Question

# A man of m = 60 kg is standing on the edge of a disc of...

A man of m = 60 kg is standing on the edge of a disc of m = 100 kg and R = 2m on the ice. A pole goes through the center of the disc about which the disc can rotate freely. He is throwing a ball of m = .5 kg at a speed of v = 3.5 m/s and an angle of theta.

(Everything is measured in the rest frame. Assume frictionless ice surface for parts a and b)

1. Initially at rest, he throws the ball at 90 degrees. What is the angular velocity of the disc + man? (0.008 rad/s)

2. He then walks to the center of the disc, what is the new angular velocity? (0.018 rad/s)

3. By grabbing onto the pole he stops the rotation. Then he yanks out the pole (m = 1.5 kg) and throws it at a speed of 1.5 m/s. Taking into account now the small kinetic friction between the ice and the disk (u = 0.02), how long does it take for the disc + man to come to a complete stop? (0.07s)

1] The angular momentum before and after the throw is conserved.  => => 2] The person is now at the center.

so, the contribution from the person to the moment of inertia will be zero.  => 