# Descent time on cycloid

1. Nov 1, 2011

### brainpushups

I was trying to derive the time it takes a particle to reach the bottom of a cycloid. I know the result is π√(a/g) where g is the acceleration due to gravity and a is the constant in the following parametric equations that give the cycloidal path

x(θ) = a(θ-sin[θ])

y(θ)= a(1-cos[θ])

but I am having trouble with the derivation. Here's what I have:

The time is found by t = ∫ds/v where ds is the arc length and v is the speed. The speed is easily found using energy conservation

v = √(2g(y(0)-y(t))

In terms of the parameter θ, we have

v = √(2ga(cos[ψ]-cos[θ]) where ψ gives the initial position of the particle.

The arc length ds = √(dx^2+dy^2), but since the curve is parametrized in terms of θ we can write

ds = (ds/dθ)dθ = √((dx/dθ)^2+(dy/dθ)^2)dθ

We have

dx/dθ = a(1-cos[θ])
dy/dθ=a sin[θ]

After some simplification (squaring dx/dθ and dy/dθ, adding the squares and simplifying)

ds = a √(1-2 cos[θ])

Finally we get

t = ∫a√(1-2 cos[θ])/√(2ga(cos[ψ]-cos[θ]) with limits from ψ to π. Not sure how to approach this integral. I did find an online homework assignment that suggests to write

ds = 2 R sin[θ/2] where I assume R is what I was calling 'a', but I don't see how to apply any identity(ies) to get this result.

2. Nov 2, 2011

### brainpushups

Got it.

For starters, I made an algebraic mistake. ds should be a √(2(1-cos[θ]))

From this the half angle identity can be applied to give the 2 a sin[θ/2] result.

Also, by applying the identity cos[θ]=2 cos^2[θ/2]-1 to both terms in the denominator you can get

√(2 g a(cos^2[ψ/2]-cos^2[θ/2]). From this point its a straightforward substitution.