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Potential energy of a pendulum and where you place the datum. 
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#1
Mar1814, 02:37 AM

P: 104

So I've always been confused about this. Suppose you have your normal pendulum: length L, mass m, and angle Θ.
When you describe the potential energy PE = mgh, you must decide where to measure your h from. Throughout my years I've seen it measured from the mass to the 0 equilbrium point where you'd get that PE = mgL*(1cosΘ) and also measured from the mass to the horizontal position where Θ=π/2 where you would get PE = mgLcosΘ. the signs are with respect to the positive yaxis pointing up. These are clearly not the same number, so whats the distinction? what is the actual potential energy? why have i seen it done both ways? 


#2
Mar1814, 02:53 AM

PF Gold
P: 377

It does not matter where you chose the reference point for potential energy.
Try to solve th problem with and arbitrary point of reference, and observe that you always end up with the same solution. Can you see why? 


#3
Mar1814, 03:08 AM

P: 104

suppose i wanted to know the energy of the system?



#4
Mar1814, 04:32 AM

PF Gold
P: 377

Potential energy of a pendulum and where you place the datum.
You would only like to know how much energy could be released if the pendulum falls from one place to another. 


#5
Mar1814, 07:28 AM

P: 1,991




#6
Mar1814, 07:33 AM

P: 104




#7
Mar1814, 07:36 AM

Mentor
P: 17,543




#8
Mar1814, 07:42 AM

Sci Advisor
Thanks
PF Gold
P: 1,911

When you calculate forces from a potential it goes: F =  gradient(potential energy).
You will note that shifting the potential energy by any constant amount does not change the force ... hence the dynamics is not affected by the choice of origin for a potential. Energy is still conserved ... just don't change your origin partway through a calculation! 


#9
Mar1814, 07:55 AM

PF Gold
P: 377

But it depend on the reference point chosen. When you change the system of coordinate, the coordinates of the reference point are also changed. The coordinates used do not matter. 


#10
Mar1814, 08:25 AM

Mentor
P: 17,543

I understand your point. You are distinguishing between coordinate system and reference point. It is a tenuous distinction since you can always consider h to be a coordinate, however, even accepting the distinction the fact remains that energy does depend on the coordinate system. Consider kinetic energy. If you are sitting in a car then in a coordinate system attached to the car your KE is 0, but in a coordinate system attached to the ground your KE is non0. Energy therefore does depend on the coordinates. 


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