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Homework Help: Angular velocity question

  1. Sep 17, 2006 #1
    A solid uniform disk of mass m and radius R is pivoted about a horizontal axis through its center, and a small body of mass m is attached to the rim of the disk. If the disk is released from rest with the small body at the end of a horizontal radius, find the angular velocity when the body is at the bottom.
    Loss of P.E., = gain in K.E. Therefore

    m*g*R = .5*I*w^2 + .5*m*v^2.

    Where I = rotational inertia and w = angular velocity.

    w=2*(g*R - 2*v)/R^2)^2.
    Is this correct, if it isn't why or where? Thanks very much.
  2. jcsd
  3. Sep 17, 2006 #2
    Note: I= .5*m*R^2
  4. Sep 19, 2006 #3
    I was woundering what is your ideas on above. Many thanks.
  5. Sep 19, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Since the disk and the small mass constitute a single object that ends up rotating at some angular speed, you need to find the rotational inertia of that composite object. The final KE will be .5*I*w^2, where I is the total rotational inertia.
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