Answer to Question #248973 in Physics for prettygirl

Question

A 90 kg physics teacher stands on a rotating platform with a 15 kg mass in each hand. With his arms outstretched (each arm is 75 cm long) he spins at a rate of 4 rev s -1 . He brings his arms in and his speed increases to 9 rev s -1 . Compute the ratio of his rotational inertia when arms are out to his rotational inertia when arms are in.

Answer 1

The moment of inertia of both arms out is

$$L_o=I_o\omega_o=2mr^2\omega=67.5\text{ kg m}^2/\text{s}.$$

The moment of inertia of both arms brought in is

$$L_i=I_i\omega_i=2mr_i^2\omega=0\text{ kg m}^2/\text{s}.$$

Here we assume that the teacher's mass is uniformly distributed along the axis of rotation, so, it creates zero moment of inertia. Same reasonings are applied to arms brought in.