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that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them. Elastic Collision An elastic collision is one that conserves internal kinetic energy. Internal Kinetic Energy Internal kinetic energy is the sum of the kinetic energies of the objects in the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 329 Figure 8.9 An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved. Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is or 1 + 2 = ′1+ ′2 net = 0 1 1 + 22 = 1′1 + 2′2 net = 0, (8.43) (8.44) where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus, 2 = 1 2 (two-object elastic collision) 2 + 1 2 + 1 (8.45) 22 2 21 ′1 22 ′2 1 21 1 expresses the equation for conservation of internal kinetic energy in a one-dimensional collision. Making Connections: Collisions Suppose data are collected on a collision between two masses sliding across a frictionless surface. Mass A (1.0 kg) moves with a velocity of +12 m/s, and mass B (2.0 kg) moves with a velocity of −12 m/s. The two masses collide and stick together after the collision. The table below shows the measured velocities of each mass at times before and after the collision: Table 8.1 Time (s) Velocity A (m/s) Velocity B (m/s) 0 1.0 s 2.0 s 3.0 s +12 +12 −4.0 −4.0 −12 −12 −4.0 −4.0 The total mass of the system is 3.0 kg. The velocity of the center of mass of this system can be determined from the conservation of momentum. Consider
the system before the collision: ( + ) = + (3.0) = (1)(12) + (2)( − 12) = − 4.0 m/s (8.46) (8.47) (8.48) 330 Chapter 8 \| Linear Momentum and Collisions After the collision, the center-of-mass velocity is the same: ( + ) = ( + ) (3.0) = (3)( − 4.0) = − 4.0 m/s The total momentum of the system before the collision is: + = (1)(12) + (2)( − 12) = − 12 kg • m/s The total momentum of the system after the collision is: ( + ) = (3)( − 4) = − 12 kg • m/s (8.49) (8.50) (8.51) (8.52) (8.53) Thus, the change in momentum of the system is zero when measured this way. We get a similar result when we calculate the momentum using the center-of-mass velocity. Since the center-of-mass velocity is the same both before and after the collision, we calculate the same momentum for the system using this method both before and after the collision. Example 8.4 Calculating Velocities Following an Elastic Collision Calculate the velocities of two objects following an elastic collision, given that 1 = 0.500 kg, 2 = 3.50 kg, 1 = 4.00 m/s, and 2 = 0. (8.54) Strategy and Concept First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 8.9 where both objects are initially moving. We are asked to find two unknowns (the final velocities ′1 and ′2 ). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus 2 = 0. Once we simplify these equations, we combine them algebraically to solve for the unknowns. Solution For this problem, note that 2 = 0 and use conservation of momentum. Thus, or 1 = ′1 + ′2 1 1 = 1′1 + 2′2. Using conservation of internal kinetic energy and that 2 = 0
, Solving the first equation (momentum equation) for ′2, we obtain 1 21 1 2 = 1 21 ′1 2 + 1 22 ′2 2. ′2 = 1 2 1 − ′1. (8.55) (8.56) (8.57) (8.58) Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable ′2, leaving only ′1 as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are and ′1 = 4.00 m/s ′1 = −3.00 m/s. (8.59) (8.60) As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution (′1 = −3.00 m/s) is negative, meaning that the first object bounces backward. When this negative value of ′1 is used to find the velocity of the second object after the collision, we get This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions ′2 = 1 2 1 − ′1 = 0.500 kg 3.50 kg 4.00 − (−3.00) m/s or Discussion ′2 = 1.00 m/s. 331 (8.61) (8.62) The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged. The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any onedimensional elastic collision of two objects. These equations can be extended to more objects if needed. Making Connections: Take-Home Investigation—Ice Cub
es and Elastic Collision Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum. PhET Explorations: Collision Lab Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens. Figure 8.10 Collision Lab (http://cnx.org/content/m55171/1.3/collision-lab_en.jar) 8.5 Inelastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: • Define inelastic collision. • Explain perfectly inelastic collisions. • Apply an understanding of collisions to sports. • Determine recoil velocity and loss in kinetic energy given mass and initial velocity. The information presented in this section supports the following AP® learning objectives and science practices: • 4.B.1.1 The student is able to calculate the change in linear momentum of a two-object system with constant mass in linear motion from a representation of the system (data, graphs, etc.). (S.P. 1.4, 2.2) • 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) • 5.D.1.3 The student is able to apply mathematical routines appropriately to problems involving elastic collisions in one dimension and justify the selection of those mathematical routines based on conservation of momentum and restoration of kinetic energy. (S.P. 2.1, 2.2) • 5.D.1.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2
) • 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2) • 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two- object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1) • 5.D.2.3 The student is able to apply the conservation of linear momentum to a closed system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2) 332 Chapter 8 \| Linear Momentum and Collisions • 5.D.2.4 The student is able to analyze data that verify conservation of momentum in collisions with and without an external friction force. (S.P. 4.1, 4.2, 4.4, 5.1, 5.3) • 5.D.2.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum as the appropriate solution method for an inelastic collision, recognize that there is a common final velocity for the colliding objects in the totally inelastic case, solve for missing variables, and calculate their values. (S.P. 2.1 2.2) • 5.D.2.6 The student is able to apply the conservation of linear momentum to an isolated system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2) We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle. Inel
astic Collision An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). Figure 8.11 shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially 1 22 + 1 22 = 2. The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum. Perfectly Inelastic Collision A collision in which the objects stick together is sometimes called “perfectly inelastic.” Figure 8.11 An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example. Example 8.5 Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie (a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure 8.12 ) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 333 Figure 8.12 An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and sound in this inelastic collision. Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and
that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested. Solution for (a) Momentum is conserved because the net external force on the puck-goalie system is zero. Conservation of momentum is or 1 + 2 = ′1 + ′2 1 1 + 22 = 1′1 + 2′2. (8.63) (8.64) Because the goalie is initially at rest, we know 2 = 0. Because the goalie catches the puck, the final velocities are equal, or ′1 = ′2 = ′. Thus, the conservation of momentum equation simplifies to Solving for ′ yields 1 1 = (1 + 2)′. ′ = 1 1 + 2 1. Entering known values in this equation, we get ′ = 0.150 kg 70.0 kg + 0.150 kg (35.0 m/s) = 7.48×10−2 m/s. Discussion for (a) (8.65) (8.66) (8.67) This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect. Solution for (b) Before the collision, the internal kinetic energy KEint of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, KEint is initially KEint = 1 22 = 1 2 = 91.9 J. 0.150 kg (35.0 m/s)2 After the collision, the internal kinetic energy is KE′int = 1 2 ( + )2 = 1 2 = 0.196 J. The change in internal kinetic energy is thus 70.15 kg 7.48×10−2 m/s (8.68) 2 (8.69) 334 Chapter 8 \| Linear Momentum and Collisions KE′int − KEint = 0.196 J − 91.9 J = − 91.7 J (8.70) where the minus sign indicates that the energy was lost. Discussion for (b) Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. KEint is mostly converted to thermal energy and sound. During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into
internal kinetic energy during a collision. Figure 8.13 shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 8.6 deals with data from such a collision. Figure 8.13 An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in Example 8.6, the potential energy of a compressed spring is released during the collision and is converted to internal kinetic energy. Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far. The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. Take-Home Experiment—Bouncing of Tennis Ball 1. Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend’s hand during the collision. Explain your observations and measurements. 2. The coefficient of restitution () is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a of 1. For a ball bouncing off the floor (or a racquet on the floor), can be shown to be = ( / )1 / 2 which the ball bounces and is the height from which the ball is dropped.
Determine for the cases in Part 1 and where is the height to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 335 for the case of a tennis ball bouncing off a concrete or wooden floor ( = 0.85 for new tennis balls used on a tennis court). Example 8.6 Calculating Final Velocity and Energy Release: Two Carts Collide In the collision pictured in Figure 8.13, two carts collide inelastically. Cart 1 (denoted 1 carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of 2.00 m/s. Cart 2 (denoted 2 in Figure 8.13) has a mass of 0.500 kg and an initial velocity of −0.500 m/s. After the collision, cart 1 is observed to recoil with a velocity of −4.00 m/s. (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)? Strategy We can use conservation of momentum to find the final velocity of cart 2, because net = 0 (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring. Solution for (a) As before, the equation for conservation of momentum in a two-object system is The only unknown in this equation is ′2. Solving for ′2 and substituting known values into the previous equation yields 1 1 + 22 = 1′1 + 2′2. (8.71) ′2 = = 1 1 + 22 − 1 ′1 2 (2.00 m/s) + 0.350 kg 0.500 kg (−0.500 m/s) 0.500 kg − 0.350 kg (−4.00 m/s) 0.500 kg = 3.70 m/s. Solution for (b) The internal kinetic energy before the collision is 2 + 1 KEint = 1 = 1 2 21 1 0.350 kg 2 22 2 (2.00 m/s)2 + 1 2 0.500
kg ( – 0.500 m/s)2 After the collision, the internal kinetic energy is = 0.763 J. 2 + 1 KE′int = 1 = 1 2 = 6.22 J. 21 ′1 0.350 kg 2 22 ′2 (-4.00 m/s)2 + 1 2 0.500 kg (3.70 m/s)2 The change in internal kinetic energy is thus KE′int − KEint = 6.22 J − 0.763 J = 5.46 J. (8.72) (8.73) (8.74) (8.75) Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring. 8.6 Collisions of Point Masses in Two Dimensions By the end of this section, you will be able to: • Discuss two-dimensional collisions as an extension of one-dimensional analysis. Learning Objectives 336 Chapter 8 \| Linear Momentum and Collisions • Define point masses. • Derive an expression for conservation of momentum along the x-axis and y-axis. • Describe elastic collisions of two objects with equal mass. • Determine the magnitude and direction of the final velocity given initial velocity and scattering angle. The information presented in this section supports the following AP® learning objectives and science practices: • 5.D.1.2 The student is able to apply the principles of conservation of momentum and restoration of kinetic energy to reconcile a situation that appears to be isolated and elastic, but in which data indicate that linear momentum and kinetic energy are not the same after the interaction, by refining a scientific question to identify interactions that have not been considered. Students will be expected to solve qualitatively and/or quantitatively for one-dimensional situations and only qualitatively in two-dimensional situations. • 5.D.3.3 The student is able to make predictions about the velocity of the center of mass for interactions within a defined two-dimensional system. In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented
. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.14.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.14. Because momentum is conserved, the components of momentum along the - and -axes ( and ) will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of twodimensional collisions.) Figure 8.14 A two-dimensional collision with the coordinate system chosen so that 2 is initially at rest and 1 is parallel to the -axis. This coordinate system is sometimes called the laboratory coordinate system, because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are. Along the -axis, the equation for conservation of momentum is Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is 1 + 2 = ′1 + ′2. (8.76) 1 1 + 22 = 11 + 22. But because particle 2 is initially at rest, this equation becomes 1 1 = 1′1 + 2′2. (8.77) (8.78) This content is available for free at
http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 337 The components of the velocities along the -axis have the form cos. Because particle 1 initially moves along the -axis, we find 1 = 1. Conservation of momentum along the -axis gives the following equation: 1 1 = 1′1 cos 1 + 2′2 cos 2, where 1 and 2 are as shown in Figure 8.14. Conservation of Momentum along the -axis 1 1 = 1′1 cos 1 + 2′2 cos 2 Along the -axis, the equation for conservation of momentum is or 1 + 2 = ′1 + ′2 1 1 + 22 = 1′1 + 2′2. (8.79) (8.80) (8.81) (8.82) But 1 is zero, because particle 1 initially moves along the -axis. Because particle 2 is initially at rest, 2 is also zero. The equation for conservation of momentum along the -axis becomes 0 = 1′1 + 2′2. The components of the velocities along the -axis have the form sin. Thus, conservation of momentum along the -axis gives the following equation: 0 = 1′1 sin 1 + 2′2 sin 2. Conservation of Momentum along the -axis 0 = 1′1 sin 1 + 2′2 sin 2 (8.83) (8.84) (8.85) The equations of conservation of momentum along the -axis and -axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level. Making Connections: Real World Connections We have seen, in one-dimensional collisions when momentum is conserved, that the center-of-mass velocity of the system remains unchanged as a result of the collision. If you calculate the momentum and center-of-mass velocity before the collision, you will get the same answer as if you calculate both quantities after the collision. This logic also works for twodimensional collisions. For example, consider two cars of equal mass. Car A is driving east (+x-direction) with a speed of 40 m/s. Car B is driving north (+y-direction) with a speed of 80 m/s. What is
the velocity of the center-of-mass of this system before and after an inelastic collision, in which the cars move together as one mass after the collision? Since both cars have equal mass, the center-of-mass velocity components are just the average of the components of the individual velocities before the collision. The x-component of the center of mass velocity is 20 m/s, and the y-component is 40 m/s. Using momentum conservation for the collision in both the x-component and y-component yields similar answers: (40) + (0) = (2)final() final() = 20 m/s (0) + (80) = (2)final() final() = 40 m/s (8.86) (8.87) (8.88) (8.89) 338 Chapter 8 \| Linear Momentum and Collisions Since the two masses move together after the collision, the velocity of this combined object is equal to the center-of-mass velocity. Thus, the center-of-mass velocity before and after the collision is identical, even in two-dimensional collisions, when momentum is conserved. Example 8.7 Determining the Final Velocity of an Unseen Object from the Scattering of Another Object Suppose the following experiment is performed. A 0.250-kg object (1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (2). The 0.250-kg object emerges from the room at an angle of 45.0º with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (′2 and 2) of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.15 is one in which 2 is originally at rest and the initial velocity is parallel to the -axis, so that conservation of momentum along the - and -axes is applicable. Everything is known in these equations except ′2 and 2, which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the - and -directions. Solution Solving 1 1 = 1′1 cos 1 + 2
′2 cos 2 for 2′ cos 2 and 0 = 1′1 sin 1 + 2′2 sin 2 for ′2 sin 2 and taking the ratio yields an equation (in which θ2 is the only unknown quantity. Applying the identity tan = sin cos, we obtain: tan 2 = ′1 sin 1 ′1 cos 1 − 1. Entering known values into the previous equation gives tan 2 = (1.50 m/s)(0.7071) (1.50 m/s)(0.7071) − 2.00 m/s = −1.129. Thus, 2 = tan−1(−1.129) = 311.5º ≈ 312º. (8.90) (8.91) (8.92) Angles are defined as positive in the counter clockwise direction, so this angle indicates that 2 is scattered to the right in Figure 8.15, as expected (this angle is in the fourth quadrant). Either equation for the - or -axis can now be used to solve for ′2, but the latter equation is easiest because it has fewer terms. Entering known values into this equation gives ′2 = − 1 2 ′1 sin 1 sin 2 ′2 = − 0.250 kg 0.400 kg (1.50 m/s) 0.7071 −0.7485. ′2 = 0.886 m/s. Thus, Discussion (8.93) (8.94) (8.95) It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 339 Figure 8.15 A collision taking place in a dark room is explored in Example 8.7. The incoming object 1 is scattered by an initially stationary object. Only the stationary object’s mass 2 is known. By measuring the angle and speed at which 1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision
