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system : object or collection of objects whose motion is currently under investigation; however, your system is defined at the start of the problem, you must keep that definition for the entire problem
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The motion of an object depends on its mass as well as its velocity. Momentum is a concept that describes this. It is a useful and powerful concept, both computationally and theoretically. The SI unit for momentum is kg··m/s.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
When a force is applied on an object for some amount of time, the object experiences an impulse.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
This impulse is equal to the object’s change of momentum.
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Newton’s second law in terms of momentum states that the net force applied to a system equals the rate of change of the momentum that the force causes.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
The law of conservation of momentum says that the momentum of a closed system is constant in time (conserved).
https://openstax.org/books/university-physics-volume-1/pages/9-summary
A closed (or isolated) system is defined to be one for which the mass remains constant, and the net external force is zero.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
The total momentum of a system is conservedonlywhen the system is closed.
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An elastic collision is one that conserves kinetic energy.
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An inelastic collision does not conserve kinetic energy.
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Momentum is conserved regardless of whether or not kinetic energy is conserved.
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Analysis of kinetic energy changes and conservation of momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one-dimensional, two-body collisions.
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The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
Momentum is conserved in both directions simultaneously and independently.
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The Pythagorean theorem gives the magnitude of the momentum vector using thex- andy-components, calculated using conservation of momentum in each direction.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
An extended object (made up of many objects) has a defined position vector called the center of mass.
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The center of mass can be thought of, loosely, as the average location of the total mass of the object.
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The center of mass of an object traces out the trajectory dictated by Newton’s second law, due to the net external force.
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The internal forces within an extended object cannot alter the momentum of the extended object as a whole.
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A rocket is an example of conservation of momentum where the mass of the system is not constant, since the rocket ejects fuel to provide thrust.
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The rocket equation gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass.
https://openstax.org/books/university-physics-volume-1/pages/9-summary
θ = s r θ = s r
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ω = lim Δ t → 0 Δ θ Δ t = d θ d t ω = lim Δ t → 0 Δ θ Δ t = d θ d t
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v t = r ω v t = r ω
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α = lim Δ t → 0 Δ ω Δ t = d ω d t = d 2 θ d t 2 α = lim Δ t → 0 Δ ω Δ t = d ω d t = d 2 θ d t 2
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a t = r α a t = r α
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ω – = ω 0 + ω f 2 ω – = ω 0 + ω f 2
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θ f = θ 0 + ω – t θ f = θ 0 + ω – t
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ω f = ω 0 + α t ω f = ω 0 + α t
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θ f = θ 0 + ω 0 t + 1 2 α t 2 θ f = θ 0 + ω 0 t + 1 2 α t 2
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ω f 2 = ω 0 2 + 2 α ( Δ θ ) ω f 2 = ω 0 2 + 2 α ( Δ θ )
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a → = a → c + a → t a → = a → c + a → t
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K = 1 2 ( ∑ j m j r j 2 ) ω 2 K = 1 2 ( ∑ j m j r j 2 ) ω 2
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I = ∑ j m j r j 2 I = ∑ j m j r j 2
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K = 1 2 I ω 2 K = 1 2 I ω 2
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I = ∫ r 2 d m I = ∫ r 2 d m
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I parallel-axis = I center of mass + m d 2 I parallel-axis = I center of mass + m d 2
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I total = ∑ i I i I total = ∑ i I i
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τ → = r → × F → τ → = r → × F →
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| τ → | = r ⊥ F | τ → | = r ⊥ F
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τ → net = ∑ i | τ → i | τ → net = ∑ i | τ → i |
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∑ i τ i = I α ∑ i τ i = I α
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d W = ( ∑ i τ i ) d θ d W = ( ∑ i τ i ) d θ
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W A B = K B − K A W A B = K B − K A
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W A B = ∫ θ A θ B ( ∑ i τ i ) d θ W A B = ∫ θ A θ B ( ∑ i τ i ) d θ
