A solid sphere is rotating about a diameter at an angular velocity ω . If it cools so that its radius reduces to ( n 1 ) th of its original value, its angular velocity becomes

Question

A solid sphere is rotating about a diameter at an angular velocity ω . If it cools so that its radius reduces to ( n 1 ​ ) th of its original value, its angular velocity becomes

TestHub · Accepted Answer

By the principle of conservation of angular momentum, the angular momentum of the sphere remains constant since no external torque acts on it.

$I_{1} ω_{1} = I_{2} ω_{2}$

The initial moment of inertia of the solid sphere is $I_{1} = \frac{2}{5} M R^{2}$ .

When the radius reduces to $\frac{1}{n}$ of its original value, the new radius is $R^{'} = \frac{R}{n}$ .

The new moment of inertia is $I_{2} = \frac{2}{5} M (\frac{R}{n})^{2} = \frac{I _{1}}{n ^{2}}$ .

Substituting the values into the conservation of angular momentum equation:

$I_{1} ω = (\frac{I _{1}}{n ^{2}}) ω_{2}$

$ω_{2} = n^{2} ω$

Answer: $n^{2} ω$

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