First slide

Moment of inertia

Question

Four identical rods are joined end to end to form a square. The mass of each rod is M. The moment of inertia of the system about one of the diagonals is

Moderate

A

$\frac{2 M l^{2}}{3}$
B

$\frac{13 M l^{2}}{3}$
C

$\frac{M l^{2}}{6}$
D

$\frac{13 M l^{2}}{6}$

Solution

The moment of inertia about CM system perpendicular to plane is $I_{z} = 4 (\frac{m l^{2}}{12} + m {(\frac{l}{2})}^{2}) = \frac{4}{3} M l^{2}$
From perpendicular axis theorem,
$\begin{array}{l} \frac{4}{3} M l^{2} = I_{d 1} + I_{d 2} (d u e t o s y m m e t r y, I_{d 1} = I_{d 2} = I_{d}) \\ \Rightarrow I_{d} = \frac{2}{3} M l^{2} \end{array}$

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