. Elastic Collisions of Two Objects with Equal Mass Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 8.14 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 (2) is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is 1 21 2 = 1 2′1 2 + 1 2′2 2. (8.96) Because the masses are equal, 1 = 2 =. Algebraic manipulation (left to the reader) of conservation of momentum in the - and -directions can show that 2′1 (Remember that 2 is negative here.) The two preceding equations can both be true only if 1 − 2 2′2 2 + ′1′2 cos 1 21 2 = 1 2 + 1. ′1 ′2 cos 1 − 2 = 0. (8.97) (8.98) • There are three ways that this term can be zero. They are ′1 = 0 : head-on collision; incoming ball stops ′2 = 0 : no collision; incoming ball continues unaffected cos(1 − 2) = 0 : angle of separation (1 − 2) is 90º after the collision • • All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions. 340 Chapter 8 \| Linear Momentum and Collisions Connections to Nuclear and Particle Physics Two-dimensional collision experiments have revealed much of what we know about sub
atomic particles, as we shall see in Medical Applications of Nuclear Physics and Particle Physics. Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such experiments. 8.7 Introduction to Rocket Propulsion Learning Objectives By the end of this section, you will be able to: • State Newton’s third law of motion. • Explain the principle involved in propulsion of rockets and jet engines. • Derive an expression for the acceleration of the rocket. • Discuss the factors that affect the rocket’s acceleration. • Describe the function of a space shuttle. Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle—Newton’s third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. Another common example is the recoil of a gun. The gun exerts a force on a bullet to accelerate it and consequently experiences an equal and opposite force, causing the gun’s recoil or kick. Making Connections: Take-Home Experiment—Propulsion of a Balloon Hold a balloon and fill it with air. Then, let the balloon go. In which direction does the air come out of the balloon and in which direction does the balloon get propelled? If you fill the balloon with water and then let the balloon go, does the balloon’s direction change? Explain your answer. Figure 8.16 shows a rocket accelerating straight up. In part (a), the rocket has a mass and a velocity relative to Earth, and hence a momentum. In part (b), a time Δ has elapsed in which the rocket has ejected a mass Δ of hot gas at a velocity e relative to the rocket. The remainder of the mass ( − Δ) now has a greater velocity ( + Δ). The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time Δ, producing a negative impulse Δ = −Δ. (Remember that impulse is the net external force on a system multiplied by the time it acts, and it equals the change in momentum of the system.) So, the center of mass of the system is in free fall but, by rapidly expelling mass, part of the system can accelerate upward. It is a commonly held misconception that the rocket exhaust
pushes on the ground. If we consider thrust; that is, the force exerted on the rocket by the exhaust gases, then a rocket’s thrust is greater in outer space than in the atmosphere or on the launch pad. In fact, gases are easier to expel into a vacuum. By calculating the change in momentum for the entire system over Δ, and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket. “The rocket” is that part of the system remaining after the gas is ejected, and is the acceleration due to gravity. = e Δ Δ − Acceleration of a Rocket Acceleration of a rocket is = e Δ Δ − (8.99) (8.100) where is the acceleration of the rocket, e is the escape velocity, is the mass of the rocket, Δ is the mass of the ejected gas, and Δ is the time in which the gas is ejected. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 341 Figure 8.16 (a) This rocket has a mass and an upward velocity. The net external force on the system is −, if air resistance is neglected. (b) A time Δ later the system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it upward. A rocket’s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket. First, the greater the exhaust velocity of the gases relative to the rocket, e, the greater the acceleration is. The practical limit for e is about 2.5×103 m/s for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor Δ / Δ in the equation. The quantity (Δ / Δ)e, with units of newtons, is called "thrust.” The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted.
Factors Affecting a Rocket’s Acceleration • The greater the exhaust velocity e of the gases relative to the rocket, the greater the acceleration. • The faster the rocket burns its fuel, the greater its acceleration. • The smaller the rocket’s mass (all other factors being the same), the greater the acceleration. Example 8.8 Calculating Acceleration: Initial Acceleration of a Moon Launch A Saturn V’s mass at liftoff was 2.80×106 kg, its fuel-burn rate was 1.40×104 kg/s, and the exhaust velocity was 2.40×103 m/s. Calculate its initial acceleration. Strategy This problem is a straightforward application of the expression for acceleration because is the unknown and all of the terms on the right side of the equation are given. Solution Substituting the given values into the equation for acceleration yields 342 Chapter 8 \| Linear Momentum and Collisions = e − Δ Δ = 2.40×103 m/s 2.80×106 kg 1.40×104 kg/s − 9.80 m/s2 (8.101) Discussion = 2.20 m/s2. This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because decreases while e and Δ Δ show that the thrust of the engines was 3.36×107 N. remain constant. Knowing this acceleration and the mass of the rocket, you can To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth’s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is = e ln 0 r (8.102) where ln 0 / r is the natural logarithm of the ratio of the initial mass of the rocket (0) to what is left (r) after all of the fuel is exhausted. (Note that is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth’s gravity starting from rest, given that the escape velocity from Earth is about 11.2×103 m/s, and assuming an exhaust velocity e = 2.5×103 m/
s. Solving for 0 / r gives Thus, the mass of the rocket is ln 0 r = e = 11.2×103 m/s 2.5×103 m/s = 4.48 0 r = 4.48 = 88. r = 0 88. (8.103) (8.104) (8.105) This result means that only 1 / 88 of the mass is left when the fuel is burnt, and 87 / 88 of the initial mass was fuel. Expressed as percentages, 98.9% of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10%. Taking air resistance and gravitational force into account, the mass r remaining can only be about 0 / 180. It is difficult to build a rocket in which the fuel has a mass 180 times everything else. The solution is multistage rockets. Each stage only needs to achieve part of the final velocity and is discarded after it burns its fuel. The result is that each successive stage can have smaller engines and more payload relative to its fuel. Once out of the atmosphere, the ratio of payload to fuel becomes more favorable, too. The space shuttle was an attempt at an economical vehicle with some reusable parts, such as the solid fuel boosters and the craft itself. (See Figure 8.17) The shuttle’s need to be operated by humans, however, made it at least as costly for launching satellites as expendable, unmanned rockets. Ideally, the shuttle would only have been used when human activities were required for the success of a mission, such as the repair of the Hubble space telescope. Rockets with satellites can also be launched from airplanes. Using airplanes has the double advantage that the initial velocity is significantly above zero and a rocket can avoid most of the atmosphere’s resistance. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 343 Figure 8.17 The space shuttle had a number of reusable parts. Solid fuel boosters on either side were recovered and refueled after each flight, and the entire orbiter returned to Earth for use in subsequent flights. The large liquid fuel tank was expended. The space shuttle was a complex assemblage of technologies, employing both solid and liquid fuel and pioneering ceramic tiles as reentry heat shields. As a result, it permitted multiple launches as opposed to single-use rockets. (credit: NASA) PhET Expl
orations: Lunar Lander Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control. Figure 8.18 Lunar Lander (http://cnx.org/content/m55174/1.2/lunar-lander_en.jar) Glossary change in momentum: the difference between the final and initial momentum; the mass times the change in velocity conservation of momentum principle: when the net external force is zero, the total momentum of the system is conserved or constant elastic collision: a collision that also conserves internal kinetic energy impulse: the average net external force times the time it acts; equal to the change in momentum inelastic collision: a collision in which internal kinetic energy is not conserved internal kinetic energy: the sum of the kinetic energies of the objects in a system isolated system: a system in which the net external force is zero linear momentum: the product of mass and velocity perfectly inelastic collision: a collision in which the colliding objects stick together point masses: structureless particles with no rotation or spin quark: fundamental constituent of matter and an elementary particle 344 Chapter 8 \| Linear Momentum and Collisions second law of motion: physical law that states that the net external force equals the change in momentum of a system divided by the time over which it changes Section Summary 8.1 Linear Momentum and Force • Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. • In symbols, linear momentum p is defined to be where is the mass of the system and v is its velocity. • The SI unit for momentum is kg · m/s. p = v, • Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton’s second law of motion is defined to be • Fnet is the net external force, Δp is the change in momentum, and Δ is the change time. Fnet = Δp Δ, 8.2 Impulse • Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts: • Forces are usually
not constant over a period of time. Δp = FnetΔ. 8.3 Conservation of Momentum • The conservation of momentum principle is written or ptot = constant ptot = p′tot (isolated system), ptot is the initial total momentum and p′tot is the total momentum some time later. • An isolated system is defined to be one for which the net external force is zero Fnet = 0. • During projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero. • Conservation of momentum applies only when the net external force is zero. • The conservation of momentum principle is valid when considering systems of particles. 8.4 Elastic Collisions in One Dimension • An elastic collision is one that conserves internal kinetic energy. • Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions. 8.5 Inelastic Collisions in One Dimension • An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). • A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision. • Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. 8.6 Collisions of Point Masses in Two Dimensions • The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the -axis parallel to the velocity of the incoming particle. • Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the -axis), stated by 1 1 = 1′1 cos 1 + 2′2 cos 2 and along the direction perpendicular to the initial direction (the -axis) stated by 0 = 1′1 +2′2. • The internal kinetic before and after the collision of two objects that have equal masses is 2 = 1 2 + 1 2 + ′1′2 cos 1 − 2. 2′1 2′2 1 21 This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 345 • Point masses are structureless particles that cannot spin. 8.7 Introduction to Rocket Propulsion • Newton’s third
law of motion states that to every action, there is an equal and opposite reaction. • Acceleration of a rocket is = e −. Δ Δ • A rocket’s acceleration depends on three main factors. They are 1. The greater the exhaust velocity of the gases, the greater the acceleration. 2. The faster the rocket burns its fuel, the greater its acceleration. 3. The smaller the rocket's mass, the greater the acceleration. Conceptual Questions 8.1 Linear Momentum and Force 1. An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? 2. An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? 3. Professional Application Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. 4. How can a small force impart the same momentum to an object as a large force? 8.2 Impulse 5. Professional Application Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. tile floor for a day care center. 6. While jumping on a trampoline, sometimes you land on your back and other times on your feet. In which case can you reach a greater height and why? 7. Professional Application Tennis racquets have “sweet spots.” If the ball hits a sweet spot then the player's arm is not jarred as much as it would be otherwise. Explain why this is the case. 8.3 Conservation of Momentum 8. Professional Application If you dive into water, you reach greater depths than if you do a belly flop. Explain this difference in depth using the concept of conservation of energy. Explain this difference in depth using what you have learned in this chapter. 9. Under what circumstances is momentum conserved? 10. Can momentum be conserved for a system if there are external forces acting on the system? If so, under what conditions? If not, why not? 11. Momentum for a system can be conserved in one direction while not being conserved in another. What is the angle between the directions? Give an example. 12. Professional Application Explain in terms of momentum and
Newton’s laws how a car’s air resistance is due in part to the fact that it pushes air in its direction of motion. 13. Can objects in a system have momentum while the momentum of the system is zero? Explain your answer. 14. Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not. 8.4 Elastic Collisions in One Dimension 15. What is an elastic collision? 8.5 Inelastic Collisions in One Dimension 16. What is an inelastic collision? What is a perfectly inelastic collision? 17. Mixed-pair ice skaters performing in a show are standing motionless at arms length just before starting a routine. They reach out, clasp hands, and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their velocity after their bodies meet? 346 Chapter 8 \| Linear Momentum and Collisions 18. A small pickup truck that has a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system (truck plus entire load)? What is their effect on the motion of the truck? 8.6 Collisions of Point Masses in Two Dimensions 19. Figure 8.19 shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle 1 ) at which the small object can emerge after colliding elastically with the cube. How does 1 depend on, the so-called impact parameter? Ignore any effects that might be due to rotation after the collision, and assume that the cube is much more massive than the small object. (b) Answer the same questions if the small object instead collides with a massive sphere. Figure 8.19 A small object approaches a collision with a much more massive cube, after which its velocity has the direction 1. The angles at which the small object can be scattered are determined by the shape of the object it strikes and the impact parameter. 8.7 Introduction to Rocket Propulsion 20. Professional Application Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion of the center of mass affected by the explosion? How would it be affected if the pieces experienced significantly more air resistance than
the intact shell? 21. Professional Application During a visit to the International Space Station, an astronaut was positioned motionless in the center of the station, out of reach of any solid object on which he could exert a force. Suggest a method by which he could move himself away from this position, and explain the physics involved. 22. Professional Application It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 347 Problems & Exercises 8.1 Linear Momentum and Force 1. (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant? 2. (a) What is the mass of a large ship that has a momentum of 1.60×109 kg · m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s. 3. (a) At what speed would a 2.00×104-kg airplane have to fly to have a momentum of 1.60×109 kg · m/s (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of 60.0 m/s? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. 4. (a) What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 10.0 m/s? (b) At what speed would an 8.00-kg trash can have the same
momentum as the truck? 5. A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest. 6. The mass of Earth is 5.972×1024 kg and its orbital radius is an average of 1.496×1011 m. Calculate its linear momentum. 8.2 Impulse 7. A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)? 8. Professional Application A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. 9. A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not. 10. Professional Application A professional boxer hits his opponent with a 1000-N horizontal blow that lasts for 0.150 s. (a) Calculate the impulse imparted by this blow. (b) What is the opponent’s final velocity, if his mass is 105 kg and he is motionless in midair when struck near his center of mass? (c) Calculate the recoil velocity of the opponent’s 10.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer’s body. (d) Discuss the implications of your answers for parts (b) and (c). 11. Professional Application Suppose a child drives a bumper car head on into the side rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80
m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. 12. Professional Application One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×103 m/s, given the collision lasts 6.00×10 – 8 s. 13. Professional Application A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) Calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) Calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 cm. 14. Professional Application Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at 600 m/s from the gun. (b) If this part is stopped over a distance of 20.0 cm, what average force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if the bullet is accelerated to its velocity in 10.0 ms (milliseconds). 15. A cruise ship with a mass of 1.00×107 kg strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest.) 16. Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.76×104 N for 5.
