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

1327.43362 rev/s

Step-by-step 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=(I_i\omega_i)/(I_f)\\\Rightarrow \omega_f=(2.4* 500)/(0.904)\\\Rightarrow \omega_f=1327.43362\ rev/s`

The new rate of rotation is 1327.43362 rev/s