Question:

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. 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

Updated On: May 21, 2024

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\frac{1}{6}$

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The Correct Option is B

Solution and Explanation

Let centre $O_{1}$ of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at $x$ -axis. Let $CM$ of new disc is at point $O _{2} $ where, $O_{1} O _{2}=\alpha R$ (given). here mass of cut off portion, $m=\frac{\pi(R)^{2} \cdot M}{\pi(2 R)^{2}}=\frac{M}{4}$ and position of its centre of mass, $O_{1} O_{1}=R$ Hence, for remaining part (new disc) $x_{ CM }=\alpha R =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} $ $=-\frac{R}{3}$ $ \Rightarrow \alpha=\frac{1}{3}$

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Concepts Used:

Center of Mass

The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.

The formula for the Centre of Mass:

Center of Gravity

The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.