A horizontal circular platform (m = 60.1 kg, r = 3.05m) rotates about a frictionless vertical axle. A student (m = 56.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 3.5 rad/s when the student is at the rim. Find omega when the student is 2.81m from the center.
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The moment of inertia of the disk about the axle is 1/2 M R^2 . The moment if inertia of the student at the rim is mR^2. So total moment of inertia initially is
I = mR^2 + 1/2 MR^2
So the initial angular momentum is
L_initial = I w = (m R^2 +1/2MR^2) w_initial
The moment of inertia in the final position, with the student at radius r, is
I_final = (m r^2 +1/2MR^2)
and the final angular momentum is
L_final = (m r^2 +1/2MR^2) w_final
Now no external torque acts on the system, so as the student moves to a new radius r, angular momentum does not change: L_final = L_initial
Therefore
w_final = (m R^2 +1/2MR^2) / (m r^2 +1/2MR^2) w_ initial
I = mR^2 + 1/2 MR^2
So the initial angular momentum is
L_initial = I w = (m R^2 +1/2MR^2) w_initial
The moment of inertia in the final position, with the student at radius r, is
I_final = (m r^2 +1/2MR^2)
and the final angular momentum is
L_final = (m r^2 +1/2MR^2) w_final
Now no external torque acts on the system, so as the student moves to a new radius r, angular momentum does not change: L_final = L_initial
Therefore
w_final = (m R^2 +1/2MR^2) / (m r^2 +1/2MR^2) w_ initial
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angular momenum I is conserved
I depends on R, V and M
notice how a skater spins faster when pulling in their arms, or a gymnast spins when "tucking"
I depends on R, V and M
notice how a skater spins faster when pulling in their arms, or a gymnast spins when "tucking"