Double Pendulum Potential Energy

AI Thread Summary
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.
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.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?

Similar threads

Back
Top