Question

Figure shows a rectangular copper plate with is centre of mass at the origin $O$ and side $AB=2BC=2m$ . 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,
$AB=2BC=2m$

$\therefore BC=1m$
$\sigma$ be the mass per unit area
$m_1=m_2=m_3=(1\times 0.5)\sigma =0.5\sigma$
If $G\left(\bar{x},\bar{y}\right)$ be the position of centre of mass, then
$\bar{x}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{0.5\sigma \times 0.5+0.5\sigma \times \left(-0.5\right)+0.5\sigma \times 0.5}{0.5\sigma +0.5\sigma +0.5\sigma }$
$=\frac{0.5\sigma \times 0.5}{3\times 0.5\sigma }=\frac{1}{6}m$
$\bar{y}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{0.5\sigma \times 0.25+0.5\sigma \times 0.25+0.5\sigma \times \left(-0.25\right)}{0.5\sigma +0.5\sigma +0.5\sigma }$
$=\frac{0.5\sigma \times 0.25}{3\times 0.5\sigma }=\frac{1}{12}m$