Question

In the system below, the moment of inertia of the combination of thin rod of mass $m$ and solid spherical ball of mass $2m$ about the axis of rotation $A{A}^{\prime }$ is:

• A. $\frac{112}{15}m{r}^2$
• B. $\frac{302}{15}m{r}^2$
• C. $\frac{312}{15}m{r}^2$
• D. $\frac{2}{17}m{r}^2$

Answer 1

The correct option is B $\frac{302}{15}m{r}^2$

Moment of inertia of thin rod about axis $A{A}^{\prime }$ is,
$I_{rod}=\frac{m{l}^2}{3}=\frac{m(2r{)}^2}{3}=\frac{4m{r}^2}{3}$
Moment of inertia of solid spherical ball about an axis passing through its geometrical centre is, $I=\frac{2}{5}\left(2m\right)r^2=\frac{4m{r}^2}{5}$
Using parallel axis theorem, we get its moment of inertia about axis $A{A}^{\prime }$ to be,
$I_{A{A}^{\prime }}=\frac{4m{r}^2}{5}+\left(2m\right)(3r{)}^2=\frac{4m{r}^2}{5}+18m{r}^2=\frac{94m{r}^2}{5}\Rightarrow I_{total}=\frac{4m{r}^2}{3}+\frac{94m{r}^2}{5}=\left[\frac{20+282}{15}\right]m{r}^2=\frac{302m{r}^2}{15}$