Moment of inertia about an axis $AB$ 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 $AB$ axis with separation of $3$ m as shown in figure below. The value of $R$ is given as $\sqrt{\dfrac{\alpha}{2}}$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 15
The axis $AB$ passes through one end of the rod, perpendicular to it:
$$I_{rod}=\frac{ML^2}{3}=\frac{40\times9}{3}=120\ \text{kg m}^2$$
The sphere rotates about an axis parallel to $AB$ and $3$ m from its centre, so by the parallel axis theorem:
$$I_{sphere}=\frac25MR^2+Md^2=\frac25(10)R^2+10(3)^2=4R^2+90$$
Setting them equal: $4R^2+90=120\Rightarrow R^2=7.5=\dfrac{15}{2}$, so $R=\sqrt{\dfrac{15}{2}}$ and $\alpha=15$.
Solution by Sreeraj P, M.Sc Physics