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 $x_{c m}=0$
And it is $z_{c m}=0$ as the z-coordinate is zero for all particles of the semicircular ring.
Now, we will calculate its $y_{c m}$ which is given by

$$
y_{c m}=\frac{\int y \cdot d m}{\int d m}
$$

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,

$$
d m=\frac{M}{\pi R} * R d \theta=\frac{M}{\pi} d \theta
$$

$A s$,

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

So,

$$
y_{c m}=\frac{\int_\pi^0 \frac{M}{\pi R} * R * R \sin \theta d \theta}{M}=\frac{R}{\pi} \int_\pi^0 \sin \theta d \theta=\frac{2 R}{\pi}
$$