# The Ferris Wheel

## Homework Statement

The figure below shows a Ferris wheel that rotates three times each minute. It carries each car around a circle of diameter 19.0 m.

a) What is the centripetal acceleration of a rider?

(b) What force does the seat exert on a 40.0 kg rider at the lowest point of the ride?

(c) What force does the seat exert on the rider at the highest point of the ride?

(d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom, going up?
Answer in ° (measured inward from the vertical)

## The Attempt at a Solution

a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s
Radius r = 19.0 m/2 = 9.5 m
Centripetal acceleration a = w^2 * r
= 0.314^2 * 9.5
= 0.937 m/s^2

b) Mass M = 40.0 kg
Let force = F
F is upward and weight Mg is downward. Centripetal force is upward.
F - Mg = Ma
Or F = M(a+g) = 40 * (9.8 + 0.937) = 429.5 N

c) Mass M = 40.0 kg
Let force = F
F is upward and Mg is downward. Centripetal force is downward.
Mg - F = Ma
Or F = M(g - a) = 40 * (9.8 - 0.937) = 354 N

d) Let force = F at angle theta below vertical.
Vertical component of F = F cos(theta) upward.
Horizontal component of F = F sin(theta)

There is no acceleration in vertical direction.
Therefore F cos(theta) = Mg--------------------------(1)
Centripetal force is horizontal.
Therefore F sin(theta) = Ma---------------------------(2)
Dividing (2) by (1),
tan(theta) = a/g
Or theta = atan(a/g) = atan(0.937/9.8) = 5.46 deg
From (1),
F = Mg/cos(theta) = 40 * 9.8/cos(5.46 deg) = 394 N
Therefore, force by the seat is 394 N at 5.46 deg below vertical.
(Note: I do not know what exactly is meant by inward from vertical. I have calculated the angle from vertically upward. If you want the angle from vertically downward, then it is 180 - 5.46 = 174.54 deg

I only have one chance to enter this, please check to see if the way i did it is correct.