Centre of Mass of a Solid Hemisphere

Result

The CM of a uniform solid hemisphere of radius $R$ lies at:

$y_{cm} = \frac{3R}{8}$

Derivation

Divide the hemisphere into thin horizontal discs.

At height $y$ above the base, the disc has:

• Radius: $r = \sqrt{R^2 - y^2}$ (from the sphere equation $x^2 + y^2 + z^2 = R^2$ at height $y$, so $r^2 = R^2 - y^2$)
• Thickness: $dy$
• Volume: $\pi r^2 \, dy = \pi(R^2 - y^2) \, dy$
• Mass: $dm = \rho \pi(R^2 - y^2) \, dy$ where $\rho = \frac{M}{(2/3)\pi R^3} = \frac{3M}{2\pi R^3}$

$y_{cm} = \frac{1}{M}\int_0^R y \cdot \rho\pi(R^2 - y^2) \, dy$

$= \frac{\rho\pi}{M}\int_0^R (R^2 y - y^3) \, dy$

$= \frac{\rho\pi}{M}\left[\frac{R^2 y^2}{2} - \frac{y^4}{4}\right]_0^R = \frac{\rho\pi}{M}\left(\frac{R^4}{2} - \frac{R^4}{4}\right) = \frac{\rho\pi R^4}{4M}$

Substituting $\rho = \frac{3M}{2\pi R^3}$:

$y_{cm} = \frac{3M}{2\pi R^3} \cdot \frac{\pi R^4}{4M} = \frac{3R}{8}$

$\boxed{y_{cm} = \frac{3R}{8}}$

Summary of hemisphere results

The solid hemisphere's CM is lower because it contains mass throughout its volume, including near the flat base where $y$ is small.