Particle Sliding on Smooth Cycloidal Trough

[alimerzairan]

Homework Statement

A particle is free to slide along a smooth cycloidal trough whose surface is given by the parametric equations:

x = \frac{a}{4}(2 \theta + \sin{2 \theta})
y = \frac{a}{4}(1 - \cos{\theta})

where 0 <= \theta <= \pi and a is a constant.
(sorry, TeX is not working for me...doesn't work well I guess on this browser).

(a) Find the Lagrangian and equation of motion of the particle. Simplify both to the most
compact form possible using trigonometric relations.

(b) Find (or recall) a variable change that reduces both the Lagrangian and the equation of motion to the ones of a harmonic oscillator.

Homework Equations

Euler-Lagrangian

The Attempt at a Solution

What I tried was deriving the Kinetic Energy, T, and Potential Energy, U.

The Potential Energy will just be: U = m g y(theta) (the equation given).

The Kinetic Energy is:

T = 1/2 m ((x')^2 + (y')^2).

But this assumes the first derivatives: x'(t) and y'(t) while I will have x'(theta) and y'(theta). So I just use the chain rule:

dx/dt = dx/dtheta * dtheta/dt

same for y.

Then I just plug it back in. My professor says that this should compact very nicely, but I don't get that. The only place were I could make a crucial error is here where I define the Kinetic and Potential Energies. Are these definitions correct? For some reason I am thinking that I should be adding a rotational kinetic energy term, but the particle is sliding on this surface so it most likely doesn't need it.

Basically, my concern is if I have the definition for Kinetic and Potential Energies wrong.

Any help would be greatly appreciated, especially before Thursday, December 2, 2010 at 1300 PDT (GMT-7).

Thank you.

 

[gabbagabbahey]

Homework Statement

A particle is free to slide along a smooth cycloidal trough whose surface is given by the parametric equations:

x = \frac{a}{4}(2 \theta + \sin{2 \theta})
y = \frac{a}{4}(1 - \cos{\theta})

where 0 <= \theta <= \pi and a is a constant...

The Attempt at a Solution

The Potential Energy will just be: U = m g y(theta) (the equation given).

The Kinetic Energy is:

T = 1/2 m ((x')^2 + (y')^2).

Keep in mind that a surface has two degrees of freedom (2 parameters), x and y are both determined by the parameter \theta, so, without seeing a diagram of the setup, I'd say it is safe to assume that z is the other free parameter for this surface, and that gravity points in the negative z-direction and, since the particle is free to move in the z-direction, there will also be a z-compponent of velocity:

U = mgz[/itex]<br /> T = \frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)

 

[Wesleytf]

edit: gabba--make a little parametric plot in 2D and you'll get an idea of the problem. It's equivalent to a particle on a cycloid-shaped wired, if I'm not mistaken.

Ali-I did it all in theta. Think about it- if your T and your U are in {theta, theta' and t}, you can just use those to make your lagrangian. For me the T simplified down quite a bit. It's basically the same approach I used in #2, IIRC.

FWIW I'm having trouble solving for the actual equation of motion and performing the substitution that will make it all nice and pretty. Anyhoo, see you tomorrow. good luck.

 

[alimerzairan]

Hi Wesley,

I did the very exact thing that you said. The part that I am stuck is on simplifying it nicely to find this equation of motion and the substitution.

@ gabba

Wesley is right. This is only for a two dimensional surface. There is no z involved.

 

[Wesleytf]

Do you have mathematica? At least that way you could check what you need to get to... I just finished work; hopefully I can make some progress on it after dinner. I'll let you know if I do.

 

[alimerzairan]

It came with a lot of thinking until I found my mistake with the help of Mathematica.

So: T = \frac{1}{2} m a^2 \frac{d\theta}{dt}^2 (\cos{\theta})^2

U = \frac{1}{2} m g a (\sin{\theta})^2

Plugging this into the lagrangian and doing the derivation I get the final equation of motion:

\ddot{\theta} = 3 \tan{\theta} (\dot{\theta})^2 + \frac{g}{a} \tan{\theta}

where I already recognize \frac{g}{a} = \sqrt{\omega}

Any suggestions from anyone. I am trying to solve part (b) right now and possibly a correction to part (a) if my equation of motion is wrong.

 

[alimerzairan]

Btw, I meant to write (cos{\theta})^2

same is true for the sine as well

 

[Wesleytf]

The solution is actually MUCH easier to find if you rearrange your final equation to something like

Cos[theta] theta'' - Sin[theta] theta'^2 = -g/a Sin[theta]

Then, look hard at the left hand side and see if you can figure out what it's a derivative of, and then, what THAT is a derivative of. Then, you'll have a simple second-order differential equation and you'll be able to see the substitution quite easily.

hope that helps...

 

[alimerzairan]

Wesley,

There was a mistake in my derivation. What you have is right. I got the same result. But I don't understand what you said in your last post about the derivatives.

What I did was:

\ddot{\theta} = \tan{\theta} (\dot{\theta})^2 - \frac{g}{a} \tan{\theta}

I then used the hint and approximations in Problem 7.14 of the Taylor book. Consider only small angles where any power of the derivatives is very small and can be set to 0. The tangent will then become theta (approximation). So all we have is:

\ddot{\theta} = -\frac{g}{a} \theta = - \omega^2 \theta

This was my way of doing the variable change. If you make the same change to the Lagrangian, you will get the same result.

 

[gabbagabbahey]

edit: gabba--make a little parametric plot in 2D and you'll get an idea of the problem. It's equivalent to a particle on a cycloid-shaped wired, if I'm not mistaken.

That's a curve (a cycloid), not a surface. If you extend that curve in the z-direction, you get a surface which looks like a sheet of paper rolled into a spiral...that is a cycloidal trough.

 

[Wesleytf]

yes, but if you take that 2-D plot and extend it you'd have a trough, not a spiral. If the particle were sliding down just a slice of that trough, it would become the 2-D problem as described.
The solution was just posted and this was the solution.

http://www.physics.uci.edu/%7Esasha/courses_files/111A/Fall10/sols_hmw8.pdf

if you want to have a look, it's number 5. He wound up doing the substitution earlier on, which makes the problem a bit easier.