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 acceleration at the wheel's rim, over a 8.60-min interval.

Answer 1

angular speed of the Ferris wheel is given as

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

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

`\omega = 0.17 rad/min = 2.8 * 10^(-3) rad/s`

So the angle rotated by the point in 8.60 min is given as

`\theta = \omega*t`

`\theta = 0.17 * 8.60 = 1.462 rad = 83.8 degree`

now the tangential velocity is given by

`v = R\omega = (183)/(2)*2.8 * 10^(-3) = 0.26 m/s`

now the change in velocity is given as

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

`\Delta v = 2*0.26 * sin(83.8)/(2)`

`\Delta v = 0.35 m/s`

now the average acceleration is given as

`a = (\Delta v)/(\Delta t)`

`a = (0.35)/(8.6*60) = 6.73 * 10^(-3) m/s^2`

so above is the average acceleration