# Double Pendulum Potential Energy

1. May 25, 2015

### disclaimer

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?

2. May 25, 2015

### ShayanJ

It seems to me you have two origins for potential energy which is wrong!

3. May 25, 2015

### disclaimer

Thank you. I think I see what you mean. Let me recalculate it.

4. May 25, 2015

### disclaimer

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:

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

5. May 25, 2015

### nasu

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.