# Homework Help: Cable and CM

1. Nov 10, 2005

### standardflop

Hello
A cable is attached at two (nonemoving)fixpoints. Now the cables midpoint is pulled downward, thereby destroying the catenary curve.

Question: How will the cables center of mass (CM) move: op, down or stay fixed?

My thoughts:
I find it intuituitive that the CM will assume the lowest possible position in the catenary (y=cosh y) curve, but how can i "proove" this? and is it possible to describe the hight of CM as function of the downward midpoint-displacement?

Thank You

2. Nov 12, 2005

### standardflop

Is this a valid approach: ?
the y-conponent of center of mass is given by $$y_c = \frac{\int r_y dm}{m}$$ furthermore
$$dm=\rho ds$$ and $$ds=\sqrt{1+y'^2} dx$$
now we have
$$y_c = \frac{\int_{x_1}^{x_2} y \cdot \rho \sqrt{1+y'^2} dx}{\int_{x_1}^{x_2} \rho \sqrt{1+y'^2} dx}$$

now forinstance inserting (and comparing) i.e. $$y_1 =cosh(x)$$ and $$y_2=|x|$$ (zigzag-function), we see that the former has higher center of mass y-comp.
thereby concluding that pulling down the cable at the midpoint will lower the center of mass?

Last edited: Nov 12, 2005
3. Nov 12, 2005

### HallsofIvy

Since the only question is whether the center of mass will move up or down, I don't see any need for any detailed calculations.

If there were no outside force, the center of mass would not move. But there obviously is an outside force since "the cables midpoint is pulled downward". Since the force is downward, which way will the center of mass move?

4. Nov 12, 2005

### standardflop

hmm, i considered the analogy of attaching a weight to the midpoint, and ofcourse the center of mass would move down. But does external forces affect the center of mass? - i thought i needed to focus on the weightdistribution caused by change in curvature of the catenary?

furthermore, it seems odd to me that the zigzag curve has lower center of mass.. wouldn it mean that a hanging cable could minimize its potential energy by assuming this curvature?

Last edited: Nov 12, 2005
5. Nov 12, 2005

### Staff: Mentor

The catenary, the shape assumed by the hanging chain, has the minimum potential energy of any possible shape. That's equivalent to saying it's center of mass is as low as it can be. So any change in its shape can only raise the center of mass.

6. Nov 13, 2005

### Andrew Mason

I think this is the answer they are looking for. But this assumes that a chain and a cable are equivalent. A cable can stretch because of the way it is wound. So it is possible to lower the centre of mass by pulling and stretching a cable.

AM

7. Nov 13, 2005

### Staff: Mentor

Interesting point, AM. I assumed that whatever it was--cable, chain, etc.--had a fixed length and was unstretchable. There's nothing in the problem statement that implies that the cable was stretched instead of merely pulled into a different shape.

8. Nov 13, 2005

### standardflop

but can you account for this, or is it just a consideration?
Yes, this the logical conclusion - but is there a proof?, and a way to find out how it moves?

9. Nov 20, 2005

### standardflop

Could not quite deduce a conclusion from the answers to my thread..
Im still struggling with this problem of 'the center of mass of deformed catenary' ;)

The "potential-energy argument" rules out that the center of mass will move downward, when the midpoint is displaced. So it would be obvious to say, that the center of mass would move up - but i still dont see the "proof", and it would be nice, to be able to describe the movement of the center of mass as function of midpoint displacement.

Thanks

10. Nov 21, 2005

### Staff: Mentor

I'm not clear on what kind of proof you are looking for. If you accept that the chain, left to hang under its own weight, will form the shape of least PE, then it follows that the chain's CM must be its lowest in that configuration. (That shape happens to be a catenary, but who cares.) Thus any deviation from that shape must increase the PE and raise the CM.

The above is trivial. But describing the new CM position as a function of midpoint displacement is not trivial (at least as far as I can see).

11. Nov 21, 2005

### standardflop

Ofcourse..I agree with the first part. But how do you reject the possibility, that a deviation from that shape (the "new" shape is also a catenary, just with a discontinuity) wont leave the CM in the same place?

Maybe one could evaluate the mean-height of the curves? something like $$\overline{f(x)}=\frac 1{b-a} \int_a^b f(x) \ dx$$