Question

Figure shows a rectangular copper plate with is centre of mass at the origin $ O $ and side $ AB = 2BC = 2 m $ . If a quarter part of the plate (shown as shaded) is removed, the centre of mass of the remaining plate would lie at

• A. $ \frac{1}{12} m, \frac{1}{6} m $
• B. $ \frac{1}{6} m, \frac{1}{12} m $
• C. $ \frac{1}{3} m, \frac{1}{6} m $
• D. $ \frac{1}{3} m, \frac{1}{2} m $

Answer 1

Answer: (B)

Given,
$AB = 2BC = 2m$

$\therefore BC=1\,m$
$\sigma$ be the mass per unit area
$m_{1}=m_{2}=m_{3}=(1\times0.5) \sigma=0.5\sigma$
If $G\left(\bar{x}, \bar{y}\right)$ be the position of centre of mass, then
$\bar{x}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$
$=\frac{0.5\sigma\times0.5+0.5\sigma\times\left(-0.5\right)+0.5\sigma\times0.5}{0.5\sigma+0.5\sigma+0.5\sigma}$
$=\frac{0.5\sigma\times0.5}{3\times0.5\sigma}=\frac{1}{6}m$
$\bar{y}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}$
$=\frac{0.5\sigma\times0.25+0.5\sigma\times0.25+0.5\sigma\times\left(-0.25\right)}{0.5\sigma+0.5\sigma+0.5\sigma}$
$=\frac{0.5\sigma\times0.25}{3\times0.5\sigma}=\frac{1}{12}m$