- #1
freeurmind101
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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?
Answer in m/s^2
(b) What force does the seat exert on a 40.0 kg rider at the lowest point of the ride?
Answer in N
(c) What force does the seat exert on the rider at the highest point of the ride?
Answer in N
(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 N
Answer in ° (measured inward from the vertical)
Homework Equations
The Attempt at a Solution
a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s
= pi/10 rad/s
Or w = 0.314 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.
