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conservation of angular momentum |
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| Jul1-07, 07:17 AM | #1 |
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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) |
| Jul1-07, 11:07 AM | #2 |
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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.
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| Jul2-07, 04:04 AM | #3 |
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well what i meant was consider this ismple problem u hav aball rolling with speed [tex]v_{0}[/tex] 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
[tex]mv_{0}r=I\frac{v}{r}+mvr[/tex] 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 |
| Jul2-07, 07:28 AM | #4 |
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conservation of angular momentum |
| Jul2-07, 09:02 AM | #5 |
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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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| Jul4-07, 12:56 PM | #6 |
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| Jul5-07, 02:48 PM | #7 |
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@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. |
| Jul5-07, 03:20 PM | #8 |
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Recognitions:
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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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| Jul6-07, 03:15 AM | #9 |
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| Jul6-07, 07:19 AM | #10 |
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| Jul6-07, 10:31 AM | #11 |
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| Jul6-07, 12:29 PM | #12 |
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yes i think this is settled now but not so convincingly as i had thought seems too conditional thanks everyone
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| Jul7-07, 02:13 AM | #13 |
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Recognitions:
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http://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. |
| Jul7-07, 07:45 AM | #14 |
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Mentor
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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 ([itex]v = \omega r[/itex]), 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. |
| Jul7-07, 12:58 PM | #15 |
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| Jul7-07, 03:02 PM | #16 |
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Recognitions:
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http://www.physicsforums.com/showthread.php?t=159337 |
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