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 $$\overline{x}=\int_{\Omega}\overline{r}d\overline{r}.$$

half a torus

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.

  • $\begingroup$ Place it in a fixed position in $x,y,z$ coordinates. The total mass of the figure is just the volume, call it $M.$ Integrate, for example, $x$ over the figure, getting some number, call it $X.$ The $x$ coordinate of the center of mass is $X/M.$ By symmetry, two out of three of the coordinates will be completely predictable, the third will need actual triple integrals. $\endgroup$ – Will Jagy Oct 19 '13 at 0:14
  • $\begingroup$ Likely best to make the full torus a solid of revolution about the $z$ axis, slice in half with something like $x \geq 0,$ then do the actual integrals with cylindrical coordinates. $\endgroup$ – Will Jagy Oct 19 '13 at 0:16

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

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


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.

  • 1
    $\begingroup$ sorry for my comment, but can you explain me why $r∈[R_1− √R_2^2−z^2,R_1+√R_2^2−z^2]$ $\endgroup$ – Rachel May 2 '14 at 23:59
  • $\begingroup$ @ Ron Gordon: center of mass of a half torus lies outside its bulk that's why half torus can be treated as a half circle for external center of mass that's why the center of mass of half torus is $\bar{x}=\frac{2}{\pi}R_1$ $\endgroup$ – jeanne clement Dec 10 '16 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.