Simple cycloid time calculation problem

[maverick280857]

Homework Statement

Consider a wire bent into the shape of the cycloid

$$x = a(\theta - \sin\theta)$$
$$y = a(\cos\theta -1)$$

If a bead is released at the origin and slides down the wire without friction, show that [itex]\pi\sqrt{a/g}[/tex] is the time it takes to reach the point [itex](\pi a, -2a)[/tex] at the bottom.

Homework Equations

(See below)

The Attempt at a Solution

Energy conservation gives

$$\frac{1}{2}mv^{2} = mg(2a)$$
or
$$v^{2} = 4ga$$

For the point at the bottom, $\theta = \pi$. So, the arc length is

$$s = \int_{0}^{\theta}\sqrt{\left(\frac{dx}{d\theta}\right)^{2} + \left(\frac{dy}{d\theta}\right)^{2}}d\theta$$

$$v = \frac{ds}{dt}$$

How do I get rid of the $$d\theta/dt$$? I know I'm missing something here...

 

[maverick280857]

Okay I got it.

 

[m2003]

To get rid of the dθ/dt, you can use the chain rule:

ds/dt = ds/dθ * dθ/dt

Then, you can substitute in the values for ds/dθ and dθ/dt that you calculated earlier. This will give you an expression for v in terms of θ and t.

Next, you can use the fact that v^2 = 4ga to solve for t in terms of θ. This will give you an expression for the time it takes to reach the bottom in terms of θ.

Finally, you can plug in the value of θ = π to find the time it takes to reach the bottom, which will be in the form of π√(a/g).