physics question?

Question

A 5.8 m diameter merry-go-round is rotating freely with an angular velocity of 0.50 rad/s. Its total moment of inertia is 2200 kgM2. Four people standing on the ground, each of 60 kg mass, suddenly step onto the edge of the merry-go-round.
(a) What is the angular velocity of the merry-go-round now?
(b) Assume that the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round). What would be the angular velocity of the merry-go-round?

Answer 1

(a) conservation of angular momentum:

L before
= (moment of inertia before)*
(ang speed before)
=
L after
= (moment of inertia after)*(final ang speed)

They give you moment of inertia and ang speed before.
Then add mr^2 to the moment for each of the 4 people hopping on. That's your moment of inertia after.

Now you know everything in that conservation equation except final angular speed. Solve for it.

(b) If they jumped off without carrying any angular momentum (ie traveled straight out), then the merry-go-round will go back to its original speed. Note, however, that if you try to jump straight off, you will generally not fly off in a radial direction. Unless you jump backwards, your inertia will take you forward and you will carry some of the angular momentum off with you and the merry-go-round will just carry on at the same speed.

Answer 2

1st answer is good but answerer misread question b. The people jump off radially with respect to the (rotating) merry-go-round, and thus it maintains the same speed.