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Bug on a pivoted ring: rotational velocity

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi guys, this question is from Kleppner and Kolenkow, problem 6.3. A ring of mass M and radius R lies on its side on a frictionless table. It is pivoted to the 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 halfway around the pivot?

    2. Relevant equations

    3. The attempt at a solution

    We will take the bug to be moving to the left at speed v initially. I would like to ask you guys a few questions. First: which quantities are conserved? It seems obvious that total linear momentum in the x-direction is conserved. I would also argue that angular momentum is conserved, since the only external forces on the system are gravity and the normal force, but these create a zero net torque. I'm not sure whether energy is conserved. For the bug to walk around the ring, there needs to be friction between it and the ring; however, is this friction dissipative?

    I have a second question: when the bug gets to the top, does it still have a velocity v (directed the the right) with respect to the ring, or not?

    If I assume that only momentum and angular momentum are conserved, and that the bug still has speed v with respect to the ring, I get these equations:

    -mv = m(v + V) + Mv
    -mRv = -mR(v + V) + M(R^2)(omega)

    where V is the velocity of the ring and omega is the angular velocity. I solve these equations for omega, and I don't get the answer in the book.

    Where did I go wrong? Thanks in advance for all your help.
  2. jcsd
  3. Sep 18, 2009 #2
  4. Dec 14, 2011 #3
    Sorry for the late reply but I was looking for it myself just now and happened to find it!

    It's problem 1173 in the Lim Classical Mechanics problems and solutions manual.

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