Double Pendulum Potential Energy

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The potential energy of a double pendulum is commonly defined as V=-(m_1+m_2) g l_1 cosθ_1 - m_2 g l_2 cosθ_2, but an alternative equation V=(m_1+m_2) g l_1 (1-cosθ_1) + m_2 g l_2 (1-cosθ_2) is proposed. This alternative starts from the equilibrium position rather than the pinned position. Both equations yield identical partial derivatives for the angles θ_1 and θ_2, indicating they are functionally equivalent despite differing in form. Ultimately, constant terms in potential energy calculations are often disregarded in physics problems.
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Hello, everybody.

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

V=-(m_1+m_2) g l_1 cos\theta_1-m_2 g l_2 cos\theta_2

However, I came up with the following equation:

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

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

\frac{\partial V}{\partial \theta_1}=(m_1+m_2) \sin\theta_1 g l_1 \frac{\partial V}{\partial \theta_2}=m_2 g l_2 \sin\theta_2
 
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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