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Rotational and translational kinetic energy

  1. Apr 30, 2008 #1
    1. The problem statement, all variables and given/known data

    A cylinder of mass m2 and radius r rolls without slipping over a cylindrical surface. It is driven (like an inverted pendulum) by a uniform rod of mass m1 and length L. Its instantaneous position as a function of time is determined by the angular displacement
    Produce the expression for the kinetic energy of the system as a function of time.[/B]

    2. Relevant equations

    K.E(tot) = 1/2 m*v^2 + 1/2 I*omega^2
    where I(cylinder 2D) = (m*r^2)/2
    I(rod about fixed end)=(1/2)*m*L^2

    3. The attempt at a solution
    I found the velocity using the derivative of the position vector of the end of the rod which is attached to the centre of the cylinder. Now what...is this velocity considered the translational velocity of the cylinder. how do i find omega? I don't know what to do next?
    Last edited: Apr 30, 2008
  2. jcsd
  3. May 1, 2008 #2


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    Welcome to PF!

    Hi maffra! Welcome to PF! :smile:

    "without slipping" means that the point of contact is (instantaneously) stationary.

    So the velocity of the centre of the cylinder is cancelled by the rotational velocity of the point of contact (relative to the centre).

    That gives you an equation between v and ω. :smile:
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