Classical Mechanics, cycloid pendulum

[fluidistic]

Homework Statement

The cycloid pendulum consists of a particle under the effect of a constant gravitational field (\vec g = -g \hat y) that moves without friction over a curve described parametrically by x=a(\theta + \sin \theta) and y=a(1-\cos \theta ).
1)Write down the Lagrangian and the equations of motion. (Hint, use \theta as a generalized coordinate)
2)Solve the motion equation and verify that the pendulum is rigorously chronicle.

Homework Equations

L=T-V. Lagrange's equation.

The Attempt at a Solution

1)L=T-V. V=mgh=mgy.
v^2=2a^2 \dot \theta ^2 [1+\cos \theta ].
Thus the Lagrangian L=ma ^2 \dot \theta ^2 [1+ \cos \theta ] + mga [\cos \theta -1 ].
Using \theta as the only generalized coordinate, I derive the equation of motion as being 2a [\cos ( \theta )\ddot \theta + \ddot \theta - \sin (\theta )\dot \theta ^2 ]+\sin \theta (a \dot \theta ^2 +g )=0.
I must solve this... is this really serious? :/
If I consider small oscillations, \sin \theta \approx \theta and \cos \theta \approx 1. The motion equation reduces to \ddot \theta - \frac{\theta \dot \theta}{2}+ \frac{\theta \dot \theta ^2}{4}+\frac{g \theta }{4a}=0. Which still doesn't look trivial to me to solve. :/

Did I go wrong somewhere? If not, how do I tackle the equation of motion? (and which one do I choose?)
Thanking you!

 

[hikaru1221]

The reduced equation doesn't look really okay to me (check the dimensions of the 2nd and the 3rd terms). Anyway, in small angle approximation, we usually omit all the terms of high degrees of not only \theta but also d\theta/dt and d^2\theta/dt^2. So the 2nd and 3rd terms should be omitted.

That fact is, I believe, for this reason: intuitively, if the distance is short, and you know the time taken is not so small, you should expect that the speed and the acceleration are also small. Usually this argument applies well to oscillation, as when we reduce the oscillation equation to the 1st degree (to the degree of linearity) - or harmonic oscillation, the "time taken" is specified by the period which doesn't depend on how short or long the amplitude (or "distance") is and instead depends on the choice of the inherent characteristics of the system. So normally we cannot assume that the period (or "time taken") is small.

 

[fluidistic]

The reduced equation doesn't look really okay to me (check the dimensions of the 2nd and the 3rd terms). Anyway, in small angle approximation, we usually omit all the terms of high degrees of not only \theta but also d\theta/dt and d^2\theta/dt^2. So the 2nd and 3rd terms should be omitted.

That fact is, I believe, for this reason: intuitively, if the distance is short, and you know the time taken is not so small, you should expect that the speed and the acceleration are also small. Usually this argument applies well to oscillation, as when we reduce the oscillation equation to the 1st degree (to the degree of linearity) - or harmonic oscillation, the "time taken" is specified by the period which doesn't depend on how short or long the amplitude (or "distance") is and instead depends on the choice of the inherent characteristics of the system. So normally we cannot assume that the period (or "time taken") is small.

Thanks a lot for your reply.
Well the dimensions will never match for each term I think. g is in m/s while the phi's are rad, rad/s, rad/s², etc. So basically my equations is all wrong :/
You said that I should depreciate the 2nd and 3rd term. But the first term is angular acceleration I think, which should be considered small too? I mean, why not depreciate the first term too?

 

[hikaru1221]

Then you have to omit the last term too It depends on the degree of approximation. The 2nd and 3rd terms are omitted because they are of higher degree (small quantity x small quantity; while for the 1st and the last ones, it's just one small quantity).
But you'd better check the equation first.

 

[fluidistic]

Then you have to omit the last term too It depends on the degree of approximation. The 2nd and 3rd terms are omitted because they are of higher degree (small quantity x small quantity; while for the 1st and the last ones, it's just one small quantity).
But you'd better check the equation first.

Ahh thanks a lot, now I think I understand what you meant. Small quantity x small quantitie = small quantity squared = very small quantity= negligible quantity.
Ok I'll redo the algebra...

 

[fluidistic]

I've redone the algebra and I find a simpler expression, so I'm very grateful to you! :D
Equation of motion: \ddot \theta [2a + \cos \theta ]+g \sin \theta =0.
For small angles this becomes \ddot \theta (2a+1) +g \theta =0. Or written even simpler, \ddot \theta +K \theta=0.
I hope I'm right this time...

Edit: Solving this diff. eq. I get \theta (t)=c_1 \sin (\sqrt K t)+c_2 \cos (\sqrt K t). It's obvious to me now that it's a periodic function. I'll try to find the period.

