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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.

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Answer:

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

User Radu Stoenescu
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