Center of mass of semi-sphere

Question

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

Answer 1

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.

Answer 2

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:

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

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

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

Answer 3

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.

RJ

Answer 4

$a > 0$: 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}

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

Answer 5

Archimedes might have done the calculation like this: he knew the volume of the half ball equals the difference between that of a circumscribed cylinder and an inverted cone inscribed in that cylinder, i.e. 2π/3 = π - π/3, because the area of a slice of the half ball is the difference of the areas of slices at the same height of the other two figures, again by Pythagoras. (This means the moments of the slices are related the same way.) He also knew the center of mass of the cylinder is at height 1/2 and that of the cone at 3/4. Thus, if the center of mass of the half ball is at height t, the total moments are related by (2π/3)t = π(1/2) - (π/3)(3/4), so t = 3/8.