Recent content by stateofdogma

Congratulations – you have completed Periodic Table Quiz. You scored 8 points out of 12 points total. Your performance has been rated as Competent. I guess on a lot of the questions

no nevermind, i think it works if you consider the dot product a distribution over a negative differential.

sorry, one last question isn't the ds abs(ds) so its negative would be turn positive by the absolute value, since this is a dot product.

thanks i get it now

its odd to me because in the second case you are expecting the work to be positive since at the bottom of the swing there is a maximum velocity and so the intial velocity should be smaller then the final leading to positive work integral.

the first is mgR(cos(θ) - cos(θi)) and the second is -mgR((cos(θ) - cos(θi)) the second gives a negative kinetic energy for the change from a higher θ to a lower θ(like from pi/2 to 0), is that correct?

This is my Queries does it not makes sense?

it would be mgRcos(θ+pi/2)dθ for the the first and the second is mgRcos(θ-pi/2), which are equal to -mgRsin(θ)dθ and mgRsin(θ)dθ.

And taking the opposite direction of displacement and force works, but trying to derive a situation where the point mass starts at zero velocity at a some θ and going to θ = 0 at max velocity makes no sense from this integral ∫force⋅displacement = Δk, where the directions are the same. Unless...

its when you take the integral ∫mgsin(θ)⋅ds and the direction of the ds and force is the same , then the integral sin is -cos for the integral, I don't see how to get this integral through your method, but isn't it valid for work-kinetic theorem ∫force⋅displacement = Δk

If i keep with your scenario, I have no problem with anything you said here. In your same setup I have a problem with starting from the top with the θ initial greater than zero and going to θ=0, if you use the work kinetic theorem you get a negative Δk, that is the - mgR(cos(θ) -...

wait do you mean that cos(0) is cos(zero), if so doesn't that leave a negative kinetic change for a pendulum on the decline.

i get it thanks

You need the change in sign to get the right answer, - mgR(cos(θ) - cos(θ(0)) will for instance give you a negative kinetic change if you take the initial θ(0) greater than θ. Which doesn't make sense for a pendulum starting from a higher height and ending at a lower height, it should have a...

Relating the kinetic change to the potential change is an easier way, I just don't understand why I can't use the work-kinetic energy theorem.