I Predicting Motion of a Swing on a Non-Horizontal Branch

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The discussion focuses on predicting the motion of a swing suspended from a non-horizontal branch, emphasizing that sufficient information is needed to specify the problem completely. Key factors include the positions of points C and D, the nature of the initial kick, and the mass distribution of the swing. The swing's motion can be analyzed using conservation of energy principles, with specific angles defined for the ropes relative to the coordinate axes. The initial kick is crucial as it determines the swing's starting position and energy. Overall, the approach appears to be on the right track for predicting the swing's motion.
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TL;DR
Do you understand swings?
swing.png


A swing is suspended from a non-horizontal tree branch. Points C and D are fixed in space. All 4 line segments in the diagram have constant distance. After some initial "kick" imparts energy to the system the only force acting externally on the system is gravity.

Is it possible to predict the motion of the swing?
 
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Yes, if we have enough information to completely specify the problem. What are the four points and what is attached between them? Where is the “initial kick” applied and what force is it, applied for how long? How is the mass of the swing distributed?
 
AB is the seat. AC and BD are the ropes. CD is the branch. I assume the system remains under tension. The initial kick could be positioning the swing away from the minimum energy position then releasing. Mass is centered on the swing seat with some non zero moment of inertia.
 
I define ##\theta## to be the angle a rope makes relative to z and ##\phi## is the angle relative to x. The branch is in the xz plane. From conservation of energy I got
$$ \frac{r_A^2}{2} \left[\left(\frac{\partial\theta_A}{\partial t}\right)^2 + \left(\frac{\partial\phi_A}{\partial t}\right)^2\right]+ \frac{r_B^2}{2}\left[\left(\frac{\partial\theta_B}{\partial t}\right)^2 + \left(\frac{\partial\phi_B}{\partial t}\right)^2\right] - r_A\cos \theta_A - r_B\cos\theta_B = 0$$
Am I on the right track?
 
Thread 'Why higher speeds need more power if backward force is the same?'
Power = Force v Speed Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same? I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force)...

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