Question

A roller-coaster car has a mass of 660 kg when fully loaded with passengers. The car passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.2 m/s. What is the force of the track on the car at the top of the hill?

Answer 1

The forces acting on the roller coaster are:

1) Force F = mg acting downwards.

2) Force F' applied by the track on the car.

By Newton's 2nd law we know that the net force is equal to mass times acceleration.

Now at the top, the velocity is 8.2 m/s and therefore its acceleration will be:

a=v²/r

a = 8.2²/15 = 4.4826 m/s²

m = 660kg

therefore the net force F_net = 660 x 4.4826 = 2958.56 N acting vertically downwards.

And we know that the gravitational force acting downward is given by F = mg

this will be: F = mg = 660 x 9.8 = 6468 N acting vertically downwards.

But we know that the net force is must be downwards for the car to remain on the track and so the Force F' will be:

mg - F' = F_net

therefore F' = 6468 - 2958.56 = 3509.44 N upwards.