Center of mass of a uniform sphere

In summary, the center of mass of a uniform sphere is at its center.However, the formula for the center of mass in terms of coordinate choice doesn't transform simplistically as described.Therefore, one must use a coordinate transformation to get the spherical coordinates of the C.M.
  • #1
ShayanJ
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We know,by symmetry,that the center of mass of a uniform sphere is at its center.So we expect the formula [itex] r_{com}=\frac{\int r \rho d\tau}{\int \rho d\tau} [/itex] to give us zero for this case.So let's see:
[itex]
r_{com}=\frac{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^3 \sin{\theta} d\phi d\theta dr}{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^2 \sin{\theta} d\phi d\theta dr}=\frac{\int_0^R r^3 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}{\int_0^R r^2 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}=\frac{\frac{1}{4}R^4\times 2\pi \times 2}{\frac{1}{3}R^3\times 2\pi \times 2}=\frac{3}{4}R

[/itex]!

What's going on?
 
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  • #2
What's going on is that that is not the equation for the center of mass.
 
  • #3
What's going on is that the center of mass formula in terms of coordinate choice doesn't transform in that simplistic manner.

Basically, you should look at how the formula for the CM in terms of (x,y,z)-coordinates treansform to polar, and verify to yourself that a) They specify the origin as the C.M, and b) None of the formulae looks the slightest bit like your attempt.
 
  • #4
Why would you expect that formula to hold?

The denominator is just the mass of the sphere.
The numerator is supposed to be the sum of the moments due to the mass elements.
What are you computing the moments about?

[edit] arildno is telling it more directly ;)
 
  • #5
But, you might ask:
What sort of weighted average do you get then by your approach?You are, as you readily can find out, averaging r over the function (rho)*4(pi)r^2, which is the mass per length of a spherical shell centered about the origin.
 
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  • #6
Can you tell me how to gain the center of mass formulas in spherical coordinates?
Because I don't think transforming the cartesian formulas to spherical coordinates gives the right answer,because we would have e.g.
[itex]
\overline{r\sin{\theta}\cos{\phi}}=\frac{1}{M}\int r \sin{\theta}\cos{\phi} r^2 \sin{\theta}d\phi d\theta dr
[/itex]!
 
  • #7
It sure is the right formula.
Take first the integral over the cos(phi) from 0 to 2*pi; what do you get?
 
  • #8
arildno said:
It sure is the right formula.
Take first the integral over the cos(phi) from 0 to 2*pi; what do you get?

I know the formula is right!
The point is,by transforming the three cartesian formulas for center of mass coordinates to spherical coordinate,one will get three equations which give you not sphercial coordinates of the center of mass but three functions of those coordinates.Oh...so looks like then the three equations should be solved simultanously to find the spherical coordinates of the center of mass!
I got it...thanks guys!
 
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  • #9
Still another point remains.
Center of mass is in fact the weighed average of the position of mass elements of the body under consideration,with the mass of every element being the weight.So the formula [itex] r_{com}=\frac{\int r \rho d\tau}{\int \rho d\tau}[/itex] seems reasonble.Why isn't it right?
And...What if you want to write the formula in spherical coordinates in the first place,without using cartesian ones?what would you do?

Thanks everyone
 
  • #10
"Center of mass is in fact the weighed average of the position of mass elements of the body under consideration,with the mass of every element being the weight"

It sure is.
But, then again, your scalar "r" is no position vector is it?
Nor is it the scalar factor in a variable-independent vector component, either, in contrast to "x" as in [itex]x\vec{i}[/itex], where [itex]\vec{i}[/itex] is independent of x,y and z.

Thus, for position vectors [itex]\vec{r}[/itex], we have the coordinate independent formulation:
[tex]\vec{r}_{C.M}=\frac{\int_{V}\vec{r}\rho(\vec{r})dV}{\int_{V}\rho(\vec{r})dV}[/tex]

Try as much as you like, you won't be able to reduce it into what you wrote, and that is why what you write doesn't have anything to do with the position of C.M.

Therefore, the correct formula for the C.M in polar coordinates for your uniform sphere is:
[tex]\vec{r}_{C.M}=\frac{\int_{V}\vec{i}_{r}\rho{r}^{3}\sin\theta {drd\theta{d}\phi}}{\int_{V}\rho{dV}}=\frac{3}{16\pi}R\int_{\theta,\phi}\vec{i}_{r} \sin\theta{d}\theta{d}\phi[/tex]

Your fallacy can then be seen to be reduced to the following misconception:
[tex]\vec{i}_{r}=||\vec{i}_{r}||={1}[/tex]
Or, something like that anyway.
 
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  • #11
Yah - we normally "cheat" by picking a symmetry line to do the integration along.
##r_{com}## is then the distance along this line from some reference point, also on the line.
 
  • #12
Simon Bridge said:
Yah - we normally "cheat" by picking a symmetry line to do the integration along.
##r_{com}## is then the distance along this line from some reference point, also on the line.
True enough.
But what I wrote highlights the particular conceptual flaw of OP rather well, I think.
And that was the main issue to be dealt with in this thread, not particular integration techniques.
But, and I think you are right that simplification techniques, like picking out a symmetry line, and where the vectorial formulation is thereby suppressed might lead to some misunderstandings that finding C.M is "essentially" a calculation technique involving scalars.
 
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  • #13
Sure - the problem wasn't with the integration but with what it means.
 

1. What is the definition of center of mass?

The center of mass of an object is a point where the mass of the object can be considered to be concentrated. It is the point at which the object would balance if it were placed on a pivot.

2. How is the center of mass of a uniform sphere calculated?

The center of mass of a uniform sphere can be calculated by finding the average position of all the mass of the sphere. This can be done by dividing the total mass of the sphere by the volume of the sphere.

3. Why is the center of mass important in physics?

The center of mass is important in physics because it helps us understand the motion and stability of objects. It is used to calculate the overall motion of an object and to determine how it will respond to external forces.

4. Does the center of mass of a uniform sphere change if it is cut into smaller pieces?

No, the center of mass of a uniform sphere does not change if it is cut into smaller pieces. This is because the center of mass is a property of the object itself and is not affected by its shape or size.

5. How does the center of mass of a uniform sphere affect its rotational motion?

The center of mass of a uniform sphere plays a crucial role in its rotational motion. If the sphere is rotating, the center of mass will remain in a fixed position, and the sphere will rotate around it. The distance between the center of mass and the axis of rotation also determines the moment of inertia of the sphere, which affects its rotational motion.

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