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For a double pendulum what would be the gravitational energy. I am trying to work out the Lagrangian for the double pendulum. I got the kinetic energy but I am struggling on the gravitational potential energy.

I tried calculating the gravitational: $$V = g(m_1+m_2)(l_1+l_2-l_1\cos\theta_1) - gm_2(l_2\cos\theta_2)$$ but on wolfram it uses a different equation. Am I wrong?

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The expression for the potential energy is given by $V_{tot}=V_1+V_2$ where $V_1=m_1 g y_1$ and $V_2=m_2 g y_2$. The $y$-coordinate for the first mass is $y_1=-l_1 \cos(\theta)$ and for the second mass $y_2=-(l_2 \cos(\theta)+l_1 \cos(\theta))$.

Then the expression for the total potential energy is: $V_{tot}= -m_1 l_1 \cos(\theta)-m_2(l_2 \cos(\theta)+l_1 \cos(\theta))$.

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You took the zero of potential energy to be at the location of the bottom mass when both angles are zero. Wolfram took it to be at the top pivot. Your expressions simply differ by a constant, which does not affect the dynamics.

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Your expression differs from ScienceWorld's expression by a constant (namely, $g(m_1 + m_2)(l_1 + l_2)$.) Since you can always add or subtract a constant from the potential energy without changing the physics, your expression and ScienceWorld's expression are equivalent.

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