƒ = – = –––––– = 0.6 × 1015 Hz = 6.0 × 1014 Hz v 3 × 108 λ 5 × 10–7 211 Example 7.5 Fig. 7.19 shows a wave produced in a string. B A D C 4.0 cm Fig. 7.19 (i) Calculate the wavelength of the wave. (ii) If ten complete waves are produced in a duration of 0.25 seconds, calculate the speed of the waves. Solution (i) Wavelength (λ) = length of a number of waves number of waves = 4 cm 2 = 2 cm (ii) f = 10 0.25 = 40 Hz v = f λ = 0.02 × 40 = 0.8 m/s Example 7.6 Fig. 7.20 shows the displacement–time graph of a wave travelling at 200 cm/s.2 0.2 0.2 0.4 0.6 0.8 time (s) Fig. 7.20: Displacement – time graph Determine the: (a) amplitude (b) Period (c) frequency (d) wavelength 212 Solutions (a) 0.3 cm (c) f = 1 T = 2.5 Hz (b) T = 0.4 s f = 2.00 (d) λ = 2.5 v = 0.8 m Example 7.7 A spring vibrates at the rate of 20 cycles every 5 seconds (a) Calculate the frequency of the waves produced. (b) If the wavelength of the waves is 0.01 m, find the speed of the waves. Solutions (a) 20 cycles = 5 seconds 4 cycles = 1 second ∴ f = 4 Hz (b) v = f λ = 4 × 0.01 = 0.04 m/s Exercise 7.2 1. Draw a wave and mark on it the wavelength and amplitude. 2. Explain the phrase ‘a wave has a frequency of 5 Hz’. 3. A flag is fixed in an ocean. If two waves pass the flag every second, what is (a) its frequency? (b) the period of the water waves? 4. Derive the wave equation. 5. A sound wave has a frequency of 170 Hz and a wavelength of 2 m. Calculate the velocity of this wave. 6. The range of frequencies used in telecommunication varies from 1.0 × 106 to 2.0 × 107 Hz. Determine the shortest wavelength in this range. (The Speed of the wave is 3 × 103 m/s). 7. The speed of sound in air is 320 m/s. Calculate the frequency of sound when the wavelength of sound is 60 cm. 8. Define the term ‘wave’. 9. Distinguish between: (a) Mechanical wave and electromagnetic wave (b) Transverse wave and longitudinal wave. 213 (c) Compression and rarefaction. 10. The figure below shows a displacement-time graph for a certain wave ) m ( 0.25 0 Time (s0.25 2.5 × 10–4 7.5 × 10–4 Fig. 7.21: Displacement - time graph (a) Identify the type of wave. (b) State the period of the wave (c) Determine the frequency of the wave. (d) If the wave has a wavelength of 3.5 cm, what is its velocity? 11. A wave source generates 300 waves signals in a second. Each of the wave signals has a wavelength of 4.5 cm. (a) Determine the: (i) Frequency of the wave. (ii) Period of the wave. (iii) velocity of the wave. (b) Determine the time taken by the generated waves to hit a barrier that is 250 m away from the wave. 12. Using specific properties of light, explain why it is a transverse wave. 13. Define the following terms and state its S.I units: (a) Amplitude (b) Period (c) Wavelength (d) Frequency 14. Clouds FM broadcasts on a frequency of 88.5 kHz producing signals of wavelength 3389.83 m. Determine: (a) The period of its signals 214 (b) The velocity of radiowaves (c) The velocity of radio free East Africa if its signals have a wavelength of 3405.22 m. 15. (a) Give the meaning of the symbols in the equation v = f λ. (b) Calculate the wavelength of a wave if the speed is 45 m/s and the frequency is 5 Hz. 16. Radio wave travel with a speed of 3 × 108 m/s in air. If a radio station broadcasts at a wavelength 125 m, calculate the frequency of the transmitted waves. Topic summary • A wave is a periodic disturbance that transfers energy in space from one point to another in a medium. • When a rope fixed at one end is shaken up and down two waves trains are produced: transversal and longitudinal. • In longitudinal waves, motion of the medium particles are displaced in the same direction as that of the travel of wave. In the transversal waves motion of the medium particles are displaced perpendicular to the direction of the travel of the wave. • Wavelength is the distance between successive crests or troughs of a wave. • Amplitude is the displacement of a particle from its mean or rest position. • Period is the time taken for a wave to make one cycle. • Frequency is the number of waves passing at a given point per second. • Compression is a region in a longitudinal wave with high concentration of vibrating particles. • Rarefaction is a region in a longitudinal wave with low concentration of vibration particles. • A pulse is a single disturbance that moves through a medium from one point to the next point. • A ripple tank is an apparatus used to demonstrate the various properties of waves like reflection, refraction, diffraction and interference. • A wavefront is an imaginary line which joins a set of particles which are in phase in a wave motion. 215 • A ray is a line draw to show the direction of travel of wave energy and is perpendicular to the wavefront. • Water and sound waves like light waves, obey the laws of reflection. Topic Test 7 1. Two waves that are in phase, they form a type of interference called _____. A. Constructive C. Coherent B. Destructive D. Out of phases 2. When a plane waves are reflected, the reflected waves take the shape of the reflecting surface. Draw the reflected waves emerging from a: (a) Convex reflector (b) Concave reflector 3. Copy and complete this paragraph about waves. When a wave enters a shallow region, it __________ down and bends towards the _________. This change of direction is called ___________. 4. What happens to the following properties of waves after the waves move into shallow water. (a) Frequency (b) Speed (c) Wavefront direction (d) Wavelength 5. The figure 7.22 shows a displacement-time graph for a certain wave ) m ( 0.25 0.25 0 Time (s) 2.5 × 10–4 7.5 × 10–4 Fig. 7.22: Displacement - time graph Identify the type of wave. (a) (b) State the period of the wave. (c) Determine the frequency of the wave. (d) If the wave has a wavelength of 3.5 cm, what is its velocity? 6. A wave source generate 300 waves signals in a second. Each of the wave signals has a wavelength of 4.5 cm. (a) Determine the: (i) Frequency of the wave. 216 (ii) Period of the wave. (iii) velocity of the wave. (b) Determine the time taken by the generated waves to hit a barrier that is 250 m away from the wave. 7. Using specific properties of light, explain why it is a transverse wave. 8. Define the following terms and state its S.I units: (a) Amplitude (c) Wavelength (b) Period (d) Frequency 9. A radio station broadcasts on a frequency of 88.5 kHz producing signals of wavelength 3389.83 m. Determine: (a) The period of its signals (b) The velocity of radiowaves (c) The velocity of radio Africa if its signals have a wavelength of 3405.22 m braodcasting on the frequency 88.5 kHz. 10. (a) Give the meaning of the symbols in the equation v = f λ. (b) Calculate the wavelength of a wave if the speed is 45 m/s and the frequency is 5 Hz. 11. Radio wave travel with a speed of 3 × 108 m/s in air. If a radio station broadcasts at a wavelength 125 m, calculate the frequency of the transmitted waves. 217 TOPIC 8 Sound Waves Unit outlines • Production of sound waves • Sources of sound waves • Nature of sound waves • Characteristics of sound waves • Propagation of sound • Sound pollution Introduction In Topic 7, we learnt that sound is an example of longitudinal waves. Since a wave is a form of energy, sound is thus a form of energy propagated in a longitudinal manner. In this topic, we shall study the production, propagation, characteristics and applications of sound waves. 8.1 Production of sound The following activity will help us understand how sound is produced. Activity 8.1 To demonstrate sound production (Work in groups) Materials: Metallic string, rubber band, metere rule, drum, piece of wood, glass beaker, pens Steps 1. Pluck a stretched metallic string or rubber band. 2. Fix on end of a half-metre rule near the edge of one side of a table and press the free end downwards slightly and release it. 3. Blow a whistle or a flute. 4. Hit a metallic rod against another. 5. Hit the ‘skin’ of a drum gently with a piece of wood. 6. Gently tap a glass beaker with a pen. 7. Why do the objects above produce such noise? 8. How is the noise produced? In each of the activites within Activity 8.1, sound is produced as the objects vibrate. Sound is a form of wave caused by vibrating bodies. 218 Sound Waves Fig. 8.1 shows some of the sound producing instruments. Drum guitar whistle speaker Fig. 8.1: Examples of sound producing instruments Activities 8.2 (Work in groups) Materials To investigate that a vibrating source produces some energy • A tuning fork • A pith ball • • water in a container glass plate • • tooth brush bristle lamp soot Steps 1. Take a tuning fork and strike hard on a rubber pad with one of the prongs on and observe what happens. 2. Make one of the vibrating prongs of the tuning fork to touch a small pith ball suspended by a thread (Fig. 8.2) and see what happens. Prongs 3. What makes the tuning fork vibrate? Fig. 8.2: Vibrating tuning fork displaces a pith ball. 4. Dip the vibrating prongs in water in a container and observe what happens. 5. What happens to water immediately the fork is dipped in it? 6. Why does it stop vibrating after sometime? 7. Cover a glass plate with a uniform coating of lamp soot. Attach a short stiff h
air of a tooth brush (bristle) to one of the prongs of a tuning fork. Set the tuning fork into vibration and let the bristle lightly touch the soot on the glass plate. Pull the glass plate gently along a straight line and observe what happens (Fig. 8.3). 219 pull bristle smoked glass plate Fig. 8.3: Vibrating tuning fork on a glass plate. The prongs start vibrating. The pith ball is seen to be jerked to one side. Water is violently agitated. A wavy trace is seen on the glass plate. The vibrating prongs of the tuning fork produce energy. Therefore, sound is a form of energy produced by vibrating objects. 8.2 Sources of sound waves All sources of sound have some structures which vibrate. A guitar has strings, a drum has a stretched skin and the human voice has a vocal cord that vibrate and produce sound. Sound travels through the air to our ears through vibrations enabling us to hear it. The air (medium of propagation) is necessary as shown by Experiment 1.11. In the experiment, the sound disappeared though the striker can still be seen hitting the gong. Evidently sound cannot travel in a vacuum as light can do. All materials, including solids and liquids and gases transmit sound. Sound waves produced for example by a loudspeaker consists of a train of compressions and rarefactions (See Fig. 8.4). Hence, sound waves are longitudinal waves. Loud speaker Compression Rarefaction Compression Rarefaction Compression Wave λ λ Fig. 8.4: Sound waves as longitudinal waves. 220 8.3 Nature of sound waves Consider a tuning fork in a state of vibration as shown in Fig. 8.5. As prong X moves to the right, it compresses the layer of air in contact with it (Fig. 8.5(b)). The compressed layer passes the energy to the next layer of air molecules and returns to the original position. Thus a region of compressions moves to the right (Fig. 8.5(c)). As prong X moves to the left, a region of reduced pressure or a rarefaction is produced in the vicinity of (Fig. 8.5(d)). The compressed air in the next layer, moves towards the left to ‘equalise’ the reduced pressure and hence produces another rarefaction to its right and so on. Thus a region of rarefaction moves to the right. x x x x x x (a) (b) (c) (d) (e) ( Fig. 8.5: Production of rarefactions and compressions in a sound wave. 221 Note: As long as the vibrations are periodical, the number of times representing a compression must be equal to the number representing rarefaction and evenly spaced respectively. Therefore, as the prong X vibrates to and fro, a series of compressions and rarefactions are produced. Each layer of air vibrates back and forth about its mean position along the direction in which propagation of energy takes place. Thus, sound waves are longitudinal waves. The wavelength of sound waves is the distance between two succussive compressions and rarefactions. 8.4 The concept of audible range An audible range is a range of frequencies of sound which can be detected by the human ear. Human being’s hear only sounds with frequencies from about 20 Hz to 20 000 Hz (20 kHz). These frequencies are the limits of audibility. The upper limit decreases with age. Sounds with frequencies below 20 Hz are called infrasonic sounds while sounds with frequencies above 20 kHz are called ultrasonic sounds (ultrasound). An average human ear can distinguish between two simultaneous sounds if their frequencies differ by at least 7 Hz. 8.4.1 The human ear Humans and animals detect sound waves using their ears. The human ear converts sound energy to mechanical energy and then to a nerve impulse that is transmitted to the brain. Figure 8.6 shows the parts of a human ear. Hammer Outer ear Anvil Stirrup Eardrum Nerve to brain P Cochlea Pinna Ear canal Eustachian tube Outer ear Middle ear Inner ear Fig. 8.6: The Human Ear 222 The human ear has three main sections: the outer ear, middle ear and the inner ear. Sound waves enter the outer ear and travel through the ear canal to the middle ear. The ear canal channels the waves to the eardrum. The eardrum is a thin, sensitive membrane stretched tightly over the entrance of the middle ear. The waves cause the eardrum to vibrate. It passes these vibrations to the hammer which is one of the three bones in the middle ear. The hammer vibrates causing the anvil, the other small bone touching the hammer to vibrate. The anvil passes the vibrations to the stirrup, another small bone touching the anvil. From the anvil, the vibrations pass into the inner ear. The vibrating stirrup touches a liquid filled sack and the vibrations travel into the cochlea, which is shaped like a shell. Inside the cochlea, there are hundreds of special cells attached to the nerve fibres which transmit the information to the brain. The brain processes the information from the ear enabling us to distinguish the different types of sounds. 8.4.2 Human audible frequency ranges The compressions and rarefactions produced in air by sound waves reach the eardrum of a person and force the eardrum into similar vibrations. The physical movements of the eardrum are transmitted to the brain and produce a mental sensation of hearing. The human ear can detect sound waves of frequencies about 20 to 20 000 Hz (cycles per second). We cannot hear the sound waves, if the frequency is less than 20 Hz or is above 20 000 Hz. The upper limit, however varies with persons and age, it is higher in the case of children than in the old people. It is still higher in certain animals like bats. Just as other types of waves, sound waves obey the wave equation, v = λf. Therefore, the audible frequency ranges is given by; fmaximum = λ fminimum = λ v ______ minimum v ______ maximum Example 8.1 A certain animal can hear sound of wavelength in the range of 2 m to 10 m. Calculate its audible range of frequency. Take the speed of sound in air as 330 m/s. 223 Solution v fminimum = ______ λ maximum fmaximum = v ______ λ minimum = 330 m/s 10 m 330 m/s = 2 m = 33 Hz = 165 Hz Its audible frequency range is 33 Hz to 165 Hz 8.5 Ultrasonic sound Ultrasonic sound is a sound wave that have a frequency above the normal human audible frequency range. Very high frequency waves can penetrate deep sea-water without loss of energy by diffraction. Examples of sources of ultrasonic sound is ship siren and some factory sirens. Therefore, ultrasonic sound has a fundamental frequency that is above the human hearing range i.e. sound with fundamental frequency above 20 000 Hz. The reverse of ultrasonic wave is the infrasonic. Infrasonic is a wave in which the fundamental frequency is lower than the human ear hearing range (audible range). 8.5.1 Uses of ultrasonic sound waves The following are some of the uses: 1. In medical and surgical diagnosis Ultrasonic waves are used in place of X-rays during X-radiography scanning parts of the body using an ultrasonic beam. Ultrasonic is also used to sterilize surgical instruments, jewellery and cleaning medicare instruments. Ultrasonic waves are also used to monitor patient’s heart beats, kidney, growth of foetus (prenatal scanning) and destroy kidney stones. 2. In industries Ultrasonic waves is used in cleaning of the machine parts in industries. Objects or parts with dirt are placed in a fluid through which ultrasonic waves are passed. The waves are used in analysing the uniformity and purity of liquids and solid particles. 3. In fishing Ultrasonic waves are is used to locate shoals of fish in deep sea by the process called echolocation i.e. use of echo to locate an object. More interesting is that this method can detect different types of fish. This is because different fish reflect sound to different extents. 224 4. In security Ultrasonic waves are used in security systems to detect even the slightest movement. Many buildings have ultrasonic motion sensors that detect motion. 5. Estimation of distance by bats: Bats judge the distance away from an object by emitting ultrasound and interpreting the time taken by the reflected wave (echo) to return. The sound they emit is partially or totally reflected from the surface on an obstacle depending on the density of the medium at that point. This helps them to tell where to pass through or perch. 6. Mapping: Sound bounces off an object, the shorter the time lapse between the initial sound and the echo, the smaller the distance. It is particularly useful in mapping the depth of the ocean and also finding lost objects at sea. 7. Ultrasonic echoes are used to determine the shape and size of an object that is not visible such as sunken ship or a baby in the womb. Example 8.2 A ship sends out a sound wave and receives an echo after 1 second. If the speed of sound in water is 1 500 m/s, how deep is the water? Data: Time taken = 1 s; speed, v = 1 500 m/s for to and fro Solution Time taken for sound to reach the seabed = t 2 = 1 2 = 0.5 s From; v = d t or use v = 2d total time d = v × t = 1 500 m/s × 0.5 s = 750 m Exercise 8.1 1. Define the term sound. 2. Describe an experiment to show how sound is produced. 3. Explain the following terms in respect to sound wave: (a) Compression (b) Rarefaction 225 4. Distinguish between utrasonic and infrasonic waves. 5. An animal has audible frequency range of 40 Hz to 20 000 Hz. Calculate the corresponding wavelengths of the frequencies. 6. Explain why a human being cannot hear sound above 20 000 Hz. 7. Explain how ultrasonic sound is used in: (a) Industry (b) Security 8. Define: (a) Sound (c) Echo (b) Pitch (d) Reverberation 9. Distinguish between: (a) Infrasonic sound and ultrasonic sound (b) Sound and echo 10. With the aid of a well labelled diagram, describe how the human ear works. 11. A gun is fired and an echo heard at the same place 1.5 seconds later. How far is the barrier which reflected the sound from the gun? (velocity of sound = 330 m/s). 12. State four uses of echo. 13. A policeman standing between two parallel walls fires a gun. He hears an echo after 2.0 seconds and another one after 3.5 seconds. Determine the separ
