Centre of Mass of a Hollow Cone

Result

The CM of a thin uniform hollow conical shell (open base) of height $h$ lies at $\frac{h}{3}$ from the base:

$y_{cm} = \frac{h}{3}$

Derivation

Let the cone have half-angle $\beta$ (so base radius $= h\tan\beta = R$). Place apex at origin, axis along $y$.

Divide into thin horizontal rings at height $y$ from apex:

• Ring radius: $r = y\tan\beta$
• Slant height element: $dl = dy/\cos\beta$
• Ring circumference: $2\pi y\tan\beta$
• Ring area: $2\pi y\tan\beta \cdot \frac{dy}{\cos\beta}$
• Mass: $dm = \sigma \cdot \frac{2\pi y\tan\beta}{\cos\beta} \, dy$

Total slant surface area: $\pi R l = \pi R \frac{h}{\cos\beta}$ where $l = \frac{h}{\cos\beta}$ is the slant height.

So $\sigma = \frac{M}{\pi R h/\cos\beta} = \frac{M\cos\beta}{\pi Rh}$

Height above base: $y_{base} = h - y$ (since $y$ is measured from apex)

$y_{cm,base} = \frac{1}{M}\int_0^h (h-y) \cdot \sigma \cdot \frac{2\pi y\tan\beta}{\cos\beta} \, dy$

Using $\tan\beta/\cos\beta = R/h \cdot 1/\cos^2\beta$... this gets messy. Use a cleaner variable: let $y$ be height above base (0 at base, $h$ at apex).

At height $y$ above base, ring radius $= R(1 - y/h)$ (linearly decreasing to 0 at apex).

Ring slant area: $2\pi R(1-y/h) \cdot \frac{dy}{\cos\beta}$

Surface density: $\sigma = \frac{M}{\pi R \cdot h/\cos\beta} = \frac{M\cos\beta}{\pi Rh}$

$dm = \frac{M\cos\beta}{\pi Rh} \cdot \frac{2\pi R(1-y/h)}{\cos\beta} dy = \frac{2M}{h}\left(1 - \frac{y}{h}\right) dy$

$y_{cm} = \frac{1}{M}\int_0^h y \cdot \frac{2M}{h}\left(1-\frac{y}{h}\right) dy = \frac{2}{h}\int_0^h\left(y - \frac{y^2}{h}\right) dy$

$= \frac{2}{h}\left[\frac{h^2}{2} - \frac{h^2}{3}\right] = \frac{2}{h} \cdot \frac{h^2}{6} = \frac{h}{3}$

$\boxed{y_{cm} = \frac{h}{3}}$

Note on the $h/3$ result

Both the triangular lamina and the hollow cone give CM at $h/3$ from the base. This is not coincidence — both have mass distributions that decrease linearly to zero at the apex, giving the same weighted average.