Question

The wheel on an upside-down bicycle moves through 18.0 rad in 5.39 s. What is the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s? Answer in units of rad/s 2 .

Answer 1

Explanation:

Formula to calculate angular acceleration is as follows.

   $\Delta (\theta) = \frac{1}{2} \alpha \Delta t^{2} + \omega_{1} \Delta t$

or,      $\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

Putting the given values into the above formula as follows.

    $\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

                = $\frac{2(18.0 rad - 2.5 rad/s \times 5.30 s)}{(5.39)^{2}}$    

                = 0.326 $rad/s^{2}$

Thus, we can conclude that the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s is 0.326 $rad/s^{2}$.