ation of walls. (Take velocity of sound 340 m/s). 14. Winfred is standing 600 m from a cliff. She bangs two pieces of wood together and hears an echo 3.5 seconds later. Determine the velocity of sound. 15. A spectator watching athletics in a stadium sees the light from the starting gun and hears its sound after 3 seconds. How far is the spectator from the starting point? (Speed of sound in air is 330 m/s) 16. Maryanne is standing between two walls. She is 400 m from the nearest wall. The walls are “y” m apart. Each time she presses a hooter, she hears two echoes one after 2.5 seconds and the second one after 4.5 seconds. Determine: (a) The velocity of sound. (b) The separation distance “y”. 226 8.6 Characteristics of sound waves The three main characteristics of musical sounds are: (a) Pitch It is the characteristic of a musical sound which enables us to distinguish a sharp note from a hoarse one. For example, the voices of women or children, usually of high pitch than of men. Similarly, the note produced by the buzzing of a bee or the humming of a mosquito is of much higher pitch than the roaring of a lion, though the latter is much louder. Pitch is purely qualitative and cannot be measured quantitatively. The greater the frequency of a vibrating body, the higher is the pitch of sound produced and vice versa. It should be noted that pitch is not frequency; it is a characteristic depend on the frequency. Frequency is a physical quantity and can be measured. Pitch cannot be measured. The pitch of sound depends on the following two factors: 1. Frequency of the sound produced Pitch is directly proportional to the frequency. 2. Relative motion between the source and the observer When a source of sound is approaching, a listener or the listener approaches the pitch of sound appears to become higher. On the other hand, if the source is moving away from the listener or the listener moves away from the source, the pitch appears to become lower. (This effect is known as the Doppler’s effect). (b) Intensity and loudness sound Intensity of sound at any point is the quantity of energy received per second on a surface area of 1 m2 placed perpendicular to the direction of propagation at those points. Thus, the intensity of sound is purely a physical quantity, quite independent of the ear and can be measured quantitatively. It is measured in joules/second/m2. (Js–1m–2) The loudness of sound is the degree of sensation of sound produced in the ear. It depends on the intensity of sound waves producing the sound and the response of the ear. In general, the sound waves of higher intensity are louder. Intensity of sound depends on the following factors: 1. Amplitude of vibrating body The intensity or loudness I, of sound is directly proportional to the square of the amplitude of the vibrating body. If the amplitude of the vibrating body is doubled, the loudness of sound produced becomes two times greater. 227 2. Distance from the vibrating body The intensity or loudness of sound I, is inversely proportional to the square of the distance from the vibrating body. ∴ Intensity α 1 (distance)2 Surface Area of the vibrating surface If the distance from the source of sound is doubled, its intensity of sound becomes 1 and so on. 4 3. Intensity is directly proportional to surface area of the vibrating body. This is because the greater the area of the vibrating surface, the larger the energy transmitted to the medium and the greater is the loudness of the sound produced. 4. Density of the medium The intensity of sound is directly proportional to the density of the vibrating medium For example, an electric bell ringing in a jar filled with oxygen produces a much louder sound than the jar filled with hydrogen. Similarly, the intensity sound of a tuning fork is much higher when the stem of the fork is placed on the table than in air. 5. Motion of the medium If wind blows in the direction in which the sound is travelling, the intensity of sound at a point in the direction of the wind increases and vice versa. Thus, if we shout on a windy day, the sound heard is much louder at a certain distance in the direction of the wind than at the same distance in the opposite direction. (c) Quality (Timbre) of sound Quality is that characteristic of musical note which enables us to distinguish a note produced by one instrument from another one of the same pitch and intensity produced by a different instrument. If, for example, a note of a given pitch is successfully produced by a violin, a guitar or a piano, the ear can distinguish between the three notes. For example, Fig. 8.7 represents two separate waves, one of which has the frequency twice that of the other. When the resultant of these two waves fall upon the ear, the ear is able to recognise the individual waves which have given rise to the resultant wave as they have different qualities (timbre). Fig. 8.7: Waves of two different frequencies 228 Musical sounds and noises In general, sound may be roughly classified as either (a) musical sounds (b) noises. If we pluck the string of a guitar or a stretched sonometer wire or set a tuning fork into vibrations, the sound produced has a pleasant effect on our ears. If however, we listen to the slamming of a door, the sound produced by thunder clouds or the rattling sound of some parts of a car, the sounds produced have an unpleasant effect on the ears. A sound of which appears pleasant to the ear is called a musical sound whereas that which produces an unpleasant or jarring effect on the ear is called a noise. The curves shown in Fig. 8.8 (a) and (b) bring out the difference between noises and musical sounds. (a) (b) Noise Musical sound Fig. 8.8: Noise and musical sound Musical sound is regular and periodic with pulses following each other very rapidly producing the sensation of a continuous sound. Noises, on the other hand, are generally sudden and have no regular period; and are usually complex in nature. 8.7 Propagation of sound Sound waves cannot be transmitted through a vacuum. The transmission of sound waves requires at least a medium which can be a solid, liquid or a gas. Activity 8.3 (Work in groups) Materials To show that sound requires material medium to travel through • Air tight bell jar • Steps • Electric bell Power supply (battery) and connecting wire • Vacuum pump 1. Suspend the electric bell inside an air-tight bell jar as shown in the Fig. 8.9. 229 switch + – S stopper gong air tight bell jar electric bell striker vacuum rubber tube to vacuum pump valve Fig. 8.9: Electric bell 2. Switch on the bell, while there is some air in the bell jar. 3. Start the pump to take the air molecules out of the jar as you listen to the change in the intensity of sound. 4. Return the air to the bell-jar again by opening the stopper slightly as you listen the change in sound. 5. How does the bell produce sound? When there is air in the jar the bell is heard ringing. When the pump is switched on to remove the air, the sound dies down gradually and is eventually not heard at all. When air is allowed to return to the jar, the sound is heard once again. This experiment shows that a medium like air is necessary for propagation of sound. Sound cannot travel through a vacuum. 8.8 Speed of sound in solids, liquids and gases The speed of sound is different in solids, liquids and gases. The arrangement of particles in matter determines how fast sound can travel in matter. The following experiment will help us illustrate that sound require material media to travel. 230 Activity 8.4 (Work in groups) Materials To design and investigate the speed of sound in solids and fluids • A metal spoon Steps • A tuning fork 1. Tie a metal spoon or a light tuning fork to one end of a string and hold the other end near the ear, by not touching it. 2. Let someone touch the spoon with a finger or the set prongs of the fork into vibration by gently hitting the prongs with a rubber. Listen to the sound produced. 3. Remember to repeat the experiment with the free end of the string in contact with the ear. Compare the loudness of sound heard in both cases. Which sound is louder? 4. Write a report about the speed of sound in solids, liquids and gases. Present your findings in a classroom discussion. The loudness of sound heard is more when the string is in contact with the ear. The string transmits sound through it and does it better than air. This experiment shows that sound can travel through solids. Similarly if sound is produced inside water e.g in a swimming pool, it can be heard a short distance away. From the experiments, it has been established that the speed of sound in water is about 1 500 m/s and in steel about 5 500 m/s. Comparison of the speed of sound in solids, liquids and gases The speed of sound varies in solids, liquids and gases. Activity 8.5 will help us to show the speed of sound in solids, liquids and gases. Comparing speed of sound in solids and gases Activity 8.5 (Work in groups) Materials Wooden plank Steps 1. Let one learner place the ear on one end of 20 m wooden plank, while another taps the plank once with a stone on the opposite end. What do you hear? 2. Which sound is heard faster and why do you think this is? 231 In Activity 8.5, two sounds, will be heard by the listerner: one coming through the wooden plank followed by another through the air. This shows that sound travels faster in solids than in air. Several experiments proved that • The speed of sound is higher in liquids than in gases and slower than in solids. • The speed of sound is faster in solids than in liquids because the particles or atoms in solids are closely packed. This makes it easier for particles to transmit sound from one point to another. • The speed of sound in liquids is faster than in gases because the particles in liquids are relatively closer than those in gases. • Therefore the speed of sound is slowest in gases. Lightning and thunder About the middle of the 18th Century, an American
Scientist Benjamin Franklin demonstrated that charged thunder clouds in the atmosphere produce thunderstorms. These thunderstorms produce a lot of sound which we hear as thunder on the earth. Due to the spark discharge occuring between two charged clouds or between a cloud and the earth, electric spark discharge, called the lightning occurs. Though the sound due to thunder is produced first, we see the flash of lightning first and after a few seconds we hear the sound of thunder. This is due to the fact that light travels much faster than sound in air. Experiments have proved that the speed of light in air (or vacuum) is 3.0 ×108 m/s. Take care! Avoid walking in drain water or standing under tall trees when it is raining. Example 8.3 The time interval between “seeing” the flash of lightning and “hearing” the sound of thunder clouds is 5 seconds. (a) Calculate the distance between the thunder clouds and the observer on the earth. (b) Explain why the calculated distance is only approximate. (Speed of sound in air = 330 m/s) 232 Solution (a) speed of sound = v = x t distance time x = v × t = 330 × 5 = 1 650 m/s ∴ The distance between the thunder clouds and the observer is 1 650 m. (b) The clouds may be moving. 8.8.1 Factors affecting the speed of sound in gases (a) Density The higher the density of a gas, the higher the speed of sound. For example, the density of oxygen is 16 times higher than the density of hydrogen hence sound travels faster in hydrogen than in oxygen (speed of sound in hydrogen = 4 × speed of sound in oxygen). (b) Humidity Moist air containing water vapour is less dense than dry air. The density of water vapour is about 0.6 times that of dry air under the same temperature conditions. If the humidity of air increases, density of air decreases hence the speed of sound in air increases. Early in the morning the percentage of humidity of air is more and sound travels faster in the morning air. (c) Pressure The speed of sound is not affected by any change in pressure provided temperature is constant. For example, on a day when the temperature and humidity of air is the same in Lilongwe and a city at the sea level, the speed of sound is the same in the two cities, although the air pressure in Lilongwe is lower than that at the city situated at the sea level. (d) Temperature A change in the temperature of a gas changes its density and hence affects the speed of sound through it. If temperature increases, the density of air decreases and hence the speed of sound increases. If temperature decreases the reverse is the effect. 233 (e) Wind Wind “drifts” air through which the sound waves travel. If air blows in the direction of sound, then the speed of sound increases. The speed of wind is added to the speed of sound in air, to get the resultant speed of sound. If wind blows in the opposite direction to that of sound, then the sound travels more slowly. Table 8.1 summarises how the speed of sound in matter is related and their corresponding reasons. Table 8.1 Matter Solid Liquid Gas Exercise 8.2 Speed of sound Reason Fastest Medium Slowest Particles are closely packed Particles loosely packed Particles are very far apart 1. Explain why the speed of sound in solids is faster than the speed of sound in air. 2. Name two factors that affect the speed of sound in air. 3. State the characteristics of sound waves. 4. Explain why at night sound from a source is clear than during hot daytime. 5. Describe two factors that affect the pitch of sound. 6. Define the following terms: (a) Resonance (b) Quality 7. Distinguish between music and noise. 8. Explain the factors that affect the frequency of sound. 9. During thunder and lightening, there are two types of waves produced. (a) Name the two waves. (b) Which one reaches the ground first? Explain. 10. Sound is a longitudinal wave. How is it propagated? Describe an experiment to demonstrate the fact that sound is actually produced by vibrating body. 8.9 Reflection of sound waves Just like light, sound waves undergo reflection on striking plane hard surface as well as curved surfaces. 234 Activity 8.6 (Work in groups) Materials To investigate the laws of reflection of sound waves • Two tubes A and B • Hard drawing board • • Flat piece of metal Stopwatch Steps 1. Set up two tubes A and B, about 1.2 m long and 4 or 5 cm in diameter as shown in Fig. 8.10. Place a flat piece of a large metal or a hard drawing board facing the tubes about 10 cm from their ends. 2. Place a stop watch at the mouth of the tube A and place your ear at the end of the tube B. A soft board S is placed in between the two tubes to prevent the sound waves from the stopwatch to reach the ear directly. Metal plate reflector A S B Stop watch Ear Fig. 8.10: Reflection of sound waves 3. Adjust the position of tube B until the sound heard is the loudest. 4. Measure angles of incidence i and reflection r. What do you notice? Explain. 5. Are the angles the same? 6. Is the sound heard same as the one from the source? 7. What property do the sound obey from the above observations? These angles are found to be approximately equal. Both the tubes containing the incidence waves and the reflected waves lie in the same plane as the normal to the reflecting surface. We can then conclude that sound waves obey the laws of reflection as is the case with light wave. You should note that since audible sounds have large wavelengths, you need a reasonably large reflector. When sound waves meet a boundary between one medium and another, a part of it is reflected, a part is refracted and the remaining part is absorbed. The relative amounts of these parts are determined by the size and the nature of the boundary under consideration. The proportion of energy reflected is greater in the case of hard substances such as stone and metal. An echo, a reflection of sound, is 235 frequently heard in mountainous regions. There is very little reflection from cloth, wool and foam rubber. Sound which is incident on such soft materials is mainly transmitted through them or absorbed. In places where the effect of echo has to be illuminated, e.g. musical recording room and concert halls, soft materials are used to line the walls of the hall. Also the ear is a sound reflector, reflecting sound waves down the ear cancel to the ear drum. Uses of reflection of sound Sound waves 2. 