I haven't calculated center of mass before and I'd like to know how I can do it in practise.

I want to find the center of mass of a semi-sphere. Could you explain me, step by step, what I have to do?

Many thanks

  • $\begingroup$ By "semi-sphere", do you mean half a spherical solid or half a spherical shell? $\endgroup$ – dfeuer Sep 16 '13 at 16:22
  • $\begingroup$ @dfeuer a spherical solid $\endgroup$ – sunrise Sep 16 '13 at 16:40
  • $\begingroup$ @dfeuer: You could figure out the shell centre of mass from the solid centre of mass. $\endgroup$ – copper.hat Sep 16 '13 at 16:49
  • $\begingroup$ @copper.hat: you could, but it's best to be clear from the start, no? $\endgroup$ – dfeuer Sep 16 '13 at 17:35

We calculate the centre of mass of a half-ball of radius $1$. Without loss of generality we may assume that the ball is made of material with density $1$.

Imagine that the ball is sitting on a table, flat side down. By symmetry the centre of mass is on the vertical line through the centre of the ball. The only question is: How far up?

We will calculate the moment of the ball about the plane of the table, and divide by the mass of the half-ball. By a standard formula, the mass of the half-ball is $\dfrac{2\pi}{3}$.

Imagine now that the half-ball is an industrial ham. Imagine a very thin slice of that ham, sliced parallel to the table, but left in place. Let the slice be taken from height $z$ to height $z+dz$, where $dz$ is extremely small. The slice is almost a cylinder of very small height $dz$.

We first calculate the radius $r=r(z)$ of the slice. By the Pythagorean Theorem, we have $r^2+z^2=1$, so $r=\sqrt{1-z^2}$.

Thus the area of the slice is $\pi r^2=\pi(1-z^2)$. The thickness is $dz$, so the volume, and therefore the mass, of the slice is approximately $\pi (1-z^2)\,dz$.

The slice is at perpendicular distance $z$ from the table. So the moment of the slice about the plane of the table is approximately $\pi (1-z^2)(z)\,dz$.

"Add up" (integrate) from $z=0$ and $z=1$. The full moment of the ball is $$\int_0^1 \pi (1-z^2)(z)\,dz.$$ Calculate. We get $\dfrac{\pi}{4}$.

Finally, divide by the mass $\dfrac{2\pi}{3}$. We get $\dfrac{3}{8}$.

For a ball of radius $R$, just multiply by $R$. The centre of mass is $\dfrac{3 R}{8}$ above the centre of the half-ball.

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    $\begingroup$ Thanks, it would be nice to have one glitch-free post. Maybe next year. $\endgroup$ – André Nicolas Sep 16 '13 at 18:09
  • $\begingroup$ Anything non-trivial will not be glitch-free :-). $\endgroup$ – copper.hat Sep 16 '13 at 18:10
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    $\begingroup$ You are welcome. To locate the center of mass in $3$-space, we have to find the moments about $3$ planes. If we are working with coordinates, the usual choices are the $x$-$y$ plane, the $y$-$z$ plane, and the $x$-$z$ plane. But in our half-ball example, symmetry tells us that the centre of mass is on a certain line. So our task is much simplified: we only need the moment about any plane parallel to the base. Might as well use the plane that contains the base. The description with slices of infinitesimal thickness parallel to the plane is very useful for the intuition. (More) $\endgroup$ – André Nicolas Sep 16 '13 at 20:12
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    $\begingroup$ (Cont) Reasoning with slices of infinitesimal thickness can be replaced by more formal language. As a side remark, note that our calculation was needed for only one coordinate of the centroid. Symmetry took care of two, so symmetry was the more powerful tool. $\endgroup$ – André Nicolas Sep 16 '13 at 20:15
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    $\begingroup$ Yes, that's right. $\endgroup$ – André Nicolas Sep 16 '13 at 21:29

Let $f(x) = \sqrt{r^2-x^2}$ and model the hemisphere as $H=\{(x,y) | x \in [0,r], \, |y| \le f(x) \}$.

Then compute $\overline{x} = \frac{\int_0^r 2 \pi xf(x)^2 dx}{\int_0^r 2 \pi f(x)^2 dx}$.

These integrals are straightforward to evaluate:

$\int_0^r 2 \pi xf(x)^2 dx = 2 \pi \int_0^r x ( r^2-x^2) dx = 2 \pi \frac{r^4}{4}$.

$\int_0^r 2 \pi f(x)^2 dx = 2 \pi \int_0^r ( r^2-x^2) dx = 2 \pi \frac{2 r^3}{3}$.

$\overline{x} = \frac{3}{8} r$.

  • $\begingroup$ I haven't worked with similar things before. Could you show me, step by step, the entire proceeding? (I'm not a matematician!) $\endgroup$ – sunrise Sep 16 '13 at 16:43
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    $\begingroup$ This is homework. Evaluate the integral above. $\endgroup$ – copper.hat Sep 16 '13 at 16:51
  • $\begingroup$ This isn't an assignment (I don't go to school!). I'd want to understand how I have to proceed in similar cases. I said that I want to calculate the center of mass of a hemispere because I have started from this object. If you don't believe me, you can show me the proceeding step by step using another solid.. please, choose the solid that you prefer. I don't know the mathematical symbolism that you have used and I haven't "modelled" anything before. $\endgroup$ – sunrise Sep 16 '13 at 17:06
  • $\begingroup$ Evaluating the above integrals is pretty straightforward. What exactly are you having a problem with? If there is a specific problem, I can help, but I am not going to provide basic material than can be found in most relevant textbooks. $\endgroup$ – copper.hat Sep 16 '13 at 17:13
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    $\begingroup$ @RonGordon Many thanks for your tolerance! $\endgroup$ – sunrise Sep 16 '13 at 19:50

1- Assume the flat surface is located on a horizontal plane with the bowl (semi-sphere) below) 2- X = left/right 3- Z = frt/back 4- Y = vertical offset

Both X & Z coordinates will be on the axis of rotation, i.e. at 0 offset. The center of mass will be .1875 * D from the flat surface of the semi-sphere.


  • $\begingroup$ Please consider formatting your answer in LateX properly. $\endgroup$ – Vishal Gupta Sep 16 '13 at 17:06

$a > 0$: radius.

\begin{align} \vec{r}_{\rm cm} &\equiv {1 \over V}\int_{V}\vec{r}\,{\rm d}V = {1 \over \left(4\pi a^{3}/3\right)/2}\int_{V}{1 \over 2}\nabla r^{2}\,{\rm d}V = {3 \over 4\pi a^{3}}\int_{S}r^{2}\,\hat{r}\,{\rm d}S \\[3mm]&= {3 \over 4\pi a^{3}}\left[% \int_{\Huge\frown}a^{2}\,{z \over a}\,\hat{z}\,{\rm d}S\ +\ \int_{\Huge\_}\left(x^{2} + y^{2}\right)\left(-\hat{z}\right)\,{\rm d}x\,{\rm d}y \right] \\[3mm]&= {3 \over 4\pi a^{3}}\,\hat{z}\left\lbrack% a^{2}\int_{0}^{2\pi} \int_{0}^{\pi/2}\cos\left(\theta\right)\ a^{2}\sin\left(\theta\right)\,{\rm d}\theta\,{\rm d}\phi\ -\ \int_{0}^{a}\rho^{3}\,{\rm d}\rho\int_{0}^{2\pi}{\rm d}\phi \right] \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \vec{r}_{\rm cm} = {3 \over 8}\,a\;\hat{z} \quad} \\ \\ \hline \end{array} $$


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