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Hoop rolling in a pipe

  1. Sep 25, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A hoop of mass ##m## and radius ##a## rolls without slipping inside a pipe of radius ##R## (the motion is 2D). Write down the kinetic and potential energy. Hence find the frequency of small oscillations about equilibrium.

    2. Relevant equations
    Moment of inertia of a hoop: I=Mr2
    Rotational kinetic energy

    3. The attempt at a solution
    I envisaged the problem as a hoop rolling backwards and forwards in a pipe shaped like an arc of radius R.
    Kinetic energy is sum of rotational and kinetic, so $$T = \frac{1}{2}mv^2 + \frac{1}{2} (ma^2) \left(\frac{v^2}{a^2}\right) = mv^2.$$

    The path of the C.O.M of the hoop is arc shaped. Let ##\theta## be the angle between a vertical passing through the centre of the hoop and the C.O.M. Draw a horizontal at the base of the pipe. After a little time, the C.O.M is at angle ##\theta##. For small oscillations the increase in height of the C.O.M from its initial position (at a height a above the base of the pipe) is a + a(1-cosθ). So V = mg(a+a(1-cosθ)).

    Is this correct? I am wondering if this can be solved more easily using the centre of momentum frame.
     
  2. jcsd
  3. Sep 25, 2013 #2
    V doesn't look correct. The CM of hoop is at a constant distance (R-a) from the centre of pipe. When you displace the hoop by angle ##\theta##, what is the distance of CM from the base?
     
  4. Sep 25, 2013 #3

    CAF123

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    Hi Pranav-Arora,
    I believe that would be a + (R-a)(1-cosθ)
     
  5. Sep 25, 2013 #4
    Yes, looks right to me.
     
  6. Sep 25, 2013 #5

    CAF123

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    Ok, so ##E = mv^2 + mg(a+(R-a)(1-\cos \theta))##. I think to find the freq of small oscillations, I should find a form ##\ddot{\theta}+ \omega^2 \theta = g##, where g is not a function of θ. Differentiating E would give me $$\ddot{\theta} + \left(\frac{g(R-a) - ga}{2a^2}\right)\theta = 0, $$from which I can extract ##\omega## and hence f.
     
  7. Sep 26, 2013 #6
    That doesn't look right.

    Rewrite the potential energy as mg(R-(R-a)cosθ). Substitute in the energy equation and differentiate again.

    What did you substitute for v?
     
  8. Sep 26, 2013 #7

    CAF123

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    Could you please explain why? The equation is certainly dimensionally consistent.

    Because of no slip, I said ##v_{COM} = a \dot{\theta}##.
     
  9. Sep 26, 2013 #8
    No, the hoop does not rotate about CM here, it rotates about the centre of pipe so ##v=(R-a)\dot{\theta}##.
     
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