Double Pendulum Potential Energy

Click For Summary
SUMMARY

The potential energy (PE) of a double pendulum can be expressed using two different equations: V=-(m_1+m_2) g l_1 cos(θ_1)-m_2 g l_2 cos(θ_2) and V=(m_1+m_2) g l_1 (1-cos(θ_1))+m_2 g l_2 (1-cos(θ_2). The first equation is commonly used in literature, while the second equation is derived from considering the equilibrium position of the pendulum. Both equations yield identical partial derivatives with respect to the angles θ_1 and θ_2, indicating that the choice of potential energy reference point does not affect the dynamics of the system.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly potential energy.
  • Familiarity with double pendulum dynamics and its equations of motion.
  • Knowledge of trigonometric functions and their applications in physics.
  • Basic calculus, specifically partial derivatives.
NEXT STEPS
  • Explore the derivation of potential energy equations in classical mechanics.
  • Study the dynamics of double pendulums and their stability analysis.
  • Learn about the role of reference points in potential energy calculations.
  • Investigate numerical simulations of double pendulum motion using tools like MATLAB or Python.
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in advanced dynamics and energy calculations in multi-body systems.

disclaimer
Messages
25
Reaction score
0
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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K