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

`\alpha = 5.7 * 10^(-4) rad/s^2`

Step-by-step 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 = (1)/(59) * 6.283rad/s`

`w_0 = (6.283)/(59)\ rad/s`

Substitute in the required values, the expression becomes:

`w = w_0 + \alpha * t`

`0 = (6.283)/(59) + \alpha * 186`

`0 = (6.283)/(59)+ 186\alpha`

Collect Like Terms

`186\alpha = -(6.283)/(59)`

Make
`\alpha` the subject

`\alpha = (6.283)/(59 * 186)`

`\alpha = (6.283)/(10974)`

`\alpha = 0.00057253508rad/s^2`

`\alpha = 5.7 * 10^(-4) rad/s^2`