Question

A square of side '$a$' is cut from a square of side '$2a$' as shown in figure. Mass of this square with hole is M. Then its moment of inertia about an axis passing through its CM and perpendicular to its plane will be

• A. $\frac{{Ma}^2}{6}$
• B. $\frac{{2Ma}^2}{6}$
• C. $\frac{{4Ma}^2}{6}$
• D. $\frac{{5Ma}^2}{6}$

Answer 1

The correct option is D $\frac{{5Ma}^2}{6}$

Mass of remaining part = $M$

Mass per unit area = $\frac{M}{4a^2-a^2}=\frac{M}{3a^2}$

Mass of square without hole = $\frac{M}{3a^2}\times 4a^2=\frac{4M}{3}$

Mass of part is taken out =$\frac{M}{3a^2}\times a^2=\frac{M}{3}$

Moment of inertia of total without hole about center of mass or center of square is :

$\frac{1}{6}\times \frac{4M}{3}\times (2a{)}^2=\frac{16M{a}^2}{18}$

Moment of inertia of square part taken out about center of mass:

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

Moment of inertia of remaining part = $\frac{16M{a}^2}{18}-\frac{M{a}^2}{18}$

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