Question

A rotating wheel requires a time Δt = [01]____________________s to rotate 37.0 revolutions. Its angular speed at the end of the interval is ω = 75.9 rad/s. What is the constant angular acceleration (rad/s2 ) of the wheel?

Answer 1

Here is the correct question

A rotating wheel requires a time Δt = 4.28 seconds to rotate 37.0 revolutions. Its angular speed at the end of 4.28 seconds interval is ω = 75.9 rad/s. What is the constant angular acceleration (rad/s² ) of the wheel?

Answer:

10.09 rad/s²

Step-by-step explanation:

Given that :

`\theta = \ 37.0 \ revolutions`

Then since 1 revolution =
`2 \pi \ rad`

`\theta = 37.0 \ rev * (2 \pi \ rad )/( 1 \ rev)`

`\theta = 232.48 \ rad`

The first equation of motion for wheel can be expressed as :

`\omega = \omega_0t + \alpha t`

`\omega_0 = \omega- \alpha t`

where
`\omega` = 75.9 rad/s

`\omega_0 = 75.9 rad/s - \alpha (4.28 \ s)`

From the second equation of the motion

`\theta = \omega_0t + (1)/(2) \alpha t ^2`

where ;

`\omega_0 = 75.9 rad/s - \alpha (4.28 \ s)`

t = 4.28 s

`\theta = 232.48 \ rad`

Then

`232.48 \ rad= (75.9 rad/s - \alpha (4.28 \ s))(4.28 \ s)+ (1)/(2) \alpha (4.28 \ s) ^2`

`232.48 \ rad= 324.852 rad/s - 18.3184 \alpha s^2 + 9.1592 \alpha s ^2`

`324.852 rad/s - 232.48 \ rad= 18.3184 \alpha s^2 - 9.1592 \alpha s ^2`

`92.372 \ rad= 9.1592 \alpha s ^2`

`\alpha = \frac {92.372 \ rad} {9.1592 \ s ^2}`

`\alpha = 10.09 \ \ rad/s^2`

Thus, the angular acceleration of the wheel = 10.09 rad/s²