The rollercoaster is designed so the wagon and its passenger start at a height H

Question

The rollercoaster is designed so the wagon and its passenger start at a height H=50 feet above ground, roll down and up to h=20 feet at which the wagon stops abruptly and releases the passenger horizontally to fly as a simple projectile, landing in a pond distance d away. Assuming the bottom of the track is at ground level and curves at a radius of 10 feet compute the maximal (centripetal) acceleration experienced by the passenger Assuming no friction and no loss in passenger's speed during ejection (i.e. s the keeps going as fast as just prior the ejection) compute d;. Suppose there is a 20% speed loss at the time of ejection, will d be longer or shorter? Compute it. Lastly, estimate the importance of friction by assuming the coefficient of kinetic friction mu ~ 0.005. total length of the track L = 100 feet and average normal force corresponding to weight of 400 lbs. lint: include energy loss due to friction and recompute 'exit' velocity, from which d can be recomputed md compared to d in part (a) and (b)

Explanation / Answer

a) The centripetal force is given by mv^2/r so if v is high the force will be higher.

So it will happen at the bottom of the circular path.

Now to find the velocity at the bottom, we have the work energy theorem like this:

mgH = 0.5 mv^2 - 0

v = sqrt(2gH) = sqrt(2*32.2*50) = 56.745 ft/s

so the centripetal acceleration = v^2 /r = 56.745^2/10 = 322 ft/s^2

b) The velocity of the person when the wagon stops:

v = sqrt(2g(H-h)) = sqrt(2*32.2*(50-20)) = 43.95 ft/s and the time taken to reach the ground = sqrt(2h/g) = sqrt(2*20/32.2) = 1.11 sec.

hence the distance d = v*t = 43.95*1.11 = 48.78 ft.

c) In this case the time will be constant and the velocity will be lower so the distance d will be lower.

d) The friction force = 0.005*400*32.2 = 64.4 and work done will be 64.4*100 = 6440 poundal*foot

So energy loss due to friction will be 6440 poundal*foot

work done by gravity = mg(H-h) = 400*32.2*(50-20) = 386400 poundal*foot

so the energy equation will be:

386400-6440 = 0.5*400*v^2

v = 43.59 ft/s

d_new = 43.59*1.11 = 48.38 ft. which is less than that in part b

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