Question

Four identical thin rods each of mass M and length $l$ , form a square frame . Form of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is :

Answer 1

Moment of inertia for the rod $AB$ rotating about an axis through the midpoint of $AB$ perpendicular to the plane of the paper is $\frac{M{I}^2}{12}$

Moment of inertia about the axis through the centre of the square and parallel to this axis,

$I=I_0+M{d}^2=M(\frac{l^2}{12}+\frac{l^2}{4})=\frac{M{l}^2}{3}$

For all the four roads, $I=\frac{4}{3}M{I}^2$.