1. Sound waves can be used to measure the speed of sound in air by reflecting sound at hard surfaces. In public halls and churches, parabolic sound reflection is often placed behind the speaker. It reflects the sound waves back to the audience and thus increasing the loudness of the sound. 3. Sound waves undergo a total internal reflection just like light. Speaking metal tubes that are used to pass message on ships (Fig. 8.11) use total internal reflection of sound waves. Air Fig. 8.11: ‘Speaking’ metal tubes Determining speed of sound by echo method Activity 8.7 (Work in groups) To determine speed of an echo sound 1. Stand about 100 m away from a cliff or a large hard surface such as the wall of a building and clap your hands. What do you hear? 2. How can you determine the speed of the sound you hear? In Activity 8.7, you will hear two sounds; the one you produce and the reflected sound. The reflected sound produced is called an echo. An echo is a reflection of sound from a large hard surface. 236 Activity 8.8 (Work in groups) Steps To investigate how echo sound is produced 1. Stand about 100 m from an isolated, large hard surface or a stone wall. 2. Shout loudly and start a stopwatch at the same time. Stop the watch on hearing the echo. Find the time interval between the production of the loud noise and hearing the echo. Are you able to time the echo accurately? 3. Repeat this a number of times and find the average time taken. 4. How can you increase the accuracy of this experiment? Note For activities above to be more accurate: 1. A large obstacle, e.g. a cliff or a wall is needed. This is because the wavelength of sound waves is large. 2. A minimum distance between the source and the reflecting surface is required. This minimum distance, called persistence of hearing is about 17 m. In Activity 8.8, you should have noticed that an echo is heard after some time interval. During this time, the sound travels to and from the hard surface covering twice the distance. The speed of sound in air is given by the formula: Speed = Total distance travelled by sound Total time taken Here are typical results from Activity 8.8: Distance from the wall is d, metres. Average time interval between the production of sound and hearing its echo is t seconds. Total distance travelled by sound is 2d metres. Speed of sound = Total distance travelled Total time taken v = 2 × d (m) t(s) ∴ The speed of sound in air is given by v = 2d t 237 Example 8.4 A girl standing 100 m from a tall wall and bangs two pieces of wood once. If it takes 0.606 s for the girl to hear the echo, calculate the speed of sound in air. Solution Speed of sound, v = Total distance Total time taken = 200 0.606 = 2d t = 2 × 100 0.606 = 330 m/s ∴ the speed of sound in air is 330 m/s Example 8.5 A man stands in front of a cliff and makes a loud sound. He hears the echo after 1.2 s. If the speed of sound in air is 330 m/s, calculate the distance between the man and the cliff. Solution Let the distance between the man and the cliff be x. (Fig. 8.12) x man cliff Fig. 8.12. Speed of sound = 330 m/s = Total distance Total time 2x 1.2 2x = 330 × 1.2 = 396 m ∴ x = 198 m The distance between the man and the cliff is 198 m. 238 Example 8.6 A man standing between two parallel cliffs fires a gun. He hears the first echo after 1.5 s and second echo after 2.5 s. (a) What is the distance between the cliffs? (b) When does he hear the third echo? (Take speed of sound in air to be 336 m/s). Solution (a) The sketch is as shown in Fig. 8.13. cliff A cliff B x2 x2 x1 x1 man Fig. 8.13: A man between parallel cliffs From Cliff A: Speed, v, = Tot
al distance travelled Total time taken v = 2x1 1.5 ⇒ 2x1 = 1.5 × v ∴ 2x1 = 1.5 × 336 x1 = 252 m From cliff B: v = 2x2 2.5 ∴ 2 x2 = 2.5 × 336 = 840 ⇒ 2 x2 = 2.5 × v x2 = 420 m ∴ The distance between the cliffs is 252 m + 420 m = 672 m (b) The first echo (after 1.5 s) reaches cliff B and returns after 2.5 s. So the man hears the 3rd echo after 1.5 + 2.5 = 4 s. Exercise 8.3 1. How is sound propagated? 2. Define the term echo. 3. A person stands infront of a wall and makes a loud sound. She hears the echo after 1.55s. If the speed of sound is 333 m/s. Calculate the distance between the person and the cliff. 239 4. A person standing 80 m from the foot of a cliff claps and hears an echo after 0.9 s. What is the speed of sound in air? 5. A pupil, standing between two cliffs and 500 m from the nearest cliff clapped his hand, and heard the first echo after 3 s and the second echo 2 s later. Calculate: (a) The speed of sound in air, (b) The distance between the cliffs. 6. An echo of the sound produced by a whistle is heard after 0.50 s. If the speed of sound in air is 332 m/s, find the distance between the whistle and the reflecting surface. 8.10 Sound pollution Sound is a very important form of energy. Human beings and animals use sound as a way of communication. But if sound is unorganised, it becomes noise. Any unwanted sound becomes a nuisance and leads to pollution in form of noise. Therefore, sound pollution is a type of pollution caused by undesirable or unwanted sound. Sound pollution can cause damages to eardrum or hinder communication. Sources of sound pollution are: very high music from discos, concerts, celebrations, factory sirens, traffic noise, aircrafts, alarms and others. Everybody is encouraged to minimize sound pollution at all cost. The government through some agencies must prohibit sound pollution by enacting some laws to govern this. The following are some of the ways used to minimize sound pollution. 1. Factories are encouraged to use sound sirens that are environmental friendly. Most of them use the normal fire alarms. 2. During construction of musical concert halls, the constructor should use materials that absorb most of incident waves of sound to avoid reverberation (reflected multiple sound). 3. Proper laws must be enacted by the government to reduce sound pollution. 4. Proper education of the citizens on sound pollution should be done to sensitize them on the important of reducing sound pollution. Listen to moderate music! Loud music can affect your eardrum if you listen for long. 240 Exercise 8.4 1. Explain what is sound pollution? 2. Sound wave just like light wave undergo reflection. Explain two uses of reflection of sound. 3. Explain two ways in reducing sound pollution. Topic summary • All vibrating bodies produce sound. • • Sound cannot travel through a vacuum. It needs a material medium like solid, liquid or gas. Sound waves are longitudinal in nature consisting of compressions and rarefactions. • Human audible frequency range is between 20 Hz and 20 000 Hz. • • • Speed of sound in air = 332 m/s at 0ºC. Speed of light in vacuum = 3 × 108 m/s. Sound waves undergo reflection. Reflection is the bouncing back of sound wave when it strikes plan hard surface or curved surface. • Echo is the reflection of sound from a large, rigid barrier like cliff, tall wall etc. • Speed of sound in air can be determined by echo method. • Density of air, humidity, temperature and wind affect the speed of sound. • Pressure, amplitude of wave and loudness of sound do not affect the speed of sound. • Ultrasonic wave is a sound wave that have a fundamental frequency above the human audible range frequencies. • A sound which appears pleasant to the ear is called a musical sound and the one which produces a jarring effect on the ear is called noise. 241 Unit Test 8 1. Sound cannot pass through a A. solid C. air liquid B. D. vacuum 2. A normal human being can hear sound of frequency less than 20 Hz. A. B. between 20 Hz and 20 000 Hz. C. between 20 Hz and 200 Hz. D. above 20 000 Hz. 3. Which of the following is correct? Sound waves A. are transverse in nature. B. are longitudinal in nature. C. can never undergo diffraction. D. can never interfere with each other. 4. The speed of sound is NOT affected by A. pressure C. temperature B. humidity. D. wind. 5. Which statement is true about the music produced by the loudspeaker of a radio? When the music is made louder, A. the amplitude of sound decreases. B. the amplitude of sound increases. C. the speed of sound increases. D. the speed of sound decreases. 6. Suggest a simple experiment to establish each of the following: (a) Sound is produced by a vibrating body. (b) Sound cannot travel through vacuum. 7. State three factors which affect the speed of sound in air. Choose one of the factors and explain how it affects the speed of sound in air. 8. In which gas is the speed of sound greater, hydrogen or oxygen? 9. (a) Describe an experiment to show how echoes are produced. (b) The echo method can be used to determine the speed of sound in air. (i) What measurements would you take in order to do this? (ii) Show how you would calculate the speed of sound in air from your measurements. 242 (iii) State a precaution to be taken to improve your result. 10. A person standing 80 m from the foot of a cliff claps and hears an echo after 0.9 s. What is the speed of sound in air? 11. A student, standing between two cliffs and 500 m from the nearest cliff clapped his hand, and heard the first echo after 3 s and the second echo 2 s later. Calculate: (a) The speed of sound in air. (b) The distance between the cliffs. 12. A worker uses a hammer to knock a pole into the ground (Fig. 8.14). hammer cliff boy worker girl Fig. 8.14: A worker knocking hammer against the pole (a) A girl at the foot of the cliff hears the sound of the hammer after 2.0 s. Calculate the distance of the worker from the girl (speed of sound in air is 340 m/s) (b) A boy on the other side of the cliff observes that each time the hammer hits the pole, he hears two separate sounds, one after the other. Explain this observation. Given that the first sound is heard by the boy after 1.0 s, determine the: (i) Distance of the boy from the worker. (ii) Time interval between the two sounds. 13. A soldier standing between 2 cliffs fires a gun. She hears the first echo after 2 s and the next after 5 s. (a) What is the distance between the two cliffs? (b) When does she hear the third echo? (speed of sound in air = 336 m/s). (c) Why is the 3rd echo faint than the 2nd one? 14. A student makes observations of a distant thunderstorm and finds the time interval between seeing the lightning flash and hearing the thunder as 4.0 s. Given the speed of sound in air = 340 m/s and speed of light in air = 3.0 × 108 m/s, 243 (a) Explain why there is a time delay? (b) Calculate the distance between the thunder cloud and the student. (c) Explain why the speed of light is not taken into account in this calculation. (d) Calculate the frequency of the flash of light emitted if the mean wavelength of light emitted is 6.0 × 10–7 m. 15. In an athletics competition, the time keeper in a 100 m race starts the stopwatch on hearing the sound from the starter’s pistol and records the time as 10.00 s. Calculate: (a) The actual time taken by the athlete to cover the 100 m race. (b) The average speed of the athletee. (speed of sound in air = 340 m/s). 244 UNIT 6 Heat Transfer Topics in the unit Topic 9: Heat Transfer Learning outcomes Knowledge and Understanding • Understand the nature of heat • and describe its effects on matter Skills • Design tests to show the factors affecting heat transfer, distinguish between conduction and radiation of heat, and between good and bad conductors of heat. • Observing carefully. • Predict expectations. • Use appropriate measures. • Collect and present results appropriate in writing. Interpret results accurately. • • Report findings appropriately • Explain applications of heat transfer. Attitudes • Appreciate the application of modes of heat transfer. Key inquiry questions • Why is heat important? • How can heat be produced? • Why is that the expansion of material a nuisance? • Why that a rough surface is a better emitter of radiation than a polished surface? • Why that a dull black surface is a better absorber of heat than a polish one? 245 TOPIC 9 Heat Transfer Unit Outline • Heat and temperature • Heat transfer by conduction • Heat transfer by convection • Heat transfer by radiation • Applications of heat transfer • Linear expansion Introduction In our environment, most interactions between systems involve transfer of heat from one system to another. For example, when we bask in the sun, we feel warmer, when we touch a hot sauce pan, we feel the heat. In this unit, we will discuss the different modes of heat transfer through which heat is transferred from one region to another. We will begin by reviewing the difference between heat and temperature. 9.1 Heat and temperature The following activity will enable us to understand the difference between heat and temperature. Activity 9.1 (Work in groups) To investigate the difference between heat and temperature Steps 1. In secondary 1, we learnt about heat and temperature. What is the difference between heat and temperature? 2. With the help of your teacher, recall and conduct an experiment to differentiate between heat and temperature. 3. Record the observation. Draw conclusions and explain your findings in a group and class discussion. 246 Heat is a form of energy that flows from a hot to a cold body while temperature is the degree of hotness or coldness of a substance. 9.2 Methods of heat transfer Activity 9.2 To describe the methods of heat transfer (Work in pairs or in groups) Fig 9.1 below shows a person heating some liquid in saucepan over fire. Fig. 9.1: Heating some liquid in a saucepan 1. Identify the modes of heat transfer marked A, B and C. 2. Discuss how each of
the modes of heat transfer takes place, citing the states of matter through which the processes take place. 3. Describe one application of each type of the above modes of heat transfer in real life. 4. Present your findings to the rest of the class in a class discussion. There are three modes of heat transfer: conduction, convection and radiation. Conduction of heat is through solids, convection in fluids and radiation in gases. 9.2.1 Heat transfer by conduction 9.2.1.1 Demonstration of conduction of heat The following experiment will illustrate conduction of heat in solids. Activity 9.3 (Work in groups) Materials: • A metal spoon • Bunsen burner To design and investigate heat transfer in solids • A beaker full of boiling water • Wax 247 Instructions 1. This activity involves an investigation. You are required to set-up the apparatus as shown in Fig. 9.2 below. Come up with a procedure and execute it to investigate the heat transfer in solids. 2. With the help of your teacher carry out the investigation. Write a report and discuss your findings in a class presentation. 3. How can the investigations be improved? 4. Besides the materials provided which other locally available materials that can be used to carry out the investigation? waxed end metal spoon Fig. 9.2: A spoon inside boiling water 5. Why do you think the free end of the spoon gets hot after sometime? Explain. Solids transfer heat from one point to another. For instance, the free end of the spoon outside the beaker in Fig. 9.2 becomes hot. Heat energy is transferred from the inside to the outside through the metal spoon i.e. from a region of higher temperature to a region of lower temperature. This process of transfer of heat energy in solids is called conduction. Conduction is the transfer of heat from one substance to another that is in direct contact with it. In conduction there is no visible movement of the heated particles. 9.2.1.2 Mechanism of conduction of heat We have already learnt that when temperature increases, the molecules have larger vibrations. This knowledge can help us understand the mechanism of conduction of heat. When the molecules at one end of a solid receive heat energy from the heat supply, they begin to vibrate vigorously. These molecules collide against the neighbouring molecules and agitate them. The agitated molecules, in turn, agitate the molecules in the next layer and so on till the molecules at the other end of the solid are agitated. Thus, the heat is passed from one point to another till the other end becomes hot. Hence, in conduction, energy transfer takes place by vibration of the molecules. There is no actual movement of the heated particles. 