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------------------------------------------------ A rotating wheel requires 3.05 seconds to rotate through 37.0 revolutions. Its angular speed at the end of the 3.05-second interval is 97.9 rad/s. What is the constant angular acceleration of the wheel?

Answer :

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]