A uniform disc of radius R is put over another uniform disc of radius 2R of the same thickness and density. The peripheries of the two discs touch each other. Locate the centre of mass of the system.
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Let the centre of the bigger disc be the origin. 2R = Radius of bigger disc R = Radius of smaller disc $\text{m}_1=\pi\text{R}^2\times\text{T}\times\rho$ $\text{m}_2=\pi(\text{2R})^2\times\text{T}\times\rho$ where T = Thickness of the two discs $\rho=$ Density of the two discs $\therefore$ The position of the centre of mass
$\Big(\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2}{\text{m}_1+\text{m}_2},\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2}{\text{m}_1+\text{m}_2}\Big)$ $\text{x}_1=\text{R},\ \text{y}_1=0$ $\text{x}_2=0,\ \text{y}_2=0$ $\Big(\frac{\pi\text{R}^2\text{T}\rho\text{R}+0}{\pi\text{R}^2\text{T}\rho+\pi(2\text{R})^2\text{T}\rho},\frac{0}{\text{m}_1+\text{m}_2}\Big)=\Big(\frac{\pi\text{R}^2\text{T}\rho\text{R}}{5\pi\text{R}^2\text{T}\rho},0\Big)=\Big( \frac{\text{R}}{5},0\Big)$ At $\frac{\text{R}}5{}$ from the centre of bigger disc towards the centre of smaller disc.
$\Big(\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2}{\text{m}_1+\text{m}_2},\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2}{\text{m}_1+\text{m}_2}\Big)$ $\text{x}_1=\text{R},\ \text{y}_1=0$ $\text{x}_2=0,\ \text{y}_2=0$ $\Big(\frac{\pi\text{R}^2\text{T}\rho\text{R}+0}{\pi\text{R}^2\text{T}\rho+\pi(2\text{R})^2\text{T}\rho},\frac{0}{\text{m}_1+\text{m}_2}\Big)=\Big(\frac{\pi\text{R}^2\text{T}\rho\text{R}}{5\pi\text{R}^2\text{T}\rho},0\Big)=\Big( \frac{\text{R}}{5},0\Big)$ At $\frac{\text{R}}5{}$ from the centre of bigger disc towards the centre of smaller disc.
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