## conservation of angular momentum

well i saw a proof and therea re hundreds of them where conservation of angular momentum is used to solve the problems.let me state an example and then my question.
a ball with some initial velocity is put on a rough plane find the speed of the c.m when it stops pure rolling.well of course on solution is via the torque equation.but other solution is by conserving the angular momentum about the point of contact since there no net torque acts about that point.

my question is this point itself is moving how can we conserve momentum about two diff. points(since the point initially with which the floor was in contact with is in general not same as the point it contacts after the ball starts pure rolling)
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 I don't think I understand your question: the speed of the center of mass when the ball stops pure rolling is 0.
 well what i meant was consider this ismple problem u hav aball rolling with speed $$v_{0}$$ initially and u put it ona rough surface and conserve momentuma about the point of contact(???) say that no torque acts about thsi point.thenn finally when it starts pure rolling we put $$mv_{0}r=I\frac{v}{r}+mvr$$ but my question is the point about which we are conserving omentum is itself moving and finally when the sphere starts pure rolling then the point may not be the same as the one with which it was put in contact withthe ground hope i was clear

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## conservation of angular momentum

 Quote by pardesi but my question is the point about which we are conserving omentum is itself moving and finally when the sphere starts pure rolling then the point may not be the same as the one with which it was put in contact withthe ground
Yes, but the frame you're looking at is not "attached" to a particular physical part of the sphere which is rotating. The frame is located at the point of contact at all times. Now, this frame isn't inertial and is accelerating. However, if you calculate the "apparent torque" in this system about the point of contact at all times, you'll find it's zero. So, you can apply the conservation of momentum.
 yes that' true i can always conserve angular momentum about the point of contact but this point is itself alwys changing.here the point about which i conserve is itself changing with time so how do i conserve angular momentum about two apparently different points

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 Quote by siddharth Yes, but the frame you're looking at is not "attached" to a particular physical part of the sphere which is rotating. The frame is located at the point of contact at all times. Now, this frame isn't inertial and is accelerating. However, if you calculate the "apparent torque" in this system about the point of contact at all times, you'll find it's zero. So, you can apply the conservation of momentum.
No you can't. In this case, there will be a torque due to fictitious forces which arise since you're in an accelerating frame. Therefore, conservation of angular momentum is invalid in this case.
 @pardesi Hey buddy, you are not conserving angular momentum about the moving point, you are conserving angular momentum about any stationary point on the ground, which is in line with the frictional force (even the initial point will do), so that the net torque of friction about the stationary point is zero. You cannot conserve angular momentum about a moving point.
 Recognitions: Homework Help Assuming no losses, then angular moementum isn't conserved, but total energy is. So the initial condition could be 0 angular rotation, and some velocity as the ball starts off initially sliding on the surface. The initial total engery is all linear kinetic energy. The final state will be the sum of kinetic energy due to rotation and linear movement, and the rate of rotation will be proportional to the linear movement.

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 Quote by lugita15 No you can't. In this case, there will be a torque due to fictitious forces which arise since you're in an accelerating frame. Therefore, conservation of angular momentum is invalid in this case.
 Hey buddy, you are not conserving angular momentum about the moving point, you are conserving angular momentum about any stationary point on the ground, which is in line with the frictional force (even the initial point will do), so that the net torque of friction about the stationary point is zero.
Yeah, that's right. Nice catch.

 Quote by jee.anupam @pardesi Hey buddy, you are not conserving angular momentum about the moving point, you are conserving angular momentum about any stationary point on the ground, which is in line with the frictional force (even the initial point will do), so that the net torque of friction about the stationary point is zero. You cannot conserve angular momentum about a moving point.
well that won't do because there will be torque due to normal force exerted by the ground. in fact no other point except the nstantaneous point of contact would supposedly do

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 Quote by pardesi well that won't do because there will be torque due to normal force exerted by the ground. in fact no other point except the nstantaneous point of contact would supposedly do
This will cancel with the torque due to the force of gravity about the stationary point on the ground. wouldn't it?
 yes i think this is settled now but not so convincingly as i had thought seems too conditional thanks everyone

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 a ball with some initial velocity is put on a rough plane find the speed of the c.m when it stops pure rolling.
Shouldn't that be when the ball starts pure rolling? The initial condition is a sliding ball that isn't rolling and friction is converting some the linear kinetic energy into rotational energy and some into heat (losses). This thread shows the math:

My previous most mentioned "assume no losses", but this could only happen in a case where no sliding occurred. The friction surface would have to immediate grip the sphere, and flex horizontally with a spring like reaction applied to the bottom surface of the sphere to convert the linear energy into a combination of linear and angular energy with no losses.
 Mentor Blog Entries: 1 As jee.anupam explained, angular momentum is conserved about any point along the line of action of the friction force. That fact, plus the no-slipping condition ($v = \omega r$), allows you to compute the speed at which rolling without slipping takes place. (Of course, angular momentum about the cm of the ball is not conserved, otherwise the ball will never start spinning.) You can also solve this using Newton's 2nd law, which takes longer but is instructive. Energy is not conserved: Until rolling without slipping is attained, the ball scrapes along the floor, dissipating mechanical energy. Assuming energy conservation gives you the wrong answer.

 Quote by pardesi well that won't do because there will be torque due to normal force exerted by the ground. in fact no other point except the instantaneous point of contact would supposedly do
The torque due to the normal force will be canceled by the torque due to gravity. Please consider all the forces acting on the rolling body before jumping to any conclusion.Thank you.

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 Quote by Doc Al Energy is not conserved: Until rolling without slipping is attained, the ball scrapes along the floor, dissipating mechanical energy. Assuming energy conservation gives you the wrong answer.
True, I was referring to an idealized case. I saw in an older thread where you and another posted what the losses would be for this situation. In this thread the losses are taken into account.