50×10–2 s. 17. Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the magnitude of the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero. 18. A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. (a) Calculate the duration of the impact. (b) What was the average force exerted on the nail? 19. Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum. 348 Chapter 8 \| Linear Momentum and Collisions 20. A ball with an initial velocity of 10 m/s moves at an angle 60º above the -direction. The ball hits a vertical wall and bounces off so that it is moving 60º above the − -direction with the same speed. What is the impulse delivered by the wall? 21. When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball. 22. A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55º above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? 8.3 Conservation of Momentum 23. Professional Application Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity? 24. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick
together. What is their final velocity? 25. Professional Application Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer. 26. What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 12.0 m/s in the same direction? Assume the deer remains on the car. 27. A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What is their velocity after impact if the falcon’s velocity is initially 28.0 m/s and the dove’s velocity is 7.00 m/s in the same direction? which it came. What would their final velocities be in this case? 8.5 Inelastic Collisions in One Dimension 31. A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left? 32. During an ice show, a 60.0-kg skater leaps into the air and is caught by an initially stationary 75.0-kg skater. (a) What is their final velocity assuming negligible friction and that the 60.0-kg skater’s original horizontal velocity is 4.00 m/s? (b) How much kinetic energy is lost? 33. Professional Application Using mass and speed data from Example 8.1 and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c)
Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass? 34. A battleship that is 6.00×107 kg and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs. 35. Professional Application Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00×103 kg, and the second a mass of 7.50×103 kg. (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both. 8.4 Elastic Collisions in One Dimension 36. Professional Application 28. Two identical objects (such as billiard balls) have a onedimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved. 29. Professional Application Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00×103 kg, and the second a mass of 7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost? 37. Professional Application Space
probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation? 30. A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/ s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from 38. A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 349 (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifleshoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem. 39. Professional Application One of the waste products of a nuclear reactor is plutonium-239. This nucleus is radioactive and 239 Pu decays by splitting into a helium-4 nucleus and a uranium-235 nucleus, the latter of which is also 4 He + 235 U radioactive and will itself decay some time later. The energy emitted in the plutonium decay is 8.40×10 – 13 J and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is 6.68×10 – 27 kg, while that of the uranium is 3.92
×10 – 25 kg (note that the ratio of the masses is 4 to 235). (a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b) How much kinetic energy does each nucleus carry away? Note that the data given here are accurate to three digits only. 40. Professional Application The Moon’s craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of 5.00×1012 kg (about a kilometer across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is 7.36×1022 kg )? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon. (c) In October 2009, NASA crashed a rocket into the Moon, and analyzed the plume produced by the impact. (Significant amounts of water were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the rocket was 2000 kg and its speed upon impact was 9000 km/h. How does the plume produced alter these results? 41. Professional Application Two football players collide head-on in midair while trying to catch a thrown football. The first player is 95.0 kg and has an initial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of –3.50 m/s. What is their velocity just after impact if they cling together? 42. What is the speed of a garbage truck that is 1.20×104 kg and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest? 43. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 10.0 kg and the horizontal component of its velocity is 8.00 m/s when the 65.0-kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity? 44. (a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s
ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from? 8.6 Collisions of Point Masses in Two Dimensions 45. Two identical pucks collide on an air hockey table. One puck was originally at rest. (a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of 30.0º,what is the velocity (magnitude and direction) of the second puck? (You may use the result that 1 − 2 = 90º for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic. 46. Confirm that the results of the example Example 8.7 do conserve momentum in both the - and -directions. 47. A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of 20.0º above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired? 48. Professional Application A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0º to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision. 49. Professional Application Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei from 4 He 197 Au gold-197 nuclei. The energy of the incoming helium nucleus was 8.00×10−13 J, and the masses of the
helium and gold nuclei were 6.68×10−27 kg and 3.29×10−25 kg, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120º during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus? 50. Professional Application Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at 8.00 m/s due south. The second car has a mass of 850 kg and is approaching at 17.0 m/s due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that 350 Chapter 8 \| Linear Momentum and Collisions gravity. The mass of the rocket just as it runs out of fuel is 75,000-kg, and its exhaust velocity is 2.40×103 m/s. Assume that the acceleration of gravity is the same as on Earth’s surface. (b) Why might it be necessary 9.80 m/s2 to limit the acceleration of a rocket? 59. Given the following data for a fire extinguisher-toy wagon rocket experiment, calculate the average exhaust velocity of the gases expelled from the extinguisher. Starting from rest, the final velocity is 10.0 m/s. The total mass is initially 75.0 kg and is 70.0 kg after the extinguisher is fired. 60. How much of a single-stage rocket that is 100,000 kg can be anything but fuel if the rocket is to have a final speed of 8.00 km/s, given that it expels gases at an exhaust velocity of 2.20×103 m/s? 61. Professional Application (a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-N frictional force opposing the squid’s movement. (b) How much energy is lost to work done against friction? 62. Unreasonable
Results Squids have been reported to jump from the ocean and travel 30.0 m (measured horizontally) before re-entering the water. (a) Calculate the initial speed of the squid if it leaves the water at an angle of 20.0º, assuming negligible lift from the air and negligible air resistance. (b) The squid propels itself by squirting water. What fraction of its mass would it have to eject in order to achieve the speed found in the previous part? The water is ejected at 12.0 m/s ; gravitational force and friction are neglected. (c) What is unreasonable about the results? (d) Which premise is unreasonable, or which premises are inconsistent? 63. Construct Your Own Problem Consider an astronaut in deep space cut free from her space ship and needing to get back to it. The astronaut has a few packages that she can throw away to move herself toward the ship. Construct a problem in which you calculate the time it takes her to get back by throwing all the packages at one time compared to throwing them one at a time. Among the things to be considered are the masses involved, the force she can exert on the packages through some distance, and the distance to the ship. 64. Construct Your Own Problem Consider an artillery projectile striking armor plating. Construct a problem in which you find the force exerted by the projectile on the plate. Among the things to be considered are the mass and speed of the projectile and the distance over which its speed is reduced. Your instructor may also wish for you to consider the relative merits of depleted uranium versus lead projectiles based on the greater density of uranium. because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the -axis and -axis; instead, you must look for other simplifying aspects. 51. Starting with equations 1 1 = 1′1 cos 1 + 2′2 cos 2 and 0 = 1′1 sin 1 + 2′2 sin 2 for conservation of momentum in the - and -directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, 2+ 1 1 21 as discussed in the text. 2+′1′2 cos 1 − 2 2′2 2′1 2 = 1 52. Integrated Concepts A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is friction
less, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away? 8.7 Introduction to Rocket Propulsion 53. Professional Application Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196 kg of gas per second at an exhaust velocity of 2.50×103 m/s? 54. Professional Application What is the acceleration of a 5000-kg rocket taking off from the Moon, where the acceleration due to gravity is only 1.6 m/s2, if the rocket expels 8.00 kg of gas per second at an exhaust velocity of 2.20×103 m/s? 55. Professional Application Calculate the increase in velocity of a 4000-kg space probe that expels 3500 kg of its mass at an exhaust velocity of 2.00×103 m/s. You may assume the gravitational force is negligible at the probe’s location. 56. Professional Application Ion-propulsion rockets have been proposed for use in space. They employ atomic ionization techniques and nuclear energy sources to produce extremely high exhaust velocities, perhaps as great as 8.00×106 m/s. These techniques allow a much more favorable payload-to-fuel ratio. To illustrate this fact: (a) Calculate the increase in velocity of a 20,000-kg space probe that expels only 40.0-kg of its mass at the given exhaust velocity. (b) These engines are usually designed to produce a very small thrust for a very long time—the type of engine that might be useful on a trip to the outer planets, for example. Calculate the acceleration of such an engine if it expels 4.50×10−6 kg/s at the given velocity, assuming the acceleration due to gravity is negligible. 57. Derive the equation for the vertical acceleration of a rocket. 58. Professional Application (a) Calculate the maximum rate at which a rocket can expel gases if its acceleration cannot exceed seven times that of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 351 Test Prep for AP® Courses 8.1 Linear Momentum and Force 1. A boy standing on a frictionless ice rink is initially at rest.
He throws a snowball in the +x-direction, and it travels on a ballistic trajectory, hitting the ground some distance away. Which of the following is true about the boy while he is in the act of throwing the snowball? a. He feels an upward force to compensate for the downward trajectory of the snowball. b. He feels a backward force exerted by the snowball he is throwing. c. He feels no net force. d. He feels a forward force, the same force that propels the snowball. 2. A 150-g baseball is initially moving 80 mi/h in the –xdirection. After colliding with a baseball bat for 20 ms, the baseball moves 80 mi/h in the +x-direction. What is the magnitude and direction of the average force exerted by the bat on the baseball? 8.2 Impulse 3. A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in a 0.045-s time interval. What is the magnitude and direction of the average force exerted on the ball by the floor during the collision? a. 33 N, up b. 120 N, up c. 120 N, down d. 240 N, down 4. A 75-g ball is dropped from rest from a height of 2.2 m. It bounces off the floor and rebounds to a maximum height of 1.7 m. If the ball is in contact with the floor for 0.024 s, what is the magnitude and direction of the average force exerted on the ball by the floor during the collision? 5. A 2.4-kg ceramic bowl falls to the floor. During the 0.018-s impact, the bowl experiences an average force of 750 N from the floor. The bowl is at rest after the impact. From what initial height did the bowl fall? a. 1.6 m b. 2.8 m c. 3.2 m d. 5.6 m 6. Whether or not an object (such as a plate, glass, or bone) breaks upon impact depends on the average force exerted on that object by the surface. When a 1.2-kg glass figure hits the floor, it will break if it experiences an average force of 330 N. When it hits a tile floor, the glass comes to a stop in 0.015 s. From what minimum height must the glass fall to experience sufficient force to break? How would your answer
change if the figure were falling to a padded or carpeted surface? Explain. 7. A 2.5-kg block slides across a frictionless table toward a horizontal spring.As the block bounces off the spring, a probe measures the velocity of the block (initially negative, moving away from the probe) over time as follows: Table 8.2 Velocity (m/s) Time (s) −12.0 −10.0 −6.0 0 6.0 10.0 12.0 0 0.10 0.20 0.30 0.40 0.50 0.60 What is the average force exerted on the block by the spring over the entire 0.60-s time interval of the collision? a. 50 N b. 60 N c. 100 N d. 120 N 8. During an automobile crash test, the average force exerted by a solid wall on a 2500-kg car that hits the wall is measured to be 740,000 N over a 0.22-s time interval. What was the initial speed of the car prior to the collision, assuming the car is at rest at the end of the time interval? 9. A test car is driving toward a solid crash-test barrier with a speed of 45 mi/h. Two seconds prior to impact, the car begins to brake, but it is still moving when it hits the wall. After the collision with the wall, the car crumples somewhat and comes to a complete stop. In order to estimate the average force exerted by the wall on the car, what information would you need to collect? a. The (negative) acceleration of the car before it hits the wall and the distance the car travels while braking. b. The (negative) acceleration of the car before it hits the wall and the velocity of the car just before impact. c. The velocity of the car just before impact and the duration of the collision with the wall. d. The duration of the collision with the wall and the distance the car travels while braking. 10. Design an experiment to verify the relationship between the average force exerted on an object and the change in momentum of that object. As part of your explanation, list the equipment you would use and describe your experimental setup. What would you measure and how? How exactly would you verify the relationship? Explain. 11. A 22-g puck hits the wall of an air hockey table perpendicular to the wall with an initial speed of 14 m/s.The puck is in contact with
the wall for 0.0055 s, and it rebounds from the wall with a speed of 14 m/s in the opposite direction.What is the magnitude of the average force exerted by the wall on the puck? a. 0.308 N b. 0.616 N c. 56 N d. 112 N 12. A 22-g puck hits the wall of an air hockey table perpendicular to the wall with an initial speed of 7 m/s. The puck is in contact with the wall for 0.011 s, and the wall exerts an average force of 28 N on the puck during that time. Calculate the magnitude and direction of the change in momentum of the puck. 13. 352 Chapter 8 \| Linear Momentum and Collisions Which of the following will be true about the total momentum of the two cars? It will be greater before the collision. It will be equal before and after the collision. It will be greater after the collision. a. b. c. d. The answer depends on whether the collision is elastic or inelastic. 18. A group of students has two carts, A and B, with wheels that turn with negligible friction. The carts can travel along a straight horizontal track. Cart A has known mass mA. The students are asked to use a one-dimensional collision between the carts to determine the mass of cart B. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together. a. Describe an experimental procedure to determine the velocities of the carts before and after a collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure. b. There will be sources of error in the measurements taken in the experiment, both before and after the collision. For your experimental procedure, will the uncertainty in the calculated value of the mass of cart B be affected more by the error in the measurements taken before the collision or by those taken after the collision, or will it be equally affected by both sets of measurements? Justify your answer. A group of students took measurements for one collision. A graph of the students’ data is shown below. Figure 8.22 The image shows a graph with position in meters on the vertical axis and time in seconds on the horizontal axis. c. Given mA =
