1. May 11, 2014

### darksyesider

1. The problem statement, all variables and given/known data

I have a conceptual question which is not HW, although the moderators of this forum think different.

In the "hoop" example, and the example after that here: http://tonic.physics.sunysb.edu/~dteaney/F06_Ph2034/lectures/lexam3.pdf

do you ALWAYS have the change in gravitational potential energy as the difference in the height of the CENTER OF MASS?

Rewording: When you use $mgh + 1/2mv^2 + 1/2 I\omega^2 = (mgh + 1/2mv^2+1/2I\omega^2 )f$,

is Mgh ALWAYS the difference in height of the CENTER OF MASS? or can it [the center of mass] be any point on the object?

2. Relevant equations

see above

3. The attempt at a solution

This is not a homework problem.

Thanks, and hope this is answered before AP tests!

2. May 11, 2014

### BvU

Yes. Basically, when calculating potential energy, what you are doing is adding up all contributions from an extended mass distribution: PE = $\Sigma g h \, \Delta m$. With g constant and $h_{\rm c.o.m} = {\Sigma h \Delta m \over \Sigma \Delta m}$ you get PE = $g\,h_{\rm c.o.m}\,\Sigma \Delta m$.

If the orientation of the object doesn't change, $h_{\rm c.o.m}$ and the h of any point on the object differ by a constant, so you can take any point and don't need to calculate the center of mass position.

Simple example of where you do: A see-saw with one heavy and one light person. PE with the heavy one up high is higher than PE with the light one. Yet the axis of rotation stays at the same height.

Or the rotation point of the ring in the exercise: stays at the same height. But PE of solid circle is higher than that of the dotted one. It "falls".

 good luck with your tests!