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P = τ ω P = τ ω
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angular acceleration : time rate of change of angular velocity
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angular position : angle a body has rotated through in a fixed coordinate system
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angular velocity : time rate of change of angular position
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instantaneous angular acceleration : derivative of angular velocity with respect to time
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instantaneous angular velocity : derivative of angular position with respect to time
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kinematics of rotational motion : describes the relationships among rotation angle, angular velocity, angular acceleration, and time
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lever arm : perpendicular distance from the line that the force vector lies on to a given axis
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linear mass density : the mass per unit lengthλλof a one dimensional object
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moment of inertia : rotational mass of rigid bodies that relates to how easy or hard it will be to change the angular velocity of the rotating rigid body
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Newton’s second law for rotation : sum of the torques on a rotating system equals its moment of inertia times its angular acceleration
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parallel axis : axis of rotation that is parallel to an axis about which the moment of inertia of an object is known
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parallel-axis theorem : if the moment of inertia is known for a given axis, it can be found for any axis parallel to it
https://openstax.org/books/university-physics-volume-1/pages/10-key-terms
rotational dynamics : analysis of rotational motion using the net torque and moment of inertia to find the angular acceleration
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rotational kinetic energy : kinetic energy due to the rotation of an object; this is part of its total kinetic energy
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rotational work : work done on a rigid body due to the sum of the torques integrated over the angle through with the body rotates
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surface mass density : mass per unit areaσσof a two dimensional object
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torque : cross product of a force and a lever arm to a given axis
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total linear acceleration : vector sum of the centripetal acceleration vector and the tangential acceleration vector
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The angular positionθθof a rotating body is the angle the body has rotated through in a fixed coordinate system, which serves as a frame of reference.
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The angular velocity of a rotating body about a fixed axis is defined asω(rad/s)ω(rad/s), the rotational rate of the body in radians per second. The instantaneous angular velocity of a rotating bodyω=limΔt→0ΔθΔt=dθdtω=limΔt→0ΔθΔt=dθdtis the derivative with respect to time of the angular positionθθ, ...
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The angular velocityω→ω→is found using the right-hand rule. If the fingers curl in the direction of rotation about a fixed axis, the thumb points in the direction ofω→ω→(seeFigure 10.5).
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If the system’s angular velocity is not constant, then the system has an angular acceleration. The average angular acceleration over a given time interval is the change in angular velocity over this time interval,α–=ΔωΔtα–=ΔωΔt. The instantaneous angular acceleration is the time derivative of angular velo...
https://openstax.org/books/university-physics-volume-1/pages/10-summary
The tangential acceleration of a point at a radius from the axis of rotation is the angular acceleration times the radius to the point.
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The kinematics of rotational motion describes the relationships among rotation angle (angular position), angular velocity, angular acceleration, and time.
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For a constant angular acceleration, the angular velocity varies linearly. Therefore, the average angular velocity is 1/2 the initial plus final angular velocity over a given time period:ω–=ω0+ωf2.ω–=ω0+ωf2.
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We used a graphical analysis to find solutions to fixed-axis rotation with constant angular acceleration. From the relationω=dθdtω=dθdt, we found that the area under an angular velocity-vs.-time curve gives the angular displacement,θf−θ0=Δθ=∫t0tω(t)dtθf−θ0=Δθ=∫t0tω(t)dt. The results of the graphic...
https://openstax.org/books/university-physics-volume-1/pages/10-summary
The linear kinematic equations have their rotational counterparts such that there is a mappingx→θ,v→ω,a→αx→θ,v→ω,a→α.
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A system undergoing uniform circular motion has a constant angular velocity, but points at a distancerfrom the rotation axis have a linear centripetal acceleration.
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A system undergoing nonuniform circular motion has an angular acceleration and therefore has both a linear centripetal and linear tangential acceleration at a point a distancerfrom the axis of rotation.
