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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Solution

Let the centre of the bigger disc be the origin.

2R=Radius of bigger disc

R=Radius of smaller disc

$m_{1} = μ R \times T \times ρ$

$m_{2} = μ (2 R)^{2} \times T \times ρ$

where,T=Thickness of the two discs

$ρ$ =Density of the two discs

$∴$ Position of the centre of mass

$= (\frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}})$

$x_{1} = R, y_{1} = 0$

$x_{2} = 0, y_{2} = 0$

$(\frac{π R^{2} T ρ R + 0}{μ R^{2} T ρ R + μ (2 R)^{2} T ρ}, \frac{0}{m_{1} + m_{2}})$

$= (\frac{π R^{2} T ρ R}{5 μ R^{2} T ρ}, 0) = (\frac{R}{5}, 0)$

At $\frac{R}{5}$ from the centre of bigger disc towards the centre of the smaller disc.

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