14 points I Previous Answers A Ferris wheel o radius 100 feet is rotating at a c

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14 points I Previous Answers A Ferris wheel o radius 100 feet is rotating at a constant angular speed rad sec counterclockwise. Using a stopwatch, the rider finds it takes 3 seconds to go from the lowest point on the the lowest point of the ride is 3 feet above ground level. My Notes ide to a aint which s level with the top of a 44 t pole. Assume bound ael Let ae x y be the coordinates of the ride at time t seconds; i e the para metric equations. Assuming the rider begins at the owest point on the wheel, he the parametric equations will hove the fo m: cos t 2 and y -rsi at-T where r can be determined om the nformation given. Provide answers below accurate to 3 dec mal places. Note: we have mposed a coordinate system so that the center of the erris wheel is the origin. There are other ways to impose coordinates, leading to different parametric equations.) (a)/ " 100 -feet (b) -0325 rad/sec (c) During the first revolution of the wheel, find the times when the rider's height above the ground is 80 feet first time = 7667 second time- 11.63 x sec

Explanation / Answer

Solution:

As shown in fig r = 100ft

Find the change in angle for 3 seconds.

cos() = (100 + 3 - 44 ft) / (100 ft)
cos() = 0.59
() = 0.9397 rad

Calculate .
= /t
= (0.9397 rad) / (3 s)
= 0.313 rad/s

Parametric equation for y:

y(t) = 3 ft + (100 ft)sin(0.313t - /2)

Find first 2 times for y = 80 ft.

80 = 103 + 100 sin(0.313t - /2)
-0.23 = sin(0.313t - /2)
sin-1(-0.23) = (0.313t - /2)

-0.232078 + 2n = 0.313t - /2              or    ( + 0.232078) + 2n = 0.313t - /2

First time:
-0.232078 = 0.313t - /2
1.338718 = 0.313t
t = 4.277 s

Second time:
+ 0.232078 = 0.313t - /2
4.944467 = 0.313t
t = 15.797 s

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