248 To demonstrate that heat energy flows due to a temperature difference Activity 9.4 (Work in groups) Materials: • An iron bar about a metre long • Drawing pins with holes drilled at equal intervals • Wax • Water bath Steps • Stand/clamp • A bunsen burner 1. Fill the holes of the iron bar partially with wax and insert the drawing pins into them. 2. At one end of the bar put a wooden screen and insert it in a water bath. Heat the end points slowly and gradually (Fig. 9.3). Fig. 9.3: The higher the temperature difference the higher the energy transferred 4. After some time, note the temperature readings of the pins. Are the drawing pins falling at the same time? Why do you think that is so? The pin nearest to the bunsen burner registers the highest rise in temperature, and the one farthest away registers the least temperature rise. When one end of the rod was inserted into boiling water, a large temperature difference is set up between the two ends and heat energy flowed from the region of higher temperature to that of lower temperature. Hence heat energy flows due to temperature difference and the pins fall slowly. If the activity is repeated by replacing the hot water bath with a bunsen burner flame (temperature of the bluish part of the flame is about 900˚C), the rise in temperature registered by each pin is higher. Hence the higher the temperature difference, the higher the energy transfer and the pins fall this time faster. Heat energy flows in solids is due to temperature difference. The higher the temperature difference, the higher the energy flow. 249 Piniron barscreen 9.2.1.3 Comparing rates of conduction in metals Activity 9.5 To show that heat transfer in solids depends on the material (Work in groups) Materials • A copper rod • 3 match sticks • A bunsen burner Steps Iron rod • • Wax • Aluminium rod • Tripod stand 1. Take three rods, copper, aluminium and iron of the same length and thickness. Fix a matchstick (or a light metal pin) to one end of each rod using a little melted wax. 2. Place the rods on a tripod stand and heat the free ends with a burner as shown in Fig. 9.4. Observe what happens. copper aluminium matchstick bunsen burner iron Fig. 9.4: Comparing heat transfer through different conductors 3. Which rod falls first? Which one falls last? 4. Why do you think hey did not fall all of them at the same time? The matchstick falls off from the copper rod first then aluminium and finally from the iron rod. When the temperatures of the other ends of the rods reach the melting point of wax, the matchstick falls off. The matchsticks do not fall off at the same time, because the energy transferred is not equal for all the rods. The matchstick from the copper rod is the first one to fall off showing that of the three metals, copper is the best conductor of heat followed by aluminium and then iron. Good conductors and poor conductors of heat A material or substance which has the ability to transfer heat through itself is called a good conductor. Most metals are good conductors of heat e.g copper etc. 250 Substances like water, air, glass, wood, plastic, paper, etc which have a poor ability to transfer heat are called poor conductors of heat. Poor conductors of heat are sometimes refered to as insulators. 9.2.2 Heat transfer by convection 9.2.2.1 Convection in liquids To observe convection current in water Activity 9.6 (Work in groups) Materials • A long straw • A crystal of potassium permanganate • A beaker containing water • A bunsen burner Steps 1. With the help of a long straw, drop a small crystal of potassium permanganate to the right side of the bottom of a flask or a beaker containing water. What do you observe? 2. Heat the flask gently at the right side of the flask (Fig. 9.5). Observe what happens. glass flask potassium permanganate crystals Fig. 9.5: Convection currents in water 3. What do you observe in the beaker when you continously heat the water? Explain your observations. 4. Why do you think the potassium permanganate crystals behave in such manner? 251 Coloured streaks are observed to rise from the bottom to the top. The crystal dissolves and the hot water of less density starts rising displacing the cold dense water down. The streams of physically moving warm liquid are called convection currents. Heat energy is transferred by the convection currents in the liquid. The transfer of heat by this current is called convection. 9.2.2.2 Convection in gases To investigate convection current in air Activity 9.7 (Work in groups) Materials • A box with a glass window, and two chimneys • A candle • Smouldering pieces of wick Steps 1. Take a box with a glass window and two chimneys fixed at the top. 2. Place a lighted candle under one chimney and hold a smouldering piece of wick above the other chimney as shown in Fig. 9.6. What do you observe? 3. Why do you think the smouldering pieces of wick behave in such manner after heating them? smoke A B smouldering wick box glass window candle Fig. 9.6: Convection currents in air. Smoke from the smoldering wick is seen to move down through chimney B then to the candle flame and finally comes out through chimney A. Air above the candle flame becomes warm and its density decreases. Warm air rises up through chimney A and the cold dense air above chimney B is drawn 252 down this chimney and passes through the box and up the chimney A. The smoke particles from the wick enable us to see path of convection current (Fig. 9.7). Heat is transferred in air through convection currents. This convection current passes energy as shown in Activity 9.8. Activity 9.8 (Work in groups) To illustrate that convection currents possess energy Materials: • A thin circular disk • A card board • A candle flame Steps 1. Take a thin circular disk of tin or cardboard and cut out six blades all round (Fig. 9.7(a)). Pivot the disk on a bent needle (Fig. 9.7(b)). 2. Hold the disk above the candle flame for some time. Observe and explain what happens. disk pivoted disc of tin with blades cut (a) bent needle (b) Fig. 9.7: A rotating disk. 3. What makes the disk to rotate in such manner? 4. What else can be used to rotate the disk? 5. What are some of the uses of convection currents? The disk starts to rotate. The rotation is due to the convection current set up. If a powerful electric bulb is available, you can make a rotating lamp shade. thin cardboard blade thin carboard cylinder cold air lamp stand hot air thick wire wrapped around a bulb with a pointed pivot to power supply Fig. 9.8 : A rotating lamp shade 253 Convection currents possess energy. It is for this reason that steam is used to rotate the turbine in geothermal electric plants. 9.2.3 Heat transfer by radiation 9.2.3.1 The concept of radiation If you stand in front of a fireplace, you feel your body becoming warm. Heat energy cannot reach you by conduction as air is a poor conductor of heat. How about convection? The hot air molecules in and around the fireplace can only rise and do not reach you by convection. How does the energy from the fireplace then reach you? Heat energy must be transferred by a different mode other than conduction and convection. To demonstrate heat transfer by radiation Activity 9.9 (Work in groups) Materials: • Thin tin lids painted black • Thumb tacks (match
sticks) • Wax Part A Steps • A bunsen burner 1. Take a thin tin lid painted black on one side. Stick a thumb tack with melted wax on the other side. 2. Keep the bunsen burner flame close to the painted side (Fig. 9.9). What happens? Explain. 3. Why do you think the thumb tack falls off after sometime. Explain? wax lid painted black Thumb tack wooden stand Fig. 9.9: Radiation 254 Part B Steps 1. Take two thin tin lids, one with shiny inner side and the other with the inner side painted dull black. 2. Stick metal thumb tacks (or match sticks) on the outside of each lid using a little molten wax. 3. Keep a bunsen burner flame midway between the lids as shown in Fig. 9.10. Watch closely to and compare what happens to the two thumb tacks. Explain your observation. tin lids wax thumb tack dull black surface shiny surface supports for lids Fig. 9.10: Good and bad absorbers As discussed in the case of the fireplace, the energy from the flame reaches the tin lid and the wax by a different mode other than conduction and convection. This third mode of heat transfer is called radiation. Radiation is the emission or transmission of energy in the form of a wave or particles through a material or space. Heat transfer from the sun travels through empty space (vacuum) and reaches the Earth. This energy is transferred by radiation. The surfaces of all luminous bodies emit radiation. A human face also emits some mild radiations. While conduction and convection need a medium to be present for them to take place, radiation can take place without a medium. The amount of heat energy radiated depends upon the temperature of the body. In Activity 9.9, if the bunsen burner is replaced by a candle flame, it will take a longer time for the wax to melt. The temperature of the candle flame is lower than that of a bunsen burner. Heat transfer can take place without contact or in a vacuum. This method of heat transfer is called radiation. 9.2.3.2 Good and bad absorbers of heat energy by radiation If a black and shiny surface receive the same amount of heat energy by radiation, the black surface absorbs more heat than the shiny surface. 255 A dull black surface is a better absorber of heat radiation than a shiny surface. To illustrate good and bad emitters of heat Activity 9.10 (Work in groups) Materials • Three thermometers • Three identical empty cans • Three cardboards Steps 1. Take three identical empty cans of the same volume with their tops removed. Apply clean and dry paints one white and the other black on two cans (both inside and out surfaces) and leave the third can shiny. 2. Prepare three suitable cardboard covers with holes at the centre. Fill the cans to the brim with hot water at 60˚C. 3. Cover the cans with cardboards and place a thermometer in each can through the hole at the centre of the cardboard (Fig. 9.11). cardboard thermometer thermometer thermometer can can can white black shiny Fig. 9.11: Good and bad emitters 4. Record the temperature of water in the cans after a certain time interval. 5. Which can cools the water fastest? 6. Which can takes the longest time to cool the water? Explain the difference in the rate of temperate drop in the three cans. A shiny surface is a good emitter than a dull black surface 256 9.2.4 Applications of heat transfer Activity 9.11 (Work in groups) Materials • Internet Steps To find out the applications of heat transfer • Reference books 1. You have learnt about heat transfer. Referring to this book or any other source, describe three ways in which heat transfer is important in our daily lives. 2. Do a research from the internet and reference books on the applications of heat transfer. 3. In your research, highlight clearly how the modes of heat transfer are applied in vacuum flasks, construction of ventilations, domestic hot water system and solar heating. 4. What other applications of heat transfer did you come across in your research? 5. Explain to your group members how natural phenomena like sea and land breeze take place. 6. Make a presentation on your findings to the whole class through your group secretary. 1. Vacuum flask The vacuum flask popularly known as thermos flask, was originally designed by Sir James Dewar. It is designed such that heat transfer by conduction, convection and radiation between the contents of the flask and its surroundings is reduced to a minimum. A vacuum flask, Fig. 9.12 is a double-walled glass container with a vacuum in the space between the walls. The vacuum minimises the transfer of heat by conduction and convection. The inside of the glass walls, is silvered so as to reduce heat losses by radiation. The felt pads on the sides and at the bottom support the vessel vertically. The cork lid is a poor conductor of heat. 257 plastic cover cork lid vacuum felt pads vacuum seal double-walled glass container silvered surface outside case Fig. 9.12: Vacuum flask When the hot liquid is stored, the inside shiny surface does not radiate much heat. The little that is radiated across the vacuum is reflected back again to the hot liquid, by the silvering on the outer surface. There is however some heat lost by conduction through the walls and the cork. 2. Windows and ventilators in buildings As shown in Fig. 9.13, warm exhaled air of less density goes out through the ventilator and fresh air of high density enters through the windows at a lower level. This refreshes the air in a room. warm air fresh air warm air Fig. 9.13: Ventilation in building 3. Natural convection currents over the earth’s surface (a) Sea breeze During the day, the temperature of the land rises faster than the temperature of sea water and the air over the land becomes warmer than the air over the sea water. The warm air of less density rises from the land allowing the cold dense air over the sea to blow to the land. This creates a sea breeze in the daytime (Fig. 9.14). 258 warm air from the land Sun cold air from the sea Fig. 9.14 Sea breeze cold sea water (b) Land breeze During the night, the land cools faster than the sea water. Warm air from the sea rises and the dense air from the land moves to the sea. This sets up a land breeze in the sea (Fig. 9.15). cold fresh air from the land warm air above the sea water rises Fig. 9.15: Land breeze 4. Electrical devices An electric kettles has its heating coil at the bottom. A refrigerator has the freezing unit at the top. 5. Domestic hot water system A domestic hot water supply system works on the principle of convection current. A schematic diagram of a hot water supply is shown in Fig. 9.16. 259 expansion pipe C ball cock main supply of cold water cold water storage tank hot water tap pipe A hot water storage tank boiler pipe B, cold water heat Fig. 9.16: Hot water system Water is heated using fire wood, oil or electricity in the boiler. Hot water from the boiler goes up to the hot water storage tank through pipe A. Cold water flows down from the cold water storage tank into the boiler through pipe B (called return pipe). When the hot water is being drawn from the top of the hot water storage tank, it is replaced by water from the main cold water tank built at the top of the house. The expansion pipe C allows steam and dissolved air to escape. This ensures that the tank does not explode due to the pressure created by the steam produced. 6. Solar heating Flat plate collectors, called solar panels, are used to heat water. They can heat water up to 70˚C. A solar panel consists of thin copper pipes, painted black, which carry the water to be heated. These tubes are fitted in a copper collector plate which in turn is fitted on to a good thermal insulator in a metal frame. A glass plate covers the panel (Fig. 9.17). These panels can be fitted on the roof of houses. Heat radiation from the sun falls on the tubes and on the collector plate through the glass plate. The heat radiations trapped inside the panel by the glass plate heat the water. The hot water is then pumped to a heat exchange coil in a hot water tank which is connected to the domestic hot water system. 260 heat from the sun pipe 1 1 solar panel glass plate pipe 2 metal case 2 insulator thin copper tube pipe 2 pump to domestic hot water system hot water water gains energy in the exchanger cold water heat exchange coil pipe 1 cold water insulated cold water tank Fig. 9.17: Solar heating 7. Solar concentrations In some heating devices, instead of a flat plate collector, curved mirrors (concave or parabolic) are used to concentrate the heat radiations from the sun to a small area at their focus. If the boiler is placed at the point of focus, very high temperatures can be reached. Exercise 9.1 1. Distinguish between heat and temperature. 2. What are the different modes of heat transfer? Explain clearly their difference using suitable examples. 3. State three factors which affect heat transfer in metals. Explain how one of the factors you have chosen affects heat transfer. 4. Describe an experiment to show that water is a poor conductor of heat. 5. Use particle behaviour of matter to explain conduction. 6. Describe a simple experiment to demonstrate that the heat radiated from a hot body depends upon the temperature of the body. 7. With a suitable diagram, explain the working of a vacuum flask. 9.3 Thermal expansion In general, nearly all substances increase in size when heated. The process in which heat energy is used to increase the size of matter is called thermal expansion. The increase in size on heating of a substance is called expansion. On cooling, substances decrease in size. The decrease in size on cooling of a substance is called contraction. Why is this so? 261 9.3.1 Thermal expansion and contraction in solids When a solid (e.g. a metal) is subjected to heat, it: (a) Increases in length (Linear Expansion). (b) Increases in volume (Volume Expansion). (c) Increases in area (Surface Expansion). 9.3.1.1 Linear expansion (a) Demonstrations of linear expansion Activity 9.12 (Work in groups)
