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 O (0, 0) of the bigger disc be the origin.



Radius of bigger disc = 2R
Radius of smaller disc = R

Now,
m₁ = μR × T × ρ
m2 = μ (2R)² × T × ρ

where T is the thickness of the two discs, and
ρ is the density of the two discs.

Position of the centre of mass is calculated as:

$$\begin{gathered} x_1=\mathrm{R},y_1=0 \\ {x}_2=0,y_2=0 \end{gathered}$$

$$\begin{gathered} =\left(\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2},\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}\right) \\ =\left(\frac{{\mathrm{\pi R}}^2\mathrm{T\rho R}+0}{{\mathrm{\mu R}}^2\mathrm{T\rho R}+\mathrm{\mu }(2\mathrm{R}{)}^2\mathrm{T\rho }},\frac{0}{m_1+m_2}\right) \\ =\left(\frac{{\mathrm{\pi R}}^2\mathrm{T\rho R}}{5{\mathrm{\mu R}}^2\mathrm{T\rho }},0\right)=\left(\frac{\mathrm{R}}{5},0\right) \end{gathered}$$

$\Rightarrow$The centre of mass lies at distance $\frac{\mathrm{R}}{5}$ from the centre of bigger disc, towards the centre of smaller disc.

Hence, the centre of mass of the system is $\left(\frac{\mathrm{R}}{5},0\right)$.