Question

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.

Answer 1

Let the centre of the bigger disc be the origin.

2R=Radius of bigger disc

R=Radius of smaller disc

$m_1=\mu R\times T\times \rho$

$m_2=\mu (2R{)}^2\times T\times \rho$

where,T=Thickness of the two discs

$\rho$=Density of the two discs

$\therefore$ Position of the centre of mass

$=(\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2},\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2})$

$x_1=R,y_1=0$

$x_2=0,y_2=0$

$(\frac{\pi R^{2}T\rho R+0}{\mu R^{2}T\rho R+\mu (2R{)}^{2}T\rho },\frac{0}{m_1+m_2})$

$=(\frac{\pi R^{2}T\rho R}{5\mu R^{2}T\rho },0)=(\frac{R}{5},0)$

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