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Angular speed of student on merry-go-round

  1. Mar 15, 2006 #1
    A merry go round can be described as a horizontal platform in the shape of a disc which rotates on a frictionless bearing about a vertical axis through its centre. The platform has a mass of 150 kg, a radius of 2.0 m and a rotational inertia of 300 kg.m2.A 60 kg student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is 1.5 rad.s-1 when the student starts at the rim, what is the angular speed when he is 0.5 m from the centre?

    I know that w=2(pi)/T=2(pi)r and the kinetic energy= (mr^2w^2)/2, but I am not sure how these two are to be related, if these are the correct equation to begin with.
     
  2. jcsd
  3. Mar 15, 2006 #2
    There are no external torques on the system. Try conservation of angular momentum.

    -Dan
     
  4. Mar 15, 2006 #3
    Use conservation of momentum and remember that the student's mass adds mass to the moment of inertia.

    If you want to do it the long way:
    The student's weight exerts greater torque on the merry go round as it approaches the center. The torque=force*radius. The torque also equals the moment of inertia * acceleration. Solve for acceleration. Convert angular to linear because you do not know delta theta. Then use the kinematic equation to solve for your linear speed. Then divide by your new radius.

    I think that is how I would do it if you do not want to use conservation of momentum.
     
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