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I have a bar of length $L$, mass $m$, and negligible width. The bar can rotate along an axis through its center of mass (the rotation is in the $x$-$y$ plane), and also can move up and down (in the vertical axis).

My question is: What is the (gravitational) potential energy of this bar?

I've found several exercises where $PE=mgy_{CM}$, where $y_{CM}$ is the position of the bar's center of mass, though I don't know why the assumption of just taking the CM.

Since this is rigid body, I thought also in summing the potential energy of each particle (i.e, integrating the classical $mgy$ function with respect to the height $y$), but I don't know if this is correct, or necessary.

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Note that the fact that the CM shows up in the answer might not be an assumption, it might be a result, right? –  levitopher Jul 7 at 4:59

2 Answers 2

up vote 3 down vote accepted

Integrating each mass slice and just taking the center of mass yield the same result. That is because:

$$PE = g \int y \, {\rm d}m = m g y_{cm}$$

from the definition of the center of mass

$$y_{cm} = \tfrac{\int y \,{\rm d}m}{m} $$

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Good answer, but since this is basically a homework question, I think this gives away too much. –  levitopher Jul 7 at 5:00

Okay, first of all, your question is unclear.
'Potential energy of a bar' has no meaning whatsoever. PE is of a system of interacting objects or particles. We sometimes just say that the PE of a mass $m$ at a height $h$ from the ground is $mgh$. But strictly speaking, this is actually the PE of the system: Mass-$m$ + Earth.
So there you go. If you are interested in this PE, then it is simply $mgd_{CM}$, where $d_{CM}$ is the distance of the CM from the ground. But I don't think this is the answer you want. Distance from the ground is not mentioned, and you have mentioned how the rod is free to rotate about some axes etc. Potential Energy of the 'Earth + Rod' system doesn't care about this. It simply wants to know the distance of the CM of the object from the surface.
One possibility is that you wanna know the gravitational interaction energy of all the particles of the rod or 'self-energy' of the rod. Is that what you require?
Could you clarify the question? If you could post the correct answer, it would be helpful. We would know what we are actually asked to find out.

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