A horizontal platform in the shape of a circular disk rotates freely in a horizo

Question

A horizontal platform in the shape of a circular disk rotates freely in a horizontal plane about a frictionless, vertical axle (see figure). The platform has a mass M = 100 kg and a radius R = 2.2 m. A student whose mass is m = 63 kg walks slowly from the rim of the disk toward its center. If the angular speed of the system is 2.3 rad/s when the student is at the rim, what is the angular speed when she reaches a point r = 0.60 m from the center? This problem is different from previous problems because part of the moment of inertia of the system changes (that of the platform). Because the platform rotates on a frictionless axle, we identify the system of the student and the platform as an isolated system in terms of angular momentum. Let us denote the moment of inertia of the platform as I_p and that of the student as I_s. We model the student as a particle. Find the initial moment of inertia I_i of the system (student plus platform) about the axis of rotation: I_i = I_pi + I_si = 1/2 MR^2 + mR^2 Find the moment of inertia of the system when the student walks to the position r

Explanation / Answer

Angular momentum will be conserved in this example

So I*omega)i = I*omega)f

If the student was not on the rim the I would be 1/2*M*R^2 = 1/2*100*2.2^2 = 242 kg-m^2
So the initial omega would now be omega = 2.3*546.92/242 = 5.199 rad/s

The initial K = 1/2*i*omega^2 = 1/2*242*5.199^2 = 3270.58 J

After the student jumped on K = 1/2*I*omega^2 = 1/2*264.68*2.3^2 = 700.08 J

So the kinetic energy changed by 3270.58 - 700.08 = 2570.5 J

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