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Potential energy of a pendulum and where you place the datum.

by Mugged
Tags: datum, energy, pendulum, potential
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Mugged
#1
Mar18-14, 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*(1-cosΘ) 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 y-axis 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?
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maajdl
#2
Mar18-14, 02:53 AM
PF Gold
P: 354
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?
Mugged
#3
Mar18-14, 03:08 AM
P: 104
suppose i wanted to know the energy of the system?

maajdl
#4
Mar18-14, 04:32 AM
PF Gold
P: 354
Potential energy of a pendulum and where you place the datum.

Quote Quote by Mugged View Post
suppose i wanted to know the energy of the system?
You would never be interested to know that.
You would only like to know how much energy could be released
if the pendulum falls from one place to another.
nasu
#5
Mar18-14, 07:28 AM
P: 1,969
Quote Quote by Mugged View Post
suppose i wanted to know the energy of the system?
You can know it in respect to the origin you choose.
Mugged
#6
Mar18-14, 07:33 AM
P: 104
Quote Quote by nasu View Post
You can know it in respect to the origin you choose.
shouldnt the energy be invariant with respect to coordinates?
DaleSpam
#7
Mar18-14, 07:36 AM
Mentor
P: 16,985
Quote Quote by Mugged View Post
shouldnt the energy be invariant with respect to coordinates?
No, energy is definitely not invariant. It is conserved, not invariant. Those are two different concepts.
UltrafastPED
#8
Mar18-14, 07:42 AM
Sci Advisor
Thanks
PF Gold
UltrafastPED's Avatar
P: 1,908
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!
maajdl
#9
Mar18-14, 07:55 AM
PF Gold
P: 354
Quote Quote by Mugged View Post
shouldnt the energy be invariant with respect to coordinates?
It (the potential energy) is invariant with respect to the coordinate system.
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.
DaleSpam
#10
Mar18-14, 08:25 AM
Mentor
P: 16,985
Quote Quote by maajdl View Post
It is invariant with respect to the coordinate system.
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.
The coordinates do matter, energy is not invariant with respect to the coordinate system.

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 non-0. Energy therefore does depend on the coordinates.


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