2.4 Solid sphere

  Centre of Mass and Conservation of Linear Momentum

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

Volume of the solid hemisphere $(V) = \frac{{2\pi {R^3}}}{3}\quad ...(i)$

Mass per unit volume $\left( {{\lambda _V}} \right) = \frac{{3M}}{{2\pi {R^3}}}\quad ...(ii)$

Consider an infinitesimal small element in the shape of a cylinder of mass $dm$, radius $r$ and infinitesimal small thickness $dx$. Point $P$, which is the centre of the infinitesimal small cylinder and is at a vertical distance $x$ from the origin.

Co-ordinate of the centre of mass of the cylinder is at point $P$ i.e. $(0,x)$.

So, the position vector of the centre of mass $(COM)$ of point $P$ is ${\overrightarrow r _P} = x\widehat j$

Note: The small section at point $P$ is infinitesimally small, so the centre of mass of that section is assumed to be at point $P$ only.

Volume of the cylinder is, $$dV = \left( {\pi {r^2}} \right)dx\quad ...(iii)$$ Also, $$\begin{equation} \begin{aligned} dm = {\lambda _V}.\;dV \\ dm = {\lambda _V}\pi {r^2}dx\quad ...(iv) \\\end{aligned} \end{equation} $$

Therefore the centre of mass $(COM)$ of the solid hemisphere is given by, $${\overrightarrow r _{COM}} = \frac{{\int {{{\overrightarrow r }_P}dm} }}{{\int {dm} }} = \frac{{\int {xdm} }}{{\int {dm} }}\widehat j\quad ...(v)$$

From equation $(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int {x{\lambda _V}} \pi {r^2}dx}}{{\int {{\lambda _V}\pi {r^2}dx} }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{\int {x{r^2}dx} }}{{\int {{r^2}dx} }}\widehat j\quad ...(vi) \\\end{aligned} \end{equation} $$

Relation between $x$ and $r$

$x$, $r$ & $R$ are the sides of a right angled triangle $\Delta OPQ$

So, for a right angle triangle we can write, $${R^2} = {r^2} + {x^2}$$ or $${r^2} = {R^2} - {x^2}\quad ...(vii)$$

From equation $(vi)$ & $(vii)$ we get, $${\overrightarrow r _{COM}} = \frac{{\int {x\left( {{R^2} - {x^2}} \right)dx} }}{{\int {\left( {{R^2} - {x^2}} \right)dx} }}\widehat j$$

Integrating for the solid hemisphere from point $O$ to $A$,

At point $O$, $x=0$ & at point $A$, $x=R$, so the limit becomes, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int\limits_0^R {x\left( {{R^2} - {x^2}} \right)dx} }}{{\int\limits_0^R {\left( {{R^2} - {x^2}} \right)dx} }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{\left[ {\frac{{{R^2}{x^2}}}{2} - \frac{{{x^4}}}{4}} \right]_0^R}}{{\left[ {{R^2}x - \frac{{{x^3}}}{3}} \right]_0^R}}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{3R}}{8}\widehat j \\\end{aligned} \end{equation} $$

So, the centre of mass $(COM)$ of the uniform solid hemisphere is $\left( {0,\frac{{3R}}{8}} \right)$