Centre of Mass of a Solid Cone

Result

The CM of a uniform solid cone of height $h$ and base radius $R$ lies at $\frac{h}{4}$ from the base:

$y_{cm} = \frac{h}{4}$

Derivation

Divide the cone into thin horizontal discs. Taking $y$ as height above the base (0 at base, $h$ at apex):

At height $y$, the disc has radius $r = R\left(1 - \frac{y}{h}\right)$ (linear decrease from $R$ at base to $0$ at apex).

Disc volume: $\pi r^2 \, dy = \pi R^2\left(1-\frac{y}{h}\right)^2 dy$

Mass: $dm = \rho \pi R^2\left(1-\frac{y}{h}\right)^2 dy$ where $\rho = \frac{M}{\frac{1}{3}\pi R^2 h} = \frac{3M}{\pi R^2 h}$

$y_{cm} = \frac{1}{M}\int_0^h y \cdot \frac{3M}{\pi R^2 h} \cdot \pi R^2\left(1-\frac{y}{h}\right)^2 dy$

$= \frac{3}{h}\int_0^h y\left(1-\frac{y}{h}\right)^2 dy$

Expand $\left(1-\frac{y}{h}\right)^2 = 1 - \frac{2y}{h} + \frac{y^2}{h^2}$:

$= \frac{3}{h}\int_0^h \left(y - \frac{2y^2}{h} + \frac{y^3}{h^2}\right) dy$

$= \frac{3}{h}\left[\frac{h^2}{2} - \frac{2h^2}{3} + \frac{h^2}{4}\right] = \frac{3}{h} \cdot h^2\left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right)$

$= 3h\left(\frac{6 - 8 + 3}{12}\right) = 3h \cdot \frac{1}{12} = \frac{h}{4}$

$\boxed{y_{cm} = \frac{h}{4}}$

Summary: cones and analogous shapes

The solid cone's CM is at $h/4$ — lower than the hollow cone's $h/3$ — because the solid has more volume near the base (wider discs) pulling the CM down.