Answer to Question #77860 in Quantum Mechanics for Simon Hamudombe

Question

4. A boy stands at the centre of a turn table with his two arms stretched. The turn table is set rotating with an angular with an angular speed of 40 r.p.m. How much is the angular speed of the boy if he folds his hands back and thereby reduces his moment of inertia to 2/5times the initial? Assume the turn table to rotate without friction.

Answer 1

From the conservation of the angular momentum:
I_1 ω_1=I_2 ω_2
ω_2=I_1/I_2 ω_1
Thus, the angular speed of the boy if he folds his hands back is
ω_2=5/2 40=100 r.p.m.