Question

Part of a roller-coaster ride involves coasting down an incline and entering a loop 8.60 m in diameter. For safety considerations, the roller coaster's speed at the top of the loop must be such that the force of the seat on a rider is equal in magnitude to the rider's weight. From what height above the bottom of the loop must the roller coaster descend to satisfy this requirement?
m

Answer 1

Given is:-

Diameter of the loop D = 8.60m, thus radius is r = 8.60 / 2 = 4.30m

Now,

The weight of the rider is equal to the centripetal force, thus

-= mg

or

6.1 =

v2 = (4.30m)(9.8m/s2)

which gives us

v = 6.49153m/s

Now applying the conservation of energy,

Estarting = Etop of the loop

mgH = mg(2r) += mu

or

ng + (47)6 = 16

by plugging all the values we get

(9.8m/s2) H = (9.8m/s2)(2x 4.30m) + (6.49153m/s)

which gives us

H = 10.75m

This is the required height to maintain the given condition.