Question

A thin rod of length 4 $l$ , mass $4m$ is bent at the points as shown in fig. What is the moment of inertia of the rod about the axis passing point O & perpendicular to the plane of the paper.

• A. $\frac{M{l}^2}{3}$
• B. $\frac{10M{l}^2}{3}$
• C. $\frac{M{l}^2}{12}$
• D. $\frac{M{l}^2}{24}$

Answer 1

Answer: (B)

$\frac{10M{l}^2}{3}$

Distance of center of 1st rod on the left to O is $L_1=\sqrt{l^2+{\left(l/2\right)}^2}=\frac{\sqrt{5}l}{2}$
Therefore moment of inertia of 1st rod, $I_1=\frac{{ml}^2}{12}+m(\frac{\sqrt{5}l}{2}{)}^2=\frac{{4ml}^2}{3}$

Moment of Inertia of rod 2 about O (at the end point of rod) $I_2=\frac{{ml}^2}{3}$
By symmetry,
$I_3=I_2=\frac{{ml}^2}{3}{I}_4=I_2=\frac{{4ml}^2}{3}$
Therefore total Moment of Inertia:
$I_T=I_1+I_2+I_3+I_4=\frac{{10ml}^2}{3}$