Question

The centre of mass of a disc of uniform mass and radius r, when a circular portion of radius $\beta$ has been removed from it such that centre of the hole is at distance $\eta$ from the centre of disc is :

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

x be the new position of centre of mass.
mass of cut disc=$\frac{m}{\pi r^2}\left(\pi {\beta }^2\right)=\frac{{\beta }^2}{r^2}m$
since , combinational system of given figure and including disc would have its COM at centre i.e at x=0.By using that,
$\frac{(m-\frac{{\beta }^2}{r^2}m)x+\eta \left(\frac{{\beta }^2}{r^2}m\right)}{m}=0$

$\Rightarrow x=-\frac{\eta {\beta }^2}{r^2-{\beta }^2}$