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Homework Help: KE of a hoop rolling within a tube

  1. Dec 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Please see the attached
    In this example my professor is trying to calculate the total KE of the hoop,and let since the velocity at the contact point of the hoop and tube is zero we have v = vt + aω.This is equivalent to
    (R-a)Θ. = - aω.

    However,he later said that it was wrong since the no-slip condition should be RΘ. = aω but he did not give any explanation.Though it makes sense if you think about it geometrically,I think both of them are correct but I am not sure which one is correct?

    2. Relevant equations

    3. The attempt at a solution
    I am not sure,the equation T = T' + 1/2MV^2 where S and S' are an inertial frame and the centre of momentum frame,V is the speed of S' wrt S,holds under Galilean transformation.I think the set of Cartesian coordinates at CM of hoop is somewhat rotating wrt to the observer/lab frame so that it is not a Galilean transformation and we couldn't calculate the KE in the way as shown in the attached?

    Attached Files:

    • COM.png
      File size:
      120.3 KB
  2. jcsd
  3. Dec 20, 2016 #2


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    Your professor was right the first time. I think I can see how he got the second expression. If you treat the hoop as attached rigidly to an arm rotating about the centre of the cylinder, the outermost part of it slides along the cylinder wall at ##R\dot\theta##. So he thought the hoop would have to rotate about its centre at ##-R\dot\theta/a## to restore it to rolling contact. But attaching it to the rigid arm made the hoop itself rotate at rate ##\dot\theta##. If, instead, we make the hoop rotate at rate ##\omega## about its centre then the difference from the rigid arm model is a rotation rate of ##\omega-\dot\theta##. Thus the no slip condition is ##R\dot\theta+a(\omega-\dot\theta)=0##.

    The KE calculation in the attachment seems a crazy approach to me. What is the hoop's instantaneous centre of rotation? What is its moment of inertia about that point?
  4. Dec 21, 2016 #3
    I agree with haruspex. If you allow the center of mass coordinates to rotate with the arm you get the second form. If you don't allow the COM coords to rotate then the hoop rotates one fewer times and the first form is correct. The problem expressly says the COM coords do not rotate, so the first version is correct.
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