Question

From a square plate of side $8\text{m}$, a circular disc of radius $2\text{m}$ is removed as shown in the figure. The mass density of the square plate is $1{\text{kg/m}}^2$. Find the center of mass of the remaining portion.

• A. $(4.9,4.9)\text{m}$
• B. $(3.51,3.51)\text{m}$
• C. $(2,2)\text{m}$
• D. $(2.5,2.5)\text{m}$

Answer 1

The correct option is B $(3.51,3.51)\text{m}$

Mass of the square plate,
$m_1=$ mass density $\times$ area of square plate
$=1\times (8\times 8)$
$=64\text{kg}$
Mass of the circular disc,
$m_2=$ mass density $\times$ area of circular disc
$=1\times (\pi \times 2^2)$
$=4\pi \text{kg}$

Coordinates of $COM$ of square plate
$(x_1,y_1)=(4,4)\text{m}$
Coordinate of COM of circular disc
$(x_2,y_2)=(6,6)\text{m}$


Therefore, $x$-coordinate of $COM$ of remaining portion is
$x_{COM}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}=\frac{64\times 4-4\pi \times 6}{64-4\pi }$
(treating removed mass as negative)
$=3.51\text{m}$

$y$ - coordinate of $COM$ of remaining portion is
$y_{COM}=\frac{m_1{y}_1-m_2{y}_2}{m_1-m_2}=\frac{64\times 4-4\pi \times 6}{64-4\pi }$
$=3.51\text{m}$

Hence, position of $COM$ of remaining portion is $(x_{COM},y_{COM})=(3.51,3.51)\text{m}$