0.50 kg, use the graph to calculate the mass of cart B. Explicitly indicate the principles used in your calculations. d. The students are now asked to Consider the kinetic energy changes in an inelastic collision, specifically whether the initial values of one of the physical quantities affect the fraction of mechanical energy dissipated in the collision. How could you modify the experiment to investigate this question? Be sure to explicitly describe the calculations you would make, specifying all equations you would use (but do not actually do any algebra or arithmetic). 19. Cart A is moving with an initial velocity +v (in the positive direction) toward cart B, initially at rest. Both carts have equal mass and are on a frictionless surface. Which of the following Figure 8.20 This is a graph showing the force exerted by a rigid wall versus time. The graph in Figure 8.20 represents the force exerted on a particle during a collision. What is the magnitude of the change in momentum of the particle as a result of the collision? a. 1.2 kg • m/s b. 2.4 kg • m/s c. 3.6 kg • m/s d. 4.8 kg • m/s 14. Figure 8.21 This is a graph showing the force exerted by a rigid wall versus time. The graph in Figure 8.21 represents the force exerted on a particle during a collision. What is the magnitude of the change in momentum of the particle as a result of the collision? 8.3 Conservation of Momentum 15. Which of the following is an example of an open system? a. Two air cars colliding on a track elastically. b. Two air cars colliding on a track and sticking together. c. A bullet being fired into a hanging wooden block and becoming embedded in the block, with the system then acting as a ballistic pendulum. d. A bullet being fired into a hillside and becoming buried in the earth. 16. A 40-kg girl runs across a mat with a speed of 5.0 m/s and jumps onto a 120-kg hanging platform initially at rest, causing the girl and platform to swing back and forth like a pendulum together after her jump. What is the combined velocity of the girl and platform after the jump? What is the combined momentum of the girl and platform both before and after the collision? A 50-kg boy runs across a mat with a speed of 6.0 m/s and collides with a soft barrier on the
wall, rebounding off the wall and falling to the ground. The boy is at rest after the collision. What is the momentum of the boy before and after the collision? Is momentum conserved in this collision? Explain. Which of these is an example of an open system and which is an example of a closed system? Explain your answer. 17. A student sets up an experiment to measure the momentum of a system of two air cars, A and B, of equal mass, moving on a linear, frictionless track. Before the collision, car A has a certain speed, and car B is at rest. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 8 \| Linear Momentum and Collisions 353 before, − 2 statements correctly characterizes the velocity of the center of mass of the system before and after the collision? a. + 2 b. + 2 c. + 2 before, 0 after after after before, + 2 d. 0 before, 0 after 20. Cart A is moving with a velocity of +10 m/s toward cart B, which is moving with a velocity of +4 m/s. Both carts have equal mass and are moving on a frictionless surface. The two carts have an inelastic collision and stick together after the collision. Calculate the velocity of the center of mass of the system before and after the collision. If there were friction present in this problem, how would this external force affect the center-of-mass velocity both before and after the collision? 8.4 Elastic Collisions in One Dimension 21. Two cars (A and B) of mass 1.5 kg collide. Car A is initially moving at 12 m/s, and car B is initially moving in the same direction with a speed of 6 m/s. The two cars are moving along a straight line before and after the collision. What will be the change in momentum of this system after the collision? a. −27 kg • m/s b. zero c. +27 kg • m/s d. It depends on whether the collision is elastic or inelastic. 22. Two cars (A and B) of mass 1.5 kg collide. Car A is initially moving at 24 m/s, and car B is initially moving in the opposite direction with a speed of 12 m/s. The two cars are moving along a straight line before and after the collision. (a) If the two cars have an
elastic collision, calculate the change in momentum of the two-car system. (b) If the two cars have a completely inelastic collision, calculate the change in momentum of the two-car system. 23. Puck A (200 g) slides across a frictionless surface to collide with puck B (800 g), initially at rest. The velocity of each puck is measured during the experiment as follows: Table 8.3 Time Velocity A Velocity B 0 1.0 s 2.0 s 3.0 s +8.0 m/s +8.0 m/s 0 0 −2.0 m/s +2.5 m/s −2.0 m/s +2.5 m/s What is the change in momentum of the center of mass of the system as a result of the collision? a. +1.6 kg•m/s b. +0.8 kg•m/s c. 0 d. −1.6 kg•m/s 24. For the table above, calculate the center-of-mass velocity of the system both before and after the collision, then calculate the center-of-mass momentum of the system both before and after the collision. From this, determine the change in the momentum of the system as a result of the collision. 25. Two cars (A and B) of equal mass have an elastic collision. Prior to the collision, car A is moving at 15 m/s in the +x-direction, and car B is moving at 10 m/s in the –x-direction. Assuming that both cars continue moving along the x-axis after the collision, what will be the velocity of car A after the collision? a. same as the original 15 m/s speed, opposite direction b. equal to car B’s velocity prior to the collision c. equal to the average of the two velocities, in its original direction d. equal to the average of the two velocities, in the opposite direction 26. Two cars (A and B) of equal mass have an elastic collision. Prior to the collision, car A is moving at 20 m/s in the +x-direction, and car B is moving at 10 m/s in the –x-direction. Assuming that both cars continue moving along the x-axis after the collision, what will be the velocities of each car after the collision? 27. A rubber ball is dropped from rest at a fixed height.
g the collision. c. Energy was lost due to friction between the ball and the floor. d. Energy was lost due to the work done by gravity during the motion. 28. A tennis ball strikes a wall with an initial speed of 15 m/s. The ball bounces off the wall but rebounds with slightly less speed (14 m/s) after the collision. Explain (a) what else changed its momentum in response to the ball’s change in momentum so that overall momentum is conserved, and (b) how some of the ball’s kinetic energy was lost. 29. Two objects, A and B, have equal mass. Prior to the collision, mass A is moving 10 m/s in the +x-direction, and mass B is moving 4 m/s in the +x-direction. Which of the following results represents an inelastic collision between A and B? a. After the collision, mass A is at rest, and mass B moves 14 m/s in the +x-direction. b. After the collision, mass A moves 4 m/s in the –x- direction, and mass B moves 18 m/s in the +x-direction. c. After the collision, the two masses stick together and move 7 m/s in the +x-direction. d. After the collision, mass A moves 4 m/s in the +x- direction, and mass B moves 10 m/s in the +x-direction. 30. Mass A is three times more massive than mass B. Mass A is initially moving 12 m/s in the +x-direction. Mass B is initially moving 12 m/s in the –x-direction. Assuming that the collision is elastic, calculate the final velocity of both masses after the collision. Show that your results are consistent with conservation of momentum and conservation of kinetic energy. 31. Two objects (A and B) of equal mass collide elastically. Mass A is initially moving 5.0 m/s in the +x-direction prior to the collision. Mass B is initially moving 3.0 m/s in the –xdirection prior to the collision. After the collision, mass A will be moving with a velocity of 3.0 m/s in the –x-direction. What will be the velocity of mass B after the collision? a. 3.0 m/s in the +x-direction b. 5.0 m/s in the +x-direction c
. 3.0 m/s in the –x-direction 354 Chapter 8 \| Linear Momentum and Collisions d. 5.0 m/s in the –x-direction 32. Two objects (A and B) of equal mass collide elastically. Mass A is initially moving 4.0 m/s in the +x-direction prior to the collision. Mass B is initially moving 8.0 m/s in the –xdirection prior to the collision. After the collision, mass A will be moving with a velocity of 8.0 m/s in the –x-direction. (a) Use the principle of conservation of momentum to predict the velocity of mass B after the collision. (b) Use the fact that kinetic energy is conserved in elastic collisions to predict the velocity of mass B after the collision. 33. Two objects of equal mass collide. Object A is initially moving in the +x-direction with a speed of 12 m/s, and object B is initially at rest. After the collision, object A is at rest, and object B is moving away with some unknown velocity. There are no external forces acting on the system of two masses. What statement can we make about this collision? a. Both momentum and kinetic energy are conserved. b. Momentum is conserved, but kinetic energy is not conserved. a. There will be no change in the center-of-mass velocity. b. The center-of-mass velocity will decrease by 2 m/s. c. The center-of-mass velocity will decrease by 6 m/s. d. The center-of-mass velocity will decrease by 8 m/s. 40. Mass A (1.0 kg) slides across a frictionless surface with a velocity of 4 m/s in the positive direction. Mass B (1.0 kg) slides across the same surface in the opposite direction with a velocity of −8 m/s. The two objects collide and stick together after the collision. Predict how the center-of-mass velocity will change as a result of the collision, and explain your prediction. Calculate the center-of-mass velocity of the system both before and after the collision and explain why it remains the same or why it has changed. 8.5 Inelastic Collisions in One Dimension 41. Mass A (2.0 kg) has an initial velocity of 4 m/s in the +xdirection. Mass B (2.0 kg) has an initial
velocity of 5 m/s in the –x-direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision? c. Neither momentum nor kinetic energy is conserved. d. More information is needed in order to determine which a. Both will move 0.5 m/s in the –x-direction. b. Mass A will stop; mass B will move 9 m/s in the +x- is conserved. 34. Two objects of equal mass collide. Object A is initially moving with a velocity of 15 m/s in the +x-direction, and object B is initially at rest. After the collision, object A is at rest. There are no external forces acting on the system of two masses. (a) Use momentum conservation to deduce the velocity of object B after the collision. (b) Is this collision elastic? Justify your answer. 35. Which of the following statements is true about an inelastic collision? a. Momentum is conserved, and kinetic energy is conserved. b. Momentum is conserved, and kinetic energy is not conserved. c. Momentum is not conserved, and kinetic energy is conserved. d. Momentum is not conserved, and kinetic energy is not conserved. 36. Explain how the momentum and kinetic energy of a system of two colliding objects changes as a result of (a) an elastic collision and (b) an inelastic collision. 37. Figure 8.9 shows the positions of two colliding objects measured before, during, and after a collision. Mass A is 1.0 kg. Mass B is 3.0 kg. Which of the following statements is true? a. This is an elastic collision, with a total momentum of 0 kg • m/s. b. This is an elastic collision, with a total momentum of 1.67 kg • m/s. c. This is an inelastic collision, with a total momentum of 0 kg • m/s. d. This is an inelastic collision, with a total momentum of 1.67 kg • m/s. 38. For the above graph, determine the initial and final momentum for both objects, assuming mass A is 1.0 kg and mass B is 3.0 kg. Also, determine the initial and final kinetic energies for both objects. Based on your results, explain whether momentum is conserved in this collision, and state whether
the collision is elastic or inelastic. 39. Mass A (1.0 kg) slides across a frictionless surface with a velocity of 8 m/s in the positive direction. Mass B (3.0 kg) is initially at rest. The two objects collide and stick together. What will be the change in the center-of-mass velocity of the system as a result of the collision? This content is available for free at http://cnx.org/content/col11844/1.13 direction. c. Mass B will stop; mass A will move 9 m/s in the –x- direction. d. Mass A will move 5 m/s in the –x-direction; mass B will move 4 m/s in the +x-direction. 42. Mass A has an initial velocity of 22 m/s in the +x-direction. Mass B is three times more massive than mass A and has an initial velocity of 22 m/s in the –x-direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision? 43. Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (5.0 kg), initially at rest. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision? a. no change b. decrease by 225 J c. decrease by 161 J d. decrease by 64 J 44. Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (4.0 kg), initially moving 7.0 m/s in the +x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision? 45. Mass A slides across a rough table with an initial velocity of 12 m/s in the +x-direction. By the time mass A collides with mass B (a stationary object with equal mass), mass A has slowed to 10 m/s. After the collision, the two objects stick together and move as one. Immediately after the collision, the velocity of the system is measured to be 5 m/s in the +xdirection, and the system
eventually slides to a stop. Which of the following statements is true about this motion? a. Momentum is conserved during the collision, but it is not conserved during the motion before and after the collision. b. Momentum is not conserved at any time during this analysis. c. Momentum is conserved at all times during this analysis. d. Momentum is not conserved during the collision, but it is conserved during the motion before and after the collision. 46. Mass A is initially moving with a velocity of 12 m/s in the +x-direction. Mass B is twice as massive as mass A and is Chapter 8 \| Linear Momentum and Collisions 355 initially at rest. After the two objects collide, the two masses move together as one with a velocity of 4 m/s in the +xdirection. Is momentum conserved in this collision? 47. Mass A is initially moving with a velocity of 24 m/s in the +x-direction. Mass B is twice as massive as mass A and is initially at rest. The two objects experience a totally inelastic collision. What is the final speed of both objects after the collision? a. A is not moving; B is moving 24 m/s in the +x-direction. b. Neither A nor B is moving. c. A is moving 24 m/s in the –x-direction. B is not moving. d. Both A and B are moving together 8 m/s in the +x- direction. 48. Mass A is initially moving with some unknown velocity in the +x-direction. Mass B is twice as massive as mass A and initially at rest. The two objects collide, and after the collision, they move together with a speed of 6 m/s in the +x-direction. (a) Is this collision elastic or inelastic? Explain. (b) Determine the initial velocity of mass A. 49. Mass A is initially moving with a velocity of 2 m/s in the +x-direction. Mass B is initially moving with a velocity of 6 m/s in the –x-direction. The two objects have equal masses. After they collide, mass A moves with a speed of 4 m/s in the –xdirection. What is the final velocity of mass B after the collision? a. 6 m/s in the +x-direction b. 4 m/s in the +x-direction c. zero d. 4 m/
