A 491 kg roller-coaster car rolls down past point A (radius of 10m) and then up past point B (radius of 15m).

Question

(a) If the vehicle has a speed of 26.0 m/s at point A, what is the force of the track on the vehicle at this point?

(b) What is the maximum speed the vehicle can have at B for gravity to hold it on the track?

Answer 1

I don't know about other tutors, but I have a general policy to not answer multi-part questions or one of multiple questions asked on this forum.

a) In terms of centrifugal force: the normal force = weight + centrifugal force or mg + mv²/R

In terms of centripetal force (which is the norm in physics books nowadays)

The net force, F_N - F_Wt = centripetal force, so F_N = mg + mv²/R (upward was taken as positive)

b) The max velocity at B is such that only the weight supplies the centripetal force: mg = mv²/R or v = sqrt(Rg)

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