First slide

Moment of intertia

Question

The moment of inertia of a hollow cubical box of mass M and side a about an axis passing through the centres of two opposite faces is equal to

Moderate

A

$\frac{5 M a^{2}}{3}$
B

$\frac{5 M a^{2}}{6}$
C

$\frac{5 M a^{2}}{12}$
D

$\frac{5 M a^{2}}{18}$

Solution

Taking mass of plate m
Then MI of two plates through which axis is passing = $\frac{m a^{2}}{6} \times 2 = \frac{{ma}^{2}}{3}$
M.I. of 4 plates having symmetrical position from the axis.
= $4 \times [\frac{{ma}^{2}}{12} + m {(\frac{a}{2})}^{2}] = 4 \times [\frac{{ma}^{2}}{3}]$
Total MI = $\frac{4 {ma}^{2}}{3} + \frac{{ma}^{2}}{3} = \frac{5 {ma}^{2}}{3}$
using $m = \frac{M}{6} \Rightarrow MI = \frac{5 {Ma}^{2}}{18}$

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