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

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

Step-by-step explanation:

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

Moment of inertia
`I=0.4kgm^2`

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

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

As we know that angular momentum is conserved

So
`15.072=I* 7.85`

`I=1.92kgm^2`

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