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7-3: Rotational Energy and Angular Momentum

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7-3: Rotational Energy and Angular Momentum Lecture #7.3 Rotational Energy, Angular Momentum 1. Work and kinetic energy for rotational motion We have already seen that solution of many problems may become much easier if they are solved not by means of the Newton's laws, but using laws of conservation of energy and momentum. Now we have to develop similar technique for rotational problems. We have introduced kinetic energy as the energy of motion. Rotation is also example of motion. We saw that motion of the rigid body can be considered as combination of two types of motion: translational motion of its center of mass for which mv2 we know kinetic energy is KE com and rotation around the axis passing through the t 2 body's center of mass. How can we find kinetic energy associated with this rotation? Rotating object can be treated as a system of particles of masses mi moving with velocities vi , then kinetic energy of rotation is the sum of individual kinetic energies of the particles, which is 1 2 KEr m i v i , i 2 Even though speed of all the particles is not the same during the rotation, but their angular speed is the same for the same rotational axis, so vi ri and we have 12 12 2 1 2 KEr m i r i m i r i I , . (7.3.1) ii2 2 2 2 where miri I is moment of inertia for the system of particles. Equation 7.3.1 i represents rotational kinetic energy of a rigid body. It is easy to see that it has the same structure as kinetic energy of translational motion, if we replace moment of inertia by mass and angular velocity by linear velocity. As we know from the case of translational motional, the force acting on the body performs work which results in change of the body's kinetic energy. The similar situation should occur in the case of rotation. However, now we have to relate work and torque, because torque plays the same role for rotating bodies as force does for transnational motion. If we apply force F at distance r from the body's rotational axis, it performs work. In the case of rotation for a small angle , the elementary work W is going to be W F r , (7.3.2) where r is the magnitude of displacement along the circular path for this elementary rotation. F is the tangential component of force. Only this component of force (which is in the direction of displacement) needs to be taken into account, since the force which is perpendicular to displacement does not do any work. Then Fr is the torque of this force, this means that W . (7.3.3) If one rotates this body from original angle 1 to final angle 2 , the total work done in this case is W lim , (7.3.4) 0 which is the same as in the case of linear displacement, if one replaces force by torque and linear displacement by angular displacement. If the torque acting on the body is constant then equation 7.3.4 becomes even simpler W b 2 1g. (7.3.5) We also know that according to work-kinetic energy theorem, work done on the body goes to change its kinetic energy. Indeed according to equation (22.5) and the Newton's second law we have 11 W()() I I 22 I KE KE KE , (7.3.6) 2 1 2 122 2 1 2 1 which is work-kinetic energy theorem in the case of the rotation. Even though we only proved it for the case of constant angular acceleration, it is valid in general case too. 2. Rolling Let us now turn our attention back to the problem of rolling which we have considered last time. We looked at wheel rolling smoothly (without sliding) along the road. It participates in both translational and rotational motions. If the center of mass O of this wheel moves forward along the road with constant speed vcm then we proved that the relation between speed of translational motion of the center of mass and angular speed of the wheel is vRcom . (7.3.7) If the wheel accelerates along the road with acceleration acm for its center of mass then the wheel also has angular acceleration, which can be related to acm as aRcom , (7.3.8) where is the angular acceleration of this wheel. In order for the wheel to have a nonzero angular acceleration, the nonzero net torque is needed. This torque comes from force of friction acting at the bottom of the wheel. Example 7.3.1 Let us continue with the same example we started last time: rolling of the wheel down the incline plane. We proved last time that acceleration of the wheel as is goes down the ramp is g sin a . (7.3.9) I 1 mR2 To find the final linear speed of the wheel, which started from rest, we can use the equation from kinematics, which is 22 v v0 2, ad v0 0, 2gd sin v2 , I 1 mR2 2gh v I 1 mR2 h where h is the height of the incline plane and d is the distance the wheel moves sin along the plane. So, we solved this problem using Newton’s second law. Now let us see how we can apply conservation of energy towards solution of this problem. Rolling of the wheel involves both: translation of the wheel's center of mass and rotation around this center of mass. So that, the kinetic energy of the rolling wheel can be found as 11 KE I22 mv . (7.3.10) 22com com Here the first term refers to rotational kinetic energy around the axis passing through the center of mass and the second term refers to translational kinetic energy of the center of mass. To find the final speed of the wheel, which stared its motion from rest, we can use the energy conservation. The total mechanical energy at the top of the ramp equal to the total mechanical energy at the bottom, which is EE12 , PE1 KE 1 PE 2 KE 2 , 11 mgh0 0 I 22 mv , 22 where we took into account that kinetic energy at the top of the ramp was zero and the body only had potential energy. At the bottom potential energy is zero but kinetic energy has two contributions translational and rotational. In the case of rolling without slipping we also have v R , so 11 mgh I 22 mv , 22 Iv2 1 gh v2 , 22mR2 I ( 1)v2 2 gh , mR2 2gh v . I 1 mR2 This solution is much shorter compared to that using Newton's laws. Exercise: Think about solutions of this problem in the case of different objects, such as ball, cylinder or hoop. 2. Angular momentum During previous lectures we have discussed many problems involving collisions. The most powerful method to solve those problems was the law of conservation of linear momentum. Let us see if there is a similar concept in the case of rotational motion. If the particle of mass m moves with velocity v , it has linear momentum p mv . If position of this particle in space is defined by radius vector r directed from the origin of coordinate system then I shall define the magnitude of angular momentum for this particle as l rpsin rmv sin , (7.3.11) where is the angle between directions of vectors and p . (Note: If you are familiar with concept of cross product of two vectors, you might notice that, the angular momentum of the particle is a vector, which is equal to the vector product of the radius vector of particle and its momentum) The SI unit for angular momentum is kilogram times meter squared per second dkg m2 si. It is easy to see that angular momentum plays the same role with respect to momentum as torque plays with respect to force. We can also rewrite the last equation in terms of vector components, like l rp rmv r p rmv , (7.3.12) where r is the component of vector perpendicular to vector p and p is the component of vector perpendicular to vector . We saw that Newton's second law can be written as p F . (7.3.13) net t This equation can be used to describe motion on of the particle-like object. Now let us see how this equation will change if we apply it for rotational motion. Let us consider the angular momentum of the particle moving around the circle, defined by the equation (22.11), which gives l () rp p r rF , t t t net net so for a single particle we have l . (7.3.14) net t The sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle, which is an angular form of the Newton's second law for a single particle. Now let us consider the system of many particles. The total angular momentum of this system with respect to the same origin is the sum of the angular momenta of individual particles, which is n Ll i . (7.3.15) i1 Here we shall use the same convention as we used for torques: the angular momentum for the particle is positive if it is moving in counterclockwise direction and negative if it is moving in clockwise direction. This angular momentum will change with time according to n n n L li li neti net .
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