Question

A circular plate of uniform thickness has a radius, $a=10\text{cm}$. A circular portion of radius, $b=1\text{cm}$ is removed from the plate as shown in figure. Find the position of center of mass of the remaining part of the circular plate. (Take $\text{O}$ to be the origin)

• A. $(0,-\frac{1}{11})$
• B. $(0,-\frac{9}{88})$
• C. $(0,-\frac{8}{99})$
• D. $(0,-\frac{10}{99})$

Answer 1

The correct option is C $(0,-\frac{8}{99})$

Due to symmetry, $X_{com}$ of the remaining part will be zero.
Let $\rho$ be the mass per unit area of circular plate.

Let $m_1$ be the mass of circular plate of radius $a=10\text{cm}$ and $m_2$ be the mass of the removed part of circular plate. $b=1\text{cm}$.
$m_1=\rho \left(\pi a^2\right)=\rho (\pi \times (10{)}^2)$
$=100\rho \pi$
$m_2=\rho \left(\pi b^2\right)=\rho \left(\pi \right(1{)}^2)=\rho \pi$
(Mass of removed plate).

Let $\text{O}$ be the reference point.
$y_1\to$ Centre of mass of circular plate of mass $m_1$
$y_2\to$ Centre of mass of removed plate.
Then, $y_1=0,y_2=8$
(from the diagram)

Centre of mass of remaining part of circular plate :
$Y_{com}=\frac{M_1{y}_1-M_2{y}_2}{M_1-M_2}$
$=\frac{100\rho \pi \times \left(0\right)-\rho \pi \left(8\right)}{100\rho \pi -\rho \pi }$
$=\frac{-\rho \pi \left(8\right)}{99p\pi }$
$Y_{com}=-\frac{8}{99}$
$\therefore (X_{com},Y_{com})=(0,\frac{-8}{99})$