Question

Four identical rods each of mass $M$ are joined to form as square frame. The moment of inertia of the system about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is

• A. $\frac{4M{l}^2}{3}$
• B. $\frac{13M{l}^2}{3}$
• C. $\frac{M{l}^2}{6}$
• D. $\frac{13M{l}^2}{6}$

Answer 1

The correct option is A $\frac{4M{l}^2}{3}$

Moment of Inertia of a rod about it Center of Mass= $\frac{M{L}^2}{12}$

Moment of Inertia of a rod about any other point at a distance D from its COM= $\frac{M{L}^2}{12}+M{D}^2$

The distance of rod COM from point of intersection of diagonals is $D=L/2$

$\therefore$ Moment of Inertia of a rod about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is ($D=L/2$)

$I_1=\frac{M{L}^2}{12}+M{D}^2=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$

As the distance of each rod COM from point of intersection, mass & length of each rod is constant.

$\therefore$ Moment of Inertia of 4 rod about an axis passing through the point of intersection of diagonals and perpendicular to the plane of the square is

$4I_1=\frac{4M{L}^2}{3}$