Physics - Rotation Question with Solution | TestHub

The system consists of $5$ solid spheres, each of mass $m$ and radius $r$.

From the given figure, the axis of rotation AA' passes through the centers of $3$ spheres (one at the top and two at the bottom).

The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm}=\frac{2}{5}m{r}^2$.

Total moment of inertia for these $3$ spheres is:

$I_1=3\times \frac{2}{5}m{r}^2=\frac{6}{5}m{r}^2$

The remaining $2$ spheres are placed such that they touch each other, and the axis AA' passes through their point of contact. Thus, the distance of their centers from the axis AA' is $r$.

Using the parallel axis theorem, the moment of inertia of one such sphere about the axis AA' is:

$I_{off}=I_{cm}+m{r}^2=\frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2$

Total moment of inertia for these $2$ spheres is:

$I_2=2\times \frac{7}{5}m{r}^2=\frac{14}{5}m{r}^2$

The total moment of inertia of the system about the axis AA' is the sum of the moments of inertia of all $5$ spheres:

$I=I_1+I_2=\frac{6}{5}m{r}^2+\frac{14}{5}m{r}^2=\frac{20}{5}m{r}^2=4m{r}^2$

Answer: $4m{r}^2$