Question

A stationary tyre begins to rotate at constant angular acceleration to reach an angular velocity of 12.0 rad/s in 3.00 s. Find (a) the magnitude of the angular acceleration of the tyre and (b) the angle in radians that it rotates during this time interval.

Result: (a) 4.00 rad/s2 (b) 18.0 rad

Answer 1

(a) By the definition of the angular acceleration, we have:

$$\alpha=\dfrac{\omega-\omega_0}{t}=\dfrac{12.0\ \dfrac{rad}{s}}{3.0\ s}=4.0\ \dfrac{rad}{s^2}.$$

b) We can find the angle in radians that the tyre rotates in 3 s from the kinematic equation:

$$\theta=\omega_0t+\dfrac{1}{2}\alpha t^2,$$ $$\theta=\dfrac{1}{2}\cdot4.0\ \dfrac{rad}{s^2}\cdot(3.0\ s)^2=1.0\ rad.$$