justadream Full Member 10+ Year Member Joined Apr 29, 2011 Messages 2,171 Reaction score 863 Sep 14, 2014 #1 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? Members don't see this ad.

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? Members don't see this ad.

Cawolf PGY-2 10+ Year Member Joined Feb 27, 2013 Messages 3,469 Reaction score 2,287 Sep 14, 2014 #2 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. Upvote 0 Downvote

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.

justadream Full Member 10+ Year Member Joined Apr 29, 2011 Messages 2,171 Reaction score 863 Sep 14, 2014 #3 @Cawolf 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) Upvote 0 Downvote

@Cawolf 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)

Cawolf PGY-2 10+ Year Member Joined Feb 27, 2013 Messages 3,469 Reaction score 2,287 Sep 14, 2014 #4 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)] Upvote 0 Downvote

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)]