Question

Four thin rods of same mass $3\text{kg}$ and same length $2\text{m}$, form a square as shown in figure. Moment of inertia of this system about an axis passing through centre $\text{'O'}$ and perpendicular to its plane is $P{\text{kg-m}}^2$. The value of $P$ is

Answer 1

Given, mass of each rod (m) $=3\text{kg}$,
length of each rod $\left(l\right)=2\text{m}$
Moment of inertia of rod AB about axis passing through its COM and perpendicular to the plane $I=\frac{m{l}^2}{12}$
Using parallel axis theorem,
moment of inertia of rod AB about point 'O'
$=\frac{m{l}^2}{12}+m{\left(\frac{l}{2}\right)}^2$
$=\frac{m{l}^2}{3}$
As the system consists of 4 rods of similar type, by symmetry
$I_{sys}=\frac{4}{3}m{l}^2=\frac{4}{3}\times 3\times (2{)}^2=16{\text{kg-m}}^2$