Question

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's angular speed is 28.0 rad/s. What is the tire's angular acceleration (in rad/s2) during this time?

Answer 1

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$