Question

A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one complete revolution every 20 s, find an equation that gives the height above the ground of a person on the Ferris wheel as a function of time. (Let y be the height above the ground in meters and let t be the time in seconds. Assume that when t = 0 the person is 11 m above the ground and going up.)

Answer 1

Answer:

$y = 11 - 10cos(0.1\pi t)$

Step-by-step explanation:

Suppose at t = 0 the person is 1m above the ground and going up

Knowing that the wheel completes 1 revolution every 20s and 1 revolution = 2π rad in angle, we can calculate the angular speed

2π / 20 = 0.1π rad/s

The height above ground would be the sum of the vertical distance from the ground to the bottom of the wheel and the vertical distance from the bottom of the wheel to the person, which is the wheel radius subtracted by the vertical distance of the person to the center of the wheel.

$y = h_g + d_b = h_g + R - d_c$ (1)

where $h_g = 1$ is vertical distance from the ground to the bottom of the wheel, $d_b$ is the vertical distance from the bottom of the wheel to the person, R = 10 is the wheel radius, $d_c$ is the vertical distance of the person to the center of the wheel.

So solve for $d_c$ in term of t, we just need to find the cosine of angle θ it has swept after time t and multiply it with R

$d_c = Rcos(\theta(t)) = Rcos(\omega t) = 10cos(0.1\pi t)$

Note that $d_c$ is negative when angle θ gets between π/2 (90 degrees) and 3π/2 (270 degrees) but that is expected since it would mean adding the vertical distance to the wheel radius.

Therefore, if we plug this into equation (1) then

$y = h_g + R - d_c = 1 + 10 - 10cos(0.1\pi t) = 11 - 10cos(0.1\pi t)$