Question

Find the center of mass of uniform semicircular ring of radius $R$ and mass $m$

Answer 1

Step. 1 Calculation of the center of mass at x-axis,

The mass of this element is $dm=\lambda Rd\theta$.

$R$ is the ring of radius.
The center of mass will be calculated along X-axis as well as Y-axis.
For this, we need to consider the distance of the mass$dm$.

Along the X-axis the distance will be $Rsin\theta$ and the distance of this mass $dm$ along the Y-axis will be $Rcos\theta$ .
Thus, the x-coordinate of the center of mass $x_{cm}$ will be:
$x_{cm}={\displaystyle \frac{1}{m}}\int \left(R\cos \theta \right)\lambda Rd\theta$
$\Rightarrow x_{cm}={\displaystyle \frac{1}{m}}\underset{0}{\overset{\pi }{\int }}\left(R\cos \theta \right)\lambda Rd\theta$
As $\theta$ lies between $\left[0,\pi \right]$
$\Rightarrow x_{cm}={\displaystyle \frac{\lambda R^2}{m}}\underset{0}{\overset{\pi }{\int }}\cos \theta d\theta$
$\Rightarrow x_{cm}={\displaystyle \frac{\lambda R^2}{m}}{\left[\sin \theta \right]}_0^{\pi }$
$\Rightarrow x_{cm}={\displaystyle \frac{\lambda R^2}{m}}\left[\sin \pi -\sin 0\right]$
$\therefore x_{cm}={\displaystyle \frac{\lambda R^2}{m}}\left[0\right]=0$

Step 2. Calculation of the center of mass at the y-axis,
The y-coordinate of the center is:
$y_{cm}={\displaystyle \frac{1}{m}}\int \left(Rsin\theta \right)\lambda Rd\theta$
$\Rightarrow y_{cm}={\displaystyle \frac{\lambda R^2}{m}}\underset{0}{\overset{\pi }{\int }}\left(sin\theta \right)d\theta$
As $\theta$ lies between $\left[0,\pi \right]$
$\Rightarrow y_{cm}={\displaystyle \frac{\lambda R^2}{m}}{\left[-\cos \theta \right]}_0^{\pi }$
$\Rightarrow x_{cm}={\displaystyle \frac{\lambda R^2}{m}}\left[-1\left(\cos \pi -\cos 0\right)\right]$
$\Rightarrow y_{cm}={\displaystyle \frac{\lambda R^2}{m}}\left[-1\left(-1-1\right)\right]$
$\Rightarrow y_{cm}={\displaystyle \frac{\lambda R^2}{m}}\times 2$
But $\lambda$ is mass per unit length, substituting$\lambda ={\displaystyle \frac{m}{\pi R}}$, we get
$\Rightarrow y_{cm}={\displaystyle \frac{m}{\pi R}}\times {\displaystyle \frac{R^2}{m}}\times 2$
$\therefore y_{cm}={\displaystyle \frac{2R}{\pi }}$
Hence the position of the center of mass of the semi-circular ring lies at $\left(0,{\displaystyle \frac{2R}{\pi }}\right).$