How Does Pendulum Motion Affect Work and Kinetic Energy Calculations?

[stateofdogma]

Homework Statement

Trying to derive the work and kinetic energy from principle ie ∫force⋅ds, for a pendulum, with θ at the vertical.

Homework Equations

∫force⋅dx

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 don't really understand how to interpret this ? I thought the answer would be the same regardless of the direction you pick to go, you should have got an answer which was correct both ways like the first result.

 

[AlephNumbers]

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

 

[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.

 

[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

 

[AlephNumbers]

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

 

[stateofdogma]

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.

 

[AlephNumbers]

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

 

[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.

 

[haruspex]

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...

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.

 

[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 doesn't 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).

 

[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) ]

 

[stateofdogma]

i get it thanks

 

[stateofdogma]

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.

 

[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?

 

[stateofdogma]

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?

If i keep with your scenario, I have no problem with anything you said here.

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.

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 doesn't 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?

 

[haruspex]

if you use the work kinetic theorem you get a negative Δk, that is the - mgR(cos(θ) - cos(θ(initial)),

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

 

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

 

[haruspex]

its when you take the integral ∫mgsin(θ)⋅ds and the direction of the ds and force is the same

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

 

[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.

 

[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 ?

 

[stateofdogma]

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.

 

[ehild]

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.

Well done. What is the work in both cases?

 

[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.

This is my Queries does it not makes sense?

 

[ehild]

I do not see any problem with work. What work is done when the pendulum moves from 0 to some angle θ, or from some angle to zero
angle? You need to include the cosine of the angle between force and displacement which is not the same in those two cases.

 

[stateofdogma]

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?

 

[stateofdogma]

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.

 

[ehild]

In the second case $W= \int_{\theta_i}^{\theta_f}{mg\sin(\theta) ds }$ with ds= - Ldθ. So $W=- \int_{\theta_i}^{\theta_f}{mgL\sin(\theta) d\theta }=mgL(\cos(\theta_f)-cos(\theta_i))$ positive if θ_f< θ_i

 

[haruspex]

∫force⋅displacement = Δk, where the directions are the same.

If you are considering theta starting at $-\theta_{max}$ and increasing to zero then $d\theta$ is positive and the force is positive, so the integral is positive. If you take theta starting at $\theta_{max}$ and decreasing to zero then $d\theta$ is negative and the force is negative, so the integral is positive.

 

[stateofdogma]

thanks i get it now

 

[stateofdogma]

In the second case $W= \int_{\theta_i}^{\theta_f}{mg\sin(\theta) ds }$ with ds= - Ldθ. So $W=- \int_{\theta_i}^{\theta_f}{mgL\sin(\theta) d\theta }=mgL(\cos(\theta_f)-cos(\theta_i))$ positive if θ_f< θ_i

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.

 

[stateofdogma]

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

 

[haruspex]

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

Not sure what you mean by that, but to be clear, no, ds does not mean abs(ds). If the range "a to b" is such that the variable of integration, x say, decreases from 'lower' bound a to 'upper' bound b then dx is negative. E.g. $\int_a^b f.dx = -\int_b^a f.dx$.