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Pendulum Potential energy equation |
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| Feb5-13, 02:56 AM | #1 |
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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 : [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 |
| Feb5-13, 06:18 AM | #2 |
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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.
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| Feb5-13, 06:54 AM | #3 |
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[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] |
| Feb5-13, 07:21 AM | #4 |
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Pendulum Potential energy equation |
| Feb5-13, 08:55 AM | #5 |
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so:
[itex]\Delta y=[/itex](center of mass level-reference level) not the other way around , am i correct? |
| Feb5-13, 10:46 AM | #6 |
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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. |
| Feb5-13, 11:33 AM | #7 |
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Now i understand , thank you very much for you kindly help,I'm very appreciative
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