[SOLVED] Conservation of Angular Momentum

[XxBollWeevilx]

1. The problem statement, all variables and given/known data

A 60-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500 kgm^{}2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth. In what direction and with what angular speed does the turntable rotate?

2. Relevant equations

L(initial) = L(final)
I \omega(initial) = I \omega(final)

3. The attempt at a solution

My book lists this as a conservation of angular momentum problem and gives an answer of 0.360 rad/s. However, I can't seem to figure out how to apply the conservation of angular momentum, and how to incorporate the momentum of the woman into the problem. Everything is initially at rest, then the woman starts walking and everything is in motion. From where to where do I examine momentum? I'm assuming that the woman has a linear momentum since she does not have a moment of inertia, but I'm not sure. Any help would be greatly appreciated!

[Dick]

The woman is walking around the center of rotation of the wheel. She does have a moment of inertia. Treat her as a point mass M at distance R from the axis. Her moment of inertia is MR^2. Since angular momentum is conserved the wheel must have equal and opposite angular momentum to her.

[XxBollWeevilx]

Thanks so much! I was able to use that to get the right answer. Here's my final work:

moment of inertia (woman) = mr^2 = 240 kgm^2
angular speed of woman = v/r = 0.750 rad/s

Comservation of Angular momentum:
I(woman)(omega(woman)) = I(turntable)(omega(turntable))

So, Omega (turntable) = (I(woman)omega(woman)) / I(turntable)

= 0.360 rad/s

Thanks!