Question

Find the moment of inertia of a uniform rod of mass $M$ and length $L$, about $OA$ {$PQ$ makes an angle of ${30}^{\circ }$ with $OA$ as shown in figure}.

• A. $\frac{M{L}^2}{4}$
• B. $\frac{M{L}^2}{12}$
• C. $\frac{M{L}^2}{3}$
• D. $\frac{M{L}^2}{8}$

Answer 1

The correct option is B

$\frac{M{L}^2}{12}$

We can use the very basic formula i.e. $I$ = $\int m{r}^2$ in order to find the MI of rod.

Let's assume a small element dx , x units away from point P

Now $dm$ = $\frac{M}{L}dx$

Perpendicular distance of $dx$ from $OA$

$X_0$ = $xsin{30}^{\circ }$ = $\frac{x}{2}$

Also, $dI$ = $dm.x_0^2$

= $\frac{x^2}{4}\frac{M}{L}dx$

On integrating,

$I$ = $\underset{O}{\overset{L}{\int }}\frac{M}{4L}{x}^{2}dx$

= $\frac{M}{4L}\times \frac{L^3}{3}$ = $\frac{M{L}^2}{12}$