A wheel, starting from rest, rotates with a constant angular acceleration of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{rad}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}\mathbf{}$. During a certain $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{s}}\mathbf{}$interval, it turns through $\mathbf{90}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{rad}}$. (a) What is the angular velocity of the wheel at the start of the$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{s}}$interval? (b) How long has the wheel been turning before the start of the$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{s}}$interval?

$$\begin{gathered} {\theta }_f-{\theta }_0=90rad \\ \alpha =2.0\frac{\text{rad}}{{\text{s}}^{\text{2}}} \\ \text{t}=3.0\text{s} \end{gathered}$$

Using the second kinematic equation of rotational motion, we can find the angular velocity of the wheel at the start of interval. Using this angular velocity into the formula of angular acceleration, we can find the time for which the wheel has been turning before the start of the i$3.00\text{s}$nterval.

Formula:

$$\begin{gathered} {\theta }_f-{\theta }_0={\omega }_0\text{t}+\frac{1}{2}\alpha {\text{t}}^2 \\ \alpha =\frac{{\omega }_0}{\text{t}} \end{gathered}$$

The second kinetic equation in rotational motion,

$$\begin{gathered} {\theta }_f-{\theta }_0={\omega }_0\text{t}+\frac{1}{2}\alpha {\text{t}}^2 \\ \Rightarrow 90\text{rad}={\omega }_0\left(3.0\text{s}\right)+\frac{1}{2}\left(2.0\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right){\left(3.0\text{s}\right)}^2 \\ \Rightarrow 90\text{rad}={\omega }_0\left(3.0\text{s}\right)+9\text{rad} \\ \Rightarrow 90\text{rad}-9\text{rad}={\omega }_0\left(3.0\text{s}\right) \\ \Rightarrow 81\text{rad}={\omega }_0\left(3.0\text{s}\right) \\ \Rightarrow {\omega }_0=27\frac{\text{rad}}{\text{s}} \end{gathered}$$

Therefore, the angular velocity of the wheel at start of the $3.00\text{s}$interval is$27\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

For angular acceleration,

$$\begin{gathered} \alpha =\frac{{\omega }_0}{\text{t}} \\ \Rightarrow \text{t}=\frac{{\omega }_0}{\alpha } \\ \Rightarrow \text{t}=\frac{27\frac{\text{rad}}{\text{s}}}{2\frac{\text{rad}}{{\text{s}}^{\text{2}}}} \\ \Rightarrow \text{t}=13.5\text{s} \end{gathered}$$

Therefore, the time for which the wheel has been turning before the start of the interval is$13.5\text{s}$