Question

An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he reduces his rate of spin to 1.25 rev/s by extending his arms and increasing his moment of inertia.

Answer 1

Answer:

New moment of inertia will be $I=1.92kgm^2$

Explanation:

It is given initially angular velocity $\omega =6rev/sec=6\times 2\pi =37.68rad/sec$

Moment of inertia $I=0.4kgm^2$

Angular momentum is equal to $L=I\omega =37.68\times 0.4=15.072kgm^2/sec$

Now angular velocity is decreases to $\omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec$

As we know that angular momentum is conserved

So $15.072=I\times 7.85$

$I=1.92kgm^2$

So new moment of inertia will be $I=1.92kgm^2$