2.10 Hollow Cone

  Centre of Mass and Conservation of Linear Momentum

Consider a uniform solid cone of radius $R$, height $H$ & slant height $l$, whose centre is at the origin of a cartesian co-ordinate (for simplicity).

Area of the hollow cone $(A) = \pi Rl\quad ...(i)$

Mass per unit volume $({\lambda _A}) = \frac{{M}}{{\pi Rl}}\quad ...(ii)$

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

Point $P$, which is the centre of the infinitesimal small circular ring, and is at a vertical distance $y$ from the origin.

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

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

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.

Area of the circular ring is, $$dA = \left( {2\pi r} \right)dy\quad ...(iii)$$ Also, $$\begin{equation} \begin{aligned} dm = {\lambda _A}dA \\ dm = {\lambda _A}2\pi rdy\quad ...(iv) \\\end{aligned} \end{equation} $$

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

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

Relation between $r$ & $y$

As we know that $\Delta AOB$ & $\Delta APQ$ are similar.

Mathematically, $$\Delta AOB \approx \Delta APQ$$ Therefore, $$\frac{H}{R} = \frac{{H - y}}{r}$$ So, $$r = \frac{{R(H - y)}}{H}\quad ...(vii)$$

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

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

At point $O$, $y=0$ & at point $A$, $y=H$, so the limit becomes,

$$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int\limits_0^H {y\frac{{R(H - y)}}{H}dy} }}{{\int\limits_0^H {\frac{{R(H - y)}}{H}dy} }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{\left[ {\frac{{{y^2}H}}{2} - \frac{{{y^3}}}{3}} \right]_0^H}}{{\left[ {Hy - \frac{{{y^2}}}{2}} \right]_0^H}}\widehat j \\ {\overrightarrow r _{COM}} = \frac{H}{3}\widehat j \\\end{aligned} \end{equation} $$

So, the centre of mass $(COM)$ of the uniform hollow cone is $\left( {0,\frac{H}{3}} \right)$.