Centre Of Mass Of Semicircular Ring MCQ - Practice Questions with Answers

Have a look at the figure of semicircular ring.

Since it is symmetrical about $y$-axis on both sides of the origin
So we can say that its $x_{cm}=0$
And it is $z_{cm}=0$ as the z-coordinate is zero for all particles of the semicircular ring.
Now, we will calculate its $y_{cm}$ which is given by

$$y_{cm}=\frac{\int y\cdot dm}{\int dm}$$

So, Take a small elemental are of mass dm at an angle $\theta$ from the x-direction.

Its angular width $\mathrm{d}\theta$
If the radius of the ring is $R$ then its $y$ coordinate will be $R\sin \theta$

So,

$$dm=\frac{M}{\pi R}\ast Rd\theta =\frac{M}{\pi }d\theta$$

$As$,

$$y_{\mathrm{cm}}=\frac{\int y\cdot dm}{\int dm}$$

So,

$$y_{cm}=\frac{{\int }_{\pi }^0\frac{M}{\pi R}\ast R\ast R\sin \theta d\theta }{M}=\frac{R}{\pi }{\int }_{\pi }^0\sin \theta d\theta =\frac{2R}{\pi }$$