Question

A uniform thin rod of length $(4a+2\pi a)$ and of mass $(4m+2\pi m)$ is bent and fabricated to form square surrounded by semicircles as shown in the figure. The moment of inertia of this frame about an axis passing through its centre and perpendicular to its plane is

• A. $\frac{\left(4+2\pi \right)}{3}m{a}^2$
• B. $\frac{\left(4+\pi \right)}{2}m{a}^2$
• C. $\frac{\left(4+3\pi \right)}{3}m{a}^2$
• D. $\frac{\left(10+2\pi \right)}{3}m{a}^2$

Answer 1

The correct option is C $\frac{\left(4+3\pi \right)}{3}m{a}^2$

Let,

Mass of the each side of the square $=m$

Length of each side of the square $=a$

Mass of each semi circle $=\frac{\pi m}{2}$

Radius of each semi circle $=\frac{a}{2}$

Moment of inertia of the square part, $I_s=4(m\frac{a^2}{12}+m\frac{a^2}{4})=\frac{4}{3}m{a}^2$

Moment of ineria of the semi circular part, $I_{sc}=4(\frac{\pi m}{2}\frac{a^2}{4}+\frac{\pi m}{2}\frac{a^2}{4})=\pi m{a}^2$

Total moment of inertia, $I=\frac{4}{3}m{a}^2+\pi m{a}^2=\frac{(4+3\pi )}{3}m{a}^2$