Question

The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.

1.Find the magnitude of the average velocity at the wheel's rim, over a 7.80-min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.80-min interval.

3.Find the magnitude of the wheel's instantaneous acceleration.

4.Determine the ratio of the difference between the magnitude of the wheel's instantaneous acceleration and the magnitude of the average acceleration to the magnitude of the average acceleration.

Answer 1

This question is based on the concept of:

1.Circular motion

2.Average velocity

3.Average acceleration

4.Instantaneous acceleration

5.Mathematical simplification

The detailed solution is explained below:

The wheel travels through
Θ = (7.80/37.3)*360º = 75.28º
and so the length of the line segment connecting the initial and final position is
L = 2*L*sin(Θ/2) = 2 * (183m/2) * sin(75.28º/2) = 111.76 m
so the average velocity is
v = L / t = 111.76m / 7.80*60s = 0.238 m/s (Answer 1)

Initially, let's say the velocity is along the +x axis:
Vi = π * 183m / (37.3*60s) i = 0.257 m/s i
Later, it's rotated through 75.28º, so
Vf = 0.257m/s * (cos75.28º i + sin75.28º j) = [0.0653 i + 0.248 j] m/s
ΔV = Vf - Vi = [(0.0653 - 0.257) i + 0.248 j] m/s = [-0.1917 i + 0.248 j] m/s
which has magnitude
|ΔV| = √(0.1917² + 0.248²) m/s = 0.313 m/s
Then the average acceleration is
a_avg = |ΔV| / t = 0.313m/s / (7.80*60s) = 6.688*10^-4 m/s² (Answer 2)

The instantaneous acceleration is centripetal: a = ω²r
a = (2π rads / (37.3*60s)² * 183m/2 = 7.21*10^-4 m/s² (Answer 3)

Ratio = (7.21 – 6.688)*10^-4 / 6.688*10^-4 = 0.078 or 7.8% (Answer 4)

THANKS!!!