Question

A uniform rod of mass 𝑚 and length $l_0$ is rotating with a constant angular speed $\omega$ about a vertical axis passing through its point of suspension. The moment of inertia of the rod about the axis of rotation if it makes an angle $\theta$ to the vertical (axis of rotation) is

• A. $\frac{2m{l}_0^2{\sin }^2\theta }{3}$
• B. $\frac{m{l}_0^2{\sin }^2\theta }{3}$
• C. $\frac{m{l}_0^2{\sin }^2\theta }{12}$
• D. $\frac{m{l}_0^2{\sin }^2\theta }{6}$

Answer 1

The correct option is B $\frac{m{l}_0^2{\sin }^2\theta }{3}$

Step 1:Draw a labelled diagram.

Step 2:Find the moment of inertia of the rod about the axis of rotation.
Formula used:
$I=\int r^{2}dm$

We can observe each and every element of rod in rotation with different radius about the axis of rotation.
Take an elementary mass dm of the rod.

$dm=\frac{m}{l_0}dl$

The moment of inertia of the elementary mass is given as

$dI=\left(dm\right)r^2$

The moment of inertia of the rod

$=I=\int dI\Rightarrow I=\int r^{2}dm$
Substituting $r=l\sin \theta$ and
$dm=\frac{m}{l_0}\cdot dl$
"we obtain,"
$I=\int \left(l^2{\sin }^2\theta \right)\frac{m}{l_0}dl$
$\frac{m{\sin }^2\theta }{l_0}{\int }_0^{l_0}{l}^{2}dl=\frac{m{l}_0^3}{3l_0}{\sin }^2\theta$
$\Rightarrow I=\frac{m{l}_0^2{\sin }^2\theta }{3}$

Final Answer: (d)