Solving a Cycloid Equation: Finding θ(t)

[thomas91]

Hello! Nice to meet you. I'm studying a Bachelor of Science and I have a doubt:

A bead slides without friction on a frictionless wire in the shape of a cycloid with equations

x = a(θ-sin θ) y = a(1+cos θ)

where θ's range is 0 to 2π and a>0

a) Find the equation of motion:

And I resolved:

I checked in a physics book that the solution is correct

b) Make the change of variable u= cos (θ/2) and find a linear equation for u(t). Solve this equation and find the solution θ(t) which in the case of a bead at t = 0 falls from rest from the corresponding point θ= π/2

I don't know how to make the change. Can anybody help me? Thanks and nice to meet you.

 

[Orodruin]

Exactly what in the change of variables are you having problems with? It is a matter of reexpressing the differential equation in terms of u instead of theta.

 

[thomas91]

Exactly what in the change of variables are you having problems with? It is a matter of reexpressing the differential equation in terms of u instead of theta.

Thanks for answering! I do not understand how I have to make the change of variable, I'm trying to do since Wednesday and not even know where to start the change of variable

 

[Orodruin]

It is just a normal change of variables. You insert the new variable in place of the old one using the relation between them.

 

[thomas91]

It is just a normal change of variables. You insert the new variable in place of the old one using the relation between them.

I tried doing a simple change of variables. But I obtain a non linear equation..

 

[Orodruin]

Please show your work and how you got there. Your non-linear term should not be there. The other terms are correct.

 

[thomas91]

Please show your work and how you got there. Your non-linear term should not be there. The other terms are correct.

u= cos (θ/2) then:

θ = 2 arccos u

then:

and:

Then substitute:

Now:

and

and substitute:

We now that cos(arccos (a)) = a and sen(arccos(a) =

Using first and second derivates and simplifying:

Then multiplied by

and dividing by (-4) we obtain:

Which isn't linear.What is your opinion? Thanks!

 

[Orodruin]

It is much easier to help you if you stop pasting images and instead use the forum LaTeX to type in your equations (they become quotable).

Your second derivative is wrong, the second term needs to contain $(u')^2$. Also, it becomes much much easier if you do not invert the relation $u = \cos(\theta/2)$ but instead differentiate it as it stands.

 

[thomas91]

It is much easier to help you if you stop pasting images and instead use the forum LaTeX to type in your equations (they become quotable).

Your second derivative is wrong, the second term needs to contain $(u')^2$. Also, it becomes much much easier if you do not invert the relation $u = \cos(\theta/2)$ but instead differentiate it as it stands.

¿$(u')^2$? I don't see this.. I don't understand what you mean. Sorry.
Thanks Orodruin!

 

[thomas91]

Now I think that the second derivative is :

But it doesn't contains $(u')^2$

 

[Orodruin]

Now I think that the second derivative is :

But it doesn't contains $(u')^2$

It does, you did it wrong again. What is the derivative of
$$
\frac{1}{\sqrt{1-u^2}}?
$$

 

[thomas91]

$uu' ((1-u^2))^{-3/2}$
It's that right?

 

[Orodruin]

$uu' ((1-u^2))^{-3/2}$
It's that right?

So what is then the derivative of
$$
\frac{u'}{\sqrt{1-u^2}}?
$$

 

[thomas91]

So what is then the derivative of
$$
\frac{u'}{\sqrt{1-u^2}}?
$$

It's ok?

Thanks!

 

[Orodruin]

What happened to the $u$ coming out of the derivative of $1/\sqrt{1-u^2}$? But again, the best way forward is not to solve for $\theta$ before differentiating, it is to directly differentiate $u = \cos(\theta/2)$.

 

[thomas91]

What happened to the $u$ coming out of the derivative of $1/\sqrt{1-u^2}$? But again, the best way forward is not to solve for $\theta$ before differentiating, it is to directly differentiate $u = \cos(\theta/2)$.

Hello! If you do not mind, I prefer to finish the exercise by this method and later solve by the method you suggest

Then, the second derivative is:

Replacing first and second derivative:

And simplifying:

Simplifying again:

What do you think? Thanks and have a nice sunday

 

[Orodruin]

Looks fine apart from a square disappearing in your middle step (the end result is correct).

 

[Orodruin]

Ah yes, and the $a$ should be in the denominator in the end result ...

 

[thomas91]

Ah yes, and the $a$ should be in the denominator in the end result ...

Sorry! It was a mistake...
The linear equation is:

The characteristics polynomial is:

With complex roots:

The general solution:

We need to know the particular solution. True?Thanks!

 

[thomas91]

Edit: The equation is homogeneous. No particular solution. But then we can not replace, I can not find the solution

 

[Orodruin]

What do you mean by "cannot replace"? The homogenous equation has constants you have to adapt to the initial conditions. That will give you the solution you want.

 

[thomas91]

What do you mean by "cannot replace"? The homogenous equation has constants you have to adapt to the initial conditions. That will give you the solution you want.

Hello!

I think this:
a) Find the solution θ(t) which in the case of a bead at t = 0 falls from rest from the corresponding point θ= π/2

Then:

What do you think?

 

[Orodruin]

This settles A. You also need to find B.

 

[thomas91]

This settles A. You also need to find B.

But ... B is out of the equation because sin (0) = 0
Then B may be any value. Yes?

Thanks!

 

[Orodruin]

For that particular value of t it does not affect the position. But this is a second order differential equation so you need two conditions (and you have one more!).

 

[thomas91]

For that particular value of t it does not affect the position. But this is a second order differential equation so you need two conditions (and you have one more!).

Hello! I can't find the other condition. Can you give me a clue?

Thanks

 

[thomas91]

Can I set two arbitrary (Between 0 to 2π ) differentes values of θ and solve a equation system for t and B?

 

[Orodruin]

Can you give me a clue?

b) Make the change of variable u= cos (θ/2) and find a linear equation for u(t). Solve this equation and find the solution θ(t) which in the case of a bead at t = 0 falls from rest from the corresponding point θ= π/2

My highlighting.

 

[thomas91]

My highlighting.

Does the rest position mean that the velocity mean that θ' = 0 for t=0?

I tried to derivate θ and solve B for t=0 and I obtain B=0...

 

[Orodruin]

Does the rest position mean that the velocity mean that θ' = 0 for t=0?

Yes.

I tried to derivate θ and solve B for t=0 and I obtain B=0...

So now you can insert this into your solution and solve for theta to get ...?

 

[thomas91]

Yes.So now you can insert this into your solution and solve for theta to get ...?

Sorry but... If I know

and I know B = 0. What I need to get?

 

[Orodruin]

You still need to solve for $\theta(t)$. Your solution was for \cos(\theta)##...

 

[thomas91]

You still need to solve for $\theta(t)$. Your solution was for \cos(\theta)##...

Then...

Is correct? Thanks!

 

[Orodruin]

No, theta is a function of time.

 

[thomas91]

Then... Recapitulating
2nd condition = > θ' = 0 and t = 0 (rest position)
So I obtained B setting θ' (t = 0) = 0; so B = 0

And now, knowing A and B I have the solution for cos(θ/2)