Question

A man, standing on a turn-table, is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is $75\%$ of that with outstretched arms, his rotational kinetic energy will

• A. Increase by $33.3\%$
• B. Decrease by $33.3\%$
• C. Increase by $25\%$
• D. Decrease by $25\%$

Answer 1

The correct option is A Increase by $33.3\%$

Let $I_1\text{and}{\omega }_1$ be the moment of inertia and angular speed respectively when his arms are outstretched.

And $I_2\text{and}{\omega }_2$ be the moment of inertia and angular speed respectively when his arms are folded.


Given
$I_2=\frac{3}{4}{I}_1$

There is no external torque, so we can apply angular momentum conservation

$\therefore I_1{\omega }_1=I_2{\omega }_2$

$I_1{\omega }_1=\frac{3}{4}{I}_1{\omega }_2$

$\Rightarrow {\omega }_2=\frac{4}{3}{\omega }_1$

Initial Kinetic energy, $K_1=\frac{1}{2}{I}_1{\omega }_1^2$

And, final Kinetic energy,
$K_2=\frac{1}{2}{I}_2{\omega }_2^2$

$K_2=\frac{1}{2}\times \frac{3I_1}{4}\times {\left(\frac{4{\omega }_1}{3}\right)}^2=\frac{4}{3}\left(\frac{1}{2}{I}_1{\omega }_1^2\right)$

$K_2=\frac{4}{3}{K}_1$

$\therefore$ Percentage increase in Kinetic energy

$=\frac{K_2-K_1}{K_1}\times 100$

$=\frac{\frac{4}{3}{K}_1-K_1}{K_1}\times 100$

$=\frac{100}{3}=33.3$%

Hence, the correct option is $\left(\text{A}\right)$.