Pendulum: KE and PE

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Apr 29, 2011
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I have this in my notes:

PEpendulum= -mgL(1-cosθ)
L = string length

KEpendulum= sqrt(2gL(1-cosθ))

Is the KE equation wrong? Because now that I think about it, doesn't that only provide the VELOCITY?

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The PE formula is right as it is just a derived version of mgh - I am not sure about the negative sign though?

KE = .5mv^2 with your formula for the velocity.

Easier is to say that the total energy is conserved and that PEmax when θ = 90

So the PEmax = mgL (1-0) = mgL

Since the energy is conserved, KE + PE = PEmax

So KE = PEmax - PE =

KE = mgL - mgL(1-cosθ) = mgL[1 - (1-cosθ)] = mgL(cosθ)

So KE = mgLcosθ

I don't memorize them so it may be wrong, but this is how I would work it out if I had to.
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Do you have any idea what "sqrt(2gL(1-cosθ))" should represent?

It seems to be from the conversion of PE into velocity (as seen in Toricelli's result where effusion speed = sqrt(2gh)).
And h = L(1-cosθ)
Yes, you are right, it is the velocity of the bob.

It can be derived by looking at when all of the energy of the system is completely kinetic, such as when θ = 180.

mgh = 0.5mv^2

mgL(1-cosθ) = 0.5mv^2

2gL(1-cosθ) = v^2

v = sqrt[2gL(1-cosθ)]
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