Question

The disk of radius $5\text{cm}$ is cut from the uniform disc of radius $10\text{cm}$ in such a way that the edge of the hole touches the edge of the disc. What is the distance of the center of mass of the remaining portion w.r.t the point $\text{O}$ ?

• A. $2.66\text{cm}$
• B. $6\text{cm}$
• C. $1.66\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is C $1.66\text{cm}$


Let $\rho$ be the mass density of the disk.
Let $m_1$ be the mass of circular disk of radius $10\text{cm}$ and $m_2$ be the mass of circular disk of radius $5\text{cm}$.
Take $\text{O}$ to be the origin for reference.
So, $x_1=0$ (COM of circular disk of radius $10\text{cm}$)
$x_2=5\text{(COM of removed disk)}.$

Now,
$X_{COM}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$
(we treat the removed portion of disc as having negative mass)
where
$m_1=\rho \left(\pi \right(10{)}^2)=100\pi \rho$
$m_2=\rho \left(\pi \right(5{)}^2)=25\pi \rho$
$X_{COM}=\frac{100\pi \rho \times \left(0\right)-25\pi \rho \times \left(5\right)}{100\pi \rho -25\pi \rho }=\frac{-25\times 5}{100-25}$
$=-1.66$
Due to symmetry, $Y_{COM}$ will be zero.

Therefore, $(X_{COM},Y_{COM})=(-1.66,0).$
So, COM of remaining portion will be shift towards left side on the line joining the points 'O' and 'C'.