Moment of inertia about an axis A B for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass of 10 kg and radius $R$ about an axis parallel to A B axis with separation of 3 m as shown in figure below. The value of $R$ is given as $\sqrt{\frac{\alpha}{2}}$. The value of $\alpha$ is $\_\_\_\_$ .
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Solution
$$
I_{R o d}=\frac{M \ell^2}{3}
$$
$\mathrm{I}_{\text {sphere }}[$ which is at 3 m from axis AB$]=\frac{2}{5} m R^2$
$$
\begin{aligned}
& \frac{M \ell^2}{3}=\frac{2}{5} m R^2 \\
& \frac{40 \times(3)^2}{3}=\frac{2}{5} \times 10 \times R^2 \\
& R^2=30 \\
& R=\sqrt{30}=\sqrt{\frac{\alpha}{2}} \\
& \alpha=60
\end{aligned}
$$
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