Answer:
[tex]\alpha=14.2rad/s^2[/tex]
Explanation:
The formula that relates angular displacement with angular acceleration is:
[tex]\Delta \theta=\omega_i t+\frac{\alpha t^2}{2}[/tex]
We can obtain [tex]\omega_i[/tex] from the definition of angular acceleration:
[tex]\alpha=\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{t}[/tex]
[tex]\omega_i=\omega_f-\alpha t[/tex]
Putting all together:
[tex]\Delta \theta=(\omega_f-\alpha t) t+\frac{\alpha t^2}{2}=\omega_f t-\frac{\alpha t^2}{2}[/tex]
Which, since we want the angular acceleration, is:
[tex]\alpha=\frac{2(\omega_f t-\Delta \theta)}{t^2}[/tex]
And for our values is:
[tex]\alpha=\frac{2((97.9rad/s)(3.05s)-(37(2\pi rad)))}{(3.05s)^2}=14.2rad/s^2[/tex]
