Question

A square plate of edge $\frac{a}{2}$ is cut out from a uniform square plate of edge ${}^{\prime }{a}^{\prime }$ as shown in the figure. The mass of the remaining portion is $M$. The moment of inertia of the remaining portion about an axis passing through $\text{'O'}$ (centre of the square of side $a$) and perpendicular to the plane of the plate is:

• A. $\frac{9}{64}M{a}^2$
• B. $\frac{3}{16}M{a}^2$
• C. $\frac{5}{12}M{a}^2$
• D. $\frac{9}{16}M{a}^2$

Answer 1

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

Let $m_1=$ mass of the square plate of side ${}^{\prime }{a}^{\prime }$
and $m_2=$ mass of the square plate of side $\frac{a}{2}$
Then $m_1=\sigma a^2$ and $m_2=\sigma {\left(\frac{a}{2}\right)}^2$
where $\sigma$ is the areal density
and, $m_1-m_2=M$

MOI of square plate of side ${}^{\prime }{a}^{\prime }$ about perpendicular axis through $\text{O}$
$=\frac{m_1{a}^2}{6}$
MOI of square plate of side ${}^{\prime }{\frac{a}{2}}^{\prime }$ about perpendicular axis through $\text{O}$
$=\frac{m_2{a}^2}{4\times 6}+\frac{m_2{a}^2}{16}$
$=\frac{5m_2{a}^2}{48}$ (using parallel axis theorem)

Total MOI about perpendicular axis through $\text{O}$
$I=\frac{m_1{a}^2}{6}-\frac{5m_2{a}^2}{48}$
(subtracting the MOI of the removed portion)
$=\frac{a^2}{6}(m_1-\frac{5}{8}{m}_2)$
$\Rightarrow I=\sigma a^4\left\{\frac{27}{32\times 6}\right\}$

Also $M=\sigma (1-\frac{1}{4})a^2\Rightarrow \sigma =\frac{4}{3}\frac{M}{a^2}$
$\Rightarrow I=\left(\frac{4}{3}\frac{M}{a^2}\right).a^4\left\{\frac{27}{32\times 6}\right\}$
$\Rightarrow I=\frac{3M{a}^2}{16}$