A 42.0-kg child takes a ride on a Ferris wheel that rotates four times each minu

Question

A 42.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 20.0 m. (a) What is the centripetal acceleration of the child? magnitude m/s2 direction (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? magnitude N direction (c) What force does the seat exert on the child at the highest point of the ride? magnitude N direction (d) What force does the seat exert on the child when the child is halfway between the top and bottom? (Assume the Ferris wheel is rotating clockwise and the child is moving upward.) magnitude N direction

Explanation / Answer

(a)
Angular speed is 4 rpm = 4/60 revs per second = 0.0667 rps, diameter D = 24m, radius r = 12m
Tangential speed v = 0.0667 *pi*D m/s = 5.03 m/s
Centripetal acceleration a = v^2/r = (5.03)^2 / 12 m/s^2 = 2.105 m/s^2

(b)
The centripetal accelerating force is always towards the centre of the Ferris wheel
But the apparent centrifugal force experienced by the child is in the opposite direction.
So its a downward force assisting gravity at the lowest point of the ride, upward against gravity at the highest point of the ride, and horizontal at right angles to gravity at the halfway point (so the seat then only has to resist gravity).

The apparent centrifugal force is F = m*v^2/r = 33kg * (5.03m/s)^2 / 12m = 69.48N
The force of gravity is f = mg = 33kg * 9.81m/s^2 = 324N

At the lowest point of the ride, force exerted by seat = 324N + 69.48N = 393N STRAIGHT UP

(c)
At the highest point of the ride, force exerted by seat = 324N - 69.48N = 255N STRAIGHT UP

(d)
At the halfway point of the ride, force exerted by seat = 324N STRAIGHT UP (resisting gravity only)

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