A 200 g, 42.0-cm-diameter turntable rotates on frictionless bearings at 56.0 rpm. A 20.0 g block sits at the center of the turntable. A compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.
What is the turntable's rotation angular velocity when the block reaches the outer edge?
.5Iw^2 (initial) = .5Iw^2 (final)
For finding the final rotational inertia:
I(final) = Icm + Md^2 = .5MR^2 + md^2
The Attempt at a Solution
I believe that energy is conserved in this process so the equation above is valid. I calculated the final rotational inertia to be (1/2)*(0.2kg)*(0.21m)^2 + (0.02kg)*(0.21m)^2 = 0.005292
The initial energy is (.5)*(.5*0.2kg*0.21m^2)*(56rpm)^2 = 6.915
The final energy is (.5)*(0.005292)*wfinal^2
Solving for wfinal I get 53.6 rpm. I've done the calculations repeatedly and I can't come up with an alternate way of doing it but this answer is not correct. Any ideas?
Thank you so much in advance!