Question

Find the moment of inertia of two solid spherical balls of mass $m$, radius $r$ attached at the two ends of a thin straight rod of mass $m$ and length $4r$, if this system is free to rotate about an axis passing through an end of the rod and perpendicular to its length.

• A. $\frac{92m{r}^2}{15}$
• B. $\frac{166m{r}^2}{15}$
• C. $\frac{86m{r}^2}{15}$
• D. $\frac{332m{r}^2}{15}$

Answer 1

The correct option is D

$\frac{332m{r}^2}{15}$

$I=I_{rod}+I_{Lsphere}+I_{Rsphere}$
$I_{rod}$ is Moment of inertia about corner of the rod
$I_{Lsphere}$ is Moment of inertia of the leftmost sphere
$I_{Rsphere}$ is Moment of inertia of the rightmost sphere
$I_{rod}=\frac{1}{3}m{L}^2$
As Axis of rotation is passing through diameter
$I_{Lsphere}=\frac{2}{5}m{r}^2$
As Axis of rotation is at a distance of $L$ and parallel to axis passing through diameter,by parallel axis Theorem
$I_{Rsphere}=\frac{2}{5}m{r}^2+m{L}^2$
$I=\frac{2}{5}m{r}^2+\frac{1}{3}m{L}^2+\frac{2}{5}m{r}^2+m{L}^2$
As $L=4r$
$I=\frac{332m{r}^2}{15}$