A Ferris wheel of radius 100 feet is rotating at a constant angular speed omega

Question

A Ferris wheel of radius 100 feet is rotating at a constant angular speed omega rad/sec counterclockwise. Using a stopwatch, the rider finds it takes 3 seconds to go from the lowest point on the ride to a point Q, which is level with the top of a 44 ft pole. Assume the lowest point of the ride is 3 feet above ground level. Let Q(t)=(x(t), y(t)) be the coordinates of the rider at time f seconds; i.e., the parametric equations. Assuming the rider begins at the lowest point on the wheel, then the parametric equations will have the form: x(t)=rcos(omega t - pi/2) and y(t)=r sin (omega t -pi/2), where r, omega can be determined from the information given. Provide answers below accurate to 3 decimal places. During the first revolution of the wheel, find the times when the rider is height above the ground is 80 feet.

Explanation / Answer

Find the change in angle for 6 seconds.
cos(??) = (100 + 3 - 44 ft) / (100 ft)
?? = 0.9397375 rad

Calculate ?.
? = ??/?t
? = (0.9397375 rad) / (6 s)
? = 0.156623 rad/s

Parametric equation for y:
y(t) = 3 ft + (100 ft)sin(0.156623t - p/2)

Find first 2 times for y = 80 ft.
80 = 103 + 100 sin(0.156623t - p/2)
-0.23 = sin(0.156623t - p/2)
-0.232078 + 2pn = 0.156623t - p/2 or (p + 0.232078) + 2pn = 0.156623t - p/2

First time:
-0.232078 = 0.156623t - p/2
1.338718 = 0.156623t
t = 8.547 s

Second time:
p + 0.232078 = 0.156623t - p/2
4.944467 = 0.156623t
t = 31.569 s

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