2.5 Hemispherical Shell

  Centre of Mass and Conservation of Linear Momentum

1.2.5 Hemispherical shell

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

Area of the hemispherical shell $(A) = 2\pi {R^2}\quad ...(i)$

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

Consider an infinitesimal small element in the shape of a circular ring of mass $dm$, radius $r$ and infinitesimal small thickness $dx$.

Point $P$ is the centre of the infinitesimal small circular ring.

Centre of mass of the circular ring is at point $P$ i.e. $\left( {0,R\sin \theta } \right)$.

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

Note: The thickness of the ring at point $P$ is infinitesimally small, so its centre of mass is assumed to be at point $P$ only

Area of the ring, $$dA = \left( {2\pi r} \right)dx$$ As, $$\begin{equation} \begin{aligned} dx = Rd\theta \quad ,\quad r = R\cos \theta \\ dA = \left( {2\pi R\cos \theta } \right)d\theta = 2\pi {R^2}\cos \theta d\theta \quad ...(iii) \\\end{aligned} \end{equation} $$ Also, $$\begin{equation} \begin{aligned} dm = {\lambda _A}dA \\ dm = {\lambda _A}2\pi {R^2}\cos \theta d\theta \quad ...(iv) \\\end{aligned} \end{equation} $$

Therefore, the centre of mass $(COM)$ of the hemispherical shell is given by, $${\overrightarrow r _{COM}} = \frac{{\int {{{\overrightarrow r }_P}dm} }}{{\int {dm} }} = \frac{{\int {R\sin \theta dm} }}{{\int {dm} }}\widehat j\quad ...(v)$$

From equation $(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int {R\sin \theta {\lambda _A}2\pi {R^2}\cos \theta d\theta } }}{{\int {{\lambda _A}2\pi {R^2}\cos \theta d\theta } }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{\int {R\sin \theta \cos \theta d\theta } }}{{\int {\cos \theta d\theta } }}\widehat j\quad ...(vi) \\\end{aligned} \end{equation} $$

Integrating the hemispherical shell from point $B$ to $A$,

At point $B$, $\theta = 0$ & at point $A$, $\theta = \frac{\pi }{2}$, so the limit becomes, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int\limits_0^{\frac{\pi }{2}} {R\sin \theta \cos \theta d\theta } }}{{\int\limits_0^{\frac{\pi }{2}} {\cos \theta d\theta } }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{R}{2}\widehat j \\\end{aligned} \end{equation} $$

So, the centre of mass $(COM)$ of the uniform hemispherical shell is $\left( {0,\frac{R}{2}} \right)$.