A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to$\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$, what is its angular velocity at t = 2.50 s ? (b) Through what angle has the wheel turned between t = 0 and $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{s}$ ?

The angular speed is${\omega }_{0_z}=1.5rad/s$ .

The angular acceleration is${\alpha }_z=0.200rad/s^2$ .

The time is$t=2.50s$ .

Angular velocity at time t of a rigid body with constant angular acceleration is given by,

${\mathit{\omega }}_{\mathbf{z}}\mathbf{=}{\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{+}{\mathit{\alpha }}_{\mathbf{z}}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here,${\mathit{\alpha }}_{\mathbf{z}}$ is the angular acceleration,${\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}$ is the angular velocity of body at time 0, is the time.

Angular position at time t of a rigid body with constant angular acceleration is given by,

$\mathit{\theta }\mathbf{=}{\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\alpha }}_{\mathbf{z}}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Substitute${\omega }_{0_z}=1.5rad/s,t=2.5s,and{\alpha }_z=0.200rad/s^2$ in equation (1).

$$\begin{gathered} {\omega }_z=1.5rad/s+\left(0.200rad/s^2\right)\left(2.5s\right) \\ =1.5rad/s+0.5rad/s \\ =2rad/s \end{gathered}$$

Therefore, the required angular speed is 2 rad/s .

Substitute ${\mathrm{\omega }}_{0\mathrm{z}}=1.5\mathrm{rad}/\mathrm{s},\mathrm{t}=2.5\mathrm{s}$and${\mathrm{\alpha }}_{\mathrm{z}}=0.200\mathrm{rad}/{\mathrm{s}}^2$ in equation (2).

$$\begin{gathered} \theta =\left(1.5rad/s\right)\left(2.5s\right)+\frac{1}{2}\left(0.200rad/s^2\right){\left(2.5s\right)}^2 \\ =3.75rad+0.625rad \\ =4.375rad \end{gathered}$$

Therefore, the required angle is 4.375 rad .