Merry-go-rounds and changing angular speeds?

Nov 2009

I'm in introductory level mechanics and I was wondering if someone could give me a hand with this question:

A playground ride consists of a disk of mass M = 56 kg and radius R = 2.0 m mounted on a low-friction axle. A child of mass m = 24 kg runs at speed v = 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. The child on the disk then walks inward on the disk and ends up standing at a new location a distance R/2 = 1 m from the axle. Now what is the angular speed?

I calculated most, if not all, of the things that I think I need already, like the moments of inertia at R=2, the angular momentum when R=2, etc., but I don't know what expression I can use to find out the new angular speed of the child once he moves. I was thinking that if angular momentum stayed constant, I could calculate using the expression Ltotal = Ldisk + L child, since I can find all of the things needed, but that isn't giving me the right answer.

Sorry my question is a little long. Thank in advance for the help!
Oct 2009
I don't have the time to go though and try and figure it out right now, (darn mid-terms) but I think the key to figuring out the problem is that the axle is low-friction, and thus the velocity at each point on the disk remains constant. All you would have to do I would think is to change the r value, and which center of mass/moment of inertia you use, and keep all the other values constant.
Sep 2009
angular mom conservation will do the trick.
mvR=(MR^2/2 +m(1)^2) W GIVES w=0.81 rad/s
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