Question

When a circular portion of radius $\beta$ has been removed from a disc of uniform mass and radius $r$ such that centre of the hole is at a distance $\eta$ from the centre of the original disc. Find the position of centre of mass of the remaining part from the centre $\text{C}$ as shown in figure.

• A. $x=\frac{\beta {\eta }^2}{r^2+{\beta }^2}$
• B. $x=\frac{\eta \beta }{r+\beta }$
• C. $x=\frac{-\eta {\beta }^2}{r+\beta }$
• D. $x=-\frac{\eta {\beta }^2}{(r^2-{\beta }^2)}$

Answer 1

The correct option is D $x=-\frac{\eta {\beta }^2}{(r^2-{\beta }^2)}$

Let $m$ be the total mass of the disc before the portion was cut.
Mass per unit area of the uniform disc is $\frac{m}{\pi r^2}$
$\therefore$ Mass of disc with radius $\beta$ which has been removed
$m_2=\frac{m}{\pi r^2}\left(\pi {\beta }^2\right)=m\frac{{\beta }^2}{r^2}$

The complete system will have its $x-$coordinate of COM at $x=0$ (taking origin $(0,0)$ at point $\text{C}$)
Using formula for x coordinate of COM,
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=0$
where $m_1=(m-m\frac{{\beta }^2}{r^2})$, $x_1=x,x_2=\eta ,m=(m_1+m_2)$

$\Rightarrow \frac{(m-\frac{{\beta }^2}{r^2}m)x+\left(\frac{{\beta }^2}{r^2}m\right)\eta }{m}=0$
Or $x(r^2-{\beta }^2)+{\beta }^2\eta =0$
$\therefore x=-\frac{\eta {\beta }^2}{r^2-{\beta }^2}$

The $COM$ will lie along the $x-$ axis, due to the symmetry of the system about $x-$ axis.
$-ve$ sign represents that new $COM$ lies to the left of $\text{C}$