Question

Four thin uniform rods each of length L and mass m are joined to form a square. The moment of inertia of square about an axis along its one diagnol is

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

Answer 1

The correct option is B $\frac{2}{3}m{l}^2$

Moment of inertia about the centre of one rod, $I_1=\frac{m{l}^2}{12}$

Moment of inertia through the centre of the square, will be through an axis parallel to moment of inertia perpendicular to the rod,

$I_2=I_1+m(\frac{l}{2}{)}^2$

$I_2=\frac{4m{l}^2}{12}$

For four rods,

$I_2=\frac{4m{l}^2}{12}\times 4$

From perpendicular axis theorem,

$I_2=I_3+I_4$

where, $I_3$ and $I_4$ are moment of inertia through the diagonals. For a square, both diagonals are equal.

$I_3=I_4$

$I_2=2I_3$

$\frac{4m{l}^2}{12}\times 4=2I_3$

$I_3=\frac{2m{l}^2}{3}$