
#1
Feb513, 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}(1cos\alpha)[/itex] [itex]V=mgl_{p}(cos\alpha1)[/itex] Now i'am really confused which equation is the correct one? Please help me 



#2
Feb513, 06:18 AM

Mentor
P: 40,905

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
Feb513, 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}(1cos\alpha)[/itex] or [itex]P.E=mgl_{p}(cos\alpha1)[/itex] 



#4
Feb513, 07:21 AM

Mentor
P: 40,905

Pendulum Potential energy equation 



#5
Feb513, 08:55 AM

P: 4

so:
[itex]\Delta y=[/itex](center of mass levelreference level) not the other way around , am i correct? 



#6
Feb513, 10:46 AM

Mentor
P: 40,905

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 nonzero angles will be negative since they are below that point. 



#7
Feb513, 11:33 AM

P: 4

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



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