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    Hey guys, was wondering if someone can tell me the equations for the kinetic energy and gravitational energy of the pendulum bob. I've been searching on the internet for some time and have gotten confused badly.
    Right now I know that h(height) should be: L(1-cosϴ) where L is the length of the string.
    This should mean GPE is m*g*L(1-cosϴ) where m is mass and g is gravity right?
    Also v(velocity) is the derivative of displacement(am*cos(ωt))
    so surely KE is 0.5*m*v^2?
    If I'm wrong please correct me I really would appreciate it.
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    Those equations seem correct, although it slightly depends on what you want them for.

    GPE is always mgh, and KE is always mv^2/2. In this case, you might want to find them as functions of angle though, in which case you correctly say that the hight above the equilibrium point is h=L(1-\cos \theta). If you want to find the kinetic energy, then you can use conservation of energy to find it at any angle.
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    (Original post by lerjj)
    Those equations seem correct, although it slightly depends on what you want them for.

    GPE is always mgh, and KE is always mv^2/2. In this case, you might want to find them as functions of angle though, in which case you correctly say that the hight above the equilibrium point is h=L(1-\cos \theta). If you want to find the kinetic energy, then you can use conservation of energy to find it at any angle.
    Thank's for the answer, was also wondering am I allowed to do that if say damping was involved. If not, is there another way to find the energies?
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    (Original post by crazyy)
    Thank's for the answer, was also wondering am I allowed to do that if say damping was involved. If not, is there another way to find the energies?
    If damping is involved then the energies can be found in the same way (i.e. you can always find potential energy given the position of the particle).

    What changes with damping is that some energy is being converted into heat - this means you lose the ability to predict the kinetic energy given the position of the pendulum and the initial velocity - I'll explain why:

    Say we have a pendulum and it's in the equilibirium position. We give it a kick and it starts to move with speed u - in the undamped case we know that the particle's energy is  \frac{1}{2}mu^2 and since we can work out the potential energy at any point we can find the kinetic energy at that point by saying  \frac{1}{2}mv^2+mgL(1-\cos\theta) = \frac{1}{2}mu^2 . However, if the system is damped then we have lost some unknown amount of energy as heat. In this case there is an extra term - 

 \frac{1}{2}mv^2+mgL(1-\cos\theta) = \frac{1}{2}mu^2 - \Delta H and so we can no longer solve the equation.

    If you knew how strong the damping was you could in principle calculate the rate at which energy was lost and still do the calculation using energies, but since damping will be velocity dependant, you're probably going to need to do it numerically.
 
 
 
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