2.3 Semicircular disc

  Centre of Mass and Conservation of Linear Momentum

Consider a uniform semicircular disc of radius $R$ and mass $M$, whose centre is at the origin of a cartesian co-ordinate (for simplicity).

Area of the semicircular disc $(A) = \frac{{\pi {R^2}}}{2}\quad ...(i)$

Mass per unit area $\left( {{\lambda _A}} \right) = \frac{{2M}}{{\pi {R^2}}}\quad ...(ii)$

Consider a ring of radius $s$ and infinitesimal small thickness $ds$ having its centre at the origin and mass $dm$ at point $P$ on the disc.

Co-ordinate of the centre of mass of the ring passing through point $P$ is $\left( {0,\frac{{2s}}{\pi }} \right)$.

Co-ordinate of point $P$ is $(R\sin \theta ,\;R\cos \theta )$

So, the position vector of the centre of mass $(COM)$ of ring passing through point $P$ is ${\overrightarrow r _{CO{M_P}}} = \frac{{2s}}{\pi }\widehat j$

Area of the ring is, $$dA = \left( {\pi s} \right)ds\quad ...(iii)$$ Also, $$\begin{equation} \begin{aligned} dm = {\lambda _A}.\;dA \\ dm = {\lambda _A}\pi sds\quad ...(iv) \\\end{aligned} \end{equation} $$

Therefore, the centre of mass $(COM)$ of the semicircular disc is given by, $${\overrightarrow r _{COM}} = \frac{{\int {{{\overrightarrow r }_{CO{M_P}}}dm} }}{{\int {dm} }} = \frac{{\int {\frac{{2s}}{\pi }dm} }}{{\int {dm} }}\widehat j\quad ...(v)$$

From equation $(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}}\frac{{\int {\frac{{2s}}{\pi }{\lambda _A}\pi sds} }}{{\int {{\lambda _A}\pi sds} }}\widehat j \\ {\overrightarrow r _{COM}}\frac{2}{\pi }\frac{{\int {{s^2}ds} }}{{\int {sds} }}\widehat j \\\end{aligned} \end{equation} $$

Integrating for the semicircular disc from point $O$ to $A$,

At point $O$, $s=0$ & at point $A$, $s=R$, so the limit becomes, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}}\frac{2}{\pi }\frac{{\int {{s^2}ds} }}{{\int {sds} }}\widehat j \\ {\overrightarrow r _{COM}}\frac{2}{\pi }\frac{{\int\limits_0^R {{s^2}ds} }}{{\int\limits_0^R {sds} }}\widehat j \\ {\overrightarrow r _{COM}}\frac{2}{\pi }\frac{{\left[ {\frac{{{s^3}}}{3}} \right]_0^R}}{{\left[ {\frac{{{s^2}}}{2}} \right]_0^R}}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{4R}}{{3\pi }}\widehat j \\\end{aligned} \end{equation} $$

So, the centre of mass $(COM)$ of the uniform semicircular disc is $\left( {0,\frac{{4R}}{{3\pi }}} \right)$.