Question

Surface mass density of a semicircular disk varies with position as $\sigma ={\sigma }_0\frac{r}{R}$ where ${\sigma }_0$ is constant, $R$ is the radius of the disc and $r$ is measured from the centre of the disc. Find the COM of the disc.

• A. $(0,\frac{R}{2})$
• B. $(0,\frac{2R}{\pi })$
• C. $(0,\frac{3R}{2\pi })$
• D. $(0,\frac{4R}{3\pi })$

Answer 1

The correct option is C $(0,\frac{3R}{2\pi })$

Cut an elementary ring of radius ${}^{\prime }{r}^{\prime }$ and thickness ${}^{\prime }d{r}^{\prime }$.


Mass of elementary ring
$dm=\sigma \left(\pi rdr\right)$
$={\sigma }_0\frac{r}{R}\pi rdr$
$=\frac{{\sigma }_0\pi }{R}{r}^{2}dr$
Centre of mass of elementary ring $=(0,\frac{2r}{\pi })$
So, x coordinate of COM of elementary ring, $\overline{x}=0$
& $\overline{y}=\frac{\int ydm}{\int dm}=\frac{{\int }_0^R\frac{2r}{\pi }\frac{{\sigma }_0\pi }{R}{r}^{2}dr}{{\int }_0^R\frac{{\sigma }_0\pi }{R}{r}^{2}dr}$
$=\frac{\frac{2{\sigma }_0}{R}{\int }_0^R{r}^{3}dr}{\frac{{\sigma }_0\pi }{R}{\int }_0^R{r}^{2}dr}=\frac{2}{\pi }\frac{\frac{R^4}{4}}{\frac{R^3}{3}}=\frac{3R}{2\pi }$
So, coordinates of COM of semicircular disc
$(x_{com},y_{com})=(0,\frac{3R}{2\pi })$