Register to reply

Pendulum Potential energy equation

by cres222
Tags: energy, equation, pendulum, potential
Share this thread:
cres222
#1
Feb5-13, 02:56 AM
P: 4
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 :
[itex]V=mgl_{p}cos\alpha[/itex]
[itex]V=-mgl_{p}cos\alpha[/itex]
[itex]V=mgl_{p}(1-cos\alpha)[/itex]
[itex]V=mgl_{p}(cos\alpha-1)[/itex]
Now i'am really confused which equation is the correct one?
Please help me
Phys.Org News Partner Physics news on Phys.org
Scientists uncover clues to role of magnetism in iron-based superconductors
Researchers find first direct evidence of 'spin symmetry' in atoms
X-ray laser probes tiny quantum tornadoes in superfluid droplets
Doc Al
#2
Feb5-13, 06:18 AM
Mentor
Doc Al's Avatar
P: 41,440
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.
cres222
#3
Feb5-13, 06:54 AM
P: 4
[itex]P.E=0[/itex] at the vertical position ([itex]\alpha=0[/itex] from the vertical line in the picture above), now which formula should i use :
[itex]P.E=mgl_{p}(1-cos\alpha)[/itex]
or
[itex]P.E=mgl_{p}(cos\alpha-1)[/itex]

Doc Al
#4
Feb5-13, 07:21 AM
Mentor
Doc Al's Avatar
P: 41,440
Pendulum Potential energy equation

Quote Quote by cres222 View Post
[itex]P.E=0[/itex] at the vertical position ([itex]\alpha=0[/itex] from the vertical line in the picture above), now which formula should i use :
[itex]P.E=mgl_{p}(1-cos\alpha)[/itex]
or
[itex]P.E=mgl_{p}(cos\alpha-1)[/itex]
Using the vertical position as your reference, Δy will be negative for any nonzero angle. The second of those is the one you want.
cres222
#5
Feb5-13, 08:55 AM
P: 4
so:
[itex]\Delta y=[/itex](center of mass level-reference level)
not the other way around , am i correct?
Doc Al
#6
Feb5-13, 10:46 AM
Mentor
Doc Al's Avatar
P: 41,440
Quote Quote by cres222 View Post
so:
[itex]\Delta y=[/itex](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.
cres222
#7
Feb5-13, 11:33 AM
P: 4
Now i understand , thank you very much for you kindly help,I'm very appreciative


Register to reply

Related Discussions
Pendulum Problem: Potential energy equals? kinetic energy Introductory Physics Homework 2
Pendulum (Conservative Forces and Potential Energy Introductory Physics Homework 4
Ballistic Pendulum - Kinetic Energy vs Gravitational Potential Energy General Physics 13
Potential energy and a pendulum Introductory Physics Homework 10
Potential energy of a pendulum when the angle is small Introductory Physics Homework 13