Question

A uniform disc of radius R/2 is put over another uniform disc of radius R of same thickness and density. The peripheries of the two disc touch each other. The centre of mass of the system is from centre of big disc

Answer 1

Let σ is mass per unit area.

Mass of disc of radius R is $M_1=\sigma \times A_1=\sigma \times \pi R^2$

Mass of disc of radius R/2 is $M_2=\sigma \times A_2=\sigma \times \frac{\pi R^2}{4}$

Let the centre of the disc of radius R be at the origin. The arrangement is shown in the figure below.

We know that,
Centre of mass = $x_{cm}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

$\Rightarrow x_{cm}=\frac{\sigma \pi R^2\times 0+\frac{\sigma \pi R^2}{4}\times \frac{R}{2}}{\sigma \times \pi R^2+\frac{\sigma \times \pi R^2}{4}}$

$\Rightarrow x_{cm}=\frac{R}{10}$

Hence, $x_{cm}=\frac{R}{10}$ right to the centre of the big disc.