How to verify centre of mass of a solid hemisphere?

Question

I know that the centre of mass for a solid hemisphere is 3/8 R. Whereas for a hollow hemisphere it is 1/2 R. By intuition that a solid hemisphere is made of infinite number of hemispheres filling out the largest one, I am tempted to think that the CoM of a solid hemisphere would remain at R/2 (though it's not correct). So I want to verify the answer by slicing the hemisphere at 3/8 R from the flat surface leaving just a point in the centre and then hanging this system of two chunks of masses connected at the CoM point by a thread. My calculations tell me that the two masses would be $0.357 \pi$ and $0.31\pi$ respectively. However their moments $0.357\pi * 3/8$ and $0.31\pi * 5/8$ are not equal, indicating that this mass system will not balance out. What mistake am I making in the verification?

Answer 1

No, not like that. Your hollow hemispheres filling the solid one have a smaller radius $r$, the height of their center of mass is $r/2$ and their mass is proportional to their surface, i.e. proportional to $r^2$. The weighted average is $$\frac{\int^R_0\frac{r}2\cdot r^2\,dr}{\int^R_0r^2\,dr}=\frac38R,$$ as it should be.

Answer 2

First method

We can express the center of mass as $$ z_c=\frac{\iiint_V \rho(x,y,z) z\,\mathrm dV}{\iiint_V \rho(x,y,z) \,\mathrm dV} $$ assuming that the hemisphere is of uniform density, so we can take the constant function out of the integral and we can then cancel out the density factor from the mass and plug in the volume of a hemisphere $$ z_c=\frac{\rho}{M}\iiint_V z\,\mathrm dV=\frac{3}{2\pi R^3}\iiint_V z\,\mathrm dV $$ In spherical coordinates,where $r$ represents the radius,$\phi$ is the angle between the point and the $z-$axis, and $\theta$ is the azimuthal angle, the integral becomes $$ \begin{align} z_c&=\frac{3}{2\pi R^3}\int_0^{\pi/2}\int_0^{2\pi}\int_0^R r^3\cos\phi\sin\phi \,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi\\ &=\frac{3}{16\pi R^3}\int_0^{\pi/2}\int_0^{2\pi} R^4\sin(2\phi) \,\mathrm d\theta\,\mathrm d\phi\\ &=\frac{3}{8 R^3}\int_0^{\pi/2} R^4\sin(2\phi) \,\mathrm d\phi\\ &=\frac{3}{16 R^3} 2R^4\\ &=\frac{3}{8 } R \end{align} $$

Second Method

The center of mass of a hemispherical shell of constant density and inner radius $R_i$ and outer radius $R$ can be found as before $$ \begin{align} z_c&=\frac{\int \rho z \,\mathrm dV}{\int \rho \,\mathrm dV} =\frac{\displaystyle\int_{0}^{\pi/2}\int_{0}^{2\pi} \int_{R_i}^{R} r^3\cos\phi \sin\phi\,\mathrm dr\,\mathrm d\theta \,\mathrm d\phi }{\displaystyle\int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{R_i}^{R} r^2 \sin\phi\,\mathrm dr \,\mathrm d\theta \,\mathrm d\phi} \end{align} $$

and we obtain $$ z_c=\frac{3}{8}\frac{R^4-R_i^4}{R^3-R_i^3} $$ and for $R_i\to 0$ we have $z_c=\frac{3}{8}R$.

Answer 3

You are calculating the moment incorrectly. Yes, you got the masses of the pieces OK, but not the moments. Why would you multiply by the length of the radius contained in the piece?