The New York Wheel is the world\'s largest Ferris wheel. It\'s 183 meters in dia

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. Find the magnitude of the average velocity at the wheel's rim, over a 8.60-min interval. Express your answer with the appropriate units. Part B Find the magnitude of the average acceleration at the wheel's rim, over a 8.60-min interval. Express your answer with the appropriate units. Part C Find the magnitude of the wheel's instantaneous acceleration. Express your answer with the appropriate units. Part D 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.

Explanation / Answer

a)

The wheel travels through
theta = (8.6/37.3)*360 = 83 degrees
and so the length of the line segment connecting the initial and final position is
L = 2Lsin(theta/2) = 2 * (183/2) * sin(83/2) = 121.26m
so the average velocity is
v = L / t = 121.26 / 8.6*60 = 0.235 m/s

b)
Initially, let's say the velocity is along the +x axis:
Vi =pi * 183m / (37.3*60) i = 0.257 m/s i
Later, it's rotated through 83 degrees, so
Vf = 0.257m/ * (cos83i + sin83 j) = [0.0819 i + 0.244 j] m/s
dV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] = [-0.175 i + 0.244 j] m/s
magnitude
|dV| = sqrt[(0.1752+ 0.2442)] = 0.300 m/s
Then the average acceleration is
aavg = |dV| / t = 0.300m/s / (8.6*60) = 5.814*10-4 m/s2
c)

The instantaneous acceleration is centripetal:

a inst=W2r
ainst = (2/ (37.3*60))2 * (183/2) = 7.21*10-4 m/s

d)

ainst/aavg=7.21/5.814=1.24

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