s in the –x-direction 50. Mass A is initially moving with a velocity of 15 m/s in the +x-direction. Mass B is twice as massive and is initially moving with a velocity of 10 m/s in the –x-direction. The two objects collide, and after the collision, mass A moves with a speed of 15 m/s in the –x-direction. (a) What is the final velocity of mass B after the collision? (b) Calculate the change in kinetic energy as a result of the collision, assuming mass A is 5.0 kg. 8.6 Collisions of Point Masses in Two Dimensions 51. Two cars of equal mass approach an intersection. Car A is moving east at a speed of 45 m/s. Car B is moving south at a speed of 35 m/s. They collide inelastically and stick together after the collision, moving as one object. Which of the following statements is true about the center-of-mass velocity of this system? a. The center-of-mass velocity will decrease after the collision as a result of lost energy (but not drop to zero). b. The center-of-mass velocity will remain the same after the collision since momentum is conserved. c. The center-of-mass velocity will drop to zero since the two objects stick together. d. The magnitude of the center-of-mass velocity will remain the same, but the direction of the velocity will change. 52. Car A has a mass of 2000 kg and approaches an intersection with a velocity of 38 m/s directed to the east. Car B has a mass of 3500 kg and approaches the intersection with a velocity of 53 m/s directed 63° north of east. The two cars collide and stick together after the collision. Will the center-of-mass velocity change as a result of the collision? Explain why or why not. Calculate the center-of-mass velocity before and after the collision. 356 Chapter 8 \| Linear Momentum and Collisions This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 357 9 STATICS AND TORQUE Figure 9.1 On a short time scale, rocks like these in Australia's Kings Canyon are static, or motionless relative to the Earth. (credit: freeaussiestock.com) Chapter Outline 9.1. The First Condition for Equilibrium
9.2. The Second Condition for Equilibrium 9.3. Stability 9.4. Applications of Statics, Including Problem-Solving Strategies 9.5. Simple Machines 9.6. Forces and Torques in Muscles and Joints Connection for AP® Courses What might desks, bridges, buildings, trees, and mountains have in common? What do these objects have in common with a car moving at a constant velocity? While it may be apparent that the objects in the first group are all motionless relative to Earth, they also share something with the moving car and all objects moving at a constant velocity. All of these objects, stationary and moving, share an acceleration of zero. How can this be? Consider Newton's second law, F = ma. When acceleration is zero, as is the case for both stationary objects and objects moving at a constant velocity, the net external force must also be zero (Big Idea 3). Forces are acting on both stationary objects and on objects moving at a constant velocity, but the forces are balanced. That is, they are in equilibrium. In equilibrium, the net force is zero. The first two sections of this chapter will focus on the two conditions necessary for equilibrium. They will not only help you to distinguish between stationary bridges and cars moving at constant velocity, but will introduce a second equilibrium condition, this time involving rotation. As you explore the second equilibrium condition, you will learn about torque, in support of both Enduring Understanding 3.F and Essential Knowledge 3.F.1. Much like a force, torque provides the capability for acceleration; however, with careful attention, torques may also be balanced and equilibrium can be reached. The remainder of this chapter will discuss a variety of interesting equilibrium applications. From the art of balancing, to simple machines, to the muscles in your body, the world around you relies upon the principles of equilibrium to remain stable. This chapter will help you to see just how closely related these events truly are. 358 Chapter 9 \| Statics and Torque The content in this chapter supports: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F A force exerted on an object can cause a torque on that object. Essential Knowledge 3.F.1 Only the force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis. 9.1 The First Condition for Equilibrium Learning Objectives By the end of this section, you will be
able to: • State the first condition of equilibrium. • Explain static equilibrium. • Explain dynamic equilibrium. The first condition necessary to achieve equilibrium is the one already mentioned: the net external force on the system must be zero. Expressed as an equation, this is simply net F = 0 Note that if net is zero, then the net external force in any direction is zero. For example, the net external forces along the typical x- and y-axes are zero. This is written as net = 0 and = 0 Figure 9.2 and Figure 9.3 illustrate situations where net = 0 for both static equilibrium (motionless), and dynamic equilibrium (constant velocity). (9.1) (9.2) Figure 9.2 This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. Figure 9.3 This car is in dynamic equilibrium because it is moving at constant velocity. There are horizontal and vertical forces, but the net external force in any direction is zero. The applied force app between the tires and the road is balanced by air friction, and the weight of the car is supported by the normal forces, here shown to be equal for all four tires. However, it is not sufficient for the net external force of a system to be zero for a system to be in equilibrium. Consider the two situations illustrated in Figure 9.4 and Figure 9.5 where forces are applied to an ice hockey stick lying flat on ice. The net external force is zero in both situations shown in the figure; but in one case, equilibrium is achieved, whereas in the other, it is not. In Figure 9.4, the ice hockey stick remains motionless. But in Figure 9.5, with the same forces applied in different places, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 359 the stick experiences accelerated rotation. Therefore, we know that the point at which a force is applied is another factor in determining whether or not equilibrium is achieved. This will be explored further in the next section. Figure 9.4 An ice hockey stick lying flat on ice with two equal and opposite horizontal forces applied to it. Friction is negligible, and the gravitational force is balanced by the support of the ice (a normal force). Thus, net = 0. Equilibrium is achieved, which is static equilibrium in this case. Figure 9.5 The
same forces are applied at other points and the stick rotates—in fact, it experiences an accelerated rotation. Here net = 0 but the system is not at equilibrium. Hence, the net = 0 is a necessary—but not sufficient—condition for achieving equilibrium. PhET Explorations: Torque Investigate how torque causes an object to rotate. Discover the relationships between angular acceleration, moment of inertia, angular momentum and torque. Figure 9.6 Torque (http://cnx.org/content/m55176/1.2/torque_en.jar) 9.2 The Second Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: • State the second condition that is necessary to achieve equilibrium. • Explain torque and the factors on which it depends. • Describe the role of torque in rotational mechanics. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) 360 Torque Chapter 9 \| Statics and Torque The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. Several familiar factors determine how effective you are in opening the door. See Figure 9.7. First of all, the larger the force, the more effective it is in opening the door—obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular
to the door—we push in this direction almost instinctively. Figure 9.7 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) Counterclockwise torque is produced by this force, which means that the door will rotate in a counterclockwise due to F. Note that ⊥ is the perpendicular distance of the pivot from the line of action of the force. (b) A smaller counterclockwise torque is produced by a smaller force F′ acting at the same distance from the hinges (the pivot point). (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) The same force as in (a), but acting in the opposite direction, produces a clockwise torque. (e) A smaller counterclockwise torque is produced by the same magnitude force acting at the same point but in a different direction. Here, is less than 90º. (f) Torque is zero here since the force just pulls on the hinges, producing no rotation. In this case, = 0º. The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). In equation form, the magnitude of torque is defined to be = sin (9.3) where (the Greek letter tau) is the symbol for torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between the force and the vector directed from the point of application to the pivot point, as seen in Figure 9.7 and Figure 9.8. An alternative expression for torque is given in terms of the perpendicular lever arm ⊥ as shown in Figure 9.7 and Figure 9.8, which is defined as so that ⊥ = sin = ⊥. (9.4) (9.5) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 361 Figure 9.8 A force applied to an object can produce a torque, which depends on the location of the pivot point
. (a) The three factors,, and for pivot point A on a body are shown here— is the distance from the chosen pivot point to the point where the force is applied, and is the angle between F and the vector directed from the point of application to the pivot point. If the object can rotate around point A, it will rotate counterclockwise. This means that torque is counterclockwise relative to pivot A. (b) In this case, point B is the pivot point. The torque from the applied force will cause a clockwise rotation around point B, and so it is a clockwise torque relative to B. The perpendicular lever arm ⊥ is the shortest distance from the pivot point to the line along which F acts; it is shown as a dashed line in Figure 9.7 and Figure 9.8. Note that the line segment that defines the distance ⊥ is perpendicular to F, as its name implies. It is sometimes easier to find or visualize ⊥ than to find both and. In such cases, it may be more convenient to use τ = r ⊥ F rather than = sin for torque, but both are equally valid. The SI unit of torque is newtons times meters, usually written as N · m. For example, if you push perpendicular to the door with a force of 40 N at a distance of 0.800 m from the hinges, you exert a torque of 32 N·m(0.800 m×40 N×sin 90º) relative to the hinges. If you reduce the force to 20 N, the torque is reduced to 16 N·m, and so on. The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque, since both and depend on the location of the pivot. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen “pivot point.” Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, as illustrated for points B and A, respectively, in Figure 9.8. If the object can rotate about point A, it will rotate counterclockwise, which means that the torque for the force is shown as counterclockwise relative to A. But if the object can rotate about point B, it will
rotate clockwise, which means the torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer. Making Connections: Pivoting Block A solid block of length d is pinned to a wall on its right end. Three forces act on the block as shown below: FA, FB, and FC. While all three forces are of equal magnitude, and all three are equal distances away from the pivot point, all three forces will create a different torque upon the object. FA is vectored perpendicular to its distance from the pivot point; as a result, the magnitude of its torque can be found by the equation τ=FAd. Vector FB is parallel to the line connecting the point of application of force and the pivot point. As a result, it does not provide an ability to rotate the object and, subsequently, its torque is zero. FC, however, is directed at an angle ϴto the line connecting the point of application of force and the pivot point. In this instance, only the component perpendicular to this line is exerting a torque. This component, labeled F⊥, can be found using the equation F⊥=FCsinθ. The component of the force parallel to this line, labeled F∥, does not provide an ability to rotate the object and, as a result, does not provide a torque. Therefore, the resulting torque created by FC is τ=F⊥d. 362 Chapter 9 \| Statics and Torque Figure 9.9 Forces on a block pinned to a wall. A solid block of length d is pinned to a wall on its right end. Three forces act on the block: FA, FB, and FC. Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero. An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space—but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame of reference.) The second condition necessary to achieve equilibrium is stated in equation form as net = 0
(9.6) where net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative. When two children balance a seesaw as shown in Figure 9.10, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely. Figure 9.10 Two children balancing a seesaw satisfy both conditions for equilibrium. The lighter child sits farther from the pivot to create a torque equal in magnitude to that of the heavier child. Example 9.1 She Saw Torques On A Seesaw The two children shown in Figure 9.10 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the pivot? (b) What is p, the supporting force exerted by the pivot? Strategy Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 363 Solution (a) The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be = sin. (9.7) Here = 90º, so that sin = 1 for all three forces. That means ⊥ = for all three. The torques exerted by the three forces are first, second, and third, 1 = 11 2 = – 22 p = pp = 0 ⋅ p = 0. (9.8) (9.9) (
9.10) Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since p acts directly on the pivot point, the distance p is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore or 2 = – 1, 2 2 = 11. Weight is mass times the acceleration due to gravity. Entering for, we get Solve this for the unknown. The quantities on the right side of the equation are known; thus, 2 is 2 = (1.60 m) 26.0 kg 32.0 kg = 1.30 m. As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw. Solution (b) This part asks for a force p. The easiest way to find it is to use the first condition for equilibrium, which is The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as net = 0 net F = 0. (9.11) (9.12) (9.13) (9.14) (9.15) (9.16) (9.17) where we again call the vertical axis the y-axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that p – 1 – 2 = 0. This equation yields what might have been guessed at the beginning: p = 1 + 2. So, the pivot supplies a supporting force equal to the total weight of the system: p = 1 + 2. (9.18) (9.19) (9.20) 364 Chapter 9 \| Statics and Torque Entering known values gives p = 26.0 kg 9.80 m/s2 + 32.0 kg 9.80 m/s2 (9.21) Discussion = 568 N. The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the
location of the seesaw's actual pivot! Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since p is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force p is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem. Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case. Always enter the correct forces—do not jump ahead to enter some ratio of masses. Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances 1 and 2 are the distances to points directly below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point. Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter. Take-Home Experiment Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? 9.3 Stability Learning Objectives By the end of this section, you will be able to: • State the types of equilibrium. • Describe stable and unstable equilibriums. • Describe neutral equilibrium. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.
F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man's hand in Figure 9.11, for example, is not in stable equilibrium. There are three types of equilibrium: stable, unstable, and neutral. Figures throughout this module illustrate various examples. Figure 9.11 presents a balanced system, such as the toy doll on the man's hand, which has its center of gravity (cg) directly over the pivot, so that the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 365 Figure 9.11 A man balances a toy doll on one hand. A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For another example of stable equilibrium, see the pencil in Figure 9.12. Figure 9.12 This pencil is in the condition of equilibrium. The net force on the pencil is zero and the total torque about any pivot is zero. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly. An obvious example is a ball resting on top of a hill. Once displaced, it accelerates away from the crest. See the next several figures for examples of unstable equilibrium. Figure 9.
