# The Ferris Wheel

by freeurmind101
Tags: ferris, wheel
 P: 5 1. The problem statement, all variables and given/known data 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) 2. Relevant equations 3. 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.
 PF Patron HW Helper P: 3,394 All calcs look good. For (d) it would be the 5.44 degree answer because the Fg is much larger than the Fc, so only a little off vertical.

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