Question

A uniform semi-circular ring of mass $M$ and radius $R$ is lying as shown. The coordinates of the centre of mass is

• A. $(0,\frac{R}{\pi })$
• B. $(0,R\pi )$
• C. $(0,\frac{R}{2})$
• D. $(0,\frac{2R}{\pi })$

Answer 1

The correct option is D $(0,\frac{2R}{\pi })$


Let us choose an element of length $Rd\theta$ on the circumference of the semicircle.

As the wire is uniform, the mass per unit length of the wire is $\frac{M}{\pi R}$.

Thus, the mass of the element is,

$dm=\left(\frac{M}{\pi R}\right)\left(Rd\theta \right)=\frac{M}{\pi }d\theta$

So, the $x-$coordinates of the centre of mass is given by

$x=\frac{1}{M}\int xdm=\frac{1}{M}\underset{0}{\overset{\pi }{\int }}\left(R\cos \theta \right)\left(\frac{M}{\pi }\right)d\theta =0$

Similarly, for $y-$ coordinate,

$y=\frac{1}{M}\int ydm=\frac{1}{M}\underset{0}{\overset{\pi }{\int }}\left(R\sin \theta \right)\left(\frac{M}{\pi }\right)d\theta$

$y=\frac{R}{\pi }\underset{0}{\overset{\pi }{\int }}\sin \theta =\frac{2R}{\pi }$

Hence, position of center of mass is $(0,\frac{2R}{\pi })$.

Therefore, option $\left(D\right)$ is correct.

Why this question? This question will help the students to have a sound grip over the calculation of COM for mass distribution.