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Coservation of angular momentum

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data
    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


    2. Relevant equations
    initial angular momentum=final angular momentum


    3. 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
     
  2. jcsd
  3. Oct 14, 2012 #2

    ehild

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    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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  4. Oct 14, 2012 #3
    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?
     
  5. Oct 14, 2012 #4

    ehild

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    The centre of mass of the ring moves along a circle. So the frame of reference fixed to it is not inertial.

    ehild
     
  6. Oct 14, 2012 #5
    ok thanks
     
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