13 If the pencil is displaced slightly to the side (counterclockwise), it is no longer in equilibrium. Its weight produces a clockwise torque that returns the pencil to its equilibrium position. 366 Chapter 9 \| Statics and Torque Figure 9.14 If the pencil is displaced too far, the torque caused by its weight changes direction to counterclockwise and causes the displacement to increase. Figure 9.15 This figure shows unstable equilibrium, although both conditions for equilibrium are satisfied. Figure 9.16 If the pencil is displaced even slightly, a torque is created by its weight that is in the same direction as the displacement, causing the displacement to increase. A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is an example. Combinations of these situations are possible. For example, a marble on a saddle is stable for displacements toward the front or back of the saddle and unstable for displacements to the side. Figure 9.17 shows another example of neutral equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 367 Figure 9.17 (a) Here we see neutral equilibrium. The cg of a sphere on a flat surface lies directly above the point of support, independent of the position on the surface. The sphere is therefore in equilibrium in any location, and if displaced, it will remain put. (b) Because it has a circular cross section, the pencil is in neutral equilibrium for displacements perpendicular to its length. When we consider how far a system in stable equilibrium can be displaced before it becomes unstable, we find that some systems in stable equilibrium are more stable than others. The pencil in Figure 9.12 and the person in Figure 9.18(a) are in stable equilibrium, but become unstable for relatively small displacements to the side. The critical point is reached when the cg is no longer above the base of support. Additionally, since the cg of a person's body is above the pivots in the hips, displacements must be quickly controlled. This control is a central nervous system function that is developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger base of support. Stability is also increased by lowering one's center of gravity by bending the knees, as when a football player prepares to receive a ball or braces themselves for a tackle
. A cane, a crutch, or a walker increases the stability of the user, even more as the base of support widens. Usually, the cg of a female is lower (closer to the ground) than a male. Young children have their center of gravity between their shoulders, which increases the challenge of learning to walk. Figure 9.18 (a) The center of gravity of an adult is above the hip joints (one of the main pivots in the body) and lies between two narrowly-separated feet. Like a pencil standing on its eraser, this person is in stable equilibrium in relation to sideways displacements, but relatively small displacements take his cg outside the base of support and make him unstable. Humans are less stable relative to forward and backward displacements because the feet are not very long. Muscles are used extensively to balance the body in the front-to-back direction. (b) While bending in the manner shown, stability is increased by lowering the center of gravity. Stability is also increased if the base is expanded by placing the feet farther apart. Animals such as chickens have easier systems to control. Figure 9.19 shows that the cg of a chicken lies below its hip joints and between its widely separated and broad feet. Even relatively large displacements of the chicken's cg are stable and result in 368 Chapter 9 \| Statics and Torque restoring forces and torques that return the cg to its equilibrium position with little effort on the chicken's part. Not all birds are like chickens, of course. Some birds, such as the flamingo, have balance systems that are almost as sophisticated as that of humans. Figure 9.19 shows that the cg of a chicken is below the hip joints and lies above a broad base of support formed by widelyseparated and large feet. Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well as for side-to-side displacements. Figure 9.19 The center of gravity of a chicken is below the hip joints. The chicken is in stable equilibrium. The body of the chicken is supported from above by the hips and acts as a pendulum between them. Engineers and architects strive to achieve extremely stable equilibriums for buildings and other systems that must withstand wind,
earthquakes, and other forces that displace them from equilibrium. Although the examples in this section emphasize gravitational forces, the basic conditions for equilibrium are the same for all types of forces. The net external force must be zero, and the net torque must also be zero. Take-Home Experiment Stand straight with your heels, back, and head against a wall. Bend forward from your waist, keeping your heels and bottom against the wall, to touch your toes. Can you do this without toppling over? Explain why and what you need to do to be able to touch your toes without losing your balance. Is it easier for a woman to do this? 9.4 Applications of Statics, Including Problem-Solving Strategies Learning Objectives By the end of this section, you will be able to: • Discuss the applications of statics in real life. • State and discuss various problem-solving strategies in statics. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. Problem-Solving Strategy: Static Equilibrium Situations 1. The first step is to determine whether or not
the system is in static equilibrium. This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur. 2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known. 3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net = 0 and net = 0, depending on the list of known and unknown factors. If the second condition is involved, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 369 choose the pivot point to simplify the solution. Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then = 0 ), or along a line through the pivot point (then = 0 )). Always choose a convenient coordinate system for projecting forces. 4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience. Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure 9.20, the pole's cg lies halfway between the vaulter's hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net = 0). The second condition (net = 0) is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight. In Figure 9.20, a pole vaulter holding a
pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, = = / 2. (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 9.20. If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand. Similar observations can be made using a meter stick held at different locations along its length. Figure 9.20 A pole vaulter holds a pole horizontally with both hands. Figure 9.21 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand. 370 Chapter 9 \| Statics and Torque Figure 9.22 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter. If the pole vaulter holds the pole as shown in Figure 9.20, the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If =, then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces and is straightforward, as the next example shows. If the pole vaulter holds the pole from near the end of the pole (Figure 9.22), the direction of the force applied by the right hand of the vaulter reverses its direction. Example 9.2 What Force Is Needed to Support a Weight Held Near Its CG? For the situation shown in Figure 9.20, calculate: (a), the force exerted by the right hand, and (b), the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand. Strategy Figure 9.20 includes a free
body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net = 0 ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net = 0) if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand. Solution for (a) There are now only two nonzero torques, those from the gravitational force ( w ) and from the push or pull of the right hand ( ). Stating the second condition in terms of clockwise and counterclockwise torques, or the algebraic sum of the torques is zero. Here this is net cw = –net ccw. = –τw (9.22) (9.23) since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, = sin, noting that = 90º, and substituting known values, we obtain (0.900 m) = (0.600 m)(). Thus, Solution for (b) = (0.667) = 32.7 N. 5.00 kg 9.80 m/s2 (9.24) (9.25) The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton's second law: + – = 0 (9.26) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque From this we can conclude: Solving for, we obtain + = = = − = − 32.7 N 371 (9.27) (9.28) = 5.00 kg 9.80 m/s2 − 32.7 N Discussion FL is seen to be exactly half of, as we might have guessed, since is applied twice as far from the cg as. = 16.3 N If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 9.22, the forces change again. Both are considerably greater, and one force reverses direction. Take-Home Experiment This is an experiment to perform while standing in a bus
or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity! PhET Explorations: Balancing Act Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game. Figure 9.23 Balancing Act (http://phet.colorado.edu/en/simulation/balancing-act) 9.5 Simple Machines By the end of this section, you will be able to: • Describe different simple machines. • Calculate the mechanical advantage. Learning Objectives The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we apply the force. The word for “machine” comes from the Greek word meaning “to help make things easier.” Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA
). MA = o i (9.29) 380 Chapter 9 \| Statics and Torque Figure 9.31 This figure shows that large forces are exerted by the back muscles and experienced in the vertebrae when a person lifts with their back, since these muscles have small effective perpendicular lever arms. The data shown here are analyzed in the preceding example, Example 9.5. What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess. There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body—a few of these are the subject of endof-chapter problems. Glossary center of gravity: the point where the total weight of the body is assumed to be concentrated dynamic equilibrium: velocity are zero a state of equilibrium in which the net external force and torque on a system moving with constant mechanical advantage: the ratio of output to input forces for any simple machine neutral equilibrium: a state of equilibrium that is independent of a system's displacements from its original position perpendicular lever arm: the shortest distance from the pivot point to the line along which F lies SI units of torque: newton times meters, usually written as N·m stable equilibrium: displacement a system, when displaced, experiences a net force or torque in a direction opposite to the direction of the static equilibrium: a state of equilibrium in which the net external force and torque acting on a system is zero static equilibrium: equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur torque: turning or twisting effectiveness of a force unstable
equilibrium: from equilibrium a system, when displaced, experiences a net force or torque in the same direction as the displacement Section Summary 9.1 The First Condition for Equilibrium • Statics is the study of forces in equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 381 • Two conditions must be met to achieve equilibrium, which is defined to be motion without linear or rotational acceleration. • The first condition necessary to achieve equilibrium is that the net external force on the system must be zero, so that net F = 0. 9.2 The Second Condition for Equilibrium • The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be = sin where is torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between F and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm ⊥ is defined to be so that ⊥ = sin = ⊥. • The perpendicular lever arm ⊥ is the shortest distance from the pivot point to the line along which acts. The SI unit for torque is newton-meter (N·m). The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: By convention, counterclockwise torques are positive, and clockwise torques are negative. net = 0 9.3 Stability • A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the direction of the displacement. • A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium. • A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. 9.4 Applications of Statics, Including Problem-Solving Strategies • Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. 9.5 Simple Machines
• Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we have to apply the force. • The ratio of output to input forces for any simple machine is called its mechanical advantage • A few simple machines are the lever, nail puller, wheelbarrow, crank, etc. 9.6 Forces and Torques in Muscles and Joints • Statics plays an important part in understanding everyday strains in our muscles and bones. • Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to joints. • Someone with good posture stands or sits in such as way that their center of gravity lies directly above the pivot point in their hips, thereby avoiding back strain and damage to disks. Conceptual Questions 9.1 The First Condition for Equilibrium 1. What can you say about the velocity of a moving body that is in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to represent all external forces on the body. 2. Under what conditions can a rotating body be in equilibrium? Give an example. 9.2 The Second Condition for Equilibrium 3. What three factors affect the torque created by a force relative to a specific pivot point? 4. A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base. 382 Chapter 9 \| Statics and Torque 5. Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also hazardous since it can break the bolt.) 9.3 Stability 6. A round pencil lying on its side as in Figure 9.14 is in neutral equilibrium relative to displacements perpendicular to its length. What is its stability relative to displacements parallel to its length? 7. Explain the need for tall towers on a suspension bridge to ensure stable equilibrium. 9.4 Applications of Statics, Including Problem-Solving Strategies 8. When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs
to be directly above the person's neck vertebrae. 9.5 Simple Machines 9. Scissors are like a double-lever system. Which of the simple machines in Figure 9.24 and Figure 9.25 is most analogous to scissors? 10. Suppose you pull a nail at a constant rate using a nail puller as shown in Figure 9.24. Is the nail puller in equilibrium? What if you pull the nail with some acceleration – is the nail puller in equilibrium then? In which case is the force applied to the nail puller larger and why? 11. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 12. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces (see previous Question)? 9.6 Forces and Torques in Muscles and Joints 13. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 14. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces? 15. Certain types of dinosaurs were bipedal (walked on two legs). What is a good reason that these creatures invariably had long tails if they had long necks? 16. Swimmers and athletes during competition need to go through certain postures at the beginning of the race. Consider the balance of the person and why start-offs are so important for races. 17. If the maximum force the biceps muscle can exert is 1000 N, can we pick up an object that weighs 1000 N? Explain your answer. 18. Suppose the biceps muscle was attached through tendons to the upper arm close to the elbow and the forearm near the wrist. What would be the advantages and disadvantages of this type of construction for the motion of the arm? 19. Explain one of the reasons why pregnant women often suffer from back strain late in their pregnancy. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 383 Problems & Exercises 9.2 The Second Condition for Equilibrium 1. (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a
distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? 2. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton × meters (relative to the center of the bolt)? (b) Convert this torque to footpounds. 3. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. 4. Use the second condition for equilibrium (net τ = 0) to calculate p in Example 9.1, employing any data given or solved for in part (a) of the example. 5. Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. 9.3 Stability 6. Suppose a horse leans against a wall as in Figure 9.32. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits. 8. (a) Calculate the magnitude and direction of the force on each foot of the horse in Figure 9.32 (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal. 9. A person carries a plank of wood 2 m long with one hand pushing down on it
at one end with a force F1 and the other hand holding it up at 50 cm from the end of the plank with force F2. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2? 10. A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure 9.33. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall. Figure 9.33 11. (a) What force must be exerted by the wind to support a 2.50-kg chicken in the position shown in Figure 9.34? (b) What is the ratio of this force to the chicken's weight? (c) Does this support the contention that the chicken has a relatively stable construction? Figure 9.32 7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance? Figure 9.34 12. Suppose the weight of the drawbridge in Figure 9.35 is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg. 384 Chapter 9 \| Statics and Torque Figure 9.35 A small drawbridge, showing the forces on the hinges ( F ), its weight ( w ), and the tension in its wires ( T ). 13. Suppose a 900-kg car is on the bridge in Figure 9.35 with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and
magnitude of the force exerted by the hinges on the bridge. 14. A sandwich board advertising sign is constructed as shown in Figure 9.36. The sign's mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge? Figure 9.36 A sandwich board advertising sign demonstrates tension. 15. (a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.36 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge? 16. A gymnast is attempting to perform splits. From the information given in Figure 9.37, calculate the magnitude and direction of the force exerted on each foot by the floor. This content is available for free at http://cnx.org/content/col11844/1.13 Figure 9.37 A gymnast performs full split. The center of gravity and the various distances from it are shown. 9.4 Applications of Statics, Including ProblemSolving Strategies 17. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom? 18. In Figure 9.22, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 9.20, show that the second condition for equilibrium (net = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above. 9.5 Simple Machines 19. What is the mechanical advantage of a nail puller—similar to the one shown in Figure 9.24 —where you exert
a force 45 cm from the pivot and the nail is 1.8 cm on the other side? What minimum force must you exert to apply a force of 1250 N to the nail? 20. Suppose you needed to raise a 250-kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 2.0-m long lever, where would you place the fulcrum if your force was limited to 300 N? 21. a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure 9.25, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground? 22. A typical car has an axle with 1.10 cm radius driving a tire with a radius of 27.5 cm. What is its mechanical advantage assuming the very simplified model in Figure 9.26(b)? 23. What force does the nail puller in Exercise 9.19 exert on the supporting surface? The nail puller has a mass of 2.10 kg. Chapter 9 \| Statics and Torque 385 equivalent lever system. Calculate the force exerted by the upper leg muscle to lift the mass at a constant speed. Explicitly show how you follow the steps in the ProblemSolving Strategy for static equilibrium in Applications of Statistics, Including Problem-Solving Strategies. 24. If you used an ideal pulley of the type shown in Figure 9.27(a) to support a car engine of mass 115 kg, (a) What would be the tension in the rope? (b) What force must the ceiling supply, assuming you pull straight down on the rope? Neglect the pulley system's mass. 25. Repeat Exercise 9.24 for the pulley shown in Figure 9.27(c), assuming you pull straight up on the rope. The pulley system's mass is 7.00 kg. 9.6 Forces and Torques in Muscles and Joints 26. Verify that the force in the elbow joint in Example 9.4 is 407 N, as stated in the text. 27. Two muscles in the back of the leg pull on the Achilles tendon as shown in Figure 9.38. What
total force do they exert? Figure 9.40 A mass is connected by pulleys and wires to the ankle in this exercise device. 30. A person working at a drafting board may hold her head as shown in Figure 9.41, requiring muscle action to support the head. The three major acting forces are shown. Calculate the direction and magnitude of the force supplied by the upper vertebrae FV to hold the head stationary, assuming that this force acts along a line through the center of mass as do the weight and muscle force. Figure 9.38 The Achilles tendon of the posterior leg serves to attach plantaris, gastrocnemius, and soleus muscles to calcaneus bone. 28. The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure 9.39. Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur). Figure 9.39 The knee joint works like a hinge to bend and straighten the lower leg. It permits a person to sit, stand, and pivot. 29. A device for exercising the upper leg muscle is shown in Figure 9.40, together with a schematic representation of an Figure 9.41 31. We analyzed the biceps muscle example with the angle between forearm and upper arm set at 90º. Using the same numbers as in Example 9.4, find the force exerted by the biceps muscle when the angle is 120º and the forearm is in a downward position. 32. Even when the head is held erect, as in Figure 9.42, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head 386 Chapter 9 \| Statics and Torque erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head? Figure 9.44 A child being lifted by a father's lower leg. 35. Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in Figure 9.45, is attached relatively far from the joint, enabling large forces to be exerted by the back teeth.
