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Central Forces

  1. Mar 6, 2005 #1
    Could someone please help me with this question:

    Q. Two particles of mass m are connected by a light inextensible string of length l. One of the particles moves on a smooth horizontal table in which there is a small hole. The string passes through the hole so that the second particle hangs vertically below the hole. Use the conservation of energy and angular momentum to show that:

    (dr/dt)^2 = A - (B/r^2) - gr

    where r(t) is the distance of the first particle from the hole, A and B are constants and g is the acceleration due to gravity. [Hint: Use polar coordinates in the plane of the table with the origin at the hole].

    I know that the conservation of energy is given by:

    1/2.m(dr/dt)^2 + L^2/2mr^2 + V(r) = E

    where L=angular momentum and E is the total energy which is constant. V(r) is the integral of the force wrt r. I'm assuming that mg is equal to the tension in the string and it is this that is providing the centripetal force? So surely V(r) = int mg dr or V(r) = mgr? Am I right so far? How do I proceed now? Thanks :rofl:
  2. jcsd
  3. Mar 7, 2005 #2
    Any one have any ideas?
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