# How to verify centre of mass of a solid hemisphere?

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?

• Considering the hemisphere to be of unit radius Commented Oct 8, 2017 at 11:28

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.

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 $z_c=\frac{3}{8}R$.

• Thank you @alexjo for detailing out nicely the 3-D integration. I know the answer is 3/8 R. But the whole idea is to find a method to verify it. Presently, I am unable to correctly calculate the moments of the two halves of hemisphere, if it were to be suspended by a thin thread at its CoM. Commented Oct 10, 2017 at 18:18

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?

• I know that the answer is $3/8R$. But I want to verify it (practically). I would therefore make a very thin cut to the hemisphere in a plane parallel to the flat surface at a distance 3/8 R from the centre of the bottom surface. If I leave just a small part near the CoM, so that I can suspend this hemisphere at that point (CoM). If the CoM is correct, then the suspended hemisphere's axis will remain parallel to ground plane. To verify this mathematically, the moments of the two parts must be same. Question is how to calculate the two moments? Commented Oct 10, 2017 at 18:09
• @J Pet: if I understand center of mass correctly, you can attach a wire to that point, drill a straight hole in any direction and suspend it through it, and the body will not want to rotate in any way. the moment of a piece of a body wr to any point is the same as the moment of a point of same mass, located at the center of mass of that piece of body. The moments of the pieces cut by a plane through the center of mass balance out. Commented Oct 10, 2017 at 18:18
• Tks @orangeskid. Presently I am trying to correctly calculate the moments. I have tried to find the new CoMs of the two parts and then using their position vectors from the original CoM for calculating moments. But still they do not turn out to be equal. Can you help in calcs? Commented Oct 10, 2017 at 18:26
• @J. Pet: did you factor in the masses of the pieces too? Commented Oct 10, 2017 at 18:28
• Yes of course. As you said before, I've calculated their individual masses correctly. Commented Oct 10, 2017 at 18:29