Question

Calculate the moment of inertia of a rectangular frame formed by uniform rods having mass $m$ each as shown in figure about an axis passing through its centre and perpendicular to the plane of frame? Also find moment of inertia about an axis passing through $PQ$?

Answer 1

Moment of inertia of rod along length $l$ about it's centre of mass is given by:

$I_{com}=\frac{m{l}^2}{12}$

So moment of inertia about an axist at a distance $\frac{1}{2}$ from this axist and parallel to it is:

$I_1=I_{(}com)+m{\left(\frac{1}{2}\right)}^2$ (Using parallel axis theorem)

$\Rightarrow I_1=\frac{m{l}^2}{12}+\frac{m{l}^2}{4}=\frac{4m{l}^2}{12}=\frac{m{l}^2}{3}\to \left(1\right)$

Similarly the moment of Inertia of othet rods about the same axis is calxulate as:

$I_2=\frac{m{l}^2}{3}\to \left(2\right)$

$I_3=\frac{m{b}^2}{3}\to \left(3\right)$

$I_4=\frac{m{b}^2}{3}\to \left(4\right)$

The moment of inertia of the whole system about the axis is given by sum of moment of inertias of the rods about the axis i.e,

$I=I_1+I_2+I_3+I_4$

$\Rightarrow I=\frac{2m{l}^2}{3}+\frac{2m{b}^2}{3}=\frac{2m}{3}(l^2+b^2)\to \left(Answer\right)$