To demonstrate expansion and contraction using the bar and gauge apparatus Materials • A bar and gauge apparatus Steps • Bunsen burner 1. Move the metal bar in and out of the gauge at room temperature (Fig. 9.18). Observe what happens. wooden handle gauge metal bar Fig. 9.18: Bar and gauge apparatus 2. Keep the metal bar away from the gauge and heat the bar for sometime. 3. Try to fit the bar into the gauge. Does it fit or not? Explain your observation. 4. Allow the bar to cool and try to fit it into the gauge. Does the bar now fit into the gauge? Explain. A bar and gauge apparatus consists of a metal bar with a suitable wooden handle and a gauge. When both the metal bar and the gauge are at room temperature, the bar just fits into the gauge. On heating, the metal bar expands. There is an increase in length. It hence expands linearly and therefore, the bar cannot fit into the gauge. 262 On cooling the bar easily fits into the gauge due to contraction. Solids expand i.e increase in length on heating and contract i.e reduced in length on cooling. Activity 9.13 (Work in groups) To demonstrate the bending effect of expansion and contraction Materials: • A bimetallic strip Steps • Bunsen burner 1. Observe a bimetallic strip at a room temperature (Fig. 9.19). wooden handle brass iron Fig. 9.19: A bimetallic strip 2. Take the bimetallic strip with the brass strip at the top and heat it with a bunsen burner flame for sometime. Observe what happens. Explain the observation. Sketch the shape of the bimetallic strip after heating. 3. Remove the flame and allow the bar to cool to a room temperature. What happens to the bimetallic strip? Sketch its shape after cooling. 4. Discuss with your friend what will happen if the bar is cooled below room temperature. Sketch the strip at that temperature. When the bimetallic strip is heated, it bends downwards with the brass strip on the outer surface of the curvature, as shown in Figure 9.20(a). Why does this happen? When the flame is remove and the bar left to cool to room temperature, the bar returns back to its initial state (straight) as shown in Figure 9.19 above. If the bar is cooled below room temperature, it bends upwards with the iron strip underneath as shown in Figure 9.20 (b). What has happened? brass Bunsen burner iron brass iron (a) Heating the bimetallic strip (b) Cooling the bimetallic strip below room temperature Fig. 9.20: Bending effect of expansion and contraction 263 As the bimetallic strip is heated, brass expands more than iron. The large force developed between the molecules of brass forces the iron strip to bend downwards. On cooling below a room temperature, the brass contracts more than iron and the iron strip is forced to bend upwards. The force developed during expansion or contraction causes a bending of the metals. (b) Comparison of rates of expansion of different solids As we know from the kinetic theory of matter, the different states of matter expands when heated but at different rates. The following activity shows that different solids have different rates of expansion. Activity 9.14 To compare rates of expansion and contraction of different solids (Work in groups) Materials: • Thin metal rods of different metals • Source of heat • Rollers connected to a pointer • G - clamp Steps 1. Clamp one end of a long thin metal rod tightly to a firm support, with the end of the rod resting on a roller fitted with a thin pointer (See Fig. 9.21). clamp pointer fixed to roller thin copper rod deflection of the pointer roller table heat Fig. 9.21: Expansion and contraction of thin metal rods. 2. Heat the metal rod for sometime. Observe and explain what happens to the rod. 3. Remove the burner and allow the rod to cool. What happens to the rod again after cooling. Does it reduce in size? Explain why. 4. Repeat the activity with thin rods of different materials. Observe and explain what happens, accounting for any differences. The pointer deflects in the clockwise direction on heating and in the anticlockwise direction on cooling. 264 The pointer deflects to different extents depending on the material. On heating, there is an increase in length (linear expansion) of the rods. The expanding rod moves the roller to the right making the pointer attached to the roller to deflects in a clockwise direction. On cooling, the rod contracts and decreases in length. The contracting rod moves the roller to the left hence the pointer deflects in the opposite direction (anticlockwise direction). When a different material e.g lead is used, the pointer deflects more to the right (clockwise). When cooled, the pointer deflects more to the left (anticlockwise). Different solids (e.g metals) expand and contract to different extents when heated by the same quality of heat. (c) Coefficient of linear expansion Consider a thin metal of length l0 in Fig. 9.22. l0 Δl Fig. 9.22: A thin rod showing increase in length. When the rod is heated, a temperature change of Δθ occurs and its length increases by Δl. The ratio of increase or decrease in length to original length ( Δl proportional to the change in temperature Δθ. l0 ) is directly Δl l0 ∝ Δθ ⇒ Δl l0 = α Δθ and α = Δl lθ Δθ where α is a constant called the coefficient of linear expansion. It is the value of the increase in length per unit rise in temperature for a given material. The SI units of α is K–1 Suppose: The temperature change = Δθ, l0 represents the original length of the rod l represents the new length for a temperature rise of θ Then, Δl = l – l0 The above expression may be expressed in terms of l0, lθ, θ and α as follows. α = Δl l Δθ = l – l0 l Δθ Re-arranging l – l = l0 αΔθ l = l0 + l0 αΔθ l = l0(1 + αΔθ) 265 Example 9.1 A copper rod of length 2 m, has its temperature changed from 15 °C to 25 °C. Find the change in length given that its coefficient of linear expansion α = 1.7 × 10–6 K–1. Solution Δθ = (25 - 15) oC = 10 oC Δl = l0 = 3.4 × 10–5 m α Δθ = 2 × 1.7 × 10–6 × 10 = 0.000034 mm 9.3.1.2 Area and volume expansion (a) Demonstrations of area and volume expansions Activity 9.15 (Work in groups) Materials: To demonstrate volume and surface expansion and contraction using the ball and ring apparatus • A ball and a ring • Bunsen burner • A bowl of cold water Steps 1. Move the ball in and out of the metal ring at room temperature (see Fig. 9.23). What do you observe? 2. Keep the metal ball away from the ring and heat it for sometime. 3. Try to pass the ball through the ring. Does the ball pass through the ring this time? Why?. 4. Cool the metal ball in a bowl of cold water and try to pass the ball through the ring again. Does it pass through the ring? Explain why. metal ball metal ring Fig. 9.23: Ball and ring apparatus 266 A ball and ring apparatus consists of a ball and ring both made of the same metal. At a room temperature, the ball and the metal ring have approximately the same diameter, thus the ball just passes through the ring. On heating, the metal ball expands. There is an increase in volume and surface area of the ball. As a result, the ball cannot pass through the ring. On cooling, contraction occurs and the original volume is regained. The ball can now pass through the ring again. This activity shows volume and surface area expansion and contraction in solids. Most solids expand on heating and contract on cooling. Why solids expand on heating? In Secondary I, we learnt that molecules of a solid are closely packed and are continuously vibrating about their fixed positions. When a solid is heated, the molecules vibrate with larger amplitude about the fixed position. This makes them to collide with each other with larger forces which pushes them far apart. The distance between the molecules increases and so the solid expands. The reverse happens during cooling. (b) Coefficients of area expansion of solids Consider a solid whose surface area is A0. When the surface of the solid is heated or cooled to a temperature change of Δθ, its surface area increases or decreases by ΔA to a new value A. Experiments have proved that the ratio of the change in surface area to original area i.e ΔA is directly proportional to the change in temperature (Δθ). A0 ∝ Δθ ⇒ ΔA ΔA A0 A0 Hence β = = βΔθ (β is a constant called coefficient of area expansivity) A – A0 A Δθ ⇒ A –A0 = AθβΔθ ΔA A0Δθ Or β = (since ΔA = A – A0) A = A0 – A0βΔθ ∴ A = A0(1 - βΔθ) Note: Coefficient of area expansivity = 2 × coefficient of linear expansivity β = 2α Example 9.2 A round hole of diameter 4.000 cm at 0 °C is cut in a sheet of brass (coefficient of linear expansion is 19 × 10-6k-1(Co)-1. Find the new diameter of the hole at 40 °C. 267 Solution (θ 2 ) θ 2 - ΔA = βA0 Given: α = 19 × 10-6k-1, (θ2 – θ1) = 40°C, D = 4.000 cm so r = 2.000 cm, β = 2 α then Area (A0) = πr2 = 22 7 × 2.000 × 2.000 cm2 = 12.971 cm2 New area A = A0 + ΔA = (12.971 + 0.0197) cm2 = 12.991 cm2 = 12.987 Since A = πr2, the new radius r = A 3.141 π = 2.033 cm (c) Coefficients of volume expansion in solids When a solid is heated or cooled to a temperature change of Δθ, its volume increases or decreases by ΔV to a new value V. The ratio of the change in volume to original volume i.e ΔV V0 to the change in temperature (Δθ). is directly proportional α Δθ ⇒ ΔV V0 ΔV V0 Hence ϒ = = ϒΔθ (ϒ is a constant called coefficient of volume expansivity) ΔV V0Δθ Or ϒ = V – V0 V Δθ (since ΔV =V – V0) ⇒ V –V0 = VθϒΔθ V =V0 – V0ϒΔθ ∴ V = V0(1 - ϒΔθ) Note: Coefficient of volume expansivity = 3 × coefficient of linear expansivity ϒ = 3α Example 9.3 A metal vessel has a volume of 800.00 cm3 at 0 °C. If its coefficient of linear expansion is 0.000014/K, what is its volume at 60 °C? Solution Given: V0 = 800.00 cm3, (θ2 – θ1) = 60 °C and α = 0.000014/K = 0.000014/ °C 268 Change in volume, (ΔV) = 3 α V0Δθ = 3(0.000014/°C) × 800.00 cm3 × 60°C = 2.016 cm3 New volume (at 60°C) = V0 + ΔV = (800.00 + 2.016) cm3 = 802.016 cm3 Exercise 9.2 1. What do you understand by the phrase 'coeficient of linear expansion'? 2. A vertical s
teel antenna tower is 400 m high. Calculate the change in height of the tower hence its new height that takes place when the temperature changes from –19 °C on winter day to 39 °C on a summer day. (Take α = 0.00000649/K 3. A 8 m long rod is heated to 90 °C. If the rod expands to 10 m after some time, calculate its coefficient of linear expansion given that the room temperature is 32 °C. 4. A rectangular solid of Brass has a coefficient of volume expansion of 96 × 10-6 /°C. The dimensions of the rectangle are 9 cm × 6 cm × 8 cm at 10 °C. What is the change in volume and the new volume if the temperature increases to 90 °C? 5. A solid plate of lead of linear expansion 29 × 10-6 /°C is 8 cm × 12 cm at 19 °C. What is the change in area and the new area of the lead if the temperature increases to 99 °C? 9.3.2 Thermal expansion and contraction in liquids Like solids, liquids expand on heating i.e volume increases and contract i.e Volume reduces on cooling. But liquids expand more than solids since they have relatively weak intermolecular forces. Activity 9.17 will help us to understand expansion and contraction in liquids. 269 Activity 9.16 To demonstrate expansion and contraction in liquids (Work in pairsor in groups) Materials: • A glass flask • A rubber stopper • Long glass tubing Steps • Coloured water • Bunsen burner • Tripod stand • Wire guaze 1. Fill a glass flask with coloured water. 2. Fit the flask with a rubber stopper carrying a long narrow glass tubing. 3. Note the initial level of water in the glass tube before heating (Fig. 9.24). thin tube C A B glass flask coloured water wire gauze heat Fig. 9.24: Expansion of liquid 4. Heat the water in the flask. What happens to the level of water at A immediately the heating starts and after a few minutes? Explain your observation. In a similar activity, it was observed that at first the level of the coloured water in the tube drops to level B and then rises to level C. On heating, the glass flask is heated first and expands i.e its volume increases. The level of water immediately drops from A to B. On continuous heating, water starts to expand hence water level rises up the tube from B to C. If the setup is allowed to cool below room temperature, the water level drops to a point lower than A and B. This shows that liquids expand on heating and contract on cooling. Why liquids expand on heating? Molecules are loosely packed in liquids. The force of attraction between the molecules is weaker than in solids. The molecules move freely in the liquid. On 270 heating, the speed of the molecules increases. The collisions between the molecules increases the distance between them causing the liquid to expand. 9.3.3 Thermal expansion and contraction in gases Just like solids and liquids, gases expand on heating and contact on cooling. Gases expand more than liquids and solid because their molecules move furthest on heating. The following activity will help us to study expansion and contraction in gasses. Activity 9.17 To demonstrate expansion of gases (Work in pairs or in groups) Materials: • A thin glass flask • A long narrow glass tube Steps • A rubber stopper 1. Take a thin glass flask with an open top. 2. Close the flask with a rubber stopper carrying a long narrow glass tube. 3. Invert the flask so that the glass tube dips into water in a container. What do you observe? (Fig. 9.25). 4. Place your hands over the flask to warm it for sometime. What happens in water. Explain your observation. 5. Remove your hands from the flask and wait for some time. What happens to the level of water in the tube of the flask. Explain your observation. When the flask is warmed by the warmth of the hands, the level of water in the tube drops and some bubbles are formed due to air escaping from the flask through the tube. 271 Fig. 9.25: Expansion of airairthin glass flaskcontainertubecoloured waterbubble On removing the hands from the flask, water level rises up the glass tube again due to contraction of air i.e volume of air reduces on cooling.This shows that gases expand on heating and contract on cooling.. The volume of air increases in the flask due to expansion. Why a gas expands on heating? The force of attraction between the molecules of a gas is very small (almost negligible) and the distance between the molecules is large compared to solids and liquids. The molecules move freely in all directions. When a gas is warmed, the molecules gain more energy and move far apart hence volume increases. Different gases expand by the same amount when heated equally. Gases contract on cooling and expand on heating. 9.3.4 Applications of thermal expansion and contraction Activity 9.18 To find out the applications of expansion and contraction (Work in pairs or in groups) Materials • Internet enabled devices (lab computers or tablets) • Reference books Steps 1. You have now learnt about expansion and contraction. Suggest any three applications of expansion and contraction in our daily lives. 2. Carry out a research from the internet on the applications of expansion and contraction. 3. Report your findings to the whole class. Thermal expansion and contraction, on one hand is a nuisance and on the other hand is quite useful. The following are some of the applications of thermal expansion and contraction. 1. Electric thermostats A thermostat is a device made from a bimetallic strip that is used to maintain a steady temperature in electrical appliances such as electric iron boxes, refrigerators, electric geysers, incubators, fire alarms and the automatic flashing unit for indicator lamps of motor cars. Fig. 9.26 show two such devices. 272 cell iron D C A brass bell iron B A brass resistance wire (a) Fire alarm (b) Electric iron box Fig. 9.26: Electric appliances with thermostat The bimeltalic, as discussed earlier, bends on expansion and relaxes on cooling, connectin and disconnecting the circuit to regulate temperature. Be responsible and take care! Conserve energy by switching off the socket after using electrical appliances. Be careful when using electrical devices to avoid electric shocks. 2. Ordinary and pyrex glasses You may have observed that when boiling water is poured into a thick-walled glass tumbler it may break suddenly. This is because the inside of glass gets heated and expands even before the outside layer becomes warm. This causes an unequal expansion between the inside and the outside surfaces. The force produced by the expanding molecules on the inside produces a large strain in the glass and the tumbler breaks. This is the reason why pyrex glass tumblers are recommended for use while taking hot liquids. 3. Rivets In industries, steel plates are joined together by means of rivets. Hot rivets are placed in the rivet holes and the ends hammered flat. On cooling the force of contraction pulls the plates firmly together (Fig. 9.27). rivet rivet holes steel plates hot rivet rivet hammered flat Fig. 9.27: Rivets 273 4. Expansion of joint loops Metal pipes carrying steam and hot water are fitted with expansion joint or loops. These allow the pipes to expand or contract easily when steam or hot water passes through them or when the pipes cool down. The shape of the loop changes slightly allowing necessary movement of the pipes to take place (Fig. 9.28). Expansion Expansion Fig. 9.28: Expansion joint 5. Loosely fitted electric cables Telephone and electricity cables are loosely fitted between the poles to allow room for contraction in cold weather and expansion in hot weather. 6. Use of alloys The measuring tape used by surveyors for measuring land is made of an alloy of iron and nickel called invar. Invar has a very small change in length when temperature changes. 7. Gaps in railway tracks Gaps are left between the rails when the railway tracks are laid. The rails are joined together by fish-plates bolted to the rails. The oval shaped bolt holes allow the expansion and contraction of the rails when the temperature changes (Fig. 9.29). rail rail gap Bolts Oval shaped bolts holes fish plate rigid supports Fig. 9.29: Gaps left between rails In very hot weather, the gaps may not be enough if the expansion is large. The rails may buckle out. Modern methods use long welded lines rigidly fixed to the beds of the track so that the rails cannot expand. Expansion for the rails is provided by overlapping the plane ends (Fig. 9.30). 274 Fig. 9.30: Overlapping joints 8. Rollers on bridges The ends of steel and concrete bridges are supported on rollers. During hot or cold weather, the change in length may take place freely without damaging the structure (See Fig. 9.31). steel bridge fixed point wall rollers Fig. 9.31: Steel and concrete bridges are supported on rollers To demonstrate causes of expansion and contraction • Water 9. Breakages Activity 9.19 (Work in pairs) Materials: • A beaker • An immersion heater • A measuring cylinder • A thermometer • Stop watch 275 Steps the final temperature θ 1. Take 200 g of water in a beaker and note its initial temperature θ 2. Heat the water with an immersion heater for 10 minutes (Fig. 9.32 (a)). Note 1. 3. Repeat (2) above by taking 400 g of water in the same beaker and same initial 1 (Fig. 9.32 (b)). Note the time taken to produce the same 2 and calculate the change in temperature, Δθ = θ temperature θ change in temperature as before. Is it more or less? 2 – θ 1. 4. Compare the time taken to produce the same change in temperature in 200g and 400g of water. What is your conclusion? (a) (b) Beaker Thermometer Heating element 200 g of water Heating element Fig. 9.32: Relationship between heat energy and mass of the substance Beaker Thermometer 400 g of water Sudden expansion and contraction can lead to breakages of things like glasses and egg shells. This behaviour is mitigated against in the manufacture of glass items such as the drinking glass. They are made of thin walls to allow even expansion and contraction thus minimising chances of breakage. Exercise 9.
