Question

From a square plate of side $10\text{m}$, a right angled triangle is removed as shown in the figure. The square plate has mass density $2{\text{kg/m}}^2$. Find the centre of mass of the remaining portion. Assume the origin to be at the intersection of $x$ and $y$ axes.

• A. $(4.2,6.5)\text{m}$
• B. $(5,5)\text{m}$
• C. $(5.23,5.25)\text{m}$
• D. $(6.25,6.32)\text{m}$

Answer 1

The correct option is C $(5.23,5.25)\text{m}$

Given side of the triangle $a=4\text{m},b=3\text{m}$
Mass of the original square plate,
$m_1=$ mass density $\times$ (area of square plate)
$=2\times (10\times 10)$
$=200\text{kg}$

Mass of the right angled triangle,
$m_2=$ mass density $\times$ (area of right angled triangle)
$=2\times (\frac{1}{2}\times 4\times 3)$
$=12\text{kg}$

$COM$ of square plate,
$\left(x_1{y}_1\right)=(5,5)\text{m}$
$COM$ of right angled triangle is situated at the distance $\frac{a}{3}$ and $\frac{b}{3}$ from the side $b$ and $a$ respectively.


i.e $COM$ of right angled triangle is at
$(x_2,y_2)=(\frac{4}{3},\frac{3}{3})=(\frac{4}{3},1)\text{m}$

Hence, $x$-coordinate of $COM$ of remaining portion.
$x_{COM}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}=\frac{200\times 5-12\times \frac{4}{3}}{200-12}$
$=5.23\text{m}$

$y$-coordinate of $COM$ of remaining portion.
$y_{COM}=\frac{m_1{y}_1-m_2{y}_2}{m_1-m_2}=\frac{200\times 5-12\times 1}{200-12}$
$=5.25\text{m}$

So, position of $COM$ of remaining portion is at
$(x_{COM},y_{COM})=(5.23,5.25)\text{m}$