Question

Moment of inertia of a rectangular plate about an axis passing through $P$ and perpendicular to the plate is $I$. The moment of inertia of $PQRS$ about an axis perpendicular to the plane of the plate :

• A. about $P=\frac{I}{2}$
• B. about $R=\frac{I}{2}$
• C. about $P>\frac{I}{2}$
• D. about $R>\frac{I}{2}$

Answer 1

Answer: (C)

about $P>\frac{I}{2}$

Let the mass density be $\rho kg/m^2$
Let the side PQ=RS=$a$ & QR=PS=$b$ (From fig $a>b$)
Mass of rectangle is $m=$ $\rho ab$
Therefore Moment of Inertia of Rectangle about its center = $m\frac{a^2+b^2}{12}$
Distance of P point from center of rectangle is $\frac{\sqrt{a^2+b^2}}{2}$
Therefore Moment of Inertia of Rectangle about P, I= $m\frac{a^2+b^2}{12}+m\frac{a^2+b^2}{4}=m\frac{a^2+b^2}{3}$
Mass of triangle PQR$=\frac{m}{2}=$ $\frac{\rho ab}{2}$
Moment of Inertia of Triangle PQR about its centroid = $\rho \frac{a{b}^3+{ba}^3}{12}=m\frac{a^2+b^2}{12}$
Distance of point P from centroid = $\sqrt{(\frac{2a}{3}{)}^2+(\frac{b}{3}{)}^2}$
Moment of Inertia of Triangle PQR about P= $m\frac{a^2+b^2}{12}+\frac{m}{2}\left\{\right(\frac{2a}{3}{)}^2+\left(\frac{b}{3}{)}^2\right\}=\frac{m}{2}\frac{{11a}^2+5b^2}{18}>\frac{m}{2}\frac{{6a}^2+6b^2}{18}>\frac{I}{2}$
(As $a>b$)
Distance of point R from centroid = $\sqrt{(\frac{2b}{3}{)}^2+(\frac{a}{3}{)}^2}$
Moment of Inertia of TrianglePQR about P$=$ $m\frac{a^2+b^2}{12}+\frac{m}{2}\left\{\right(\frac{a}{3}{)}^2+\left(\frac{2b}{3}{)}^2\right\}=\frac{m}{2}\frac{{5a}^2+11b^2}{18}$
As $a>b$ therefore $\frac{m}{2}\frac{{5a}^2+11b^2}{18}$ can be less than or equal to or greater than $\frac{I}{2}$