suppose that a ferris wheel of a diameter of 50 meters compl

suppose that a ferris wheel of a diameter of 50 meters completes one revolution every 2 minutes. when you are at the lowest point on the ferris wheel you are still 5 meters above the ground.

A) make a table of vaules showing height above the ground as a function of time

Solution

Solution :

Diameter = 50 m

So radius = 25 m

The wheel turns one revolution in 2mins.

So it turns half a revolution in 1 min.

Convert revolutions per minute to degrees per second:

A wheel turning half a revolution per min will turn

(1/2 x 360^o ) / 60 = 3^o per second.

At t = 0 sec , height h = 5m

At t = 60 sec we will have turned 180^o i.e. we will be at the highest point i.e. 50 + 5 = 55 m.

A cosine function , i.e. cos is 1 when angle = 0^o and -1 when angle = 180^o.

That’s opposite to what we want as we want the most negative value when angle = 0 and the most positive value when angle = 180.

At t=60 , we want angle = 180^o , so we will take –cos(3t).

That’s -1 at t = 0 and +1 at t= 60

Multiply by 25 and we get -25 cos(3t).

That’s -25 at t=0 and +25 at t = 60

Add 30 and we get 30 – 25 cos(3t)

That gives 5 when t = 0 and 55 when t = 60

So our required formula is :

h =30 – 25 cos (3t) where cosine is in degrees and not in radians.

We can make a table for different values of h for change in values of t.