# Thread: Weighted mean mass of a ball?

1. ## Weighted mean mass of a ball?

I have a ball with the mass varying with the radius (ie. distance from the center, the closer it is to the center, the heavier it is). Let's call r the radius, f(r) is the mass at that radius distance from the center. How is the weighted mean mass of the ball calculated according to r and f(r)?

Many thanks!

2. ## Variable density

Hello cctest

Welcome to Math Help Forum!
Originally Posted by cctest
I have a ball with the mass varying with the radius (ie. distance from the center, the closer it is to the center, the heavier it is). Let's call r the radius, f(r) is the mass at that radius distance from the center. How is the weighted mean mass of the ball calculated according to r and f(r)?

Many thanks!
May I change your question a little, because this is what I think you mean. Calculate the total mass of a sphere of radius $\displaystyle a$, if its density (mass per unit volume) at a given point is given by $\displaystyle f(r)$ where $\displaystyle r$ is the distance from the centre of the sphere. Is that it?

If so, then consider a thin spherical shell of radius $\displaystyle r$, thickness $\displaystyle \delta r$. Its volume is its surface area multiplied by its thickness; i.e. $\displaystyle 4\pi r^2\delta r$. If its mass per unit volume is $\displaystyle f(r)$, then its mass is $\displaystyle 4\pi r^2f(r)\delta r$. To get the total mass, therefore, we integrate between $\displaystyle r = 0$ and $\displaystyle r = a$, to get

Mass $\displaystyle = 4\pi\int^a_0 r^2f(r)\,dr$

Obviously, you can't take this any further unless you know what $\displaystyle f(r)$ is.

I hope that's what you were looking for.

The answer you've given is exactly the thing that I was afraid that people would be misunderstanding when reading my question.

It would be forever to explain my situation but what I want is the weighted mean mass of the ball (since each layer of dr distance away from the center carries a different mass), NOT the total mass, it would be much easier this way tho.

Because of my further calculations, I would need the mean mass of the layers of distances dr away from the center other than the total mass. One could initially say that this mean mass is at r0 = 1/2 rmax, but if you think about it, as r grows bigger, the layer r away from the center will contribute more (in terms of volume) to the total mass, and therefore one could NOT easily says weighted_mean = sigma(f(r)r)/sigma(f(r)) as in weighted arithmetic mean... I s'pse it should be sigma(f(r)r^3)/sigma(f(r)r^2) .but i'm not sure...

4. ## Mass and density

Hello cctest

I've been thinking how to try to answer your question, but I'm still not clear what it is that you want. I don't understand your use of the word 'mass'.

You talk about the 'weighted mean mass', but I don't think this really makes any sense. The sphere just has a mass. Period. You can talk about the mean mass of several spheres, of course, but a single sphere just has one mass, and that's it. There's no such thing as the mean mass of a sphere, weighted or otherwise.

In your original post, you referred to
the mass varying with the radius (ie. distance from the center, the closer it is to the center, the heavier it is). Let's call r the radius, f(r) is the mass at that radius distance from the center
and in my reply I talked about $\displaystyle f(r)$ as a density function; i.e. mass per unit volume. It simply doesn't make sense to talk about the mass at a particular point, distance $\displaystyle r$ from the centre. An object with a finite mass will always have a finite volume - it can't have an infinitesimally small volume. If it did, then the whole object (made up of an infinitely large number of such volumes) would have an infinite mass. You have to talk about the density varying with respect to the radius, not the mass.

So I wonder whether you really want the mean density, and not the mean mass? If so, then this is simply the total mass (as given in my first reply) divided by the total volume, $\displaystyle \tfrac43\pi a^3$.

Other things that you hint at include the radius at which this 'mean mass' occurs. Again, if you mean the radius at which the mean density occurs, then you'll have to find the mean density, $\displaystyle \rho$, say and then solve the equation $\displaystyle f(r) = \rho$, to find the distance from the centre where the density of the sphere is equal to its mean density.

Or, you could find the radius of a uniform sphere which has the same mass as this variable one, and the same mean density. In this case, you'd need to solve for $\displaystyle R$ the equation

$\displaystyle \tfrac43\pi R^3 \rho$ = total mass of original sphere = $\displaystyle 4\pi\int^a_0 r^2f(r)\,dr$

The only other quantity that I can think of that might be relevant is the radius of gyration, which gives the root mean square distance of the 'infinite number' of 'infinitesimally small' variable density 'particles' of which you can imagine the sphere to be comprised. This is, of course, a sort of weighted mean.

Other than that, I'm stumped!

Is there anything here that helps?