Question

Aferris wheel is 250 ft in diameter and us boarded from a platform 15 ft high. T=0 when the wheel is at uts maximum height. The wheel takes 5. mins to make 1 rotation.

1. graoh 1 period if the wheel

2. find a firmula for the height of the wheel as a function of time

Answer 1

Radius of the wheel is given by -

R = D / 2 = [(250 ft) / 2]

R = 125 ft

2. Height of the wheel as a function of time is given by -

h (t) = (15 ft) + [(125 ft) + y]

h (t) = (140 ft) + y                                                                                           { eq.1 }

Here, y depends on the angle of rotation ().

From a trigonometric identity, we have

sin (90 - ) = y / (125 ft)

y = (125 ft) cos                                                                                                    { eq.2 }

Here, depends on the angular speed ().

We know that, = t

Then, we have

y = (125 ft) cos ( t)

Inserting the value of 'y' in eq.1 & we get

h (t) = (140 ft) + (125 ft) cos ( t)

where, = angular frequency of the wheel = 2 / T

then, we get

h (t) = (140 ft) + (125 ft) cos ( t)

h (t) = (140 ft) + (125 ft) cos {[2 / (300 s)] t}

h (t) = (140 ft) + (125 ft) cos [ t / (150 s)]