Question

Added to Six Flags St. Louis in the Colossus is a giant Ferris wheel. Its diameter is 165 feet, it rotates at a rate of about 1.6 revolutions per minute, and the bottom of the wheel is 15 feet above the ground. Determine an equation that relates a rider's height above the ground at time . Assume the passenger begins the ride at the bottom of the wheel.

Answer 1

Answer:

The height of the rider as a function of time is $h(t) = 15 + 82.5\cdot (1-\cos 0.168t) \,[ft]$, where time is measured in seconds.

Step-by-step explanation:

Given that Ferris wheel rotates at constant rate and rider begins at the bottom of the wheel, the height of the rider as a function of time is modelled after this expression:

$h(t) = h_{bottom} + (1-\cos \omega t)\cdot r_{w}$

Where:

$h_{bottom}$ - Height of the bottom with respect to ground, measured in feet.

$\omega$ - Angular speed of the ferris wheel, measured in radians per second.

$t$ - Time, measured in seconds.

$r_{w}$ - Radius of the Ferris wheel, measured in feet.

The angular speed of the ferris wheel, measured in radians per second, is obtained from the following expression:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where:

$\dot n$ - Angular speed of the ferris wheel, measured in revolutions per minute.

If $\dot n = 1.6\,rpm$, then:

$\omega = \frac{\pi}{30}\cdot (1.6\,rpm)$

$\omega \approx 0.168\,\frac{rad}{s}$

Now, given that $h_{bottom} = 15\,ft$, $r_{w} = 82.5\,ft$ and $\omega \approx 0.168\,\frac{rad}{s}$, the height of the rider as a function of time is:

$h(t) = 15 + 82.5\cdot (1-\cos 0.168t) \,[ft]$