3 1. Use particles model to explain thermal expansion of solids. 2. Explain why: (a) Steel bridges are usually supported by rollers on one loose side. (b) Metal pipes carrying steam and hot water are fitted with loops. 3. Describe how shrink fitting is done. 4. State two applications of contraction of solids. 5. Name three physical properties that change when heating a solid. 276 Topic summary • Heat is a form of energy which is transferred from a region of higher temperature to a region of lower temperature. • The SI unit of heat energy is Joule (J). • Two substances of equal masses can be at the same temperature but contain different amounts of heat energy and vice-versa. • Heat energy can be transferred by three different modes: conduction, convection or radiation. • Solids are heated by conduction and fluids by convection. Radiation can take place through vacuum. • We get heat energy from the sun by radiation. Topic Test 9 For questions 1 – 9, select the correct answer from the choices given. 1. Radiation in a thermos flask is minimized by A. Cork C. Felt pad 2. A dull black surface is a good (i) Absorber of heat energy (ii) Emitter of heat energy (iii) Reflector of heat energy B. Vacuum D. Silvered glass water (i) only (ii) and (iii) only B. (i) and (ii) only D. (i), (ii) and (iii) A. C. 3. Radiation is the transfer of heat _______ A. B. C. D. 4. The mode of transfer of heat between the boiler and the storage tank of a hot in a liquid which involves the movement of the molecules. from one place to another by means of electromagnetic waves. through a material medium without the bulk movement of the medium. through a fluid which involves the bulk movement of the fluid itself. water supply system is A. radiation C. convention B. conduction D. evaporation 277 5. The transfer of heat by the actual movement of molecules of matter takes place B. only in gases D. A. only in liquid C. in solids and liquid 6. Match each heat transfer mechanisms to its description Conduction Evaporation Radiation Convection Electromagnetic waves. Transfer of vibrational energy from particle to particle. Escaping of particles from the surface of a liquid. Movement of particles due to changes in density. in liquids and gases 7. Explain the following statements: (a) A metallic seat seems to be hotter during the day and colder during the night than a wooden seat under the same conditions. (b) The bottom of cooking vessels are usually blackened. (c) It is safer to hold the other end of a burning match stick. 8. In a experiment requiring storage of heat energy, water is preferred to other liquids. Give two reasons for this. 9. A cup made of pyrex glass has a volume of 200 cm3 at 0 °C. If the coefficient of linear expansion is 0.000003/K, what will be its volume if it holds hot water at 92 °C? 278 UNIT 7 Magnetism Topics in the unit Topic 10: Magnetism Learning outcomes Knowledge and Understanding • Understand the theory of magnetism and explain the properties of magnets. Skills • Design investigations to determine the polarities of magnets, methods of magnetization and demagnetization, and how to distinguish between magnets and non-magnets. • Carry out accurate observation. • Recording results accurately in appropriate way. • Analysis of results in groups. • Explain analysis and consider applications. Attitudes • Appreciate the properties of magnets in construction of simple compass. Key inquiry questions • Why a compass needle does always points to the north? • Why that some magnets are classified as strong? • Why that a point is identified as neutral in magnetic field lines? • Why would you shield a small compass needle from earth’s magnetic field? • Why do we use soft iron keeper? 279 TOPIC 10 Magnetism Unit Outline • Definition of a magnet • Magnetic and non-magnetic materials • The poles of bar magnet • Test for magnetism • Types of magnets Introduction The people of Magnesia in Asia Minor observed that certain kinds of naturally occurring iron ores possessed an iron-attracting property. The ore was discovered near the city of Magnesia and hence it was named as Magnetite. Huge lumps of magnetite were often called lodestone meaning “ leading” stone or natural magnet. Chemically lodestone consists of iron oxide. Dr. William Gilbert (1540-1603) did a lot of work with the natural magnets. He published a book called De magnete in 1600 in which he gave an account of his research into the magnets and their properties. In one of his work he concluded that the earth was itself magnetic and that is why compasses point to the north of the earth. 10.1 Definition of a magnet Activity 10.1 To identify magnets (Work in groups) Materials: Cooking stick, steel nail, a bar magnet, a spanner, a cork Steps 1. Identify a magnet from the materials provided (see Fig. 10.1). Suggest a reason why you think the material you have identified is a magnet. 280 (a) `(b) (c) (d) (e) Fig. 10.1: Magnetic and non-magnetic materials 2. Discuss in your group what a magnet is. From Activity 10.1, you observed that Fig. 10.1 (d) is a magnet. A magnet is a piece of metal with either natural or induced properties of attracting another metal objects e.g. steel. The common type of a magnet used in school laboratory is a bar magnet (Fig. 10.1 (d)). We shall learn about types of magnets later. 10.2 Magnetic and non-magnetic materials Materials may be classified according to their magnetic properties. There are those that are attracted by magnets and others that are not. Identifying magnetic and non-magnetic substances Activity 10.2 To identify magnetic and non-magnetic substances (Work in groups) Materials: Iron and steel nails, bar magnet, copper metal, cobalt, wood, zinc, glass rods 281 Steps 1. Place some iron nails on the table. Bring a bar magnet close to the iron nails and observe what happens. Explain your observations. 2. Repeat the activity with other material such as copper, cobalt, steel, sulphur, brass, wood, cork, nickel, plastic, pens, wax, zinc, glass rods, carbon, aluminium, paper, chalk etc. 3. Record your observations in tabular form as shown in Table 10.1. Table 10.1: Magnetic and non-magnetic materials Substances attracted by a bar magnet Substances not attracted by a bar magnet 1. 2. 3. 4. 1. 2. 3. 4. 4. Discuss your observations in step 3 in your group and suggest the name given to substances that are attracted by a magnet and those that are not. The results from Table 10.1 shows that some materials are attracted by the bar magnet while others are not. The materials which are attracted by a magnet are called magnetic materials while those which are not attracted are called non-magnetic materials. The magnetic materials that are strongly attracted by a magnet are called ferromagnetic materials. These include nickel, iron, cobalt and steel. Materials that are not attracted by a magnet are called non-magnetic materials. Examples of non-magnetic materials include copper, brass, aluminium, wood, cork, plastic etc. When metals are mixed together, they form alloys. Some alloys are ferromagnetic materials. An example is Al-ni-co which composed of aluminium (Al), nickel (Ni) and cobalt (Co) hence the name Al-ni-co. Another example of alloys which are those composed of nickel, iron, copper, chromium or titanium; they are also ferromagnetic. 282 10.3 Properties of magnets (a) Polarity property of magnets Activity 10.3 (Work in groups) To identify the poles of a magnet Materials: A bar magnet, iron filings in a container, a paper Steps 1. Lay a bar magnet on a bench and cover it with a piece of paper. 2. Sprinkle the iron filings over the paper. What happens to the iron filings? Explain your observations. 3. Which parts have attracted more iron filings? 4. Suggest the name given to the ends of a magnet. From Activity 10.3, you must have noted that the iron filings were attracted by a bar magnet. Most iron filings remained clustered around the ends of the magnet as shown in Fig. 10.2. Bar magnet Iron filings Iron filings Fig. 10.2: Distribution of iron filings around a bar magnet. The ends of a magnet where the attraction is strongest are known as the magnetic poles. Magnetic poles are the places in a magnet where the total attractive force seems to be concentrated. A straight line drawn passing through these ends is called the magnetic axis of the magnet (see Fig. 10.3). Magnetic pole Fig. 10.3: Magnetic poles and magnetic axis of a bar magnet. A bar magnet has the strongest attraction at the poles. Magnetic pole 283 (b) Directional property of a magnet Activity 10.4 (Work in groups) To observe the directional property of a magnet Materials: A bar magnet, 1 metre long thread Steps 1. Suspend a bar magnet freely at its centre by a length of a cotton thread from a support (Fig. 10.4 (a)). Make sure there are no steel or iron objects near the magnet. N S (a) Magnetic meridian Fig. 10.4: A freely suspended magnet i n e l N - S 2. Displace the magnet slightly so that it swings in a horizontal plane. 3. Note the direction in which the magnet finally comes to rest. Suggest a reason why it rests in that direction. 4. Repeat the activity at different places and note the resting direction of the magnet. What do you observe about the resting direction of the magnet? Explain the direction of the magnet when it rests. In Activity 10.4, you observed that the bar magnet swings to and fro and finally rests in a north-south (N-S) direction of the earth. The magnet comes to rest with its axis in a vertical plane called the magnetic meridian (Fig. 10.4 (b)) i.e. a bar magnet rests in a north-south direction. The pole that points towards the north pole of the earth is called the north seeking pole or simply the north pole (N). The other pole is called the south seeking pole or south pole (S). 284 Identifying the poles of a magnet by colour Activity 10.5 (Work in groups) To identify the poles of a magnet by colour Materials: A bar magnet, 1 metre long thread Instructions 1. In this activ
ity you will conduct an investigation to identify the poles of a magnet. 2. Write a brief procedure of the investigation. Execute the procedure and conduct the investigation. After the activity answer the following questions. 3. Compare the direction shown by the compass and that of the suspended bar magnet. 4. Note the pole of the suspended bar magnet that is pointing in the same direction as north pole or south pole of the magnetic compass. Deduce the poles of the magnet. 5. Write a report on poles of magnets and present it in a class discussion. From Activity 10.5, you noted that the pole that points in the direction of the north of the compass is the north pole and the other pole is the south pole. In order to easily identify the poles of a magnet, the ends are usually painted in different colours. For example, the N-pole is painted red while the S-pole is painted blue or white Fig 10.5 (a). In other cases the whole bar is painted blue with a red dot or spot on one end to identify the north pole. (See Fig. 10.5 (b)). Red (North pole) (a) Blue (South pole) Red (North pole) (b) Fig. 10.5: Colours used to identify poles of a bar magnet Blue (South pole) Hey!! Do you know that the red colour in our national flag symbolises the blood that was shed for the independence of our country. Let us always live happily with one another and keep peace in our beautiful country. 10.4 Test for magnetism Basic law of magnetism Activity 10.6 (Work in groups) To establish the basic law of magnetism Materials: Two bar magnets, cotton thread. 285 Steps 1. Suspend a bar magnet using a light cotton thread with its north and south pole clearly marked. 2. Bring a S-pole of a second bar magnet slowly towards the S-pole of the suspended magnet (Fig. 10.6(a)). What happens to the magnets. 3. Repeat the activity using the S-pole of the suspended magnet and the N-pole of the second magnet (Fig. 10.6 (b)). What happened to the magnets? N repulsion S S N attraction N S N S (a) (b) Fig. 10.6: Action of magnets on each other. 4. Repeat using the other poles and record your observation in a tabular form as shown in table 10.2. Poles of suspended magnet Pole of second magnet Observation South South North North South North South North _______________ _______________ _______________ _______________ 5. Why does some poles attract whereas others repel each other? Table 10.2: Test for magnetism From Activity 10.6, you must have discovered that a north pole attracts a south pole, a north pole repels a north pole and a south pole repels a south pole. Therefore, unlike poles attract each other while like poles repel each other. This is called the basic law of magnetism. In the previouus class, we learnt about charges. Like charges repel whereas unlike charges attract. The same concept is applied in the basic law of magnetism. Like poles repel whereas unlike poles attract. 286 Testing the polarity of magnets using the basic law of magnetism Activity 10.7 To test for polarity of magnets using the basic law of magnetism (Work in groups) Materials: A nail, two bar magnets, a cotton thread Steps 1. Freely suspend a bar magnet as shown in Fig. 10.7. 2. Bring one pole of the magnet close to a nail placed on a table. What happens to the nail?. cotton thread nail S N Fig. 10.7: Testing the polarity of a magnet. 3. Repeat with the other pole close to the nail and record your observations. 4. Repeat steps 2 and 3 using a second bar magnet instead of the nail. What do you observe? Discuss your observations. From Activity 10.7, you must have observed the following: 1. There is attraction when the south or north pole of the suspended magnet is brought near the nail. 2. When the second bar magnet is used, there is attraction with one pole and repulsion with the other pole. Therefore, there is always attraction between a magnet and a magnetic material and also between the unlike poles of magnets. But there is repulsion only between two like poles of magnets. Repulsion is therefore, the only sure way of testing for polarity of a magnet. 287 10.5 Types of magnets Activity 10.8 (Work in groups) To magnetise a piece of iron nail Materials: A piece of soft iron nail (about 3 inches), about 1 m of thin coated copper wire, a fresh size D dry cell, iron filings, cello tape. Steps 1. Leave about 15 cm of wire loose at one end and wrap part of the remaining section of the wire around the nail. 2. Cut the wire (if needed) but ensure that there is at least 15 cm of wire loose on the other end too. 3. Remove about 2 cm of the plastic coating from each end of the wire. Attach one end of the wire to terminal of the dry cell, and the other end of the wire to the other terminal of the battery using a cello tape (Fig 10.8). Be careful though, the wire might get very hot! Soft iron nail Dry cell Copper wire Fig. 10.8: Magnetising a piece of iron nail 4. Bring one end of the nail near the iron filings. What do you observe? Explain. 5. Disconnect one end of the wire from the dry cell. What happens to the iron filings? Explain. 6. Repeat the activity by replacing soft iron nail with a steel nail and increase current by adding more new dry cells to the circuit. What do you observe? Explain your observations to other members in your group. From Activity 10.8, you must have noted that the soft iron nail attracts the iron filings only when the circuit is complete i.e., when electric current is flowing through the wire.This shows that the iron nail becomes a magnet. This kind of a magnet which is made by passing current through a coil is called an electromagnet. 