Question

Two solid spherical balls of mass $m$, radius $r$ attached at the two ends of a thin straight rod of mass $m$ and length $4r$ as shown in the figure.If this system is free to rotate about an axis passing through an end of the rod and along diameter of one of the spheres, find moment of inertia of the system about this axis.

• 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 $L=4r$
$I=\frac{332m{r}^2}{15}$