Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

Let, mass of entire disc $=M$
Mass per unit area $=\frac{M}{\pi (2R{)}^2}=\frac{M}{4\pi R^2}$
Mass of removed disc of radius $R$
$M_1=\frac{M}{4\pi R^2}\times R^2=\frac{M}{4}$
Mass of remaining disc,
$\Rightarrow M_2=M-\frac{M}{4}=\frac{3M}{4}$
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 $C(0,0)$.
$\Rightarrow \frac{M_1{x}_1-M_2{x}_2}{M_1-M_2}=0$
$\Rightarrow M_1{x}_1=M_2{x}_2$
$\frac{M}{4}\cdot R=\frac{3M}{4}(\alpha R)$
$\Rightarrow \alpha =\frac{1}{3}$