Question

A uniform rectangular lamina $\text{ABCD}$ is of mass $M$, length $a$ and breadth $b$, as shown in the figure. If the shaded portion $\text{HBGO}$ is cut-off, the coordinates of the centre of mass of the remaining portion will be

• A. $(\frac{2a}{3},\frac{2b}{3})$
• B. $(\frac{5a}{3},\frac{5b}{3})$
• C. $(\frac{3a}{4},\frac{3b}{4})$
• D. $(\frac{5a}{12},\frac{5b}{12})$

Answer 1

The correct option is D $(\frac{5a}{12},\frac{5b}{12})$

Let the mass of rectangular lamina $\text{ABCD}$ be $M$, length be $a$ and breadth be $b$
The centre of mass of rectangular lamina$\text{ABCD}$ is at point ($x_1,y_1$)=$(\frac{a}{2},\frac{b}{2})$


Centre of mass of the rectangular lamina $\text{HBGO}$ is ($x_2,y_2$)=$(\frac{3a}{4},\frac{3b}{4})$
Mass of rectangular lamina $\text{HBGO}$= $\frac{M}{4}$
$x-$coordinate of c.o.m after mass removal is given by
$x_{cm}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$
$x=\frac{M\frac{a}{2}-\frac{M}{4}\times \frac{3a}{4}}{M-\frac{M}{4}}=\frac{\frac{a}{2}-\frac{3a}{16}}{\frac{3}{4}}=\frac{5a}{12}\text{Units}$
$y-$ coordinate of c.o.m after mass removal is given by
$y_{cm}=\frac{m_1{y}_1-m_2{y}_2}{m_1-m_2}$
$y=\frac{M\frac{b}{2}-\frac{M}{4}\times \frac{3b}{4}}{M-\frac{M}{4}}=\frac{5b}{12}\text{Units}$