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

• 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 $\frac{5M{a}^2}{18}$

The moment of inertia of face that through the axis is given as,

$I=\frac{m\left(L^2+W^2\right)}{12}$

Where, the length is $a$ and the mass is $\frac{M}{6}$.

It can be written as,

$I_1=\left(\frac{M}{6}\right)\frac{a^2}{6}$

Since, there are two faces that through the axis so it can be written as,

${I_1}^{\prime }=\frac{M{a}^2}{18}$

The moment of inertia that not through the axis is given as,

$I=m\left[\frac{\left(L^2+W^2\right)}{24}+R^2\right]$

Where, the length is $a$, mass is $\frac{M}{6}$.and the distance from the face center to the axis is $\frac{a}{2}$.

It can be written as,

$I_2=\left(\frac{M{a}^2}{18}\right)$

Since, there are four faces that does not through the axis so it can be written as,

${I_2}^{\prime }=\frac{2M{a}^2}{9}$

The total moment of inertia is given as,

$I_t={I_1}^{\prime }+{I_2}^{\prime }$

$=\frac{M{a}^2}{18}+\frac{2M{a}^2}{9}$

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

Thus, the total moment of inertia is $\frac{5M{a}^2}{18}$.