At what speed does the car reach the top of the loop?

A roller coaster car (mass = 988 kg including passengers) is about to roll down a track . The diameter of the circular loop is 20.0 m and the car starts out from rest 40.0 m above the lowest point of the track. Ignore friction and air resistance and assume g = 9.81 m/s2. (a) At what speed does the car reach the top of the loop? (b) What is the force exerted on the car by the track at the top of the loop? (c) From what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop?

At what speed does the car reach the top of the loop?
a) At the highest point of the loop potential energy mgh is transformed into kinetic energy 1/2 mv^2. As the roller coaster has fallen 20 meters we use that for h and find v=sqrt (2gh)=19.8 m/s





b) The force exerted by the tracks must be equal to the centripetal force mv^2/r minus gravity mg. (Minus because there are opposite.)


mv^2/r = 988 x 19.8^2 / (20/2) = 38734 N


mg = 988 x 9.81 = 9692 N


Total 2.90 x 10^4 N (rounded of to three digits)





c) In b) we found that the centripetal force is much bigger than gravity. At what speed are they equal? We can put


mv^2/r = mg, solve for v = sqrt (g r) = sqrt (9.81 10) =9.9 m/s.


This corresponds to a kinetic energy of 1/2 mv^2 and a potential energy of mgh. Solve for h and find 5 meters. This means we have to release the car at 5 meters above the top of the loop or 25 meters above the bottom.
Reply:At what height does the car start?



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