# Conservation of angular momentum

(see attachment)

## The Attempt at a Solution

I think the angular momentum can be conserved about any point on the rod because torque is zero about any point on the rod. But that's not the correct answer. Any help is appreciated. Thanks!

SammyS
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## Homework Statement

(see attachment) [ IMG]https://www.physicsforums.com/attachment.php?attachmentid=55391&d=1360078065[/PLAIN]

## The Attempt at a Solution

I think the angular momentum can be conserved about any point on the rod because torque is zero about any point on the rod. But that's not the correct answer. Any help is appreciated. Thanks!

I'm pretty sure that you are correct. It will be difficult to calculate the angular momentum for any point that's in an accelerating reference frame (one that's not inertial).

The center of mass will travel with constant velocity -- both before and after the collision. It will be easy to calculate the angular momentum w. r. t. the center of mass. That choice was not given explicitly.

The above question is from a test paper.

The solution booklet of the test says that its D. In the booklet, a single line is mentioned "Conceptual, torque of pseudo force" with no explanation. I don't have a clue what this means. How come pseudo forces come into play here?

haruspex
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Horrible question. Angular momentum of what? If rod+ball then A. If they mean of just the rod, and the ball does not strike its mass centre, then D. I think the explanation they give refers to decomposing the impulse into a linear impulse at the mass centre of the rod plus a torque impulse. That linear impulse will conserve the angular momentum, but a nonzero torque impulse cannot.

If rod+ball then A. If they mean of just the rod, and the ball does not strike its mass centre, then D. I think the explanation they give refers to decomposing the impulse into a linear impulse at the mass centre of the rod plus a torque impulse. That linear impulse will conserve the angular momentum, but a nonzero torque impulse cannot.
Lol, that made my head spin. Can you explain it in a simpler way? haruspex
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If I jump up and down, my momentum is not conserved, but that of the system me+planet is. In this problem, the rod+ball system must have its angular momentum conserved, no matter which inertial frame of reference you choose to measure it in. That's because the only external forces (gravity, normal force from table) cancel.
If the answer is not A, then I guess they just meant the ang mom of the rod (but it's awful they don't make this clear).
In general, a force applied at some point, P, is completely equivalent to a parallel force of the same magnitude applied at some other point, Q, plus an appropriate torque. The torque is the vector product of the original force and the vector P-Q. So if the ball strikes the rod away from the rod's mass centre, we can equate the impulse to an impulse at the mass centre plus an impulsive torque.
Clearly an impulse at the rod's mass centre will not change its angular momentum, but an impulsive torque (which is an amount of angular momentum) will changes its ang mom by that amount.

If I jump up and down, my momentum is not conserved, but that of the system me+planet is. In this problem, the rod+ball system must have its angular momentum conserved, no matter which inertial frame of reference you choose to measure it in. That's because the only external forces (gravity, normal force from table) cancel.
If the answer is not A, then I guess they just meant the ang mom of the rod (but it's awful they don't make this clear).
In general, a force applied at some point, P, is completely equivalent to a parallel force of the same magnitude applied at some other point, Q, plus an appropriate torque. The torque is the vector product of the original force and the vector P-Q. So if the ball strikes the rod away from the rod's mass centre, we can equate the impulse to an impulse at the mass centre plus an impulsive torque.
Clearly an impulse at the rod's mass centre will not change its angular momentum, but an impulsive torque (which is an amount of angular momentum) will changes its ang mom by that amount.
I understand it now, thanks! But what about the pseudo forces mentioned in the booklet?

haruspex
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I understand it now, thanks! But what about the pseudo forces mentioned in the booklet?
I'm guessing, but that's all, that it's a reference to decomposing the off-centre force into a force and a torque.

I'm guessing, but that's all, that it's a reference to decomposing the off-centre force into a force and a torque.
I still don't get it. ehild
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The conservation of angular momentum applies for the system rod+ball, and you can use any fix point of reference. The rod will move, so no point on it is fixed.

ehild

haruspex
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The conservation of angular momentum applies for the system rod+ball, and you can use any fix point of reference. The rod will move, so no point on it is fixed.
This puts a different slant on the question. I was reading it in the sense of using conservation of angular momentum to determine the speed and rotation immediately after collision. For that, I believe any point on the rod can be used (since there is no difference between choosing that point on the rod and the same point in the static frame). But if the idea is to take moments sometime later, then I agree this will not do.
So now we have three or four interpretations of the question, and no sound basis to choose between them. Yuk.

D H
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Anyways, thanks everyone for your help! 