# Center of mass of a ring of variable density

1. May 6, 2009

### redtree

How would I find the center of mass (COM) for a ring of variable density, where the variation in density can be described by a continuous mathematical equation.

The density at each location along the ring is described by a function (d) whose independent variables are radius (r) and angle ($$\theta$$), i.e., polar coordinates.

The value of d at each r and $$\theta$$ is the sum of two functions (d1 and d2) where:

d1 = A sinc($$\pi$$ r)

where A is a constant
AND

d2 = B sinc($$\pi$$ z)

where B is a constant
And: z = l + r * cos($$\theta$$)

where l is a constant

NOTE: THIS IS NOT A HOMEWORK QUESTION

2. May 6, 2009

### tiny-tim

Hi redtree!

(have a theta: θ and a pi: π )

Just change to x and y coordinates, and integrate in the usual way to find the x- and - components of the c.o.m. separately …

what do you get?

3. May 6, 2009

### arildno

Remember that the value of the radial variable of the center of mass is:
$$\hat{r}=\sqrt{\hat{x}^{2}+\hat{y}^{2}}$$ where $\hat{x},\hat{y}$ are the x and y-coordinates of the center of mass. Analogously for the value of the value pf the angular variable at the mass point.

There is no simplistic way to calculate these two values except by going through the intermediary x and y values.

4. May 6, 2009

### redtree

What is the formula (in x-y coordinates) for the center of mass of a ring of variable density?

I think the general formula is as follows:

R = $$\int$$ p(r) * r dV / $$\int$$ p(r) dV

In this case p(r) = d(r) = d1(r) + d2(r)
Where
d1(r) = A sinc(pi * (x2 + y2)(1/2))
d2(r) = B sinc(pi * ((l-x)2 + y2)(1/2))

So, how do I find "dV"?

This problem can also be restated as a "barycenter" problem:

r1 = a / (1 + m1/m2) and r2 = a / (1+m2/m1)

where
m1 is a function of d1(r)
m2 is a function of d2(r)
a = l.

However, even with this restatement, it seems that finding the COM for m2 is not straightforward.

5. May 7, 2009

### tiny-tim

dV is dxdydz

(or, for a 2-dimensional mass, you'd use dA = dxdy)