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The total linear acceleration is the vector sum of the centripetal acceleration vector and the tangential acceleration vector. Since the centripetal and tangential acceleration vectors are perpendicular to each other for circular motion, the magnitude of the total linear acceleration is|a→|=ac2+at2|a→|=ac2+at2.
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The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given byK=12Iω2K=12Iω2, whereIis the moment of inertia, or “rotational mass” of the rigid body or system of particles.
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The moment of inertia for a system of point particles rotating about a fixed axis isI=∑jmjrj2I=∑jmjrj2, wheremjmjis the mass of the point particle andrjrjis the distance of the point particle to the rotation axis. Because of ther2r2term, the moment of inertia increases as the square of the distance to the fixed rot...
https://openstax.org/books/university-physics-volume-1/pages/10-summary
In systems that are both rotating and translating, conservation of mechanical energy can be used if there are no nonconservative forces at work. The total mechanical energy is then conserved and is the sum of the rotational and translational kinetic energies, and the gravitational potential energy.
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Moments of inertia can be found by summing or integrating over every ‘piece of mass’ that makes up an object, multiplied by the square of the distance of each ‘piece of mass’ to the axis. In integral form the moment of inertia isI=∫r2dmI=∫r2dm.
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Moment of inertia is larger when an object’s mass is farther from the axis of rotation.
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It is possible to find the moment of inertia of an object about a new axis of rotation once it is known for a parallel axis. This is called the parallel axis theorem given byIparallel-axis=Icenter of mass+md2Iparallel-axis=Icenter of mass+md2, wheredis the distance from the initial axis to the parallel axis.
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Moment of inertia for a compound object is simply the sum of the moments of inertia for each individual object that makes up the compound object.
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The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation|τ→|=r⊥F|τ→|=r⊥F, wherer⊥r⊥is the perpendicular distance from the axis to the line upon which the force vector lies.
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The sign of the torque is found using the right hand rule. If the page is the plane containingr→r→andF→F→, thenr→×F→r→×F→is out of the page for positive torques and into the page for negative torques.
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The net torque can be found from summing the individual torques about a given axis.
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Newton’s second law for rotation,∑iτi=Iα∑iτi=Iα, says that the sum of the torques on a rotating system about a fixed axis equals the product of the moment of inertia and the angular acceleration. This is the rotational analog to Newton’s second law of linear motion.
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In the vector form of Newton’s second law for rotation, the torque vectorτ→τ→is in the same direction as the angular accelerationα→α→. If the angular acceleration of a rotating system is positive, the torque on the system is also positive, and if the angular acceleration is negative, the torque is negativ...
https://openstax.org/books/university-physics-volume-1/pages/10-summary
The incremental workdWin rotating a rigid body about a fixed axis is the sum of the torques about the axis times the incremental angledθdθ.
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The total work done to rotate a rigid body through an angleθθabout a fixed axis is the sum of the torques integrated over the angular displacement. If the torque is a constant as a function ofθθ, thenWAB=τ(θB−θA)WAB=τ(θB−θA).
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The work-energy theorem relates the rotational work done to the change in rotational kinetic energy:WAB=KB−KAWAB=KB−KAwhereK=12Iω2.K=12Iω2.
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The power delivered to a system that is rotating about a fixed axis is the torque times the angular velocity,P=τωP=τω.
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v CM = R ω v CM = R ω
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a CM = R α a CM = R α
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d CM = R θ d CM = R θ
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a CM = m g sin θ m + ( I CM / r 2 ) a CM = m g sin θ m + ( I CM / r 2 )
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l → = r → × p → l → = r → × p →
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d l → d t = ∑ τ → d l → d t = ∑ τ →
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L → = l → 1 + l → 2 + ⋯ + l → N L → = l → 1 + l → 2 + ⋯ + l → N
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d L → d t = ∑ τ → d L → d t = ∑ τ →
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