Double Pendulum Potential Energy

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  • #1
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Hello, everybody.

This website and many others define the potential energy of a double pendulum as:

[tex]V=-(m_1+m_2) g l_1 cos\theta_1-m_2 g l_2 cos\theta_2[/tex]

However, I came up with the following equation:

[tex]V= (m_1+m_2) g l_1 (1-cos\theta_1)+m_2 g l_2 (1-cos\theta_2)[/tex]

I started from the position of what looks like equilibrium (when the pendulum is fully stretched and hanging freely). They seem to start at the point where the pendulum is "pinned."

Which equation should I use? Am I missing something here?

Thanks in advance.
 

Answers and Replies

  • #2
2,791
591
It seems to me you have two origins for potential energy which is wrong!
 
  • #3
25
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Thank you. I think I see what you mean. Let me recalculate it.
 
  • #4
25
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Okay. I guess it doesn't really matter in the end, since the partial derivatives [tex]\frac{\partial V}{\partial \theta_1}[/tex] and [tex]\frac{\partial V}{\partial \theta_2}[/tex] are identical for both equations, namely:

[tex]\frac{\partial V}{\partial \theta_1}=(m_1+m_2) \sin\theta_1 g l_1[/tex] [tex]\frac{\partial V}{\partial \theta_2}=m_2 g l_2 \sin\theta_2[/tex]
 
  • #5
3,757
424
Yes, your PE has just some constant terms extra. The constant terms are usually dropped anyway in this kind of problems, even if they result from a correct calculation.
 

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