Question

A student stands on a horizontal platform that is free to rotate around a vertical axis. He holds two weights in his outstretched hands. Another student gives him a push and starts the platform rotating at 0.500 rev/s. The student then pulls the weights in close to his chest. The moment of inertia with the weights extended is 2.40 kgrn2; the moment of inertia with the weights close to the axis is 0.904 kgm2. Ignore any frictional effects. What is his new rate of rotation?

Answer 1

Answer:

1327.43362 rev/s

Explanation:

$I_i$ = Initial moment of inertia = 2.4 kgm² (arms outstretched)

$I_f$ = Final moment of inertia = 0.904 kgm² (arms close)

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0.5 rev/s

Here the angular momentum is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\frac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\frac{2.4\times 500}{0.904}\\\Rightarrow \omega_f=1327.43362\ rev/s$

The new rate of rotation is 1327.43362 rev/s