How to find the center of mass of half a torus?

Question

Consider a halved solid torus (half a donut). The radius of the torus are $R_1$ and $R_2$. I need to find its center of mass. The hint they give is that the center of mass of a homogeneous solid object $\Omega \subset \Bbb R^3$ is calculated as $$\overline{x}=\int_{\Omega}\overline{r}d\overline{r}.$$

I really don't understand this formula, I don't know what $\overline{r}$ means and what is $\Omega$ in this case. I'd appreciate that someone explains what this formula means and how to apply it in this problem. Thanks in advance.

Answer 1

I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass. Before we do so, however, I set my coordinate system up as follows. I have positive $x$ coming out of the screen, positive $y$ going to the right, and positive $z$ up. In cylindrical coordinates $(r,\phi,z)$:

$$x = r \cos{\phi}$$ $$y=r \sin{\phi}$$

where we have the limits defining the region $\Omega$:

$$\phi \in \left [ -\frac{\pi}{2},\frac{\pi}{2} \right]$$ $$z \in [-R_2,R_2]$$ $$r \in \left [ R_1 - \sqrt{R_2^2-z^2}, R_1 + \sqrt{R_2^2-z^2}\right]$$

Also, for an object of constant mass density, the expression for the $x$ component of the center of mass is

$$\bar{x} = \frac{\displaystyle \int_{\Omega} d^3 \vec{x} \, x}{\displaystyle \int_{\Omega} d^3 \vec{x}}$$

(Note that, by symmetry, we have $\bar{y}=0$ and $\bar{z}=0$.) Let's first evaluate the denominator:

$$\begin{align} \int_{\Omega} d^3 \vec{x} &= \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r \\ &= \frac{\pi}{2}\int_{-R_2}^{R_2} dz \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^2 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^2 \right ]\\ &= 4 \pi R_1 \int_0^{R_2} dz \, \sqrt{R_2^2-z^2}\\ &= \pi^2 R_1 R_2^2 \end{align}$$

Now we evaluate the center of mass:

$$\begin{align}\bar{x} &= \frac{1}{\pi^2 R_1 R_2^2} \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r^2 \cos{\phi} \\ &= \frac{4}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \, \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^3 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^3 \right ]\\ &= \frac{8}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \,\left [3 R_1^2 \sqrt{R_2^2-z^2} + \left (R_2^2-z^2 \right )^{3/2} \right ] \\ &= \frac{8}{3 \pi^2 R_1 R_2} \left ( \frac{3 \pi}{4} R_1^2 R_2 + \frac{3 \pi}{16} R_2^3 \right )\end{align}$$

Simplifying, I get

$$\bar{x} = \frac{4 R_1^2+R_2^2}{2 \pi R_1}$$

ADDENDUM

As a quick note, in the limits as $R_2 \to 0$, we find that the center of mass becomes

$$\bar{x}=\frac{2}{\pi} R_1$$

which agrees with the center of mass of a uniform wire bent into a semicircle.

Answer 2

This problem can be solved using an extension of the theorem of Pappus.

Consider a volume that can be generated by sweeping a plane figure, defined in an $rz$-plane, about the $z$-axis.

The formula for the $x$-coordinate of the centroid of a volume of revolution is:

$$\overline{x} = \frac{\sin\alpha}{\alpha} \left(\frac{I_z}{\overline{r}A}\right)$$

where $α$ denotes the swept angle in radians relative to the $x$-axis (i.e. the starting position of the sweep), while $A$ is the area, $\overline{r}$ the centroid and $I_z$ the second moment of area of the swept plane figure relative to the $z$-axis.

For the problem at hand, the centroid will be the same as that for a quarter torus swept from the plane of symmetry. Hence:

$$\overline{x} = \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} \left[\frac{\frac{1}{4}\pi R_2^4 + R_1^2 \pi R_2^2}{R_1 \pi R_2^2}\right] = \frac{4R_1^2 + R_2^2}{2 \pi R_1}$$

For further details please see: https://doi.org/10.1177/03064190231188650