A horizontal circular platform (m = 122.1 kg, r = 3.43m) rotates about a frictionless vertical axle. A student (m = 82.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 2.9 rad/s when the student is at the rim. Find omega when the student is 2.39m from the center.

Answer 1

conservation of angular momentum applies:
initial momentum=final momentum
Studentw+wheelw=studentw' + wheelw'
1/2*82.3*3.43^2*wo+1/2*122.1*3.43^2*wo=
= 1/2*82.3*2.39^2wf+1/2*122.1*3.43^2*wf
solve for wf, knowing wo=2.39