A rollercoaster car, having a mass of 950kg when loaded, starts form rest at the

Question

A rollercoaster car, having a mass of 950kg when loaded, starts form rest at the top of a long ramp. h=50m and R=10m.

a) If the friction force is negligible, how fast is the car moving at its lowest point?

b) If the speed at this point is actually 25m/s, how much energy has been dissipated through frictional losses?

c) The car enters a vertical loop-the-loop of radius 10m. Assuming that the frictional losses from the bottom to the top of the loop equal 20% of the kinetic energy at the bottom, how fast will the car be traveling when it reaches the top of the loop? Will the car need an auxiliary rail to prevent its leaving the track?

A rollercoaster car, having a mass of 950kg when loaded, starts form rest at the top of a long ramp. h=50m and R=10m. a) If the friction force is negligible, how fast is the car moving at its lowest point? b) If the speed at this point is actually 25m/s, how much energy has been dissipated through frictional losses? c) The car enters a vertical loop-the-loop of radius 10m. Assuming that the frictional losses from the bottom to the top of the loop equal 20% of the kinetic energy at the bottom, how fast will the car be traveling when it reaches the top of the loop? Will the car need an auxiliary rail to prevent its leaving the track?

Explanation / Answer

a) conservation of energy

m g h = 1/2 mv^2

v = sqrt( 2 g h) = sqrt(2*9.81*50)= 31.3 m/s

b)

dE = E - KE = 950*9.81*50 - 0.5*950*25^2= 1.69E5 J

c) so

E top = 0.8 E bottom

m g 2 R + 1/2 mv^2 = 0.8 1/2 mv^2

9.81*2*10 + 0.5*v^2 = 0.8*0.5*25^2

v=10.37 m/s

at top min speed without rail is

m g = mv^2/r

v = sqrt( g r) = sqrt(9.81*10) = 9.9 m/s
so won't need a rail

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