288 When the circuit is disconnected, no current will flow in the wire, hence the iron nail does not attract the iron filings. The electromagnet is a temporary magnet. When the steel nail was used instead of iron nail and current was increased, the iron filings did not fall off after being attracted and the circuit disconnected. Hence, the steel nail becomes a permanent magnet. Therefore,there are two basic types of magnets; permanent and temporary magnets. (a) Permanent magnets In Activity 10.8, you should have observed that the steel nail retained its magnetism for a longer time. Permanent magnets are those magnets that retain magnetic properties for a long time. They are made from hard magnetic materials e.g steel. An example of naturally occurring permanent magnet is lodestone, which is composed of a mineral called magnetite. Other permanent magnets are made from mixing magnetic materials (such a mixture is known as an alloy). Examples of alloys commonly used to make permanent magnets are Al-ni-cos i.e iron alloys containing aluminium, nickel, and cobalt. Steel which is mixture of carbon and iron and materials containing rare-earth elements like samarium, neodymium or ferrites (an oxide of iron). Permanent magnets can be made into any shape to fit the usage. They can be made into round bars, rectangles, horse-shoes, donuts, rings, disks and other custom shapes. Fig 10.9 shows some permanent magnets named according to their shapes. N S N S (a) A bar magnet (b) U-shaped magnet N S S N (c) Horse-shoe magnet (d) A ring magnet Fig. 10.9: Shapes of permanent magnets. Fig 10.10 shows a ceramic or magnadur magnet. The poles of ceramic magnet are at its faces (Fig. 10.10). These types of magnets are stronger than other magnets 289 of comparable size. They are greyish/black in colour. Magnadur magnets consist of basically iron oxide and barium oxide. south pole north pole Fig. 10.10: Ceramic magnet (magnadur). Uses of permanent magnets Permanent magnets are used to lift heavy loads in industries. They handle loads with extreme easiness in the minimum area.This makes them efficient because they always operate from the top without compressing or deforming the load. Note: Note: A permanent magnet system for lifting loads is safe since it is not affected by any electrical power failure. Therefore, no battery or generator backup system is required. Other uses of permanent magnets include: 1. Removing of iron pieces from the eyes of patients in hospitals. 2. Setting of six’s maximum and minimum thermometer in weather stations. 3. To show the direction as in compass needles for navigation. 4. Magnetic tapes use permanent magnets in audio and video recorders. 5. They are used in Jewellery e.g earing, bracelets, necklaces to assist them clapsed (closed). (b) Temporary magnets From Activity 10.8, you should have established that temporary magnets are those magnets which act as magnets only when there is a flow of electric current or a presence of a permanent magnet. They loose magnetism when the permanent magnet is removed or electric current is cut off. Some may retain weak magnetic properties. They are made using soft magnetic materials like soft iron, iron-silicon alloys and iron-nickel alloys. An example of temporary magnet is an electromagnet. A simple electromagnet is made by winding a wire carrying current round a soft magnetic material and then connecting the wire in a simple circuit. 290 Uses of temporary magnets 1. Electromagnets are used in motors, loud speakers, telephone, earphones and among other devices. 2. To seperate materials made of magnetic metals in the scrap yard. Exercise 10.1 1. Explain the meaning of the following terms: (a) a magnet. (b) a magnetic substance/material. (c) a non-magnetic material. (d) a ferro-magnetic material. 2. Group the following materials into magnetic and non-magnetic materials: Zinc, paper, aluminium, graphite, steel and plastic. 3. Explain how you can identify the polarity of a magnet whose poles are not marked? 4. Two steel pins were attracted by a magnet. When a south pole was brought in between the two pins, the pins moved further away, as shown in Fig. 10.11. Explain why the pins moved apart. bar magnet N S pins Fig. 10.11: Steel pins attracted to a bar magnet 10.6 Magnetic field pattern around a magnet Activity 10.9 (Work in pairs) To investigate the existence of magnetic field around a magnet Ma
terials: 2 bar magnets, a magnetic compass Instructions 1. In this activity you will design and carry out an investigation to investigate magnetic field pattern around a magnet. (a) Which direction did the compass needle point? why? (b) What did you observe? Discuss your observation to your class partner. 291 2. Write a brief procedure for your investigation correctly, execute the procedure and conduct the investigation. 3. Write a report and present it in a class discussion. During the discussion answers the following questions from your class members. 4. How can the investigation be improved? In Activity 10.9, you should have observed that in steps 2 and 3 when the magnetic compass was placed near the bar magnet, its direction changes. This shows that there is a magnetic effect in the region around the magnet. In this region, there exist magnetic force of attraction and repulsion. This space or region is called magnetic field, and is represented by the lines of force called magnetic field lines. These field lines form a pattern called magnetic field pattern. 10.6.1 Drawing magnetic field pattern round a magnet Activity 10.10 To investigate magnetic field pattern around a magnet using iron filings (Work in pairs) Materials: 2 bar magnets, U-shaped magnet, iron filings , stiff paper Steps 1. Place a smooth stiff paper on top of a bar magnet. 2. Sprinkle iron filings onto the stiff paper. What happens to the iron filings? Explain your observation to your class partner. 3. Tap the paper gently and draw the pattern displayed by the iron filings. 4. Repeat the activity with north poles of two bar magnets close together and then south to north poles. Observe and draw the pattern displayed by iron filings. 5. Repeat steps 1 to 3 by using a U-shaped magnet. 6. Compare and discuss your patterns with other pairs in your class before presenting your findings to the whole class. In Activity 10.10, you observed that the iron filings are attracted by the magnet, since there is a magnetic effect in the region around the magnet. The pattern displayed by the iron filings represents the magnetic lines of force. Fig.10.12 shows the photograph and magnetic field lines of the iron filings arrangement around a bar magnet. Note that the lines do not cross each other. 292 (a) (b) Fig. 10.12: Magnetic field lines around a bar magnet When the two north poles or north and south pole are placed close to each other and the steps 1 to 3 repeated, the pattern displayed is as shown in Fig 10.13(a) and (b) respectively. N N S N S N Neutral point (a) (b) Fig. 10.13: Magnetic field patterns Note that magnetic field lines originate from the north pole of a magnet to the south pole. In step 5 of Activity 10.12, the pattern displayed when the U-shaped magnet was used is as shown in Fig. 10.14. Fig. 10.14: Magnetic field patterns of U-shaped magnet 293 10.7 Magnetisation and demagnetisation 10.7. 1 Structure of an atom Activity 10.11 (Work in groups) To demonstrate the composition of an atom Materials required: A model of an atom, internet, reference books Steps 1. From the knowledge in Chemistry, tell your group members what an atom is. 2. Draw a diagram of an atom showing the nucleus and electrons. Discuss each part of the atom acknowledging the spinning of electrons around the orbit. 3. Now, using the model provided, show your group members the nucleus, the electrons and the orbit where the electron spins. 4. Hold a discussion on the existence of magnetism in an atom. In Secondary 1, and in Chemistry we learnt about atoms. An atom is the smallest part of an element that can take part in chemical reactions. In this section we will learn about how atoms take part in magnetism. Magnetism appears to be due to the spinning of electron about the nucleus of an atom. When thinking about magnetism consider the smaller units of an atom have even smaller sub units: protons, neutrons and electrons. An atom has a nucleus. Around the nucleus there are electrons spinning around it along their orbits. Orbit Electron Spin Nucleus Fig. 10.15: The orbit of a spinning electron about the nucleus of an atom. 10.7.2 Domain theory of magnetism The domain theory of magnetism developed by a scientist called Wilhelm Weber helps to explain the phenomenon of magnetism. 294 Activity 10.12 To show the existence of molecular magnets (dipoles) (Work in groups) Material: A bar magnet Steps 1. Cut the the bar magnet provided into two halves. Test the polarities of the ends of each half. 2. Cut one of the halves into two halves and again test the polarities. 3. Continue cutting one piece until you are not able to cut it any more. Each time test the polarity of the halves (Fig. 10.16). Fig. 10.16: Effects of cutting a bar magnet. 4. Do the small pieces behave like magnets? Why? The polarity test for the first halves shows that each piece is a magnet in its own right. Further cutting of the pieces still yields a smaller magnet. The smallest portion of any matter is an atom. Thus if we were able to cut the magnet further we would see that the smallest magnets. These small magnets are called dipoles. A dipole is the smallest particle of a magnetic material. It is equal to an atom in electric conductor. In ferromagnetic materials these small magnets or dipoles (with two poles) occupy tiny regions called domains. The magnetism of each domain is aligned. However, the domain points are in different and random directions (Fig. 10.17). Domains Molecular magnets (dipoles) Fig. 10.17: Unmagnetised ferromagnetic material. The domain theory may be used to explain the processes of magnetisation and demagnetisation. The domain theory states that inside a magnet there are small regions in which the magnetic direction of all dipoles are aligned in the same directions. 295 10.7.3 Magnetisation Activity 10.13 (Work in groups To find out what magnetisation is Materials: Reference books, internet Steps 1. Tell your group member what you understand by magnetisation. 2. List to them four method of magnetisation. 3. Now carry a research from Physics reference books and internet to verity your points Magnetisation is the process of making a magnet from a magnetic materials. In the second century A.D, Chinese found a method of making magnets by rubbing pieces of common iron against lodestone. Nowadays magnets are made using various methods. Some of these methods include stroking or touching method, electric method, hammering and induction method. (a) Stroking or touching method (i) Single stroking A piece of steel e.g. steel needle placed near a magnet becomes magnetised. However, the magnetism acquired usually disappears quickly when the magnet is removed. This magnetism may be enhanced by stroking. Activity 10.14 (Work in groups) To make a magnet by single stroking method Materials: A steel needle, A bar magnet Steps 1. Place a steel needle on the bench. 2. Stroke the steel needle with the south pole of a bar magnet along the whole length of the steel needle. 3. Once at the end, lift the magnet well away from the steel needle i.e. make a wide sweep as shown in Fig. 10.18. 296 N magnet wide sweep A S steel needle B Fig. 10.18: Making magnets by single stroke method. 4. Repeat the process in step 2 and 3 several times. Test the polarity of the steel needle by the repulsion method. What is the polarity of end A and B? The steel needle becomes magnetised with end B becoming a S-pole and end A becoming a N-pole. Note that the end of the magnetic material last touched by the magnet acquires a polarity opposite to the one touching it. (ii) Double stroking Activity 10.15 To make a magnet by double stroking (Work in groups) Materials: A steel needle, two bar magnets Steps 1. Stroke a steel needle using two magnets as shown in Fig. 10.19. 2. The stroking should begin at the middle of the steel needle each time making sure that the two bar magnets are lifted far away from the steel needle once you reach the ends. N S C S N D Fig. 10.19: Making magnets by double or divided stroke method. 3. Test the polarity at the end of the needle. What is the polarity at the end C and D? 4. Suggest another name of this method i.e. double stroking of making a magnet. 297 It is observed that end C becomes a N-pole while end D becomes a S-pole. The double stroke method is also called the divided stroke method. (iii) Consequent poles Activity 10.16 To make a magnet with consequent poles (Work in groups) Materials: Two bar magnets, A steel pin Steps 1. Repeat Activity 10.14 as shown in Fig. 10.20 using similar poles in stroking. N N S S Fig. 10.20: Making a magnet with consequent poles 2. Test for the polarities of the magnet made. What do you notice? Discuss. Similar poles appear at each end of the steel needle. The middle of the needle becomes a south pole (Fig. 10.21). N S S N Fig. 10.21: Consequent poles It is possible to have a magnet with like poles on both ends. These are called consequent poles. (b) Electrical method Activity 10.17 Work in groups To make a magnet by electrical method Materials: A wire, a hollow rod, a steel needle, a d.c source Steps 1. Wind a number of turns around a hollow rod. This is called a solenoid. Place a steel knitting needle in the solenoid and pass a direct current (d.c) through the turns of the solenoid as shown in Fig. 10.8. 298 2. Switch off the current and remove the needle after some minutes. Test for polarity of the needle. What is the polarity of the needle? 3. Repeat the experiment but with the electric current direction reversed. Test for polarity of the needle. low d.c supply switch ammeter A steel needle inside a solenoid N S Fig. 10.22: Making a magnet by electrical method. The steel needle attracts iron fillings after the process. The polarity of the field is easily seen by examining the path of the conventional current in the coil. If looking at the end of the coil through current is going clockwise it will produce a south seeking pole. A capital 'S' has the ends following the clockw
ise rotation. Similarly, the other end will be anticlockwise. This produces a north seeking pole. A capital 'N' has the ends following anticlockwise rotation. It is observed that the polarity of the magnet produced depends on the direction of the electric current. The steel needle is magnetized by electric method. (c) Hammering method Activity 10.18 To make a magnet by hammering method (Work in groups) Materials: A hammer, a steel bar, iron fillings Steps 1. Hammer one end of a steel bar fixed in north-south direction several times. (Fig. 10.23). 2. Place the steel bar into ironing filings. What is your observation? Discuss. S Hammer Steel bar e N-S lin N Fig. 10.23: Magnetisation by hammering. 299 The steel bar attracts iron filings after hammering becoming a weak magnet. The lower end becomes a weak north pole for countries in the northern hemisphere and a south pole for countries in the southern hemisphere. In this method the influence of earth’s magnetic field is used to magnetise the steel bar being hammered. (d) Induction method Activity 10.19 To make a magnet by induction method (Work in groups) Materials: Two bar magnets, Two steel pins Steps 1. Place a magnet near an unmagnetised steel pin and note what happens. What do you obeserve? 2. Bring another pin next to the first pin and note what happens. 3. Remove the bar magnet and observe what happens to the second pin. S N Bar magnet pins Fig. 10.24: Verifying induction method 4. Separate the first pin from the magnet and note what happens to the second pin. 5. Repeat the activity using two pins placed side by side. Introduce a north pole between the two pins and observe what happens. When a steel pin is brought near a magnet, the pin is attracted by the bar magnet. The second pin is attracted to the first pin. It is observed that the second pin falls off when the bar magnet was removed. This shows that the presence of the bar magnet sustains the magnetism between the first and second pin. The first pin becomes magnetised by the magnet through a process called induction and then gets attracted to the magnet. The second pin gets magnetised by the first pin through the same process. 