Solving a Cycloid Equation: Finding θ(t)

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Homework Help Overview

The discussion revolves around the motion of a bead sliding on a frictionless wire shaped like a cycloid, with specific equations governing its path. Participants are exploring how to derive the equation of motion and subsequently apply a change of variable to find a linear equation for a new variable related to the angle θ.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to perform a change of variable from θ to u = cos(θ/2) and are discussing the implications of this change on the differential equation. There are questions about the correct application of derivatives and the nature of the resulting equations.

Discussion Status

There is ongoing dialogue about the correctness of derivatives and the formulation of the linear equation. Some participants are providing guidance on how to approach the differentiation and the implications of initial conditions, while others are expressing confusion about specific steps in the process.

Contextual Notes

Participants are working within the constraints of a second-order differential equation and are discussing initial conditions necessary for finding a particular solution. There is mention of the need for two conditions due to the nature of the equation.

  • #31
Orodruin said:
Yes.So now you can insert this into your solution and solve for theta to get ...?

Sorry but... If I know
proxy.php?image=http%3A%2F%2Fi.imgur.com%2FhrPNb6A.png
and I know B = 0. What I need to get?
 
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  • #32
You still need to solve for ##\theta(t)##. Your solution was for \cos(\theta)##...
 
  • #33
Orodruin said:
You still need to solve for ##\theta(t)##. Your solution was for \cos(\theta)##...

Then...
dfgslvm.png

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PHmdjzD.png

5OxTaE5.png


Is correct? Thanks!
 
  • #34
No, theta is a function of time.
 
  • #35
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)
 
  • #36
thomas91 said:
And now, knowing A and B I have the solution for cos(θ/2)
So theta becomes ...?
 
  • #37
Hello!

¿
vz0R0jK.png
?

Thanks!
 
  • #38
thomas91 said:
Hello!

¿
vz0R0jK.png
?

Thanks!
There we go!
 
  • #39
Finally! How can I calculate the oscillation period of the bean?
 
  • #40
How long does it take until it returns to the original position?
 
  • #41
It will be
WLpKf4Q.png
?

Simplified:
6vODwyt.png


Thanks!
 
  • #42
Bingo!

Or even simpler
$$
T= 4\pi \sqrt{\frac{a}{g}}
$$
 
  • #43
Thanks for everything Orodruin! Have a nice day :D
 

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