Question

A Ferris wheel is moving at an initial angular velocity of 1.0 rev/59 s. If the operator then brings it to a stop in 3.1 min, what is the angular acceleration of the Ferris wheel

Answer 1

Answer:

$\alpha = 5.7 * 10^{-4} rad/s^2$

Explanation:

Given

$w_0 = 1.0/59 \ rev/s$ initial angular velocity

$t = 3.1\ min$ -- Initial time

$w = 0$ --- final angular velocity (when the wheel stops)

Required:

Determine the angular acceleration ($\alpha$)

The angular is calculated using the following formula

$w = w_0 + \alpha * t$

Convert time to seconds:

$t = 3.1\ min$

$t = 3.1 * 60s$

$t = 186s$

Convert angular velocity to rad/s

$w_0 = 1.0/59\ rev/s$

$w_0 = \frac{1}{59} * 6.283rad/s$

$w_0 = \frac{6.283}{59}\ rad/s$

Substitute in the required values, the expression becomes:

$w = w_0 + \alpha * t$

$0 = \frac{6.283}{59} + \alpha * 186$

$0 = \frac{6.283}{59}+ 186\alpha$

Collect Like Terms

$186\alpha = -\frac{6.283}{59}$

Make $\alpha$ the subject

$\alpha = \frac{6.283}{59 * 186}$

$\alpha = \frac{6.283}{10974}$

$\alpha = 0.00057253508rad/s^2$

$\alpha = 5.7 * 10^{-4} rad/s^2$