Conversion of angular momentum to linear momentum


by NANDHU001
Tags: angular, conversion, linear, momentum
NANDHU001
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#1
Feb4-13, 09:59 AM
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Suppose a rod is rotating about its center of mass, for simplicity consider it to be its geometric center(uniform density). At an instant a rigid ball is introduced near the rod so that it collides with the rod and the stick comes to rest(rotationally and translationally). Then can the ball be said to posses the equivalent linear momentum of the rotational(angular) momentum possessed by the rod initially. Also if the parameters are as below please tell me how to calculate the final linear momentum of the ball.
mass of rod= x ,mass of ball=y,angular velocity of rod=v,arm distance from center where collision takes place=d.
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Feb4-13, 01:36 PM
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At an instant a rigid ball is introduced near the rod so that it collides with the rod and the stick comes to rest(rotationally and translationally).
The rod has to be attached to something in some way to do that, otherwise linear momentum is not conserved.
Then can the ball be said to posses the equivalent linear momentum of the rotational(angular) momentum possessed by the rod initially.
No, as some momentum has to be exchanged with the environment.
Also if the parameters are as below please tell me how to calculate the final linear momentum of the ball.
Probably with conservation of angular momentum, assuming the rod is fixed to rotate around its center of mass.
sophiecentaur
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Feb4-13, 05:15 PM
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The rod could be spinning in space with no linear momentum (in the observer's frame). A situation can arise where a ball of the right mass and velocity (linear momentum but with angular momentum about the axis of spin of the rod, on impact) can strike the end of the rod and leave the rod not rotating but moving off in one direction, with the mass bouncing back - or sticking to the rod, depending on the details. Linear momentum and angular momentum would be preserved.

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Feb4-13, 06:42 PM
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Conversion of angular momentum to linear momentum


Quote Quote by sophiecentaur View Post
A situation can arise where a ball of the right mass and velocity (linear momentum but with angular momentum about the axis of spin of the rod, on impact) can strike the end of the rod and leave the rod not rotating but moving off in one direction, with the mass bouncing back - or sticking to the rod, depending on the details. Linear momentum and angular momentum would be preserved.
The OP's question says the rod has no linear momentum before and after the impact.

I don't think that is possible unless the rod is rotating about a fixed pivot. The rod has no linear momentum before the impact. If it has no linear momentum after impact, the total impulse on the rod must be zero. The ball hitting the rod gives a non-zero impulse, so you another impulse on the rod to balance it - for example the reaction at the pivot.

So the only solution is the "trick answer" that the rotation speed of the rod before the impact was zero, and you can put the ball anywhere you like.
NANDHU001
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#5
Feb5-13, 09:50 AM
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Okay, I'll provide a slight clarification. The ball is placed so that it collides with the rod. I'll ask the question in a different manner.
A rod is spinning around its center of mass with no linear velocity with respect to the observer.
Is it possible for a ball (placed so that it collides with the rod) to make the linear as well as angular momentum of the rod ZERO( The collision is neither elastic nor the ball sticks to the rod). If so how can the linear momentum of the ball be calculated. And can the final linear momentum of the ball be said to be the equivalent linear momentum of the angular momentum possessed by the rod.
DrZoidberg
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#6
Feb5-13, 11:10 AM
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It's possible if the rod collides with two balls at the same time.
Say the rod is floating in space and rotating around it's center but otherwise isn't moving relative to the observer.
If both ends of the rod collide with a ball at the same moment and the two balls have the right mass, the rod will stop moving alltogether. It's angular momentum will be 0 and the two balls will each have gained a linear momentum.
However - the angular momentum is still conserved. Remember the definition of angular momentum. L = r x p. An object doesn't need to rotate to have an angular momentum. r is the position of the object and p is it's linear momentum. It doesn't matter relative to what point in the universe you measure r, as long as you use the same point for all measurements the total angular momentum of all particles combined will stay the same, as will the linear momentum. And yes, the magnitude of the linear momentum of both balls combined will be equal to the magnitude of the angular momentum the rod originally had.
sophiecentaur
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Feb5-13, 12:46 PM
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Quote Quote by NANDHU001 View Post
Okay, I'll provide a slight clarification. The ball is placed so that it collides with the rod. I'll ask the question in a different manner.
A rod is spinning around its center of mass with no linear velocity with respect to the observer.
Is it possible for a ball (placed so that it collides with the rod) to make the linear as well as angular momentum of the rod ZERO( The collision is neither elastic nor the ball sticks to the rod). If so how can the linear momentum of the ball be calculated. And can the final linear momentum of the ball be said to be the equivalent linear momentum of the angular momentum possessed by the rod.
If linear and angular momentum are to be conserved then the answer has to be NO. The ball starts with some linear momentum and it can hardly interact with the rod without transferring some of that.
NANDHU001
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#8
Feb7-13, 04:43 AM
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Please note the word 'placed' in my clarification of the question. What I have meant is that a ball of zero linear velocity is placed by an external agency such as a human being at a later time so that the rotating rod strikes it off and the rod thus comes to rest both translationally and rotationally. Is such a situation possible.
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#9
Feb7-13, 06:27 AM
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If the rod's axis of rotation (center of mass) is not held effectively stationary by being fastened to the earth or some other very massive object, the answer is "no."

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.
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Feb7-13, 12:43 PM
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Quote Quote by jtbell View Post
If the rod's axis of rotation (center of mass) is not held effectively stationary by being fastened to the earth or some other very massive object, the answer is "no."

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.
I agree. The problem seems to keep shifting in subtle ways, I think but the basic conservation laws must apply. Changing the model slightly, it would be easy to bring a rotating bat to a halt by striking an appropriate ball. This was what I thought we were heading for aamof.
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Feb8-13, 01:21 AM
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Quote Quote by jtbell View Post
If the rod's axis of rotation (center of mass) is not held effectively stationary by being fastened to the earth or some other very massive object, the answer is "no."

The total linear momentum of the system is zero before the collision, so it must also be zero after the collision. If the ball has linear momentum after the collision, the rod must also have linear momentum, equal in magnitude and opposite in direction.
Thanks, will there be any change in the rotational velocity of the rod after the event.
If so, tell me how to calculate it.
sophiecentaur
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Feb8-13, 10:36 AM
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Quote Quote by NANDHU001 View Post
Thanks, will there be any change in the rotational velocity of the rod after the event.
If so, tell me how to calculate it.
I'd suggest you write out two equations, one for linear and one for angular momentum. I would imagine that there will be a suitable ratio of masses that will give you whatever outcome you specify. The variables are the position of the impact and the ratio of masses, which is two unknowns, which two equations should enable you to find.


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