Question

$\frac{1}{12}M{L}^2$ is the moment of inertia of a rod about an axis through its centre and perpendicular to it. The rod is bent from the middle at an angle $60^{\circ }$ . Find the moment of inertia of the bent rod about the same axis. ( $M$ is the mass and $L$ is the length of the rod)

• A. $\frac{1}{48}M{L}^2$
• B. $\frac{1}{12}M{L}^2$
• C. $\frac{1}{24}M{L}^2$
• D. $\frac{M{L}^3}{8\sqrt{3}}$

Answer 1

Answer: (A)

Given $I_1=\frac{1}{12}M{L}^2$
now as per the question

Both half rodes are in same plane.
now
$I=\frac{2m{l}^2}{3}si{n}^2\theta$
where $m$ is mass of each rod, $l$ is its length and $\theta$ is half of angle between them.
So
$I=\frac{2\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2}{3}\times {\left(sin\right)}^{2}30^{\circ }$
$I=\frac{M{L}^2}{48}$