Question

A uniform metal disc of diameter $24cm$ is taken and out of it a disc of diameter $8cm$ is cut off from the right side end. The centre of mass of the remaining part will be at:

• A. left side, $2cm$ from the centre
• B. right side, $1cm$ from the centre
• C. right side, $2cm$ from the centre
• D. left side, $1cm$ from the centre

Answer 1

Answer: (D)

left side, $1cm$ from the centre

Let the surface density of material be $\rho$.

Mass of complete disc, $M=\rho A=\rho \pi r^2$

Then, mass of the removed disc is $M_2=\rho A_2=\rho \pi r_2^2$

Mass of the remaining disc is $M_1=M-M_2=\rho \pi (r^2-r_2^2)$

Center of mass of removed disc is $x_2=r-r_2$

Since the center of mass of the remaining piece and that of the removed portion combined must lie at the center of disc (0,0).

$x_{cm}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

$\left(\rho \right(\pi {12}^2)-\rho (\pi 4^2\left)\right)x+\rho \left(\pi 4^2\right)(12-4)=0$

$x=-\frac{\left(16\right)\left(8\right)}{144-16}=-1cm$

Hence,$1cm$ to the left of center.