Coservation of angular momentum

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Homework Help Overview

The discussion revolves around the conservation of angular momentum in a system involving a ring and a bug moving on it. The ring is pivoted at its rim on a frictionless table, and the problem asks for the rotational velocity of the ring at two specific points as the bug moves around it.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and question the factors involved in the calculations, particularly the factor of 2mrω. There is uncertainty about why the velocity is considered with respect to the ring rather than the ground.

Discussion Status

Some participants have provided insights into the moment of inertia and the nature of the frame of reference, suggesting that the angular momenta of different rotations need to be considered. However, there is no explicit consensus on the interpretation of the problem or the calculations involved.

Contextual Notes

Participants note that the pivot's position affects the moment of inertia and that the center of mass of the ring moves along a circular path, raising questions about the inertial nature of the reference frame used in the analysis.

shivam jain
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Homework Statement


a ring of mass M and radius r lies on its side on a frictionless table.it is pivoted to table at its rim.a bug of mass m walks around the ring with speed v,starting at the pivot.what is the rotational velocity of the ring when the bug is a) halfway around b)back at the pivot


Homework Equations


initial angular momentum=final angular momentum


The Attempt at a Solution


conserving angular momentum about pivot
correct answer of a) comes like this
initial L=0
final L=2Mr^2ω-2mr(v-2rω)
then initial L=final L
i am unable to understand factor of 2mrω
it should come only when ring is rotating and translating both.but here it is fixed to pivot and only rotating.further why we are taking velocity with respect to ring not ground.please explain me this solution
 
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shivam jain said:

Homework Statement


a ring of mass M and radius r lies on its side on a frictionless table.it is pivoted to table at its rim.a bug of mass m walks around the ring with speed v,starting at the pivot.what is the rotational velocity of the ring when the bug is a) halfway around b)back at the pivot


Homework Equations


initial angular momentum=final angular momentum


The Attempt at a Solution


conserving angular momentum about pivot
correct answer of a) comes like this
initial L=0
final L=2Mr^2ω-2mr(v-2rω)
then initial L=final L
i am unable to understand factor of 2mrω
it should come only when ring is rotating and translating both.but here it is fixed to pivot and only rotating.further why we are taking velocity with respect to ring not ground.please explain me this solution

The pivot is on the rim, and the ring rotates about it. The moment of inertia has to be calculated with respect to the pivot, which is twice the one with respect to the centre. Or you can consider that the centre of the ring travels along a circle with radius r around the pivot and the ring rotates about its centre at the same time, both with the same angular velocity. The angular momenta of both rotations add up.
The bug moves with speed v with respect to the ring, so its velocity with respect to the ground is the velocity of the point it is on the ring + its relative velocity.

ehild
 

Attachments

  • ringbug.JPG
    ringbug.JPG
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thank you the figure was indeed very helpful in understanding what actually is happening in question.can we solve it about the origin taken as centre of mass of the ring also?
 
The centre of mass of the ring moves along a circle. So the frame of reference fixed to it is not inertial.

ehild
 
ok thanks
 

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