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Homework Help: A case of Rolling

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform cylinder of mass m of radius r(smaller) rolls on a cylindrical surface of radius R(larger). At a certain instant, the line OC has an angular velocity [tex]\omega[/tex] and angular acceleration [tex]\alpha[/tex].
    http://img82.imageshack.us/img82/5380/cylinerssm0.jpg [Broken]

    Find
    a) the acceleration of point of contact P (of the 2 cylinders) with respect to the surface.
    b) the kinetic energy of the cylinder



    3. The attempt at a solution
    First of all since the line joining the centres goes through an angular velocity of [tex]\omega[/tex]. The velocity of the centre of mass of the small cylinder with respect to the centre of fixed cylinder is [tex]\omega(R+r)[/tex]. The angular velocity about its centre would be [tex]\frac{\omega(R+r)}{r}[/tex]. I am stuck now.

    For
    a) I think the acceleration of this point= r * [tex]\beta[/tex] where [tex]\beta[/tex]is the angular acceleration of the small cylinder about its centre.
    [tex]\beta=\frac{d(\frac{\omega(R+r)}{r})}{dt}[/tex]

    Since it is given that d[tex]\omega[/tex]/dt=[tex]\alpha[/tex]
    [tex]\beta=\frac{(R+r)\alpha}{r}[/tex]

    But the answer provide is not r * [tex]\beta[/tex]. I am confused.

    b) For the kinetic energy of the cylinder, I used
    KE= [tex]0.5*m*v^{2}+0.5*I (\frac{\omega(R+r)}{r})^2[/tex]

    where I=mr*r
    and v=[tex]\omega(R+r)[/tex]

    But I get wrong results for both. Please Help@!
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 17, 2008 #2

    tiny-tim

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    Hi ritwik06! :smile:
    You're misunderstanding the question … "with respect to the surface" means with respect to the larger cylinder … in other words, with respect to the ground. :wink:

    Try again! :smile:
     
  4. Nov 17, 2008 #3
    Hi tim,
    I did not have a confusion regarding that. What I have tried to calculate is in accordance with ground frame. Is it not?
     
  5. Nov 18, 2008 #4

    tiny-tim

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    Hi ritwik06! :smile:

    (have an alpha: α and a beta: β and an omega: ω and a squared: ² :wink:)
    No, the acceleration of point of contact P (of the 2 cylinders) with respect to the surface (the ground frame) is rα. :wink:

    (rβ is the acceleration with respect to the smaller cylinder)
     
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