Centre Of Mass Of Semicircular Disc - Practice Questions & MCQ

Have a look at the figure of semicircular disc

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

$
y_{\mathrm{cm}}=\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 for semicircular ring $y_{c m}=\frac{2 R}{\pi}$
So, for 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{4 R}{3 \pi}
\end{aligned}
$
 

Have a look at the figure of  semicircular annular ring

It has inner radius as $R_1$ and outer radius as $R_2$ and centre as O
Since it is symmetrical about $y$-axis on both sides of the origin
So we can say that its $x_{c m}=0$
And its $z_{c m}=0$ as z-coordinate is zero for all particles of 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 $\mathrm{dr}=\mathrm{dx}$

So y - coordinate of COM of an elemental ring is equal to $y_{c m_r}=\frac{2 r}{\pi}$
So, $^{y_{\mathrm{cm}}}=\frac{\int y d m}{\int d m}=\frac{\int \frac{2 r}{\pi} * d m}{M}$

As

$$
\begin{aligned}
& \sigma=\frac{\text { mass }}{\text { area }}=\frac{M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)} \\
& \text { So } d m=\sigma d A=\frac{M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)} *(\pi r * d r) \\
& y_{c m}=\frac{\int \frac{2 r}{\pi} * d m}{M}=\frac{\int \frac{2 r}{\pi} \sigma *(\pi r * d r)}{M}=\frac{\int 2 r * \sigma * r * d r}{M} \\
& y_{c m}=\frac{\frac{2 M}{\frac{\pi}{2} *\left(R_2^2-R_1^2\right)}}{M} * \int_{R_1}^{R_2} r^2 d r=\frac{4}{\pi\left(R_2^2-R_1^2\right)} * \frac{\left(R_2^3-R_1^3\right)}{3} \\
& y_{c m}=\frac{4}{3 \pi} * \frac{\left(R_2^3-R_1^3\right)}{\left(R_2^2-R_1^2\right)}
\end{aligned}
$$