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For my own knowledge and to understand why. I am trying to convince myself that the center of mass for a rigid solid sphere is at the origin (0,0,0).

I begin with the basic definition of CM

$$\overrightarrow{r}_{cm}=\frac{\sum_i m_ir_i}{\sum_i m_i}$$

Turning this into an integral form where the body is continuous:

$$\overrightarrow{r}_{cm}=\frac{1}{M}\int r\space dm$$

To convert dm into something we can integrate:

I know that $dm=\rho \space dV$

I know the the volume of a sphere is: $V=\frac{4}{3} \pi r^3$

So $dV=4\pi r^2 \space dr$

Ultimately giving me an integral that I can integrate over the the entire slice of the sphere where the origin is at 0,0,0:

$$r_{cm}= \frac{4\pi \rho}{m}\int_{-r}^{r} r^3 \space dr$$

If you integrate this and try to solve it you find that $r_{cm}=0$ which is exactly what I want, but I would like to know if this is the proof that I need to convince myself that the center of mass on solid sphere is located at the center of the structure or did i stumble upon this by coincidence and I made a mistake somewhere along the line and just got lucky to get to the answer?

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    $\begingroup$ Your vector definition is inconsistent, what is the vector on the right side of the first two equations? It should be $\vec{r}_i$ in the first equation, but what about the when you get to the integral? What is $r^3$? $\endgroup$ – tmwilson26 Dec 9 '15 at 2:39
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Let's call the solid sphere $B$. We know intuitively that the center of mass has to be at zero because it is symmetric. For example, it is symmetric under reflection through the origin. So, if $-B$ is the reflection of $B$ through the origin then we know $-B=B$.

But how can we use this to prove that the center of mass is zero? Well lets call the center of mass $\newcommand{rcm}{\vec{r}_{cm}}\rcm$. Then $$\rcm = \int_B \vec{r} dm.$$ Now we know we should get our result from doing the symmetry identity so let's use that. We get $$\rcm = \int_{-B} \vec{r} dm.$$ But now we can do a change of variable ($\vec{r} \to -\vec{r}$) to get $$\rcm = \int_{B} -\vec{r} dm = -\int_{B} \vec{r} dm = -\rcm.$$ So since $\rcm = -\rcm$, the center of mass must be zero.

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If you're going to use spherical coordinates, you cannot integrate from $-r$ to $r$ and must integrate from $0$ to $r$. Also, the definition of center of mass should be a vector quantity, so you need to think carefully about how this integral can actually be done.

I don't think that spherical coordinates is a good idea, and your best bet is to solve for the center of mass in each of the cartesian components $x$, $y$, and $z$ separately. You can find this more easily for each component in cylindrical coordinates, where each mass that you choose is a disk of radius $R$ (not the same $r$ as the sphere) and thickness $dx$ and distance $x$ away from the $x$-axis. Find the mass of that disk and integrate over $x$.

As an aside, there are many different ways to have something come out to be zero, so it may be hard to know if you are really correct, especially because the symmetry of the situation will allow you to ignore things like the mass term in the denominator as long as you find equal mass components on both sides of all of the axes. If you want to prove this to yourself, you should consider moving the origin of the sphere to the point $(1,1,1)$ and see if you can solve for it to have the center of mass at that location. This will make the problem a bit more complicated, but if you show that you can be confident that your calculations are correct.

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The ball is completely symmetric around the centre. So the centre of mass cannot possibly be anywhere else. All the integrals will cancel since they go from positve R to negative R.

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  • $\begingroup$ I know that it has to be at the center of the sphere. I want to know how can I convince myself of that mathematically. $\endgroup$ – Uys of Spades Dec 9 '15 at 2:34
  • $\begingroup$ symmetry principles are mathematical. The integrand is invariant under reflection, but the integral goes from -R to +R so it has to cancel even without computing it. Oh, and, although the centre of mass is vectorial, each of its coordinates is computed separately, so the vectorial nature can be ignored. $\endgroup$ – joseph f. johnson Dec 9 '15 at 2:36
  • $\begingroup$ I have to disagree. $\int_{-r}^{r} r^2 \space dr \neq 0$ $\endgroup$ – Uys of Spades Dec 9 '15 at 2:38
  • $\begingroup$ To get the centre of mass, you have to multiply by the coordinate, x. /because of leverage). But what it multiplies is mass, which is always positive. So the integrand can not be an even function. As the other poster said, don't use spherical coordinates, use Cartesian coordinates $\endgroup$ – joseph f. johnson Dec 9 '15 at 2:39
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Method 1: Cartesian coordinates: Let $r$ be the radius of the solid sphere,

consider an elementary solid disc of thickness $dx$ at distance $x$ from the origin $(0, 0)$ now, the mass of the elementary disc $$dm=\rho\pi(r^2-x^2)\ dx$$ hence the distance of center ($\bar x$) of mass of sphere from the origin $(0, 0)$ is given as $$\bar x=\frac{\int x\ dm}{\int dm}=\frac{\int_{-r}^{r} x\rho\pi(r^2-x^2)\ dx}{\int_{-r}^{r} \rho\pi(r^2-x^2)\ dx }=0$$

Method 2: Polar coordinates:

consider an elementary solid disc subtending an angle $2\theta$ at the origin $(0, 0)$ & having small axial thickness $Rd\theta \sin \theta$. Now, the mass of the elementary disc $$dm=\rho\pi(r\sin\theta)^2rd\theta\sin\theta=\pi\rho r^3\sin^3\theta \ d\theta$$ hence the distance of center ($\bar x$) of mass of sphere from the origin $(0, 0)$ is given as $$\bar x=\frac{\int x\ dm}{\int dm}=\frac{\int_{0}^{\pi} (r\cos\theta)\pi\rho r^3\sin^3\theta \ d\theta}{\int_{0}^{\pi} \pi\rho r^3\sin^3\theta \ d\theta}=0$$

Hence, the C.M. of sphere lies at the center

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