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{\bf Question}

Show that the centre of mass of a right circular cone of radius
$r$ and height $h$ is situated at a distance $\ds\frac{h}{4}$ from
the base.

A missile is to be constructed from two parts. The propulsion unit
is effectively a cylinder of radius $r$, height $H$ and density
$\ds\frac{\omega}{2}$. The warhead is situated in the nose cone,
which is a right circular cone of radius $r$, height $h$ and
density $\omega$.

Show that the centre of mass of the missile is at a distance

$$\ds\frac{(£H^2+4Hh+h^2)}{2(3H+2h)}$$

from the base of the propulsion unit.

Hence show that if $H=h=h\ds\frac{5r}{4}$ and the missile is
placed on a perfectly rough slope inclined at an angle of less
than $\ds\frac{\pi}{4}$ between its axis of symmetry and the
horizontal, it is liable to topple over.

\medskip

{\bf Answer}

PICTURE \vspace{1in}

Density=$\omega$

By symmetry  centre of mass is along axis of symmetry of cone at
$(\bar{x},0)$

Elemental disc: mass$=\pi r(x)^2 \omega \partial{x}$

By similar triangles $\ds\frac{r}{h}=\ds\frac{r(x)}{x}$

So mass$=\pi x^2\ds\frac{r^2}{h^2}\omega \partial{x}$

moment of disc about $y$ axis\un{$=\pi\ds\frac{x^3r^2}{h^2}\omega
\partial{x}$}

Balance total moments:

$$M\bar{x}=\ds\int_0^h \ds\frac{\pi
x^3r^2\omega}{h^2}\partial{x}$$

where $M=$mass of come $=\ds\frac{\pi r^2\omega h}{3}$

so $\bar{x}=\ds\frac{\frac{\pi
r^2}{h^2}\omega[\frac{x^4}{4}]_0^h}{\frac{\pi r^2 \omega h}{3}}$

$$\un{\bar{x}=\ds\frac{3h}{4};\ \ \rm{i.e.}, \ds\frac{h}{4}\
\rm{from\ base}}$$

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\newpage
Let centre of mass of composite body be at $(\bar{x}_c,0)$ (by
symmetry).

Take moments about $y$ axis:

\begin{enumerate}
\item
moment cylinder=$\underbrace{\pi r^2 H \ds\frac{\omega}{2}} \times
\underbrace{\ds\frac{H}{2}}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ mass \ \ \ \ \ \ \ \ c
of m

\item
moment warhead=$\ds\frac{1}{3}\pi r^2 h \omega
\times\left(H+\ds\frac{h}{4}\right)$

\item
moment missile=$\left(\underbrace{\ds\frac{1}{3} \pi r^2 h
\omega}+\underbrace{\ds\frac{\pi r^2 H
\omega}{2}}\right)\bar{x}_c$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ mass warhead \ \ \ \ \
\ \ \ \ mass cylinder

\end{enumerate}
Now $(3)=(1)+(2)$

\begin{eqnarray*} \Rightarrow \left(\ds\frac{1}{3} \pi r^2 h
\omega+\ds\frac{\pi r^2 H \omega}{2}\right)\bar{x}_c & = &
\ds\frac{\pi r^2 H^2 \omega}{4}+\ds\frac{\pi r^2 h
\omega}{3}\left(H+\ds\frac{h}{4}\right)\\ \Rightarrow \bar{x}_c &
= & \ds\frac{\pi r^2
\omega(\frac{H^2}{4}+\frac{hH}{3}+\frac{h^2}{12})}{\pi r^2
\omega(\frac{h}{3}+\frac{H}{2})}\\ & = &
\un{\ds\frac{3H^2+4Hh+h^2}{2(2h+3H)}} \end{eqnarray*}

as required.

Take $H=h=\ds\frac{5r}{4} \Rightarrow
\bar{x}_c=\ds\frac{3h^2+4h^2+h^2}{2(2h+3h)}=\ds\frac{4h}{5}=r$

PICTURE \vspace{1in}

Assuming centre of mass $\approx$ centre of gravity, we have that
the critical point of toppling occurs when centre of mass lies
just outside base of missile,

i.e., $\tan\theta=\ds\frac{r}{r}=1 \Rightarrow
\un{\theta=\ds\frac{\pi}{4}}$

Any angle greater than this and the missile topples over.

\end{document}