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.

Answer 1

Answer:

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

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

Answer 2

Answer:

$v = 0.24 m/s$

$a = 6.75 \times 10^{-4} m/s^2$

Explanation:

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

so angular speed of the wheel is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{37.3} rad/min$

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

$\theta = \omega t$

$\theta = (\frac{2\pi}{37.3})(7.40) = 0.4\pi$

PART 1)

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

here displacement in given time interval is

$d = 2Rsin\frac{\theta}{2}$

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

$v = \frac{d}{t} = \frac{106.8}{444}$

$v = 0.24 m/s$

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

$v = R\omega$

$v = (91.5)(\frac{2\pi}{37.3\times 60})$

$v = 0.257 m/s$

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

$\Delta v = 2vsin\frac{\theta}{2}$

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

$\Delta v = 0.3 m/s$

time interval is given as

$t = 7.40 min = 444 s$

now average acceleration is given as

$a = \frac{0.3}{444}$

$a = 6.75 \times 10^{-4} m/s^2$