Question

Two uniform circular thin rods of mass $m$ and length $l$ are connected with each other at their mid points and perpendicular to each other. A ring is also connected to both of the rods as shown in figure. Masses of the ring and both rods are the same. Find the moment of inertia of the system about the axis passing through the mid point of the rods and perpendicular to the plane of ring.

• A. $\frac{7}{12}m{l}^2$
• B. $\frac{9}{12}m{l}^2$
• C. $\frac{3}{12}m{l}^2$
• D. $\frac{5}{12}m{l}^2$

Answer 1

The correct option is D $\frac{5}{12}m{l}^2$

Given mass of each rod $=m$
mass of ring $=m$
length of rod $=l$
So, radius of circular ring $=\frac{l}{2}.$
Moment of inertia of rod AB, $I_1=\frac{m{l}^2}{12}$
about $\text{OO}$'
Moment of inertia of rod CD, $I_2=\frac{m{l}^2}{12}$
about $\text{OO}$'
Moment of inertia of circular ring about $\text{OO}$', $I_3=m{\left(\frac{l}{2}\right)}^2=\frac{m{l}^2}{4}$

Total moment of inertia of system about the axis passing through the center and perpendicular to its plane. $I=I_1+I_2+I_3$
$I=\frac{m{l}^2}{12}+\frac{m{l}^2}{12}+\frac{m{l}^2}{4}=\frac{5m{l}^2}{12}$
$\therefore I=\frac{5}{12}m{l}^2$