Question

A square section of side $20\text{cm}$ is removed from the bigger square of side $40\text{cm}$ as shown in the figure. Find the center of mass of the remaining portion of the square.

• A. $(\frac{50}{4},\frac{50}{4})\text{cm}$
• B. $(25,25)\text{cm}$
• C. $(\frac{50}{3},\frac{50}{3})\text{cm}$
• D. $(20,20)\text{cm}$

Answer 1

The correct option is C $(\frac{50}{3},\frac{50}{3})\text{cm}$


$COM$ of square $ABCD$ is $(x_1,y_1)=(20,20)\text{cm}$
$COM$ of square $EFGC$ is $\left(x_2{y}_2\right)=(30,30)\text{cm}$

Let the mass of square $ABCD$ be $M$.
Since, the area of square $EFGC$ is one fourth of square $ABCD$, the mass of the removed square $EFGC$ is $\frac{M}{4}$.
So, take $m_1=M$ and $m_2=-\frac{M}{4}$

$X$-coordinate of $COM$ of remaining portion,
$x_{COM}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}=\frac{M\left(20\right)-\frac{M}{4}\left(30\right)}{M-\frac{M}{4}}$
$=\frac{20-\frac{30}{4}}{\frac{3}{4}}=\frac{50}{3}\text{cm}$

$Y$-coodinate of $COM$ of remaining portion,
$y_{COM}=\frac{m_1{y}_1-m_2{y}_2}{m_1-m_2}=\frac{M\left(20\right)-\frac{M}{4}\left(30\right)}{M-\frac{M}{4}}$
$=\frac{20-\frac{30}{4}}{\frac{3}{4}}=\frac{50}{3}\text{cm}$

So, position of $COM$ of remaining portion is $(\frac{50}{3},\frac{50}{3})\text{cm}$