Question

A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.

Answer 1

Let O be the origin of the system (smaller disc plus bigger disc).

The density of the rods is ρ.
The thickness of rods is T.

Let m₁ be the mass and R be the radius of the smaller disc, and
m₂ be the mass and 2R be the radius of the bigger disc.

According to the question, the smaller disc is cut out from the bigger disc.

From the figure given below, we can write:



$$\begin{gathered} m_1=\pi R^{2}T\rho \mathit{} \\ \mathrm{Position}\mathrm{of}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{smaller}\mathrm{disc}, \\ {x}_1=R,y_1=0 \\ {m}_2=\pi \mathit{(}2R{\mathit{)}}^{\mathit{2}}T\rho \\ \mathrm{Position}\mathrm{of}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{bigger}\mathrm{disc}, \\ {x}_2=0,y_2=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{system}, \\ {X}_{\mathrm{cm}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2},Y_{\mathrm{cm}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2} \\ \mathrm{Since}\mathrm{the}\mathrm{smaller}\mathrm{disc}\mathrm{is}\mathrm{removed}\mathrm{from}\mathrm{the}\mathrm{bigger}\mathrm{disc},\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{disc}\mathrm{is}\mathrm{taken}\mathrm{as}\mathrm{negative}(-m_{\mathit{1}}). \\ {X}_{\mathrm{cm}}=\frac{-\pi R^{2}T\rho R+0m_2}{-\pi R^{2}T\rho +\pi (2R{)}^{2}T\rho },Y_{\mathrm{cm}}=\frac{0+0}{m_1+m_2} \\ {X}_{\mathrm{cm}}=\frac{-\pi R^{2}T\rho R}{3\pi R^{2}T\rho },Y_{\mathrm{cm}}=0 \\ {X}_{\mathrm{cm}}=\frac{-R}{3},Y_{\mathrm{cm}}=0 \end{gathered}$$
Hence, the centre of mass of the system lies at distance $\frac{R}{3}$ from the centre of bigger disc, away from centre of the hole.