Given:
D=165 feet and the frequency of the motion is 1.6 revolutions per minute.
Solution:
The radius is half of the diameter.
The radius of the wheel is 82.5 feet.
[tex]T=\frac{1}{1.6} \text{ minutes}[/tex]
As we know: [tex]\omega=\frac{2\pi}{T}[/tex]
Substitute the value of T in the above formula.
[tex]\omega=\frac{2\pi}{\frac{1}{1.6}}\\\omega=3.2\pi[/tex]
If the center of the wheel is at the origin then for [tex]t=0[/tex] the rest position is [tex]-a[/tex].
This can be written as:
[tex]h(t)=-a\cos(\omega t)\\h(t)=-82.5cos(32.\pi t)[/tex]
The actual height of the rider from the ground is:
[tex]h(t)=\text{ Initial height from bottom}+\text{ radius}-82.5\cos(3.2\pi t)\\h(t)=15+82.5-82.5\cos(3.2\pi t)\\h(t)=97.5-82.5\cos(3.2\pi t)[/tex]
The required equation is [tex]h(t)=97.5-82.5\cos(3.2\pi t)[/tex].
