Question

Consider the carnival ride in which the rider stands against the wall inside a large cylinder. The ride then starts to spin which accelerates the rider to speed v such that the centripetal force supplied by the normal force of the cylinder wall on the rider's back supplies enough friction to support the rider's weight. Assume the rider has a mass of 65 kg and the radius of the spinning cylinder is 6.0 m. What speed does the rider have to be spinning when the floor drops out beneath her, if the normal force must be at least half their weight? Answer in m/s.

Answer 1

The speed at which the rider is spinning when the floor drops out beneath her is 5.42 m/s.

Acceleration of the rider

The acceleration of the rider when the normal force is half their weight is calculated as follows;

Fn = 0.5(W)

ma = 0.5 x mg

65 x a  = 0.5 x 65 x 9.8

a =  0.5 x 9.8

a = 4.9 m/s²

Speed of the rider

a = v²/r

v² = ar

v = √ar

v = √(4.9 x 6)

v = 5.42 m/s

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