# Center of mass of a sphere with cavity removed

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1. Oct 16, 2016

### 1v1Dota2RightMeow

1. The problem statement, all variables and given/known data
A solid sphere of density $ρ$ and radius $R$ is centered at the origin. It has a spherical cavity in it that is of radius $R/4$ and which is centered at $(R/2, 0, 0)$, i.e. a small sphere of material has been removed from the large sphere. What is the the center of mass $R_{cm} = (x_{cm}, y_{cm}, z_{cm})$ of the large sphere, including the cavity?

2. Relevant equations
$R=1/M \int \rho r dV$, where $dV=dxdydz$ and $\rho = dm/dV$ and $M=$ total mass

3. The attempt at a solution
$R=(1/M) \int r dm = (1/M) \int \rho r dV$

$M$= total mass = $\rho V = \rho(V_{total} - V_{cavity})$

$R=(1/(\rho(V_{total} - V_{cavity})) \int \rho r dV$

Here, I see that the $\rho$'s cancel. But now I'm stuck wondering what $r$ is.

2. Oct 17, 2016

### BvU

Hi DRM,

Like before, $\vec r$ is the position of $dV$. The $\rho$ do not cancel; they aren't even the same thing: the first one is the $\rho$ of the material only. The second one (within the integral) is zero where the cavity is and equal to the other one where the material is.

However, the integral you are left with in this approach is cumbersome, to say the least. Personally I'm not in favour of using tricks, but for this exercise some lateral thinking might save you a lot of work. Do you know how to calculate the center of mass of two solid bodies with mass $m_1$ and $m_2$ centered at $\vec r_1$ and $\vec r_2 \ \$ ?

3. Oct 17, 2016

### 1v1Dota2RightMeow

I do, and I've seen this trick done elsewhere. But in regards to it - why is this allowed, mathematically? Why is this equivalent to finding the CM via the integrals?

For reference: http://physics.stackexchange.com/qu...-in-finding-the-center-of-mass-of-this-sphere

4. Oct 17, 2016

### BvU

Hedging your bets with the competition, eh ?

Nothing wrong with that; never mind, all for the good cause. So:

Anything unclear about the answer by Prasad ? All I can do is rephrase, unless you give us a clue what's the step you aren't comfortable with ...

If you have to integrate the value 0 over the small sphere you can first integrate $\rho$ (and that's the same $\rho$ as in the solid part of the big sphere) and then correct by integrating $-\rho$ over the small sphere (the cavity).

5. Oct 17, 2016

### PeroK

Another way to look at it is to take the object X as having a mass $M_X$ and a centre of mass, $(x,0,0)$ then add the smaller sphere, which has a known mass $m$ and centre of mass $(R/2, 0, 0)$ and the result is the large sphere, with a known mass $M$ and centre of mass.

This gives an equation in terms of positive masses. If you move the terms for the small sphere to the other side of the equation, this is a negative term. You could interpret this as adding a negative mass, but that sounds a bit dramatic to me.