Trigonometry/Worked Example: Ferris Wheel Problem

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The Problem[edit]

Wiener Riesenrad DSC02378.JPG

Exam Question[edit]

"Jacob and Emily ride a Ferris wheel at a carnival in Billings, Montana. The wheel has a 16 meter diameter, and turns at three revolutions per minute, with its lowest point one meter above the ground. Assume that Jacob and Emily's height h above the ground is a sinusoidal function of time t, where t=0 represents the lowest point on the wheel and t is measured in seconds."

"Write the equation for h in terms of t."

[For those interested the picture is actually of a Ferris wheel in Vienna. We couldn't find one of a Ferris wheel in Billings Montana.]

Video Links[edit]

The Khan Academy has video material that walks through this problem, which you may find easier to follow:

Solution[edit]

Diameter to Radius

A 16m diameter circle has a radius of 8m.

Revolutions per Minute to Degrees per Second

A wheel turning at three revolutions per minute is turning

\displaystyle \frac{3 \times 360^\circ}{60}

per second. Simplifying that's

\displaystyle 18^\circ

per second.

Formula for height

At t=0 our height h is 1. At t =10 we will have turned through 180o, i.e. half a circle and will be at the top most point which has height 16 + 1= 17.

A cosine function, i.e. \displaystyle \cos \theta is 1 at \displaystyle \theta=0^\circ and -1 at \displaystyle \theta=180^\circ. That's almost exactly opposite to what we want as we want the most negative value at 0 and the most positive at 180. So let's start with negative cosine as our function.

At t=10 we want \theta=180^\circ, so we will take \displaystyle -\cos( 18 t ). That's -1 at t=0 and +1 at t=10. Multiply by 8 and we get:

\displaystyle -8\cos( 18 t ). That's -8 at t=0 and +8 at t=10

Add 9 and we get

\displaystyle 9-8\cos( 18 t ). Which is 1 at t=0 and +17 at t=10

Our required formula is

\displaystyle h = 9 - 8\cos( 18 t ).

with the understanding that cosine is of an angle in degrees (not radians).