# Center of Mass of compound cylinder

1. Apr 14, 2008

### corochena

[SOLVED] Center of Mass of compound cylinder

I have this problem. I have a cylinder filled with two liquids of different densities and not able to mix up (like oil and water). I have to fill the cilynder in such a way so that the center of mass will be located as low as possible. For example, mercury and water, cylinder radius = 1cm, height = 10cm. How much mercury and water should I use? What would be the location of the center of mass. Note: The cylinder must be filled in completely. Give a response in function of the densities of the fluids.

2. Apr 14, 2008

### Hootenanny

Staff Emeritus
Hi corochena and welcome to PF,

What have you attempted thus far?

3. Apr 14, 2008

### corochena

Solution

Well, I realized that the heavier substance must be located at the bottom of the cylinder and the ligther top. Then I computed the position of the center of mass of each sub-cylinder top and bottom.
Let it be:
d1: density of the heavier substance
d2: density of the lighter substance
h: height of the heavier substance
H: height of the cylinder

With those I computed the center of mass of the whole cylinder and then I derivate to find the maximum/minimum and after some algebraic manipulation I came to the equation h = ( (d2 +/- sqr2(d1*d2) ) / (d2 - d1) ) * H

I realized that in order that the center of mass to be located as low as possible the center of mass must be located in the surface of contact of both substances.

I would like to know if my results are correct and other ways to solve it or any comments.

4. Apr 14, 2008

### Hootenanny

Staff Emeritus
Could you be a little more descriptive with your working? Could you show all your steps?

5. Apr 15, 2008

### corochena

It's a bit difficult because the simbology but I'll try:

The center of mass of cylinder 1 and 2 (heavier and lighter) are located in the middle because they are homogeneus. So I am able to find the CM of the whole cylinder:

By definition of CM:
m1 x = m2 (H/2 - x), x is the distance between CM1 and CM of the whole cylinder

m = d . V (density times volume)

m1 = d1 . pi . R^2 . h
m2 = d2 . pi . R^2 . (H-h)

so

d1 . pi . R^2 . h . x = d2 . pi . R^2 . (H-h) . (H/2 - x)

simplifying

d1 . h . x = d2 . (H-h) . (H/2 - x)

here I will make a shorcut stating that in order for the CM be as low as possible the CM must be located on the surface of contact of both substances. (The demonstration of such statement I will post it later). This statement makes the demonstration much shorter and simpler.

So x = h/2 then

d1 . h . h/2 = d2 . (H - h) . (H/2 - h/2)

rearranging

(d2 - d1)h^2 - 2d2.H. h + d2 . H^2 = 0

and solving the cuadratic we get

h = H . ( d2 +/- sqr(d1.d2)) / (d2 - d1)

I solved this problem a few months ago and I thought it was pretty interesting and I wanted to share my results and get feedback if they are correct or so.

I tried to put a picture here but I get the error that I am allowed to post URLs to other sites after I have made 15 posts of more.

#### Attached Files:

• ###### planteo.JPG
File size:
4.5 KB
Views:
103
6. Apr 15, 2008

### Hootenanny

Staff Emeritus
Thanks corochena. Unfortunately I haven't got time to look at it now, but I'll have a good look at it later this evening.

In order to prevent spamming, users with less than fifteen posts are not allowed to post urls. However, I'm sure a mentor will approve your attachment soon.

7. Apr 15, 2008

### corochena

Thanks a lot! And... how were you able to answer so quickly? Is there some kind of alarm or something?

8. Apr 15, 2008

### Hootenanny

Staff Emeritus
There is a link in the top left corner called My PF, if one clicks it then it displays a list of subscribed threads.

Technically, you also make this assumption,
At the very beginning of your derivation, so you should really prove your lemma first.
I assume that your taking moments about the CM of the cylinder ($C_c$), in which case the above expression is,

$$m_1x = m_2\left(\frac{H-h}{2}\right) = m_2\left(\frac{H}{2}-\frac{h}{2}\right)$$

Now if one assumes that $C_c = h$, then the equation becomes,

$$m_1x = m_2\left(\frac{H}{2}-x\right)$$

As you originally had. However, you need to prove that $C_c = h$ before you can use it.

9. Apr 15, 2008

### corochena

Yes, I am taking moments about the CM of the cylinder Cc.

But I think m1 . x = m2 (H/2 - h/2) is wrong. The correct expression is as I wrote before (hopefully!) m1 . x = m2 . (H/2 - x). I think the new picture will explain better.

#### Attached Files:

• ###### planteo2.JPG
File size:
12.3 KB
Views:
148
10. Apr 15, 2008

### corochena

I know I have to prove that x must be equal to h/2 (that is equivalent to say that the lowest CM possible must be located at the surface of contact of the two substances).

There is another way to do it and it is not necessary to make the assumption x = h/2, but you need to find the derivative of x=f(h) to find the maxima/minima, the algebraic burden is bigger although. Result is same of course.

Also regardless the procedure and the technical details (rigor) I would like to know if somebody gets the same result and how he/she came to his/her result.

11. Apr 15, 2008

### Hootenanny

Staff Emeritus
I'm afraid that I still can't see where you're deriving the following relationship from,

$$m_1x = m_2\left(\frac{H}{2}-x\right)$$

Perhaps I am missing something, would someone else like to chime in?

12. Apr 16, 2008

### Staff: Mentor

I agree. This follows from the fact that the distance between the centers of m1 and m2 must equal H/2. (The center of m1 is h/2 from the bottom; the center of m2 is (H-h)/2 from the top.)

13. Apr 16, 2008

### Hootenanny

Staff Emeritus
Thanks Doc!

corochena, the rest of your derivation and your final expression for h is correct. Nicely done .

14. Apr 16, 2008

### corochena

Thanks both of you. What a nice site is this! It was difficult to get someone interested in this kind of stuff, ... not anymore!