1. A uniform disk of mass 1 kg is spinning on its (vertical) axis of symmetry (r

Question

1. A uniform disk of mass 1 kg is spinning on its (vertical) axis of symmetry (remember rotational inertia of a disk is given by 0.5mr^2), with an angular velocity of 6 rad/s. A piece of mud of mass 1 kg is dropped on the disk halfway between the center and the edge and the disk slows down. What is it's angular velocity now? Answer in rad/s.

2. A small child is on a merry go round which is spinning (no friction) freely. He crawls from the center to towards the rim. As he does this, angular momentum conservation will dicate that:

a) rate of turning stays the same

b) merry go round slows down

c) merry go round speeds up

Explanation / Answer

1. Problem is quite simple. Just apply conservation of angular momentum in both cases.

Intial anglular momentum = final angular momentum

Given data : initial angular velocity (w0) = 6 rad/s , inertia of disk = .5 mr2 and mass of mud = 1 kg

What we need to find: final angular velocity (wf)

L0 = Idisk w0 and Lf = (Idisk + Imud) wf

wf = Idisk w0 / (Idisk + Imud)

Imud = mmud (r/2)2 as the mud is fallen between the edge and center. So under assumption that mud is a point mass at distance r/2 from the center. if mud mass has fallen at the edge Imud would have been mmudr2.

solving gives us wf = 4 rad/sec.

2. when child is at the center of merry go round, child's inertia is zero (as r from center is 0). when he moves to the edge inertia of child becomes mchild r2 (r is the radius of the merry go round).

applying angular momentum conservation like in part a wf decreases as child starts moving towards edge therefore rate of turning will decrease as child starts moving towards edge.

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