A uniform disc of radius R is placed over ano | vedclass

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(B) Let $\rho$ be the surface mass density of the discs. Since they have the same thickness and material,$\rho$ is constant for both.Mass of the large

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$\frac{R}{5}$

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(B) Let $\rho$ be the surface mass density of the discs. Since they have the same thickness and material,$\rho$ is constant for both.Mass of the large disc $(M_1)$ = $\pi(2R)^2 \rho = 4\pi R^2 \rho$.The center of mass of the large disc is at the origin,so $x_1 = 0$.Mass of the small disc $(M_2)$ = $\pi R^2 \rho$.The center of the small disc is at a distance $R$ from the origin,so $x_2 = R$.The center of mass of the system $(X_{cm})$ is given by:$X_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$$X_{cm} = \frac{(4\pi R^2 \rho)(0) + (\pi R^2 \rho)(R)}{4\pi R^2 \rho + \pi R^2 \rho}$$X_{cm} = \frac{\pi R^3 \rho}{5\pi R^2 \rho} = \frac{R}{5}$.Thus,the center of mass is at a distance $\frac{R}{5}$ from the origin towards the smaller disc.

$A$ uniform disc of radius $R$ is placed over another uniform disc of radius $2R$ made of the same material and having the same thickness. The peripheries of the two discs touch each other. Locate the centre of mass of the system,taking the center of the large disc at the origin.

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