Physics - Rotation Question with Solution | TestHub

The moment of inertia of a solid sphere of mass $m$ and radius $r$ about its diameter is given by $I_{cm}=\frac{2}{5}m{r}^2$.

For sphere 1, the axis ${\mathrm{YY}}^1$ passes through its diameter. Therefore, its moment of inertia about the axis is:
$I_1=\frac{2}{5}m{r}^2$

For spheres 2 and 3, the axis ${\mathrm{YY}}^1$ is tangent to their surfaces. The perpendicular distance from the center of each of these spheres to the axis ${\mathrm{YY}}^1$ is equal to their radius $r$.

Using the parallel axis theorem, the moment of inertia of sphere 2 about the axis ${\mathrm{YY}}^1$ is:
$I_2=I_{cm}+m{r}^2=\frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2$

Similarly, the moment of inertia of sphere 3 about the axis ${\mathrm{YY}}^1$ is:
$I_3=I_{cm}+m{r}^2=\frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2$

The total moment of inertia of the system about the axis ${\mathrm{YY}}^1$ is the sum of the moments of inertia of the three spheres:
$I_{total}=I_1+I_2+I_3$
$I_{total}=\frac{2}{5}m{r}^2+\frac{7}{5}m{r}^2+\frac{7}{5}m{r}^2$
$I_{total}=\frac{16}{5}m{r}^2$

Answer: $\frac{16}{5}{\mathrm{mr}}^2$