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Question

A Ferris wheel is 40 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 34 meters above the ground?

Answer 1

Please draw a picture:

Let

h = 34m be the height of interest,

p be a horizontal line h = 34m above ground,

P be the intersection of p and the vertical axis of the wheel,

Q₁, Q₂ be the two intersections between the wheel and the line p,

C be the center of the wheel,

T be the top of the wheel,

r = 20m be the radius of the wheel,

b = 2m be the distance between the ground and bottom of the wheel,

α be the angle Q₁CQ₂.

We are to calculate what portion of the 10 min ride is spent above

the line p, i.e., along the arc of the angle α.

The fraction of time spent along the arc of α is same as the

ration between the angle α and the full angle of 360°.

Therefore we just need to calculate the angle α.

The angle α/2 is the angle Q₁CP in that right triangle.

cos(α/2) = CP/Q₁C = (r - TP)/r = (r - (2r + b - h))/r = (h - r - b)/r =

(34 - 20 - 2)/20 = 0.6

α/2 = 53.13°

α = 106.26°

The time spent above 34m is then

10 min × 106.26/360 = 2.95 min