Question

In the figure shown below, a quarter ring of radius $r$ is placed in the first quadrant of a cartesian co-ordinate system, with centre at origin. Find the co-ordinates of $COM$ of the quarter ring.

• A. $(\frac{2R}{\pi },\frac{2R}{\pi })$
• B. $(\frac{2R}{\pi },0)$
• C. $(0,\frac{2R}{\pi })$
• D. $(0,0)$

Answer 1

The correct option is A $(\frac{2R}{\pi },\frac{2R}{\pi })$

Let us consider a small mass ${}^{\prime }d{m}^{\prime }$ of elemental length ${}^{\prime }d{s}^{\prime }$ which is at ${}^{\prime }{x}^{\prime }\text{m}$ from $y-$ axis and ${}^{\prime }{y}^{\prime }\text{m}$ from $x-$ axis.


Mass per unit length $\lambda =\frac{dm}{ds}$
$\Rightarrow dm=\lambda ds$
$\Rightarrow dm=\lambda rd\theta$ $(\because ds=rd\theta )$

To find $COM$:
$x_{cm}=\frac{\int xdm}{\int dm}=\frac{\underset{0}{\overset{\pi /2}{\int }}\left(r\cos \theta \right)\left(\lambda rd\theta \right)}{\underset{0}{\overset{\pi /2}{\int }}\lambda rd\theta }$
$=\frac{r\underset{0}{\overset{\pi /2}{\int }}\cos \theta d\theta }{\underset{0}{\overset{\pi /2}{\int }}d\theta }=\frac{2R}{\pi }$
and $y_{cm}=\frac{\int ydm}{\int dm}=\frac{\underset{0}{\overset{\pi /2}{\int }}\left(r\sin \theta \right)\left(\lambda rd\theta \right)}{\underset{0}{\overset{\pi /2}{\int }}\lambda rd\theta }$
$=\frac{r\underset{0}{\overset{\pi /2}{\int }}\sin \theta d\theta }{\underset{0}{\overset{\pi /2}{\int }}d\theta }=\frac{2R}{\pi }$
$\Rightarrow (x_{cm},y_{cm})=(\frac{2R}{\pi },\frac{2R}{\pi })$