Edit 2: Done, period is \frac{2 \pi}{\sqrt {\frac{g}{2a+1}}}. I'd love if someone could verify my answer. (It's not even homework, I'm preparing myself for the final exam on mid/end July. I just want to be sure I'm doing things the right way :) )

 

[hikaru1221]

Check the dimension of the equation of motion
I guess it should be \frac{d^2\theta}{dt^2}2a(1+cos\theta)+gsin\theta = 0

 

[ideasrule]

Equation of motion: \ddot \theta [2a + \cos \theta ]+g \sin \theta =0.

Are you sure? I got 2a\ddot \theta [1 + \cos \theta ]+g \sin \theta =0

Edit: Solving this diff. eq. I get \theta (t)=c_1 \sin (\sqrt K t)+c_2 \cos (\sqrt K t). It's obvious to me now that it's a periodic function. I'll try to find the period.

Yes, that's right, assuming you use the right value of K. Typically, solutions of differential equations that look like \ddot x + w^2x =0 are written as Asin(wx+phi). This is mathematically identical to your solution, but it's easier to understand intuitively.

 

[ehild]

Using \theta as the only generalized coordinate, I derive the equation of motion as being 2a [\cos ( \theta )\ddot \theta + \ddot \theta - \sin (\theta )\dot \theta ^2 ]+\sin \theta (a \dot \theta ^2 +g )=0.

It is right, but you made a mistake in the next line, it should be:

\ddot \theta - \frac{\theta \dot \theta^2}{2}+ \frac{\theta \dot \theta ^2}{4}+\frac{g \theta }{4a}=0.

ehild

 

[fluidistic]

Thanks guys for your help. I've redone the algebra 3 times now, starting from the Lagrangian. My equation of motion is 2a \ddot \theta [1+ \cos \theta]-a\dot \theta ^2 \sin \theta +g \sin \theta =0.
It seems like I have an extra "-a \dot \theta ^2 \sin \theta" term. I don't know how you could have got rid of it.
\frac{\partial L}{\partial \theta}=-ma\sin \theta [a \dot \theta ^2 -g].
While \frac{\partial L}{\partial \dot \theta}=2ma^2 [1+\cos \theta ]\dot \theta.
Deriving this with respect to time gave me\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \theta} \right ) =2ma^2 \{ \ddot \theta [1+\cos \theta ] - \dot \theta ^2 \sin \theta \}. Hence my equation of motion... do you spot any error in my derivation?
And I don't really know what is meant by "check out the dimension of the equation of motion"
Some term has rad/s², another has m/s² because of the g term so the dimension seems like ill defined in my (our?!) equation of motion.
Edit: For small oscillations, our motion equation is the same, namely \ddot \phi+\frac{g \phi}{4a}=0. Solving it gives \theta (t)=A\sin \left ( \frac{1}{2} \sqrt{\frac{g}{a}} t + \phi \right ) with period \pi \sqrt{ \frac{a}{g}} if I didn't make any error. So yes, the pendulum has a period, it's chronicle.

 

[ideasrule]

It seems like I have an extra "-a \dot \theta ^2 \sin \theta" term. I don't know how you could have got rid of it.

You can assume it's negligible because it's the product of sin(theta), a small quantity, with -a \dot \theta ^2, an even smaller quantity.

\frac{\partial L}{\partial \theta}=-ma\sin \theta [a \dot \theta ^2 -g].
While \frac{\partial L}{\partial \dot \theta}=2ma^2 [1+\cos \theta ]\dot \theta.
Deriving this with respect to time gave me\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \theta} \right ) =2ma^2 \{ \ddot \theta [1+\cos \theta ] - \dot \theta ^2 \sin \theta \}. Hence my equation of motion... do you spot any error in my derivation?

Nope, it's correct.

And I don't really know what is meant by "check out the dimension of the equation of motion"
Some term has rad/s², another has m/s² because of the g term so the dimension seems like ill defined in my (our?!) equation of motion.

This isn't possible. In any equation, all of the terms must have exactly the same units, or else the equation is guaranteed to be wrong. However, your equation of motion has consistent units, because a^2 has units of m^2.

Edit: For small oscillations, our motion equation is the same, namely \ddot \phi+\frac{g \phi}{4a}=0. Solving it gives \theta (t)=A\sin \left ( \frac{1}{2} \sqrt{\frac{g}{a}} t + \phi \right ) with period \pi \sqrt{ \frac{a}{g}} if I didn't make any error. So yes, the pendulum has a period, it's chronicle.

I think the period should have an extra factor of 4 in front of it. Otherwise, your answer is correct.

 

[fluidistic]

Thank you very much for all guys. I didn't know a had units. Now everything is clear.
Problem solved.