Coservation of angular momentum

AI Thread Summary
The discussion centers on a physics problem involving the conservation of angular momentum for a ring pivoted at its rim, with a bug walking around it. The initial angular momentum is zero, and the final angular momentum is expressed as a combination of the ring's and the bug's contributions. The factor of 2mrω arises from considering both the rotation of the ring and the motion of the bug relative to the ring. The moment of inertia must be calculated about the pivot point, which is different from the center of the ring. The conversation concludes with an acknowledgment that the center of mass of the ring moves along a circular path, making the frame of reference non-inertial.
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
 

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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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