Question

A wheel has a constant angular acceleration of 3.3 rad/s2. During a certain 6.0 s interval, it turns through an angle of 91 rad. Assuming that the wheel started from rest, how long had it been in motion before the start of the 6.0 s interval?

Answer 1

The kinematic equation of the angular motion is the following:

$$\Delta\varphi = \omega_0 t + \dfrac{\varepsilon t^2}{2}$$

where $\Delta\varphi$ is changing in the angle of revolution, $\omega_0$ is the initial angular speed, $\varepsilon = 3.3 rad/s^2$ is the angular acceleration and $t$ is the interval, during which the motion is considered.

If the wheel revolve by an angle of 91 rad during 6s interval, then, according to this equation, it's initial angular speed at the start of this interval will be:

$$\omega_0 = \Delta\varphi/t - \dfrac{\varepsilon t}{2} = 91/6 - 3.3\cdot 6/2 = 75.1 rad/s$$

The changing in angular velocity and the constant angular acceleration, by definition, are connected in the following manner:

$$\varepsilon = \dfrac{\Delta\omega}{t}$$

Assuming that the wheel started from rest, obtain the time, when it reached the velocity $\omega_0$:

$$\varepsilon = \dfrac{\omega_0 - 0}{t}\Rightarrow t = \dfrac{\omega_0}{\varepsilon} = \dfrac{75.1rad/s}{3.3 rad/s^2} \approx 22.8s$$

Answer. t = 22.8 s.