# Trouble with work derivation

1. Feb 16, 2015

### stateofdogma

1. The problem statement, all variables and given/known data
Trying to derive the work and kinetic energy from principle ie ∫force⋅ds, for a pendulum, with θ at the vertical.

2. Relevant equations
∫force⋅dx

3. The attempt at a solution
I toke the ∫mgsin(θ)⋅ds as the displacement and direction of the gravitation is opposite leading to the dot product - ∫mgsin(θ)ds,
ds=Rdθ
then the answer is mgR(cos(θ) - cos(θ(0)) , θ(0) is the initial, which makes sense since the path downwards leads to a positive number which corresponds to a increase in kinetic energy at the bottom.
But its when you take the path where the direction of the displacement and force are the same that you
get the integral - mgR(cos(θ) - cos(θ(0)). Which corresponds to opposite of the correct answer, ie loss of kinetic energy at the bottom of the pendulum...
So i dont really understand how to interpret this ? I thought the answer would be the same irregardless of the direction you pick to go, you should have got an answer which was correct both ways like the first result.

2. Feb 16, 2015

### AlephNumbers

What is the problem, word for word? I don't think I understand what they want you to do.

3. Feb 16, 2015

### stateofdogma

Im just trying to find the equation of the kinetic energy for a ball on a string using the equation ∫force⋅dx = Δk, and I find when I take the force as the force of gravity tangent to the curve of the string in the same direction as the path taken along the arc of the string the answer is - mgR(cos(θ) - cos(θ(0)) , which gives the kinetic energy as negative for a path that starts from θ greater than 0 and ends at 0. Which would imply that the kinetic energy is lowest and θ=0.

4. Feb 16, 2015

### AlephNumbers

The left side of the equation ∫force⋅dx = Δk can be re-written in terms of the change in gravitational potential energy to make this much simpler. Hint: ΔU = -w

5. Feb 16, 2015

### AlephNumbers

Also, if it's a "simple" pendulum the string should be thought of as a rigid, massless spindle.

6. Feb 17, 2015

### stateofdogma

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

7. Feb 17, 2015

### AlephNumbers

They are the same thing. ∫force⋅dx = Δk = w = -ΔU. By definition, -dU(x)/dx = f(x)

8. Feb 17, 2015

### AlephNumbers

I think I see what you are saying. This is a bit over my head. I'll need some time to think about this.

9. Feb 17, 2015

### haruspex

Why do you think you have to change the sign on the algebraic expression? The expression should be the same, though its value may change sign.

10. Feb 17, 2015

### stateofdogma

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 doesnt make sense for a pendulum starting from a higher height and ending at a lower height, it should have a positive kinetic change. You get that equation if you take the force in the same direction as the dispacement (ds).

11. Feb 17, 2015

### lightgrav

when you integrate downward (inward), the initial angle is θ , and the final angle is 0 : - mgR [cos(0) - cos(θ) ] = mgR [cos(θ) - cos(0) ]

12. Feb 17, 2015

### stateofdogma

i get it thanks

13. Feb 17, 2015

### stateofdogma

wait do you mean that cos(0) is cos(zero), if so doesnt that leave a negative kinetic change for a pendulum on the decline.

14. Feb 17, 2015

### haruspex

For concreteness, let's say theta is measured anticlockwise from the vertical. The tangential force is $mg\sin(\theta)$, and it always opposes the displacement, so $r\ddot\theta=-g\sin(\theta)$. (There's no Coriolis term because r is constant.)
Integrating, $\frac r2 {\dot \theta}^2 =g\cos(\theta)-g\cos(\theta_0)$, where $\theta_0$ is the angle at which the speed is zero, i.e. $\theta_{max}$.
Now, where do you see a problem?

15. Feb 17, 2015

### stateofdogma

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) - cos(θ(initial)), which is not the answer. Which implies that the velocity starting from some value and ended at zero which doesnt make sense. Do we just cross out the negatives of the -Δk and - mgR(cos(θ) - cos(θ(0)) and get the same answer as your above derivation?

16. Feb 17, 2015

### haruspex

I still don't see how you get that minus sign in front. It's not in the integral I posted.

17. Feb 17, 2015

### stateofdogma

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 dont see how to get this integral through your method, but isn't it valid for work-kinetic theorem ∫force⋅displacement = Δk

18. Feb 17, 2015

### haruspex

But I took the integral of -mgsin(θ). The directions of displacement and force are always opposite for a pendulum.

19. Feb 17, 2015

### stateofdogma

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 you just cancel out the negative on the left side and the supposed negative on the right side, which would give you the same formula as yours.

Last edited: Feb 17, 2015
20. Feb 17, 2015

### ehild

The change of KE is equal to the work done, and work is displacement times force times the cosine of angle between them.

See attachment:: the red arrow is the force (gravity) and the blue arrows represent the direction of infinitesimal displacement. What is the angle between displacement and force when the pendulum moves in the direction θ>θ0 and when it moves in the direction θ<θ0 ?