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 $x_{cm}=0$
And its $z_{cm}=0_{z\text{-coordinate is zero for all particles of the semicircular ring. }}$
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 ring of mass dm of radius x on the disc.

$$dm=\frac{2M}{\pi R^2}\pi x(dx)$$

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

So,

$$\begin{array}{rl}& y_{cm}=\frac{1}{M}{\int }_0^R\left(\frac{2x}{\pi }dm\right)\\ & y_{cm}=\frac{1}{M}{\int }_0^R\left(\frac{4M}{\pi R^2}{x}^{2}dx\right)\\ & y_{cm}=\frac{3R}{4\pi }\end{array}$$
 

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 $x_{cm}=0$
And it's $z_{cm}=0$ 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 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 $y_{c{m}_r}=\frac{2r}{\pi }$
${\mathrm{So}}_{,}{y}_{cm}=\frac{\int ydm}{\int dm}=\frac{\int \frac{2r}{\pi }\ast dm}{M}$
As $\sigma =\frac{\text{ mass }}{\text{ area }}=\frac{M}{\frac{\pi }{2}\ast (R_2^2-R_1^2)}$
So $dm=\sigma dA=\frac{M}{\frac{\pi }{2}\ast (R_2^2-R_1^2)}\ast (\pi r\ast dr)$
$y_{cm}=\frac{\int \frac{2r}{\pi }\ast dm}{M}=\frac{\int \frac{2r}{\pi }\sigma \ast (\pi r\ast dr)}{M}=\frac{\int 2r\ast \sigma \ast r\ast dr}{M}$
$y_{cm}=\frac{\frac{2M}{\frac{\pi }{2}\ast (R_2^2-R_1^2)}}{M}\ast {\int }_{R_1}^{R_2}{r}^{2}dr=\frac{4}{\pi (R_2^2-R_1^2)}\ast \frac{(R_2^3-R_1^3)}{3}$

$$y_{cm}=\frac{4}{3\pi }\ast \frac{(R_2^3-R_1^3)}{(R_2^2-R_1^2)}$$