# Pendulum Potential energy equation

1. Feb 5, 2013

### cres222

hello every one
i have this pendulum:

i need to stabilize the pendulum in the inverted position , i need to know the potential energy for the pendulum , i read several articles in each one i have a different equation :
$V=mgl_{p}cos\alpha$
$V=-mgl_{p}cos\alpha$
$V=mgl_{p}(1-cos\alpha)$
$V=mgl_{p}(cos\alpha-1)$
Now i'am really confused which equation is the correct one?

2. Feb 5, 2013

### Staff: Mentor

What are you taking as your reference level? (Where PE = 0.) The pivot? The gravitational PE is given by mgΔy, where Δy is measured from your chosen reference level. Using that you should be able to pick the correct formula.

3. Feb 5, 2013

### cres222

$P.E=0$ at the vertical position ($\alpha=0$ from the vertical line in the picture above), now which formula should i use :
$P.E=mgl_{p}(1-cos\alpha)$
or
$P.E=mgl_{p}(cos\alpha-1)$

4. Feb 5, 2013

### Staff: Mentor

Using the vertical position as your reference, Δy will be negative for any nonzero angle. The second of those is the one you want.

5. Feb 5, 2013

### cres222

so:
$\Delta y=$(center of mass level-reference level)
not the other way around , am i correct?

6. Feb 5, 2013

### Staff: Mentor

Δy is measured from the PE = 0 reference point. Another way to write the PE is mgy, where y is the vertical position of the center of mass and y = 0 is the PE = 0 point.

Since you are choosing the reference level to be where the pendulum is vertical (the angle is zero) and thus the mass is at its highest point, all values for y and thus PE for non-zero angles will be negative since they are below that point.

7. Feb 5, 2013

### cres222

Now i understand , thank you very much for you kindly help,I'm very appreciative