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 7.40-
min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.40-
min interval.

Answer 1

`v = 0.24 m/s`

`a = 6.75 * 10^(-4) m/s^2`

Step-by-step explanation:

Given that wheel completes one round in total time T = 37.3 min

so angular speed of the wheel is given as

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/(37.3) rad/min`

now the angle turned by the wheel in time interval of t = 7.40 min

`\theta = \omega t`

`\theta = ((2\pi)/(37.3))(7.40) = 0.4\pi`

PART 1)

Now the average velocity is defined as the ratio of displacement and time

here displacement in given time interval is

`d = 2Rsin(\theta)/(2)`

R = radius = 91.5 m

`d = 183sin(0.2\pi) = 106.8 m`

Now time to turn the wheel is given as

`t = 7.40 min = 444 s`

now we have

`v = (d)/(t) = (106.8)/(444)`

`v = 0.24 m/s`

PART 2)

Now average acceleration is defined as ratio of change in velocity in given time interval

here velocity of a point on its rim is given as

`v = R\omega`

`v = (91.5)((2\pi)/(37.3* 60))`

`v = 0.257 m/s`

now change in velocity when wheel turned by the above mentioned angle is given as

`\Delta v = 2vsin(\theta)/(2)`

`\Delta v = 2(0.257)sin(0.2\pi)`

`\Delta v = 0.3 m/s`

time interval is given as

`t = 7.40 min = 444 s`

now average acceleration is given as

`a = (0.3)/(444)`

`a = 6.75 * 10^(-4) m/s^2`

Answer 2

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

Step-by-step explanation:

The wheel travels through

Θ = (7.40/37.3)*360º = 71.42º

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(71.42º/2) = 107 m

so the average velocity is

v = L / t = 107m / 7.40*60s = 0.241 m/s

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 71.42º, so

Vf = 0.257m/s * (cos71.42º i + sin71.42º j) = [0.0819 i + 0.244 j] m/s

ΔV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] m/s = [-0.175 i + 0.244 j] m/s

which has magnitude

|ΔV| = √(0.175² + 0.244²) m/s = 0.300 m/s

Then the average acceleration is

a_avg = |ΔV| / t = 0.300m/s / (7.40*60s) = 6.76e-4 m/s²

The instantaneous acceleration is centripetal: a = ω²r

a = (2π rads / (37.3*60s)² * 183m/2 = 7.21e-4 m/s²