## Pendulum Potential energy equation

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?
 Mentor Blog Entries: 1 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.
 $P.E=0$ 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 $P.E=mgl_{p}(cos\alpha-1)$

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## Pendulum Potential energy equation

 Quote by 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 $P.E=mgl_{p}(cos\alpha-1)$
Using the vertical position as your reference, Δy will be negative for any nonzero angle. The second of those is the one you want.
 so: $\Delta y=$(center of mass level-reference level) not the other way around , am i correct?

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 Quote by cres222 so: $\Delta y=$(center of mass level-reference level) not the other way around , am i correct?
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.
 Now i understand , thank you very much for you kindly help,I'm very appreciative