Question

A uniform rod of length $L$ and mass $M$ is bent at the middle point $\text{O}$ as shown in figure. Consider an axis passing through middle point $\text{O}$ and perpendicular to the plane of the bent rod. The moment of inertia about this axis is

• A. $\frac{1}{48}M{L}^2$
• B. $\frac{1}{24}M{L}^2$
• C. $\frac{1}{12}M{L}^2$
• D. depends on $\theta$

Answer 1

The correct option is C $\frac{1}{12}M{L}^2$

As we know,
Moment of inertia of a rod of mass $M$ and length $L$ about a perpendicular axis passing through its center is $\frac{M{L}^2}{12}$.
Here, there are two rods of length $\frac{L}{2}$ each and total mass $M$ uniformly distributed. Hence, mass of each will be $\frac{M}{2}$.
MOI of rod of length $\frac{L}{2}$ and mass $\frac{M}{2}$ about perpendicular axis through it centre is
$I_1=\frac{\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2}{12}$

By parallel axis theorem, MOI of both rods about axis through $\text{O}$ will be
$I_{total}=[\frac{\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2}{12}+\frac{M}{2}{\left(\frac{L}{4}\right)}^2]\times 2$
{multiplied by $2$ because there are two rods}
$\therefore I_{total}=\frac{M{L}^2}{12}$