Question

Moment of inertia of a uniform rod $AB$ about axis $Y{Y}^{\prime }$ as shown in figure is

• A. $\frac{m{L}^2{\sin }^2\alpha }{4}$
• B. $\frac{m{L}^2{\cos }^2\alpha }{3}$
• C. $\frac{m{L}^2{\sin }^2\alpha }{3}$
• D. $\frac{m{L}^2{\cos }^2\alpha }{4}$

Answer 1

The correct option is C $\frac{m{L}^2{\sin }^2\alpha }{3}$

Let us take a small mass $dm$ of width $dx$ at distance $x$ along the rod.


Moment of inertia of mass $dm$ about $Y{Y}^{\prime }$
$dI=dm{r}^2$
$\Rightarrow dI=\left(\frac{m}{L}dx\right)(x\sin \alpha {)}^2$
[$\because$ Linear mass density of rod is, $\lambda =\frac{dm}{dx}=\frac{m}{L}$
& $r=x\sin \alpha$]

$\Rightarrow dI=\left(\frac{m{\sin }^2\alpha }{L}\right)x^{2}dx$
On integrating both sides,
$\underset{0}{\overset{I}{\int }}dI=\underset{0}{\overset{L}{\int }}\frac{m{\sin }^2\alpha }{L}{x}^{2}dx$
$\Rightarrow I=\frac{m{\sin }^2\alpha }{L}{\left[\frac{x^3}{3}\right]}_0^L$
$\Rightarrow I=\frac{m{L}^2{\sin }^2\alpha }{3}$
Moment of inertia of rod $AB$ about axis $Y{Y}^{\prime }$ is $\frac{m{L}^2{\sin }^2\alpha }{3}$