# Finding linear density at a given location along a rod

1. Dec 11, 2012

### ravanbak

I hope someone can point me in the right direction on how to solve this, and hopefully I can explain it properly.

Given a rod with a known length, mass and center of mass, how can I find the linear density at a given location (x) along the length of the rod. I want to say 'instantaneous' density at location 'x', but not sure if that's the correct term.

My question is for the case where the center of mass is not at the center of the rod, so the linear density is non-uniform, but the change in density from one end of the rod to the other is linear, if that makes sense.

My goal is to represent a mass as a distribution along a length by calculating the linear density at the two end points.

Thanks in advance for any help.

2. Dec 11, 2012

### Staff: Mentor

Step 1: Express the linear density ρ as a function of position x along the rod by:

ρ = mx + b

Solve for m and b under the conditions that ρ = ρ0 at x = 0 and ρ = ρL at x = L

Step 2:

Find the mass of the rod in terms of ρ0 and ρL by integrating the density with respect to x from x = 0 to x = L. Without doing the integration, what do you think the answer will be?

Step 3: Find the moment of the rod weight distribution about x = 0.

Can you figure out what the remaining steps are?

Chet

3. Dec 12, 2012

### ravanbak

Hi Chet,

Thanks very much for your help. It's been a long time since I've done any calculus, but here's what I have so far:

Step 1:

m = (ρL - ρ0) / L
b = ρ0

So, ρ = (ρL - ρ0) / L * x + ρ0

Step 2:

After integration, mass = (ρL + ρ0)L / 2

(which is the result I was expecting)

Step 3:

Ok, here's where I'm a little lost. I don't fully understand the connection between moment and linear density.

And I'm not sure what the formula for the moment would be. For a rod, I = mL2/3, but I would expect that that's only for a rod with uniform density.

Since I know the center of mass, should I treat the rod as a shorter rod with length xc, where xc is the location of the center or mass?

In any case, I don't know where to go from here.

Thanks again,
Dave

Last edited: Dec 12, 2012
4. Dec 12, 2012

### Staff: Mentor

Hi Dave,

You've done really well so far. To take moments of forces about a point, you multiply the force by the moment arm. The force you are dealing with is the downward differential amount of weight in each differential section of rod: dF = ρgdx. The moment arm with respect to the point x = 0 is the location x itself. So the differential moment resulting from the section of rod between x and x + dx is dM = ρgxdx . The total moment of the rod weight around the point x = 0 is the integral of dM. This total moment M must be equal to the total weight of the rod times the distance from x = 0 to the center of mass xc. With this value for xc, the two sections of rod on either side of xc must balance like a see-saw.

It is my understanding that you do not know where the center of mass is located, and you are trying to solve for it. The above approach is how to do it.

5. Dec 12, 2012

### ravanbak

Thanks, I'm trying. :) But it looks like I didn't explain it properly.

I already know where the center of mass is located.

I'm trying to find a way to solve for the linear density at a given location along the rod.

Particularly, what is the linear density at x0 and at xL.

Thanks,
Dave

6. Dec 12, 2012

### Staff: Mentor

You can't determine them both independently. You can only determine their ratio. Start with the moment relationship I described, and, set ρL = k ρ0. Then solve for k.

Chet

7. Dec 12, 2012

### ravanbak

Sorry, but can you clarify: are you saying there is no way to solve for actual values of linear density?

My understanding is that even if I determine the ratio between ρ0 and ρL, I still won't be able to get density values without already knowing the density at some location.

8. Dec 12, 2012

### Staff: Mentor

I may have misspoken. If you know the mass and the location of the center of mass, you will have two equations in two unknowns for the linear densities at the ends.

9. Dec 12, 2012

### ravanbak

I find it hard to believe that knowing everything else (mass, length and center of mass) about the rod, the density at a given location cannot be calculated.

Remember that I said the density changes linearly from one end to the other, it's not just random. Given the mass and center of mass, I would expect there to be a unique solution for density at any given location.

10. Dec 12, 2012

### Staff: Mentor

Right. I stand corrected. See my post #8. So set up the equation for the total moment M, and lets review what you get.

11. Dec 12, 2012

### Staff: Mentor

Aside from a factor of g, here's what I got for the moment around x = 0:

M = (ρ0 + 2 ρL) L2/6

Chet

12. Dec 14, 2012

### Staff: Mentor

I sent you a private message giving the remaining details. I just didn't want to bore everyone else with the details.

Chet