At t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of$\mathbf{30}\mathbf{\hspace{0.33em}}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ until a circuit breaker trips at t =2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between t = 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

The angular speed at t=0 is ${\omega }_{0_z}=24\hspace{0.33em}rad/s$.

The angular acceleration is ${\alpha }_z=30\hspace{0.33em}rad/s^2$.

The angular displacement is $\Delta \theta =432\hspace{0.33em}rad$.

The time is t=2 s.

The angular acceleration is constant. So, apply the equations of rotation with constant angular acceleration.

$\mathit{\Delta }\mathit{\theta }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{+}{\mathit{\omega }}_{\mathbf{z}}\mathbf{)}\mathit{t}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

${\mathbf{\omega }}_{\mathbf{z}}\mathbf{=}{\mathbf{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{+}{\mathbf{\alpha }}_{\mathbf{z}}\mathbf{t}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

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

$\mathbf{\Delta \theta }\mathbf{=}{\mathbf{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\alpha }}_{\mathbf{z}}{\mathbf{t}}^{\mathbf{2}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here,${\mathit{\theta }}_{\mathbf{0}}$ is Angular position of body at time 0.

Substitute ${\omega }_{0_z}=24\hspace{0.33em}rad/s$, ${\alpha }_z=30\hspace{0.33em}rad/s^2$, and t =2 s in equation (2).

$$\begin{gathered} {\omega }_z=24rad/s+\left(30rad/s^2\right)\left(2s\right) \\ =84rad/s \end{gathered}$$

Substitute ${\omega }_{0_z}=24r\mathrm{ad}/\mathrm{s},{\omega }_z=84\mathrm{rad}/\mathrm{s},\mathrm{and}\mathrm{t}=2$in equation (1).

$$\begin{gathered} ∆{\theta }_1=\frac{1}{2}\left(24\mathrm{rad}/\mathrm{s}+84\mathrm{rad}/\mathrm{s}\right)\left(2\sec \right) \\ =108\mathrm{rad} \end{gathered}$$

Find the total angle as follows.

$$\begin{gathered} {\theta }_T=108\mathrm{rad}+432\mathrm{rad} \\ =540\mathrm{rad} \end{gathered}$$

Therefore, the total angle the when turned between t=0 and the time it stopped is 540 rad.

Take the interval from when the circuit breaker trips until the wheel comes to a stop. So, ${\omega }_{0_z}=84\hspace{0.33em}rad/s$.

At the end the wheel is stopped, so take ${\omega }_z=0$.

Substitute ${\omega }_{0_z}=84\hspace{0.33em}rad/s$, $\Delta \theta =432\hspace{0.33em}rad$, and ${\omega }_z=0$in equation (1).

$$\begin{gathered} 432\mathrm{rad}=\frac{1}{2}(84\mathrm{rad}/\mathrm{s}+0)t \\ t=\frac{2\times 432\mathrm{rad}}{84\mathrm{rad}/\mathrm{s}} \\ =10.3\mathrm{s} \end{gathered}$$

Find the total time as follows.

$$\begin{gathered} t_T=2s+10.3s \\ =12.3s \end{gathered}$$

Therefore, the wheel will stop after 12.3 s.

Substitute ${\omega }_{0_z}=84\hspace{0.33em}rad/s$,$t=10.3\hspace{0.33em}s$, and${\omega }_z=0$in equation (2).

$$\begin{gathered} 0=84\mathrm{rad}/\mathrm{s}+{\mathrm{\alpha }}_{\mathrm{z}}\left(10.3\mathrm{s}\right) \\ {\mathrm{\alpha }}_{\mathrm{s}}=-\frac{84\mathrm{rad}/\mathrm{s}}{10.3\mathrm{s}} \\ =-8.16\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the required angular acceleration is $-8.16rad/s^2$.