Question

There are four solid balls with their centres at the four corners of a square of side $a$. The mass of each sphere is $m$ and radius $r$. Find the moment of inertia of the system about one of the sides of the square.

• A. $\frac{8}{5}m{r}^2+2m{a}^2$
• B. $\frac{5}{8}m{r}^2+2m{a}^2$
• C. $m{r}^2+\frac{8}{5}m{a}^2$
• D. $m{r}^2+\frac{5}{8}m{a}^2$

Answer 1

The correct option is A $\frac{8}{5}m{r}^2+2m{a}^2$


Moment of Inertia$(M.I)$ of two sphere lying on the axis $I_1=2\times \frac{2}{5}m{r}^2=\frac{4}{5}m{r}^2$

$(M.I)$ of two sphere lying on the edge parallel to the axis of distance $a$ $($By using parallel axis theorem$)$
$I_2=2(\frac{2}{5}m{r}^2+m{a}^2)$

Total $M.I$ of the system about one of the sides of the square. $I=I_1+I_2=2\times \frac{2}{5}m{r}^2+2(\frac{2}{5}m{r}^2+m{a}^2)$
$=\frac{8}{5}m{r}^2+2m{a}^2$

Hence option A is the correct answer.