A uniform disc of radius R is placed upon another uniform disc of...

Let the mass of the smaller disc be $m$. Since both discs have the same thickness and density, their masses are proportional to their areas.

Area of the smaller disc $= \pi R^2$

Area of the larger disc $= \pi (2R)^2 = 4\pi R^2$

Mass of the larger disc, $M = 4m$

Let the center of the larger disc be at the origin $(0,0)$. Since the smaller disc is placed upon the larger disc and their edges touch, the smaller disc is internally tangent to the larger disc.

The distance between their centers is $2R - R = R$. Let the center of the smaller disc be at $(R, 0)$.

The position of the center of mass of the two-disc system is given by:

$x_{cm} = \dfrac{M x_1 + m x_2}{M + m}$

Substituting the values:

$x_{cm} = \dfrac{4m(0) + m(R)}{4m + m}$

$x_{cm} = \dfrac{mR}{5m} = \dfrac{R}{5}$

Thus, the center of mass is at a distance of $\dfrac{R}{5}$ from the center of the larger disc towards the center of the smaller disc.