A circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2 R$. The centre…

A circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2 R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is
  1. $\frac{1}{2}$
  2. $\frac{1}{3}$
  3. $\frac{1}{4}$
  4. $\frac{1}{6}$

Solution


Let, mass of entire disc $=M$ Mass per unit area $=\frac{M}{\pi(2 R)^2}=\frac{M}{4 \pi R^2}$ Mass of removed disc of radius $R$, $ M_1=\frac{M}{4 \pi R^2} \times R^2=\frac{\mathrm{M}}{4} $ Mass of remaining disc, $ \Rightarrow \quad M_2=M-\frac{M}{4}=\frac{3 M}{4} $ Center of mass of removed disc is $C_1$ and centre of mass of remaining new disc is $C_2$. And centre of mass of combination of $M_1$ and $M_2$ will be at $C(0,0)$. $ \begin{aligned} \Rightarrow \quad & \frac{M_1 x_1-M_2 x_2}{M_1-M_2}=0 \\ \Rightarrow \quad & M_1 x_1=M_2 x_2 \\ & \frac{M}{4} \cdot R=\frac{3 M}{4}(\alpha R) \quad \Rightarrow \alpha=\frac{1}{3} \end{aligned} $

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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