A roller coaster car (m = 750kg with passengers) starts with near zero velocity at the top of a 80m section of track that is inclined at -38 degrees from the horizontal. During the descent the average drag force acting on the car is 550 N. At the bottom of the descent is a horizontal turn to the right. In the turn the track is banked at a 33 degree angle with respect to the horizontal.
a) If the occupants are to experience a maximum centripetal acceleration of 1.5 times Earth gravity at the start of the turn, what is the minimum turn radius allowable? Assume the only forces acting on the car are its weight, drag and normal forces.
b) Assuming the same drag forces act on the car during the turn through what angle does the car travel before the centripetal acceleration decreases to 1g?
PEo + KEo = PEf + KEf
ƩF = mV^2/r
Eo + Ef = W(non conservative)
PE = mgh
KE = 1/2 mv^2
The Attempt at a Solution
So what I did is, during the descent I used the equation Eo + Ef = W(non conservative) where I set initial kinetic energy is 0, initial potential energy would be mgh where h = Length(sinθ) subtracted by the non conservative work which is (drag force x length of track) and set that equal to the final kinetic energy. I then solved for Vf which I would be using for the ascent. Now during the ascent of part a, if the occupants are to experience a maximum centripetal acceleration of 1.5 times the gravity I assumed that it would mean that the normal force would equal to 1.5g, so with that said i drew a free body diagram of the ascent and applied the Forces( in the radial direction) where Fn - Fg = mv^2/r, which I then solved for R, using the V final I obtained from earlier.
ok now for part b..i'm a little stuck. I assume I just draw the same Free body diagram but this time add the drag force and use the same radius I found in part a and then just solve for the new θ?
thanks for all the suggestions and input!!