Question
Consider a uniform square plate of side $$'a'$$ and mass $$'m'.$$ The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
A.
$$\frac{5}{6}m{a^2}$$
B.
$$\frac{1}{{12}}m{a^2}$$
C.
$$\frac{7}{{12}}m{a^2}$$
D.
$$\frac{2}{3}m{a^2}$$
Answer :
$$\frac{2}{3}m{a^2}$$
Solution :
$${I_{nn'}} = \frac{1}{{12}}M\left( {{a^2} + {a^2}} \right) = \frac{{M{a^2}}}{6}$$
Also, $$DO = \frac{{DB}}{2} = \frac{{\sqrt 2 a}}{2} = \frac{a}{{\sqrt 2 }}$$
According to parallel axis theorem
$$\eqalign{ & {I_{mm'}} = {I_{nn'}} + M{\left( {\frac{a}{{\sqrt 2 }}} \right)^2} = \frac{{M{a^2}}}{6} + \frac{{M{a^2}}}{2} \cr & = \frac{{M{a^2} + 3M{a^2}}}{6} \cr & = \frac{2}{3}M{a^2} \cr} $$
$${I_{nn'}} = \frac{1}{{12}}M\left( {{a^2} + {a^2}} \right) = \frac{{M{a^2}}}{6}$$
Also, $$DO = \frac{{DB}}{2} = \frac{{\sqrt 2 a}}{2} = \frac{a}{{\sqrt 2 }}$$
According to parallel axis theorem
$$\eqalign{ & {I_{mm'}} = {I_{nn'}} + M{\left( {\frac{a}{{\sqrt 2 }}} \right)^2} = \frac{{M{a^2}}}{6} + \frac{{M{a^2}}}{2} \cr & = \frac{{M{a^2} + 3M{a^2}}}{6} \cr & = \frac{2}{3}M{a^2} \cr} $$
