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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.

Respuesta :

Answer:

New moment of inertia will be [tex]I=1.92kgm^2[/tex]

Explanation:

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

Moment of inertia [tex]I=0.4kgm^2[/tex]

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

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

As we know that angular momentum is conserved

So [tex]15.072=I\times 7.85[/tex]

[tex]I=1.92kgm^2[/tex]

So new moment of inertia will be [tex]I=1.92kgm^2[/tex]

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