Ferris Wheel Problem

Question

A Ferris wheel is 50 meters in diameter and boarded from a platform that is 3 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 6 minutes. How many minutes of the ride is spent higher than 40 meters above the ground?

Answer 1

The height of some part of the ferris wheel above ground can be found as below.

Let height above ground by H, then a general equation modeling this is H = A cos θ + B. The angle is measured from the lowest point to position of a rider at any time.

When a rider is at the bottom of the wheel, the rider is 3 meters above the ground, so at this point we have

3 = A cos 0 + B. = A + B

At the very top, the rider is 53 m above ground, and the angle is 180 degrees.

53 = A cos 180 +B = -A + B.

Adding 3 = A + B

53 = -A + B gives 56 = 2B or B = 28 m

from 3 = A + B , A = -25 m

So, H = -25 cos θ +28

At 40 m above the ground, 40 = -25 cos θ + 28

cos θ = -0.48 or θ = 118.7 degrees from the bottom. Also, H = 40 at θ = 241.3 degrees.

So, the rider on the ferris wheel is above 40 m for 241.3 - 118.7 = 122.6 degrees.

One revolution in 6 minutes is equal to 6 min / 360 deg x 122.6 deg = 2.0 minutes above 40 m.

Please check all math.

Answer 2

d = 50m p = 3m wheel rotates at 360deg/6minutes

We need to figure out at what angles is the wheel above 40 m.

50m + 3m = 53m total height

37m of ferris wheel will be 40m above ground

Now we can create a triangle starting at the middle of the wheel with a hypotenuse of 25m (radius of wheel) ending at the point where the wheel is 37m tall.

37m - 25m = 12m this is the height of our triangle. now we can find the angle that the hypotenuse and

height make

tan(θ) = 12/25

θ₁ = arctan(12/25)

θ₁ = 25.6410° this is the first point where the wheel is 40m above ground

θ ₂ = 180° - 25.6410° = 154.3589° this is the second point where the wheel is 40m above ground

θ = θ ₂ - θ₁ = 154.3589° - 25.6410° = 128.7179° this is the total amount of "angle" that the wheel

is 40m above ground

360°/6min = 128.7179°/xmin ratio used to find the amount of time the wheel is above 40m

x = (128.7179° x 6min)/ 360°

x = 2.1453 minutes