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
Explanation:
Given that :
[tex]\theta = \ 37.0 \ revolutions[/tex]
Then since 1 revolution = [tex]2 \pi \ rad[/tex]
[tex]\theta = 37.0 \ rev * \frac{2 \pi \ rad }{ 1 \ rev}[/tex]
[tex]\theta = 232.48 \ rad[/tex]
The first equation of motion for wheel can be expressed as :
[tex]\omega = \omega_0t + \alpha t[/tex]
[tex]\omega_0 = \omega- \alpha t[/tex]
where [tex]\omega[/tex] = 75.9 rad/s
[tex]\omega_0 = 75.9 rad/s - \alpha (4.28 \ s)[/tex]
From the second equation of the motion
[tex]\theta = \omega_0t + \frac{1}{2} \alpha t ^2[/tex]
where ;
[tex]\omega_0 = 75.9 rad/s - \alpha (4.28 \ s)[/tex]
t = 4.28 s
[tex]\theta = 232.48 \ rad[/tex]
Then
[tex]232.48 \ rad= (75.9 rad/s - \alpha (4.28 \ s))(4.28 \ s)+ \frac{1}{2} \alpha (4.28 \ s) ^2[/tex]
[tex]232.48 \ rad= 324.852 rad/s - 18.3184 \alpha s^2 + 9.1592 \alpha s ^2[/tex]
[tex]324.852 rad/s - 232.48 \ rad= 18.3184 \alpha s^2 - 9.1592 \alpha s ^2[/tex]
[tex]92.372 \ rad= 9.1592 \alpha s ^2[/tex]
[tex]\alpha = \frac {92.372 \ rad} {9.1592 \ s ^2}[/tex]
[tex]\alpha = 10.09 \ \ rad/s^2[/tex]
Thus, the angular acceleration of the wheel = 10.09 rad/s²