(a) Using the information in the figure, calculate the force exerted by the lower teeth on the bullet. (b) Calculate the force on the joint. Figure 9.42 The center of mass of the head lies in front of its major point of support, requiring muscle action to hold the head erect. A simplified lever system is shown. 33. A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 9.43. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint. Figure 9.45 A person clenching a bullet between his teeth. 36. Integrated Concepts Suppose we replace the 4.0-kg book in Exercise 9.31 of the biceps muscle with an elastic exercise rope that obeys Hooke's Law. Assume its force constant = 600 N/m. (a) How much is the rope stretched (past equilibrium) to provide the same force B as in this example? Assume the rope is held in the hand at the same location as the book. (b) What force is on the biceps muscle if the exercise rope is pulled straight up so that the forearm makes an angle of 25º with the horizontal? Assume the biceps muscle is still perpendicular to the forearm. 37. (a) What force should the woman in Figure 9.46 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) Figure 9.43 The muscles in the back of the leg pull the Achilles tendon when one stands on one's toes. A simplified lever system is shown. 34. A father lifts his child as shown in Figure 9.44. What force should the upper leg muscle exert to lift the child at a constant speed? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 387 How much work does she do if her center of mass rises
0.240 m? (d) What is her useful power output if she does 25 pushups in one minute? Figure 9.46 A woman doing pushups. 38. You have just planted a sturdy 2-m-tall palm tree in your front lawn for your mother's birthday. Your brother kicks a 500 g ball, which hits the top of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only 20 cm. Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the 20 cm length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily? 39. Unreasonable Results Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 40. Construct Your Own Problem Consider a method for measuring the mass of a person's arm in anatomical studies. The subject lies on her back, extends her relaxed arm to the side and two scales are placed below the arm. One is placed under the elbow and the other under the back of her hand. Construct a problem in which you calculate the mass of the arm and find its center of mass based on the scale readings and the distances of the scales from the shoulder joint. You must include a free body diagram of the arm to direct the analysis. Consider changing the position of the scale under the hand to provide more information, if needed. You may wish to consult references to obtain reasonable mass values. Test Prep for AP® Courses 9.2 The Second Condition for Equilibrium 1. Which of the following is not an example of an object undergoing a torque? a. A car is rounding a bend at a constant speed. b. A merry-
go-round increases from rest to a constant rotational speed. c. A pendulum swings back and forth. d. A bowling ball rolls down a bowling alley. 2. Five forces of equal magnitude, labeled A–E, are applied to the object shown below. If the object is anchored at point P, which force will provide the greatest torque? Figure 9.47 Five forces acting on an object. a. Force A 388 b. Force B c. Force C d. Force D e. Force E 9.3 Stability 3. Using the concept of torque, explain why a traffic cone placed on its base is in stable equilibrium, while a traffic cone placed on its tip is in unstable equilibrium. 9.4 Applications of Statics, Including ProblemSolving Strategies 4. A child sits on the end of a playground see-saw. Which of the following values is the most appropriate estimate of the torque created by the child? a. 6 N•m b. 60 N•m c. 600 N•m d. 6000 N•m 5. A group of students is stacking a set of identical books, each one overhanging the one below it by 1 inch. They would like to estimate how many books they could place on top of each other before the stack tipped. What information below would they need to know to make this calculation? Figure 9.48 3 overlapping stacked books. I. The mass of each book II. The width of each book III. The depth of each book a. b. c. d. e. I only I and II only I and III only II only I, II, and III 6. A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order to prevent the string from breaking, a person must place an upward force of 7 N at a position along the bottom surface of the board. At what distance from its left edge would they need to place this force in order for the board to be in static equilibrium? Chapter 9 \| Statics and Torque a. If a 1000 kg car comes to rest at a point 5 meters from the left pier, how much force will the bridge provide to the left and right piers? b. How will FL and FR change as the car drives to the right side of the bridge? 8. An object of unknown mass is provided to a student. Without using a scale, design an experimental procedure detailing how the magnitude of this mass could be
experimentally found. Your explanation must include the concept of torque and all steps should be provided in an orderly sequence. You may include a labeled diagram of your setup to help in your description. Include enough detail so that another student could carry out your procedure. 9.5 Simple Machines 9. As a young student, you likely learned that simple machines are capable of increasing the ability to lift and move objects. Now, as an educated AP Physics student, you are aware that this capability is governed by the relationship between force and torque. In the space below, explain why torque is integral to the increase in force created by a simple machine. You may use an example or diagram to assist in your explanation. Be sure to cite the mechanical advantage in your explanation as well. 10. Figure 9.24(a) shows a wheelbarrow being lifted by an applied force Fi. If the wheelbarrow is filled with twenty bricks massing 3 kg each, estimate the value of the applied force Fi. Provide an explanation behind the total weight w and any reasoning toward your final answer. Additionally, provide a range of values over which you feel the force could exist. 9.6 Forces and Torques in Muscles and Joints 11. When you use your hand to raise a 20 lb dumbbell in a curling motion, the force on your bicep muscle is not equal to 20 lb. a. Compare the size of the force placed on your bicep muscle to the force of the 20 lb dumbbell lifted by your hand. Using the concept of torque, which force is greater and explain why the two forces are not identical. b. Does the force placed on your bicep muscle change as you curl the weight closer toward your body? (In other words, is the force on your muscle different when your forearm is 90° to your upper arm than when it is 45° to your upper arm?) Explain your answer using torque. m m a. b. 3 7 5 2 25 7 30 7 e. 5 m d. c. m m 7. A bridge is supported by two piers located 20 meters apart. Both the left and right piers provide an upward force on the bridge, labeled FL and FR respectively. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 389 10 ROTATIONAL MOTION AND ANGULAR MOMENTUM Figure 10.1 The mention of a tornado conjures up images
of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Chapter Outline 10.1. Angular Acceleration 10.2. Kinematics of Rotational Motion 10.3. Dynamics of Rotational Motion: Rotational Inertia 10.4. Rotational Kinetic Energy: Work and Energy Revisited 10.5. Angular Momentum and Its Conservation 10.6. Collisions of Extended Bodies in Two Dimensions 10.7. Gyroscopic Effects: Vector Aspects of Angular Momentum Connection for AP® Courses Why do tornados spin? And why do tornados spin so rapidly? The answer is that the air masses that produce tornados are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner, as seen in Figure 10.2. The skater starts her rotation with outstretched limbs and increases her rate of spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado. We will find that this is another example of the importance of conservation laws and their role in determining how changes happen in a system, supporting Big Idea 5. The idea that a change of a conserved quantity is always equal to the transfer of that quantity between interacting systems (Enduring Understanding 5.A) is presented for both energy and angular momentum (Enduring Understanding 5.E). The conservation of angular momentum in relation to the external net torque (Essential Knowledge 5.E.1) parallels that of linear momentum conservation in relation to the external net force. The concept of rotational inertia is introduced, a concept that takes into account not only the mass of an object or a system, but also the distribution of mass within the object or system. Therefore, changes in the rotational inertia of a system could lead to changes in the motion (Essential Knowledge 5.E.2) of the system. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogues in linear motion. Clearly, therefore, force,
energy, and power are associated with rotational motion. This supports Big Idea 3, that interactions are described by forces. The ability of forces to cause torques (Enduring Understanding 3.F) is extended to the interactions between objects that result in nonzero net torque. This nonzero net torque in turn causes changes in the rotational motion of an object (Essential Knowledge 3.F.2) and results in changes of the angular momentum of an object (Essential Knowledge 3.F.3). 390 Chapter 10 \| Rotational Motion and Angular Momentum Similarly, Big Idea 4, that interactions between systems cause changes in those systems, is supported by the empirical observation that when torques are exerted on rigid bodies these torques cause changes in the angular momentum of the system (Enduring Understanding 4.D). Again, there is a clear analogy between linear and rotational motion in this interaction. Both the angular kinematics variables (angular displacement, angular velocity, and angular acceleration) and the dynamics variables (torque and angular momentum) are vectors with direction depending on whether the rotation is clockwise or counterclockwise with respect to an axis of rotation (Essential Knowledge 4.D.1). The angular momentum of the system can change due to interactions (Essential Knowledge 4.D.2). This change is defined as the product of the average torque and the time interval during which torque is exerted (Essential Knowledge 4.D.3), analogous to the impulse-momentum theorem for linear motion. The concepts in this chapter support: Big Idea 3. The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F. A force exerted on an object can cause a torque on that object. Extended Knowledge 3.F.2. The presence of a net torque along any axis will cause a rigid system to change its rotational motion or an object to change its rotational motion about that axis. Extended Knowledge 3.F.3. A torque exerted on an object can change the angular momentum of an object. Big Idea 4. Interactions between systems can result in changes in those systems. Enduring Understanding 4.D. A net torque exerted on a system by other objects or systems will change the angular momentum of the system. Extended Knowledge 4.D.1. Torque, angular velocity, angular acceleration, and angular momentum are vectors and can be characterized as positive or negative depending upon whether they give rise to or correspond to counterclockwise or clockwise rotation with respect
to an axis. Extended Knowledge 4.D.2. The angular momentum of a system may change due to interactions with other objects or systems. Extended Knowledge 4.D.3. The change in angular momentum is given by the product of the average torque and the time interval during which the torque is exerted. Big Idea 5. Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A. Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Extended Knowledge 5.A.2. For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Enduring Understanding 5.E. The angular momentum of a system is conserved. Extended Knowledge 5.E.1. If the net external torque exerted on the system is zero, the angular momentum of the system does not change. Extended Knowledge 5.E.2. The angular momentum of a system is determined by the locations and velocities of the objects that make up the system. The rotational inertia of an object or system depends upon the distribution of mass within the object or system. Changes in the radius of a system or in the distribution of mass within the system result in changes in the system's rotational inertia, and hence in its angular velocity and linear speed for a given angular momentum. Examples should include elliptical orbits in an Earth-satellite system. Mathematical expressions for the moments of inertia will be provided where needed. Students will not be expected to know the parallel axis theorem. Figure 10.2 This figure skater increases her rate of spin by pulling her arms and her extended leg closer to her axis of rotation. (credit: Luu, Wikimedia Commons) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 391 10.1 Angular Acceleration Learning Objectives By the end of this section, you will be able to: • Describe uniform circular motion. • Explain nonuniform circular motion. • Calculate angular acceleration of an object. • Observe the link between linear and angular acceleration. Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity was defined as the time rate of change of
angle : = Δ Δ, (10.1) where is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity and linear velocity was also defined in Rotation Angle and Angular Velocity as = (10.2) or =, where is the radius of curvature, also seen in Figure 10.3. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative (10.3) Figure 10.3 This figure shows uniform circular motion and some of its defined quantities. Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer's hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: where Δ is the change in angular velocity and Δ is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. If increases, then is positive. If decreases, then is negative. = Δ Δ, (10.4) Example 10.1 Calculating the Angular Acceleration and Deceleration of a Bike Wheel Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of – 87.3 rad/s2, how long does it take the wheel to stop? Strategy for (a) The angular acceleration can be found directly from its definition in = Δ Δ are given. We see that Δ is 250 rpm and Δ is 5.00 s. Solution for (a) Entering known information into the definition of angular acceleration, we get because the final angular velocity and time 392 Chapter 10 \| Rotational Motion and Angular Momentum = Δ Δ 250 rpm 5.00 s =. (10.5) Because Δ is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert Δ from rpm to rad/s: Δ = 250 rev min = 26.2rad s. ⋅ 2π rad rev ⋅
1 min 60 sec Entering this quantity into the expression for, we get = Δ Δ = 26.2 rad/s 5.00 s = 5.24 rad/s2. (10.6) (10.7) Strategy for (b) In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δ, yielding Δ = Δ. (10.8) Solution for (b) Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Δ is – 26.2 rad/s, and is given to be – 87.3 rad/s2. Thus, Δ = – 26.2 rad/s – 87.3 rad/s2 = 0.300 s. (10.9) Discussion Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval. If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 10.4. Thus, linear acceleration is called tangential acceleration t. Figure 10.4 In circular motion, linear acceleration, occurs as the magnitude of the velocity changes: is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration t. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 393 Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, c, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular
motion experiences centripetal acceleration, as seen in Figure 10.5. Thus, t and c are perpendicular and independent of one another. Tangential acceleration t is directly related to the angular acceleration and is linked to an increase or decrease in the velocity, but not its direction. Figure 10.5 Centripetal acceleration c occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other. Now we can find the exact relationship between linear acceleration t and angular acceleration. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be t = Δ Δ (10.10). For circular motion, note that =, so that t = Δ() Δ. The radius is constant for circular motion, and so Δ() = (Δ). Thus, By definition, = Δ Δ. Thus, or 10.11) (10.12) (10.13) (10.14) These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car's drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration. Example 10.2 Calculating the Angular Acceleration of a Motorcycle Wheel A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure 10.6.) 394 Chapter 10 \| Rotational Motion and Angular Momentum Figure 10.6 The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels. Strategy We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration t. Then, the expression = t can be used to find the angular acceleration. Solution The linear acceleration is t = Δ Δ = 30.0 m/s 4.20 s = 7.14 m/s2. We also know the radius of the wheels. Entering the values for t and into = t, we get = t = 7.14 m/s2 0.320 m = 22.3 rad
/s2. (10.15) (10.16) Discussion Units of radians are dimensionless and appear in any relationship between angular and linear quantities. So far, we have defined three rotational quantities—,, and. These quantities are analogous to the translational quantities,, and. Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them. Table 10.1 Rotational and Translational Quantities Rotational Translational Relationship = = = Making Connections: Take-Home Experiment Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities. Check Your Understanding Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 395 Solution The magnitude of angular acceleration is and its most common units are rad/s2. The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis. PhET Explorations: Ladybug Revolution Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. Figure 10.7 Ladybug Revolution (http://cnx.org/content/m55183/1.2/rotation_en.jar) 10.2 Kinematics of Rotational Motion Learning Objectives By the end of this section, you will
be able to: • Observe the kinematics of rotational motion. • Derive rotational kinematic equations. • Evaluate problem solving strategies for rotational kinematics. Just by using our intuition, we can begin to see how rotational quantities like,, and are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration is large for a long period of time, then the final angular velocity and angle of rotation are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating,, and. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion: = 0 + (constant ) (10.17) Note that in rotational motion = t, and we shall use the symbol for tangential or linear acceleration from now on. As in linear kinematics, we assume is constant, which means that angular acceleration is also a constant, because =. Now, let us substitute = and = into the linear equation above: The radius cancels in the equation, yielding = 0 + (constant ) = 0 +. (10.18) (10.19) where 0 is the initial angular velocity. This last equation is a kinematic relationship among,, and —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Making Connections Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts): 396 Chapter 10 \| Rot
