Question

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

Answer 1

Answer:

$\alpha=14.2rad/s^2$

Explanation:

The formula that relates angular displacement with angular acceleration is:

$\Delta \theta=\omega_i t+\frac{\alpha t^2}{2}$

We can obtain $\omega_i$ from the definition of angular acceleration:

$\alpha=\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{t}$

$\omega_i=\omega_f-\alpha t$

Putting all together:

$\Delta \theta=(\omega_f-\alpha t) t+\frac{\alpha t^2}{2}=\omega_f t-\frac{\alpha t^2}{2}$

Which, since we want the angular acceleration, is:

$\alpha=\frac{2(\omega_f t-\Delta \theta)}{t^2}$

And for our values is:

$\alpha=\frac{2((97.9rad/s)(3.05s)-(37(2\pi rad)))}{(3.05s)^2}=14.2rad/s^2$