Question

An isosceles triangle piece is cut from a square plate of side $l$. The piece is one-fourth of the square and mass of the remaining plate is $M$. The moment of inertia of the plate about an axis passing through O and perpendicular to its plane is

• A. $\frac{M{l}^2}{6}$
• B. $\frac{M{l}^2}{12}$
• C. $\frac{M{l}^2}{24}$
• D. $\frac{M{l}^2}{3}$

Answer 1

The correct option is A $\frac{M{l}^2}{6}$

Mass of remaining three-fourth of plate is M, thus mass of the piece cut out should be $\frac{M}{3}$ because density is uniform.

Now moment of inertia for a rectangular plate about an axis perpendicular to the plane and passing through its centre is given by $I=\frac{m(a^2+b^2)}{12}$ which becomes $I=\frac{2m{l}^2}{12}$ or $I=\frac{m{l}^2}{6}$, where m=M+M/3=4M/3.

This moment of inertia is the sum of moment of inertias of 4 triangular parts about their vertex at centre. Also, these 4 moments will be equal because of symmetry.

Thus, $I_1+I_2+I_3+I_4=4I_{triangle}=\frac{\left(\frac{4M}{3}\right)L^2}{6}$

or, $I_{triangle}=\frac{M{L}^2}{18}$

Since after cutting one piece, moment of inertia becomes ${3.I}_{triangle}$, we get

$I={3.I}_{triangle}=\frac{M{L}^2}{6}$