Question

The moment of inertia of a hollow cubical box of mass $M$ and side ${}^{\prime }{a}^{\prime }$ about an axis passing through the centers of two opposite faces is equal to

• A. $\frac{5M{a}^2}{3}$
• B. $\frac{5M{a}^2}{6}$
• C. $\frac{5M{a}^2}{12}$
• D. $\frac{5M{a}^2}{18}$

Answer 1

The correct option is D $\frac{5M{a}^2}{18}$

Hollow cubical box = 6 identical square plates of mass ${}^{\prime }{m}^{\prime }$ joined to form a cube of side ${}^{\prime }{a}^{\prime }$.
Assume each plate of mass $m$.
Hence $M=6m$
MOI of square plate about axis through COM and perpendicular to its plane $=\frac{m{a}^2}{6}$


Axis of rotation passes through center of mass of two square plates (1) and (2) facing opposite to each other.
Therefore,
$I_2=I_1=\frac{m{a}^2}{6}$ ... (1)

MOI of a square plate about axis passing through COM and parallel to its plane is $I=\frac{m{a}^2}{12}$
and the remaining 4 plates have a distance of $\frac{a}{2}$ from the cube axis.

Hence by parallel axis theorem, MOI of plate (3), (4), (5) and (6) about cube axis is
$I_3=I_4=I_5=I_6=\frac{m{a}^2}{12}+m{\left(\frac{a}{2}\right)}^2=\frac{m{a}^2}{3}$

Therefore, total MOI about cube axis is
$I=I_1+I_2+I_3+I_4+I_5+I_6$
$I=2I_1+4I_3$
$I=2\times \frac{m{a}^2}{6}+4\times \frac{m{a}^2}{3}=\frac{5m{a}^2}{3}$
Here, $m=\frac{M}{6}$
Hence, $I=\frac{5M{a}^2}{18}$