300 The two pins separate further when a north pole is placed in between them (Fig. 10.25). S N S N S N N S Fig. 10.25: Identiying polarity of the pins This shows that the induced pole nearest to the magnet is of opposite polarity to that of the inducing magnet. Induced magnetism is a process in which unmagnetized magnetic material is made a magnet by touching or bringing it near the pole of a permanent magnet. The following activity illustrates magnetic induction. Activity 10.20 To illustrate magnetic induction (Work in groups) Materials: Bar magnet, iron bar, iron filings Steps 1. Place a magnetic material e.g. iron bar near a bar magnet for some time. 2. Sprinkle some iron filings near one end of the iron bar. 3. Observe what happens on the iron fillings. Discuss. From Activity 10.20, we notice that after sometime, the magnetic material attracts the iron filings. The magnetic material is magnetized by a permanent magnet by induction. Fig. 10.26 shows magnetic induction of a magnetic material. Before attraction, the north pole of a magnet repels the north poles of the magnetic dipoles in a magnetic material thereby creating a south pole thereafter attraction occurs as shown below. Magnet Magnetic material S N S N Fig 10.26: Magnetic induction of a magnetic material 301 Facts about magnetism Magnetization is the process of making a magnet from a magnetic material. Since the domains are aligned in all possible directions in an unmagnetised material, the net magnetism in the material is zero. In a partially magnetised material the domains align themselves as shown in Fig. 10.27(a) below. Note that the domains in Fig.10.27(a) are not all aligned in the same direction. When the material is fully magnetised the domain walls move and the molecular magnets align themselves in one particular direction as shown in Fig. 10.27 (b). S N S N (a) Partially magnetised (b) Fully magnetised Fig. 10.27: Magnetised ferromagnetic material. A ferromagnetic material is said to be magnetically saturated when the walls are swept out and the molecular magnets point in the same direction. A resultant north pole is produced at one end and a south pole at the other end as shown in Fig. 10.27(b). 10.7.4 Demagnetisation (a) Hammering method Activity 10.21 To demonstrate how a magnet can lose its magnetism (Work in groups) Materials: A hammer, A bar magnet, Iron nail Steps 1. Place a bar magnet in East-West direction and hammer it severally using the hammer. 2. Using the iron nail provided, test whether it can attract the iron vigil. What do you observe? Explain. 3. Tell your members what demagnetisation is and list the methods of demagnetization. When a bar magnet is placed in East-West direction and hammered severally, it loses it magnetism. The process through which magnets loses its magnetism is called demagnetisation. 302 (b) Heating method Activity 10.22 To investigate demagnetisation of a magnet by heating method (Work in groups) Materials: A Bar magnet, source of heat, iron filings Steps 1. Place a magnet in east-west direction and heat it for some time. 2. Cool it and put iron filings near the magnet. Observe what happens. From Activity 10.22, we notice the magnet does not attract the iron filings. This is because of demagnetisation of the magnet. Heating a magnetised material until it becomes red-hot and cooling it suddenly when resting in East-West direction makes it lose its magnetism. This method is called heating method. (c) Electrical method Activity 10.23 To investigate demagnetization by electrical method (Work in groups) Materials: A solenoid, AC supply, A magnetized needle, Iron rod Steps 1. Placing a magnetised needle in a coil placed in East-West direction and passing an alternating current (a.c) demagnetises the needle (Fig. 10.28). 2. Switch off the current supply and remove the needle from the solenoid. Test its magnetism using iron filings. What do you notice? Explain. The needle loses its magnetism and thus does not attract the iron rod. Therefore, the needle is demagnetised by electrical method a.c supply switch ammeter A steel needle inside a solenoid W E Fig. 10.28: Demagnetisation by electrical method 303 Facts about demagnitisation During demagnitisation, the walls of the domain slowly return to their original state with time as this is a more stable state; hence the material becomes demagnetised. This kind of demagnetisation is called self-demagnetisation. This is due to the poles at the end which tends to reverse the direction of the molecular magnets. The demagnetisation process can also be influenced externally by giving the molecular magnets enough energy to overcome the forces holding them in a particular direction. The energy may be provided by heating, hammering or dropping on a hard surface or by using an alternating current. Exercise 10.2 1. State the domain theory of a magnet. 2. Define: (a) Magnetisation. (b) Demagnetisation. (c) Dipoles. 3. Explain the term induced magnetism. 4. List four methods of making a magnet. 5. Describe how a single stroking is done during making of a magnet. 6. Name two methods of demagnetisation. 10.7.5 Magnetic properties of steel and iron Magnetic materials are classified according to how well they retain their magnetism once they are magnetised. Activity 10.24 (Work in groups) To distinguish between hard and soft magnetic materials Materials: Four new dry cells, iron filings in two identical dishes, insulated copper wire, switch, two identical iron and steel rods Steps 1. Make equal turns of the insulated copper wire around the iron and steel rods. 2. Connect the circuit as shown in Figure 10.29. Observe and compare the quantity of iron fillings attracted to iron and steel rods. Explain your observation. 304 Switch Dry cells Steel Iron Iron filings Fig. 10.29: Steel is more retentive than iron 3. Open the switch after a short while and compare the quantities of iron filing left attracted to the iron and steel rods. Explain your observation. When the switch is closed, iron gets magnetised easily than steel by the electric current (d.c), hence it attracts more iron filings. When the switch is opened, iron gets demagnetised easily than steel hence drops most of the filings. Steel retains most of iron filings attracted to it. Iron is easily magnetised or demagnetised. Steel is hard to magnetise than iron but once magnetised it retains its magnetism for a longer time. 10.8 The effect of magnetic materials on the magnetic field Activity 10.25 (Work in groups) To investigate the effect of magnetic materials on the magnetic field Material: Magnets with magnetic keepers, Circular magnet with a shield e.g. that of radio loudspeakers, internet, reference books Steps 1. Remind your group members the meaning of magnetic field and how to draw them. 2. Draw magnetic field lines when two bar magnets are brought near each other with North pole facing the south pole of the other magnet. What will happen if the soft iron ring is placed in between the bar magnets? Draw the magnetic field. 3. Now, discuss the meaning of magnetic shield and magnetic keepers. In your discussion, point out the importance of magnetic shielding and magnetic keepers. (You may refer to physics reference books or internet). 305 Magnetic materials affect any magnetic field in the position which it is placed. Fig. 10.30 shows the effect of soft iron in magnetic field. N S N S soft iron soft iron (a) Soft iron (b) soft iron N S soft iron ring (c) Soft iron ring Fig. 10.30: Lines of force concentrate into magnetic materials The magnetic field lines do not penentrate inside a soft iron ring as shown in Fig. 10.30(c) above. This is why a compass placed inside the ring does not point in any fixed direction. Hence a soft iron ring may be used to shield instruments from magnetic effects.
Magnetic shielding (screening) Magnetic shielding (screening) is the process that limits the coupling of a magnetic field between two locations. It is done using a sheet of metal, metal mesh, ionized gas or plasma. Magnetic keepers Magnetic keepers are bars made from soft iron or steel placed across the poles of permanent magnet to preserve the strength of the magnets. 10.9 Methods of storing magnets Activity 10.26 (Work in groups To investigate the methods of storing magnets Materials: Magnets with iron keepers, reference books, internet Steps 1. Take a close look on the magnets with iron keepers provided to you. From the knowledge learnt on magnetic keepers, suggest the function of iron keepers. 306 2. Suggest other methods of storing magnets. 3. Now carry out a research from physics reference books or internet on methods of storing magnets (a) Using keepers A bar magnet loses its magnetism with time due to self-demagnetisation. The process of self-demagnetisation starts at the ends of a magnet in which the free like poles repel each other and slowly upsetting the alignment of the molecular magnets inside it. To minimise this, soft iron bars called keepers are placed across their ends as shown in Fig. 10.31. The dipoles find themselves in a closed chain or loops round the magnet and the keepers, with no free poles, available to upset the domains. The soft iron keepers are used since they are easily magnetised by induction method. keeper S N N S S N SN keeper Fig. 10.31: Soft iron keepers used to store magnets. (b) Storing magnets away from heat and metal objects Heat can cause poles of the atoms in a magnet to randomize and this will eventually destroy a magnet. It is therefore, recommended to store magnets in a cool place, in room temperature and out of the sun. Weak magnets will easily be ruined by stronger magnets and electromagnetic field so you must make sure to keep them out of touch with other stronger magnets. (c) Keeping magnets out of reach of children Magnets are extremely dangerous to swallow and therefore should be stored far away from the reach of kids or pots. Just like you store medicine high up or behind lock, you need to do the same with magnets. Especially for magnetic jewelry that looks attractive to play with. Children might also have magnets as toys and it is therefore important parents to teach them how to store the magnets properly. 307 (d) Protecting magnets against mechanical shocks Many magnets can lose their magnetism through improper storage. It is important to understand how to store magnets if you want to prolong their life time. Don’t throw the magnets to the storage box but rather place them gently. Avoid striking them with a hammer or any other material. It will lose its magnetism or even crack and break. Uses of magnets As learnt in Activity 10.14, electromagnets are made from soft magnetic materials e.g. iron. The following are some of the uses of magnets. 1. Magnets are widely used in electric and electromechanical devices such as electric ball, motors, generators, in transformers, relays, magnetic lifts in industries to lift heavy metals in a microwave oven in kitchen etc. 2. The most important use of magnets is the magnetic compass that is used to find the geographical directions. 3. They are also used in the speakers that can convert the electrical energy into sound energy. 4. They are used in the electrical bells. 5. They are also used to sort out the magnetic and non magnetic substances from the scrap. 6. They are used in TV screens, computer screens, telephones and in tape recorders. 7. They are used in the candy or cold drink vendors to separate the metallic cap from the lots. 8. They are used in cranes. 9. They are used in the refrigerators to keep the door close. 10. They are used in the Maglev trains. In the Maglev trains, the super conducting magnets are used on the tracks on which the train floats. These types of the trains are working on the repulsion force of the magnets. 308 Topic summary • A magnet is a piece of metal with either natural or induced properties of attracting other metal objects. • Materials which are attracted by a magnet are called magnetic materials while those which are not attracted are called non-magnetic materials. The ends of a magnet where the attraction is strongest are known as the magnetic poles. There are two types of poles. North pole and south pole. • • Unlike poles attract each other while like poles repel each other. This is called the basic law of magnetism or the first law of magnetism. • Repulsion is the only sure way of testing for polarity of a magnet. • The space or region around a magnet is called magnetic field. It is represented by the lines of force called magnetic field lines. • Magnetism is due to the spinning of electrons about the nucleus of an atom. • Domain is tiny regions in a magnetic material that is occupied by the dipoles. • Dipole is the smallest particle of a magnetic material. It is equal to an atom in electric conductor. • Domain theory states that inside a magnet there are small regions in which the magnetic direction of all dipoles are aligned in the same direction. • Magnetisation is the process of making a magnet from a magnetic materials. • The methods of magnetization include: Stroking method. 1. 2. Electrical method by passing a.c current. 3. Hammering method where a magnetic material is hammered facing N-S direction. Induction method. 4. • Demagnetisation is the process through which a magnet loses its magnetism. • The method of demagnetization include: 1. Heating a magnet till rod-hot and cool it suddenly while facing East- West direction. 2. Hammering a magnet while facing East-West direction. 3. Electrical method by passing AC current. • Magnetic keepers are used to store magnets. They help to preserve the strength of the magnets by completing the magnetic circuit. 309 Topic Test 10 1. State two properties of a magnet. 2. State the basic law of magnetism. 3. Name four types of magnets according to shapes. 4. Describe an experiment to explain the existence of magnetic poles. 5. Explain what would happens to a U-shaped magnet if it is freely suspended as shown in Fig. 10.32 below. N S S Fig. 10.32: U-shaped magnet 6. What is the main difference between a ceramic magnet and a bar magnet? 7. You have been provided with the following; (a) a rod labelled S, which is a magnetic material. (b) a rod labelled N, which is a non-magnetic material. Explain how you would identify them. 8. The magnets shown in Fig. 10.33, pole B attracts pole P and pole Q attracts pole X. If pole Y is South pole:. A B P Q X Y Fig. 10.33: Bar magnet (a) What is the polarity of P? (b) What would happen if the following poles are brought close together: (i) pole B and X (iii) pole B and pole Y (ii) pole A and pole Q (c) Draw a magnetic field pattern when pole B and P are placed near each other. 9. Define the following: (a) Dipole (c) Self-demagnetisation (b) Domain theory (d) Consequent poles 310 10. Differentiate between a magnet and magnetic material using the domain theory. 11. (a) Explain the following terms using domain theory of magnetisation (i) Magnetisation (ii) Demagnetisation (b) Explain three methods of magnetising and demagnetising magnetic materials and a magnet respectively. 12. Fig. 10.34 shows what is observed when two steel pins AB and CD, hang from the north pole of a magnet. S N A C B D Fig. 10.34 (a) Explain why the distance BD is greater than AC. (b) State the polarity of ends B and D. (c) Explain what is meant by magnetic induction. 13. With the aid of a diagram, explain how bar magnets are stored so as to minimize self-demagnetisation. 14. Using the domain theory of magnetism, explain why. (a) The strength of a magnet cannot be increased beyond a certain point. (b) The temperature increase weakens or destroys the magnetism of a magnet. 15. Describe all experiment to show how to make a magnet by (a) Single stroke method (b) Electrical method 16. Discuss three applications of magnets in our society today. 311 S o u t h S u da n S o u t h S u da n Secondary Secondary Student’s Book 22 Secondary Physics has been written and developed by Ministry of General Education and Instruction, Government of South Sudan in conjunction with Subjects experts. This course book provides a fun and practical approach to the subject of Physics, and at the same time imparting life long skills to the students. The book comprehensively covers the Secondary 2 syllabus as developed by Ministry of General Education and Instruction. Each year comprises of a Student’s Book and teacher’s Guide. The Student’s Books provide: Full coverage of the national syllabus. A strong grounding in the basics of Physics. Clear presentation and explanation of learning points. A wide variety of practice exercises, often showing how Physics can be applied to real-life situations. It provides opportunities for collaboration through group work activities. Stimulating illustrations. All the courses in this secondary series were developed by the Ministry of General Education and Instruction, Republic of South Sudan. The books have been designed to meet the secondary school syllabus, and at the same time equiping the students with skills to fit in the modern day global society. Secondary Secondary Student’s Book 22 This Book is the Property of the Ministry of General Education and Instruction. This Book is not for sale. Any book found on sale, either in print or electronic form, will be confiscated and the seller prosecuted. Funded by: Published by: This Book is the Property of the Ministry of General Education and Instruction. This Book is not for sale. Funded by:

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