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Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
A
$\frac{1}{3}Ml^2$
B
$\frac{4}{3}Ml^2$
C
$\frac{2}{3}Ml^2$
D
$\frac{13}{3}Ml^2$
Detailed Solution
Moment of inertia of one rod about an axis through its own centre and perpendicular to its length: $I_{cm} = \frac{Ml^2}{12}$
The centre of the square is at a perpendicular distance $\frac{l}{2}$ from the centre of each rod.
By the parallel axis theorem, for one rod about the axis through the centre of the square: $I_1 = \frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2}{3}$
For the four rods: $I = 4\times\frac{Ml^2}{3}$
$I = \frac{4}{3}Ml^2$
The centre of the square is at a perpendicular distance $\frac{l}{2}$ from the centre of each rod.
By the parallel axis theorem, for one rod about the axis through the centre of the square: $I_1 = \frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2}{3}$
For the four rods: $I = 4\times\frac{Ml^2}{3}$
$I = \frac{4}{3}Ml^2$