Question

The London Eye is a large Ferris wheel that has diameter 135 meters and revolves continuously. Passengers enter the cabins at the bottom of the wheel and complete one revolution in about 27 minutes. One minute into the ride a passenger is rising at 0.06 meters per second. How fast is the horizontal motion of the passenger at that moment?

Answer 1

0.253 m/s

Explanation:

You want to know the horizontal component of motion of a passenger riding a Ferris wheel when they are 1/27 of the way around the circle and their rate of rise is 0.06 m/s.

Angle of elevation

The wheel makes one revolution in 27 minutes, so the angular displacement is changing at the rate of (360°)/(27 min) = 13 1/3°/min.

After 1 minute, the passenger is following a path that has an angle of elevation of 13 1/3°.

Horizontal component

The ratio of vertical speed to horizontal speed will be the tangent of the angle of elevation:

Vv/Vh = tan(13 1/3°)

Then the horizontal speed will be ...

Vh = Vv/tan(13 1/3°) = (0.06 m/s)/0.237004

Vh ≈ 0.253 m/s

The passenger's horizontal motion is about 0.253 m/s.