How Long Does It Take to Complete a Loop on a Rollercoaster with Friction?

[jaumzaum]

Calculate the time t to complete a loop of a rollercoaster of radius R by a car that has initial velocity Vo, with friction u.

We can calculate the velcity V if we have alpha, the angle that the car is in the rollercoaster

acceleration in alpha = g.sen alpha - acceleration by atrictFc=N+g.cos alpha$A(\alpha ) = g.sen\alpha - (v²/R + g.cos \alpha)u$and
$V = \sqrt{Vo² + \int a.ds}$
The problem is now, I don't know how to solve the equation. I've tried to to this, but I don't know if it's right.
$A(\alpha ) = g. (sen\alpha +cos \alpha u) -(v²/R) U$

$A(\alpha ) = g. (sen\alpha +cos \alpha u) -(v²/R) u$
$v² = g. (sen\alpha +cos \alpha u) - A (\alpha) .u/R$
$Vo² + \int a.ds =( g. (sen\alpha +cos \alpha u) - A (\alpha)) .u/R$$a.ds = d(Vo²)- d(g. (sen\alpha +cos \alpha u)) .u/R - d(A (\alpha) .u/R$
$a.ds = d(Vo²)- d(g. (sen\alpha +cos \alpha u)) .u/R - d(A (\alpha) .u/R)$Now we have a equation with da and ds, alpha = S/R so alpha depends on dS and da depends on a, ok, how do we integrate this?

 

[jbsteele]

Let's assume that a = k.u, then da = k .du so a.ds = k .du .dS/R = k .dS .du/R Integrating Vo² + \int a.ds = Vo² + k .\int dS .du/R = Vo² + k .S .du/R V = \sqrt{Vo² + k .S .du/R } Now we add the two equations and solve for t V = \sqrt{Vo² + \int a.ds} t= \sqrt{Vo² + \int a.ds} /(k.S.du/R) So the time t to complete one loop of a rollercoaster of radius R by a car that has initial velocity Vo and friction u is:t= \sqrt{Vo² + \int a.ds} /(k.S.du/R)