Question

A Ferris wheel is a vertical, circular amusement ride with radius 9 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 10 s. Consider a rider whose mass is 54 kg.

At the bottom of the ride, what is the perpendicular component of the rate of change of the rider's momentum?

Answer 1

The rate of change of momentum at the bottom of a Ferris wheel is the centripetal force, calculated using the mass of the rider, the velocity from the rotational period of the wheel, and the radius of the wheel.

Step-by-step explanation:

In answering the question about the Ferris wheel, the perpendicular component of the rate of change of the rider's momentum at the bottom of the ride can be found by calculating the centripetal force, since momentum changes direction at this point. Given the information, the mass of the rider is 54 kg and the Ferris wheel completes one rotation in 10 seconds. The radius is 9 m. Therefore, the velocity (v) of the rider can be calculated using the circumference of the circle (C = 2πr) and the period (T), which is v = C/T. The force (F) responsible for the change in momentum (rate of change of momentum) is the centripetal force, given by F = mv²/r, where m is the mass, v is the velocity, and r is the radius of the circle. By plugging in the values, we can calculate this force which is also the rate of change of momentum per unit of time.