Conservation of angular momentum proof

[pardesi]

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)

 

[WMGoBuffs]

I don't think I understand your question: the speed of the center of mass when the ball stops pure rolling is 0.

 

[pardesi]

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

 

[siddharth]

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.

 

[pardesi]

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

 

[lugita15]

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.

 

[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.

 

[rcgldr]

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.

 

[siddharth]

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.

 

[pardesi]

@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

 

[siddharth]

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?

 

[pardesi]

yes i think this is settled now but not so convincingly as i had thought seems too conditional thanks everyone

 

[rcgldr]

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:

https://www.physicsforums.com/showthread.php?t=159337

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.

 

[Doc Al]

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.

 

[jee.anupam]

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.

 

[rcgldr]

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.

https://www.physicsforums.com/showthread.php?t=159337

 

[Doc Al]

True, I was referring to an idealized case.

You must have kinetic friction in order to end up with rolling without slipping; thus you must have energy loss. You can't assume energy conservation as an ideal case for this problem: you end up creating an essentially different problem.

 

[rcgldr]

You must have kinetic friction in order to end up with rolling without slipping; thus you must have energy loss. You can't assume energy conservation as an ideal case for this problem: you end up creating an essentially different problem.

I corrected my previous post. It was too late to edit my first post. Regarding the essentially different problem:

The idealized case would be one where the friction surface flexes elastically and without any slippage. A close approximation of this would be a marble sliding at a reasonably slow speed from a very slick surface onto a surface composed of table tennis rubber which is very sticky and can flex significantly along the surface with a very high energy retention factor (it would spring back, temporarily increasing angular velocity, but then settle), assume this is done in a vacuum chamber so that aerodynamic drag isn't a factor.

Another idealized case would be a rolling gear sliding onto a geared plane. The result would be similar as obove, spring like flexing of the surfaces (gears), and no sliding. The more generalized case as I pointed out would be a very high coefficient of friction combined with a horizontally flexible surface.


In real life, dynamic friction varies with speed (coefficient of friction is reduced as speed differences in the surfaces increase). I'm not sure what affect this would have on the results. The transition from dynamic to static friction isn't instantaneous, but assuming the flex in the friction surface can be ignored, this shouldn't matter.

A more resonable example of a lossless case involving linear and angular energy would be to have a sphere resting on a sticky plane and then accelerate the sticky plane horizontally with a constant rate of acceleration. As velocity of the plane increases, what is the ratio of the velocity of the sphere versus the velocity of the plane the sphere is rolling on?

Highlight for answer: 5/9

 

[rcgldr]

Corrections and clarifications to above:

Constant acceleration of horizontal plane:
What is the ratio of the acceleration of a non-sliding sphere versus the acceleration of the plane the sphere is rolling on?

Highlight for answer: 2/7

What is the ratio of the acceleration of frictionless block versus the acceleration of the plane the block is sliding on?

Highlight for answer: 0

Constant force:
On an inclined plane what is the ratio of acceleration of a sphere rolling on the plane versus the acceleration of a frictionless block sliding on the plane?

Highlight for answer: 5/7

 

[pardesi]

@jee.anupam
yes i will try to be careful next time