Question

Find the centre of mass of a uniform:
(a) Half-disc
(b) Quarter-disc

Answer 1

a)

Given,

Mass of half disc = $M$

Surface charge density be $\left(\sigma \right)=\frac{2M}{\pi {\alpha }^2}$

Mass of the element $\left(dM\right)=\sigma dA=\sigma \pi rdr$

As disc is symmetric about $x-axis,so{x}_{CM}=0$

Here $y-$ coordinate of centre of mass of ring of radius $risy=\frac{2r}{\pi }$

$y-$ coordinate of half-disc:

$y_{CM}=\frac{\int ydM}{\int dM}$

$y_{CM}=\frac{{\int }_0^{\alpha }\left(\frac{2r}{\pi }\right)\left(\sigma \pi rdr\right)}{{\int }_0^{\alpha }\left(\sigma \pi rdr\right)}$

$y_{CM}=\left(\frac{2}{\pi }\right)\frac{{\int }_0^{\alpha }\left(r^{2}dr\right)}{{\int }_0^{\alpha }\left(rdr\right)}$

$y_{CM}=\frac{4\alpha }{3\pi }$

Final Answer: $0,\frac{4\alpha }{3\pi }$



(b)



Given,

Mass of quarter disc $=M$

Surface charge density be $\left(\sigma \right)=\frac{4M}{\pi {\alpha }^2}$

Mass of the element $\left(dM\right)=\sigma dA=\sigma \left(\frac{\pi r}{2}\right)dr$

Here centre of mass of ring element, $x=y=\frac{2r}{\pi }$

$x-$ coodinate of quarter -disc:

$x_{CM}=\frac{\int xdM}{\int dM}$

$x_{CM}=\frac{{\int }_0^{\alpha }\left(\frac{2r}{\pi }\right)\sigma \left(\frac{\pi r}{2}\right)dr}{{\int }_0^{\alpha }\left(\sigma \left(\frac{\pi r}{2}\right)dr\right)}$

$x_{CM=\left(\frac{2}{\pi }\right)}\frac{{\int }_0^{\alpha }\left(r^{2}dr\right)}{{\int }_0^{\alpha }\left(rdr\right)}$

$x_{CM}=\frac{4\alpha }{3\pi }$

Same procedure will follow for $y-$ coordinate of mass of disc

$y_{CM}=\frac{\int ydM}{\int dM}=\frac{{\int }_0^{\alpha }\left(\frac{2r}{\pi }\right)\sigma \left(\frac{\pi r}{2}\right)dr}{{\int }_0^{\alpha }\left(\sigma \left(\frac{\pi r}{2}\right)dr\right)}=\frac{4\alpha }{3\pi }$

Final Answer: $\frac{4\alpha }{3\pi },\frac{4\alpha }{3\pi }$