## Position of Point on London Eye Relative to Ground

The London Eye is a giant Ferris wheel with a radius of 67.5m and takes 30 minutes to make a complete rotation. If we take the point on the ground below the centre has the origin, the the coordinates of the centre are  {jatex options:inline}(0,67.5){/jatex}.
A point on the wheel rotates at the rate  {jatex options:inline}2 \pi{/jatex}  per 30 minutes, or  {jatex options:inline}\frac{2 \pi}{30 \times 60}= \frac{\pi}{900} rads/sec{/jatex}. In a times  {jatex options:inline}t{/jatex}  the wheel will rotate through an angle  {jatex options:inline}\theta = \frac{\pi}{900}t{/jatex}. The London eye appears to rotate in the clockwise as seen from the other side of the Thames. By convention, clockwise is negatives, so the angle of rotation is  {jatex options:inline}- \frac{\pi}{900}t{/jatex}.
By convention also the horizontal line is taken as the zero angle, so at any times  {jatex options:inline}t{/jatex}, the position of a point that started from the ground at  {jatex options:inline}t=0{/jatex}  is  {jatex options:inline}(67.5sin(- \frac{\pi}{900}t), 67.5-67.5cos(- \frac{\pi}{900}t){/jatex}.