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

Have a look at the figure of the semicircular disc

Since it is symmetrical about the $y$-axis on both sides of the origin
So, we can say that its $x_{c m}=0$
And its $z_{c m}=0_{z \text {-coordinate is zero for all particles of the semicircular ring. }}$
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 ring of mass dm of radius x on the disc.

$$
d m=\frac{2 M}{\pi R^2} \pi x(d x)
$$

As we know semicircular ring $y_{c m}=\frac{2 R}{\pi}$
So, for the elemental ring $y$-coordinate is $y_{c m}=\frac{2 x}{\pi}$

So,

$$
\begin{aligned}
& y_{c m}=\frac{1}{M} \int_0^R\left(\frac{2 x}{\pi} d m\right) \\
& y_{c m}=\frac{1}{M} \int_0^R\left(\frac{4 M}{\pi R^2} x^2 d x\right) \\
& y_{c m}=\frac{3 R}{4 \pi}
\end{aligned}
$$
 

Have a look at the figure of  semicircular annular ring

It has an inner radius of $R_1$ an outer radius of $R_2$ and centre as 0
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's $z_{c m}=0$ 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 an elemental ring of radius r  and it has mass dm and thickness as dr=dx

So $y$-coordinate of COM of an elemental ring is equal to $y_{c m_r}=\frac{2 r}{\pi}$
$\mathrm{So}_{,} y_{c m}=\frac{\int y d m}{\int d m}=\frac{\int \frac{2 r}{\pi} * d m}{M}$
As $\sigma=\frac{\text { mass }}{\text { area }}=\frac{M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)}$
So $d m=\sigma d A=\frac{M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)} *(\pi r * d r)$
$y_{c m}=\frac{\int \frac{2 r}{\pi} * d m}{M}=\frac{\int \frac{2 r}{\pi} \sigma *(\pi r * d r)}{M}=\frac{\int 2 r * \sigma * r * d r}{M}$
$y_{c m}=\frac{\frac{2 M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)}}{M} * \int_{R_1}^{R_2} r^2 d r=\frac{4}{\pi\left(R_2^2-R_1^2\right)} * \frac{\left(R_2^3-R_1^3\right)}{3}$

$$
y_{c m}=\frac{4}{3 \pi} * \frac{\left(R_2^3-R_1^3\right)}{\left(R_2^2-R_1^2\right)}
$$