ational Motion and Angular Momentum Table 10.2 Rotational Kinematic Equations Rotational Translational ¯ = = 0 + = = 0 + (constant, ) = constant, ) 2 + 2 (constant, ) In these equations, the subscript 0 denotes initial values ( 0, 0, and 0 are initial values), and the average angular velocity - and average velocity - are defined as follows: ¯ = 0 + 2 and ¯ = 0 + 2. (10.20) The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which and are constant. Problem-Solving Strategy for Rotational Kinematics 1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. 3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). 4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. 5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. 6. Check your answer to see if it is reasonable: Does your answer make sense? Example 10.3 Calculating the Acceleration of a Fishing Reel A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in Figure 10.8. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time? Strategy In each part of this example, the strategy is the same as it was for
solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown. Solution for (a) Here and are given and needs to be determined. The most straightforward equation to use is = 0 + because the unknown is already on one side and all other terms are known. That equation states that We are also given that 0 = 0 (it starts from rest), so that = 0 +. = 0 + 110 rad/s2 (2.00s) = 220 rad/s. Solution for (b) Now that is known, the speed can most easily be found using the relationship =, This content is available for free at http://cnx.org/content/col11844/1.13 (10.21) (10.22) (10.23) Chapter 10 \| Rotational Motion and Angular Momentum 397 where the radius of the reel is given to be 4.50 cm; thus, = (0.0450 m)(220 rad/s) = 9.90 m/s. (10.24) Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m×rad = m. Solution for (c) Here, we are asked to find the number of revolutions. Because 1 rev = 2π rad, we can find the number of revolutions by finding in radians. We are given and, and we know 0 is zero, so that can be obtained using = 0.500) 110 rad/s2 (2.00 s)2 = 220 rad. Converting radians to revolutions gives = (220 rad) 1 rev 2π rad = 35.0 rev. Solution for (d) The number of meters of fishing line is, which can be obtained through its relationship with : = = (0.0450 m)(220 rad) = 9.90 m. Discussion (10.25) (10.26) (10.27) This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90
m, about right for when the big fish bites. Figure 10.8 Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel. Example 10.4 Calculating the Duration When the Fishing Reel Slows Down and Stops Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of – 300 rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is 0 = 220 rad/s and the final angular velocity is zero. The angular acceleration is given to be = −300 rad/s2. Examining the available equations, we see all quantities but t are known in = 0 +, making it easiest to use this equation. Solution The equation states 398 Chapter 10 \| Rotational Motion and Angular Momentum = 0 +. We solve the equation algebraically for t, and then substitute the known values as usual, yielding = − 0 = 0 − 220 rad/s −300 rad/s2 = 0.733 s. (10.28) (10.29) Discussion Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration. Example 10.5 Calculating the Slow Acceleration of Trains and Their Wheels Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s2. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train? Strategy In part (a), we are asked to find, and in (b) we are asked to find and. We are given the number of revolutions, the radius of the wheels, and the angular acceleration. Solution for (a)
The distance is very easily found from the relationship between distance and rotation angle: Solving this equation for yields =. = Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities: = (200 rev)2π rad 1 rev = 1257 rad. Now we can substitute the known values into = to find the distance the train moved down the track: = = (0.350 m)(1257 rad) = 440 m. (10.30) (10.31) (10.32) (10.33) Solution for (b) We cannot use any equation that incorporates to find, because the equation would have at least two unknown values. 2 + 2 will work, because we know the values for all variables except : The equation 2 = 0 Taking the square root of this equation and entering the known values gives (0.250 rad/s2)(1257 rad) 1 / 2 = 25.1 rad/s. We can find the linear velocity of the train,, through its relationship to : = = (0.350 m)(25.1 rad/s) = 8.77 m/s. Discussion The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h). (10.34) (10.35) (10.36) There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 399 Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly). Example 10.6 Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.) Strategy ¯ First, find the total
number of revolutions, and then the linear distance traveled. = because - is given to be 6.0 rpm. can be used to find Solution ¯ Entering known values into = gives = = 6.0 rpm (2.0 min) = 12 rev. As always, it is necessary to convert revolutions to radians before calculating a linear quantity like from an angular quantity like : = (12 rev) 2 rad 1 rev = 75.4 rad. Now, using the relationship between and, we can determine the distance traveled: = = (0.15 m)(75.4 rad) = 11 m. Discussion (10.37) (10.38) (10.39) Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics. Check Your Understanding Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.) Solution Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause. 400 Chapter 10 \| Rotational Motion and Angular Momentum 10.3 Dynamics of Rotational Motion: Rotational Inertia Learning Objectives By the end of this section, you will be able to: • Understand the relationship between force, mass, and acceleration. • Study the turning effect of force. • Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. The information presented in this section supports the following AP® learning objectives and science practices: • 4.D.1.1 The student is able to describe a representation and use it to analyze a situation in which several forces exerted on a rotating system of rigidly connected objects change the angular velocity and angular momentum of the system. (S.P. 1.2, 1.4) • 4.D.1.2 The student is able to plan data collection strategies designed to establish that torque, angular velocity, angular acceleration, and angular momentum can be predicted accurately
when the variables are treated as being clockwise or counterclockwise with respect to a well-defined axis of rotation, and refine the research question based on the examination of data. (S.P. 3.2, 4.1, 5.1, 5.3) • 5.E.2.1 The student is able to describe or calculate the angular momentum and rotational inertia of a system in terms of the locations and velocities of objects that make up the system. Students are expected to do qualitative reasoning with compound objects. Students are expected to do calculations with a fixed set of extended objects and point masses. (S.P. 2.2) If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 10.10. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton's second law of motion. There are, in fact, precise rotational analogs to both force and mass. Figure 10.10 Force is required to spin the bike wheel. The greater the force, the greater the angular acceleration produced. The more massive the wheel, the smaller the angular acceleration. If you push on a spoke closer to the axle, the angular acceleration will be smaller. To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force on a point mass that is at a distance from a pivot point, as shown in Figure 10.11. Because the force is perpendicular to, an acceleration = is obtained in the direction of. We can rearrange this equation such that = and then look for ways to relate this expression to expressions for rotational quantities. We note that =, and we substitute this expression into =, yielding Recall that torque is the turning effectiveness of a force. In this case, because F is perpendicular to, torque is simply =. So, if we multiply both sides of the equation above by, we get torque on the left-hand side. That is
, =. or = 2 = 2α. (10.40) (10.41) (10.42) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 401 This last equation is the rotational analog of Newton's second law ( = ), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and 2 is analogous to mass (or inertia). The quantity 2 is called the rotational inertia or moment of inertia of a point mass a distance from the center of rotation. Figure 10.11 An object is supported by a horizontal frictionless table and is attached to a pivot point by a cord that supplies centripetal force. A force is applied to the object perpendicular to the radius, causing it to accelerate about the pivot point. The force is kept perpendicular to. Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences. Rotational Inertia and Moment of Inertia Before we can consider the rotation of anything other than a point mass like the one in Figure 10.11, we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia of an object to be the sum of 2 for all the point masses of which it is composed. That is, = ∑ 2. Here is analogous to in translational motion. Because of the distance, the moment of inertia for any object depends on the chosen axis. Actually, calculating is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop's moment of inertia around its axis is therefore 2, where is its total mass and its radius. (We use and for an entire object to distinguish them from and for point masses.) In all other cases, we must consult Figure 10.12 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for that have been derived from integration over the continuous body. Note that has units of mass multiplied by distance squared ( kg ⋅ m2 ), as we might expect from its definition.
The general relationship among torque, moment of inertia, and angular acceleration is or net τ = = net τ, (10.43) (10.44) where net is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in =, = net τ is the rotational analog to Newton's second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis. As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge. Take-Home Experiment Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall.) Hold the circle stationary and with the number 12 positioned at the top, attach a 402 Chapter 10 \| Rotational Motion and Angular Momentum lump of blue putty (sticky material used for fixing posters to walls) at the number 3. How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle's moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times. In what direction did the
circle rotate when you added putty at the number 3 (clockwise or counterclockwise)? In which of these directions was the resulting angular velocity? Was the angular velocity constant? What can we say about the direction (clockwise or counterclockwise) of the angular acceleration? How could you change the placement of the putty to create angular velocity in the opposite direction? Problem-Solving Strategy for Rotational Dynamics 1. Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation. 2. Determine the system of interest. 3. Draw a free body diagram. That is, draw and label all external forces acting on the system of interest. 4. Apply, the rotational equivalent of Newton's second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation. 5. As always, check the solution to see if it is reasonable. Making Connections In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton's second law of motion for rotation. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 403 Figure 10.12 Some rotational inertias. Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round Consider the father pushing a playground merry-go-round in Figure 10.13. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction. Figure 10.13 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque. Strategy Angular acceleration is given directly by the expression = net τ : 404 Chapter 10 \| Rotational Motion and Angular Momentum To solve for, we must first calculate the torque (which is the same in both cases) and moment of inertia (which is greater in the second case). To find the torque, we note that the applied
force is perpendicular to the radius and friction is negligible, so that =. (10.45) τ = sin θ = (1.50 m)(250 N) = 375 N ⋅ m. Solution for (a) The moment of inertia of a solid disk about this axis is given in Figure 10.12 to be 1 22, where = 50.0 kg and = 1.50 m, so that Now, after we substitute the known values, we find the angular acceleration to be = (0.500)(50.0 kg)(1.50 m)2 = 56.25 kg ⋅ m2. = = 375 N ⋅ m 56.25 kg ⋅ m2 = 6.67rad s2. Solution for (b) (10.46) (10.47) (10.48) (10.49) We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia, we first find the child's moment of inertia c by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then, c = 2 = (18.0 kg)(1.25 m)2 = 28.13 kg ⋅ m2. (10.50) The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of : = 28.13 kg ⋅ m2 + 56.25 kg ⋅ m2 = 84.38 kg ⋅ m2. Substituting known values into the equation for gives = τ = 375 N ⋅ m 84.38 kg ⋅ m2 = 4.44rad s2. (10.51) (10.52) Discussion The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is
on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader. Making Connections: Multiple Forces on One System A large potter's wheel has a diameter of 60.0 cm and a mass of 8.0 kg. It is powered by a 20.0 N motor acting on the outer edge. There is also a brake capable of exerting a 15.0 N force at a radius of 12.0 cm from the axis of rotation, on the underside. What is the angular acceleration when the motor is in use? The torque is found by = sin = (0.300 m)(20.0 N) = 6.00 N·m. The moment of inertia is calculated as = 1 2 2 = 1 2 8.0 kg (0.300 m)2 = 0.36 kg ⋅ m2. Thus, the angular acceleration would be = = 6.00 N ⋅ m 0.36 kg ⋅ m2 = 17 rad/s2. Note that the friction is always acting in a direction opposite to the rotation that is currently happening in this system. If the potter makes a mistake and has both the brake and motor on simultaneously, the friction force of the brake will exert a torque opposite that of the motor. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 405 The torque from the brake is = sin = (0.120 m)(15.0 N) = 1.80 N⋅m. Thus, the net torque is 6.00 N⋅m − 1.80 Ν⋅m = 4.20 Ν⋅m. And the angular acceleration is = = 4.20 N ⋅ m 0.36 kg ⋅ m2 = 12 rad/s2. Check Your Understanding Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple? Solution No. Torque depends
on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors. 10.4 Rotational Kinetic Energy: Work and Energy Revisited Learning Objectives By the end of this section, you will be able to: • Derive the equation for rotational work. • Calculate rotational kinetic energy. • Demonstrate the law of conservation of energy. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.2.1 The student is able to make predictions about the change in the angular velocity about an axis for an object when forces exerted on the object cause a torque about that axis. (S.P. 6.4) • 3.F.2.2 The student is able to plan data collection and analysis strategies designed to test the relationship between a torque exerted on an object and the change in angular velocity of that object about an axis. (S.P. 4.1, 4.2, 5.1) In this module, we will learn about work and energy associated with rotational motion. Figure 10.14 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy. Figure 10.14 The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell) Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.15) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of
the force times the arc length traveled: net = (net )Δ. (10.53) To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by, and gather terms: 406 Chapter 10 \| Rotational Motion and Angular Momentum We recognize that net = net τ and Δ / =, so that net = (net τ). net = ( net )Δ. (10.54) (10.55) This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation net = (net τ) is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net τ =, so that net =. (10.56) Figure 10.15 The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus (net )Δ. The net work goes into rotational kinetic energy. Making Connections qWork and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation. Now, we solve one of the rotational kinematics equations for. We start with the equation Next, we solve for : 2 = 0 2 + 2. = 2 − 0 2 2. Substituting this into the equation for net and gathering terms yields net = 1 22 − 1 20 (10.57) (10.58) (10.59) 2. This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term 2 to be rotational kinetic energy KErot for 1 2 an object with a moment of inertia and an angular velocity : KErot = 1 22. (10.60) The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with being analogous to and to. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in Figure 10.16. This content is available for free at http
://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 407 Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into KErot. It can also convert translational kinetic energy, when the bus stops, into KErot. The flywheel's energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction. Example 10.8 Calculating the Work and Energy for Spinning a Grindstone Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º)? The force is kept perpendicular to the grindstone's 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation net = (net τ). We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in KErot = 1 22. Solution for (a) The net work is expressed in the equation net = (net τ), where net is the applied force multiplied by the radius () because there is no retarding friction, and the force is perpendicular to. The angle is given. Substituting the given values in the equation above yields net = = (0.320 m)(200 N)(1.00 rad) = 64.0 N ⋅ m. Noting that 1 N · m = 1 J, net = 64.0 J. (10.61) (10.62) (10.63) Figure 10.17 A large grindstone is given a spin by a person grasping
its outer edge. Solution for (b) To find from the given information requires more than one step. We start with the kinematic relationship in the equation 2 = 0 2 + 2. (10.64) 408 Chapter 10 \| Rotational Motion and Angular Momentum Note that 0 = 0 because we start from rest. Taking the square root of the resulting equation gives Now we need to find. One possibility is = (2)1 / 2. = net τ, where the torque is The formula for the moment of inertia for a disk is found in Figure 10.12: net τ = = (0.320 m)(200 N) = 64.0 N ⋅ m. = 1 22 = 0.5 85.0 kg (0.320 m)2 = 4.352 kg ⋅ m2. Substituting the values of torque and moment of inertia into the expression for, we obtain = 64.0 N ⋅ m 4.352 kg ⋅ m2 = 14.7rad s2. Now, substitute this value and the given value for into the above expression for : = (2)1 / 2 = 2 14.7rad s2 1 / 2 (1.00 rad) = 5.42rad s. Solution for (c) The final rotational kinetic energy is Both and were found above. Thus, KErot = 1 22. KErot = (0.5) 4.352 kg ⋅ m2 (5.42 rad/s)2 = 64.0 J. (10.65) (10.66) (10.67) (10.68) (10.69) (10.70) (10.71) (10.72) Discussion The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples. Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid

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