Question

From a uniform square plate of side a and mass m, a square portion $DEFG$ of side $\frac{a}{2}$ is.removed. Then, the moment of inertia of remaining portion about the axis $AB$ is

• A. $\frac{7m{a}^2}{16}$
• B. $\frac{3m{a}^2}{16}$
• C. $\frac{3m{a}^2}{4}$
• D. $\frac{9m{a}^2}{16}$

Answer 1

The correct option is B $\frac{3m{a}^2}{16}$

MI of a square plate about an an edge in the plane of the plate and passing through center and parallel to an edge is:$I=\frac{m{a}^2}{12}$ using perpendicular axis theorem.
Using parallel axis theorem, MI about an edge is: $I=\frac{m{a}^2}{12}+m(\frac{a}{2}{)}^2=\frac{m{a}^2}{3}$
$I_{AB}=I_1+I_2+I_3$
$=\frac{\left(m/3\right)(a/2{)}^2}{3}+\frac{\left(m/3\right)(a/2{)}^2}{3}+[\frac{\left(m/3\right)(a/2{)}^2}{12}+(m/3\left){\left(\frac{3a}{4}\right)}^2\right]$
$=\frac{3}{16}m{a}^2$