A technician is testing a computer-controlled, variable-speed motor. She attaches a thin disk to the motor shaft, with the shaft at the center of the disk. The disk starts from rest, and sensors attached to the motor shaft measure the angular acceleration ${\mathit{\alpha }}_{\mathbf{z}}$of the shaft as a function of time. The results from one test run are shown in Fig. P9.87:

(a) Through how many revolutions has the disk turned in the first 5.0 s? Can you use Eq. (9.11)? Explain. What is the angular velocity, in rad/s, of the disk (b) at t= 5.0 s; (c) when it has turned through 2.00 rev?

Given in the question,

The angular acceleration of the disk is${\alpha }_z$ .

From the graph we can observe angular acceleration is a linear function of time

So, we can write

${\alpha }_z=\lambda t$

Where $\lambda$is a constant.

Since disk starts from rest

Therefore

The initial angular displacement of the disk,${\theta }_0=0$

The initial angular velocity of the disk, ${\omega }_0=0$

If the angular acceleration is not constant, then $\mathit{\theta }\mathbf{,}\mathbf{}\mathit{\omega }\mathbf{,}\mathit{\alpha }$are related as,

$$\begin{gathered} \mathit{\omega }\mathbf{=}\frac{\mathbf{d}\mathbf{\theta }}{\mathbf{d}\mathbf{t}} \\ \mathit{\alpha }\mathbf{=}\frac{\mathbf{d}\mathbf{\omega }}{\mathbf{d}\mathbf{t}} \end{gathered}$$

Where $\mathit{\theta }$is angular displacement,$\mathbf{}\mathit{\omega }$ is the angular velocity, and $\mathit{\alpha }$is the angular acceleration

$\begin{array}{l}6\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^{\text{2}}=\lambda \left(5\text{\hspace{0.17em}\hspace{0.17em}s}\right)\\ \lambda =\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\end{array}$From the graph

$\alpha =\lambda t$…(i)

At$t=5\text{\hspace{0.17em}\hspace{0.17em}s}$, $\alpha =6\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$

To find $\lambda$substituting the values

$\begin{array}{l}6\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^{\text{2}}=\lambda \left(5\text{\hspace{0.17em}\hspace{0.17em}s}\right)\\ \lambda =\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\end{array}$

Therefore, from equation (i)

${\alpha }_z=\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)t$

Substituting the value of${\alpha }_z$

$\begin{array}{l}\omega =\int \left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)tdt\\ \omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\int tdt\\ \omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2}+C\end{array}$

As we know at $t=0,{\omega }_0=0$

Therefore,

$\begin{array}{l}\omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2}+C\\ 0=\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{0^2}{2}+C\\ C=0\end{array}$

Hence the angular velocity is $\omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2}$

Now we know

$\begin{array}{c}\omega =\frac{d\theta }{dt}\\ d\theta =\omega dt\\ \theta =\int \omega dt\end{array}$

Now substituting the value of $\omega$and integrating

$$\begin{gathered} \begin{array}{l}\theta =\int \left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2}dt\\ \theta =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^3}{6}+C\end{array} \\ \begin{array}{l}0=\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{0^3}{6}+C\\ C=0\end{array} \end{gathered}$$

Hence the angular displacement can be given as

$\theta =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^3}{6}$

To find the number of turns at$t=5s$ , substituting the value of t in the equation of angular displacement

$\begin{array}{l}\theta =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{{\left(5\text{\hspace{0.17em}\hspace{0.17em}s}\right)}^3}{6}\\ \theta =25\text{\hspace{0.17em}\hspace{0.17em}rad}\end{array}$

Since in one revolution angular displacement is $2\pi$, so number of revolutions in $25\text{\hspace{0.17em}\hspace{0.17em}rad}$

$\begin{array}{c}\theta =\frac{25}{2\pi }\text{\hspace{0.17em}\hspace{0.17em}rev}\\ =\text{4\hspace{0.17em}\hspace{0.17em}rev}\end{array}$

Hence the disk turns 4 revolutions in 5 s

Since the angular acceleration is not constant, therefore, we cannot use the equations (9.11)

We know that

Angular velocity

$\omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2}$

To find the angular velocity at $t=5s$substituting the value in the equation of angular velocity

$\begin{array}{l}\omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{{\left(5\text{\hspace{0.17em}s}\right)}^2}{2}\\ \omega =15\text{\hspace{0.17em}rad}/{\text{s}}^2\end{array}$

The angular velocity at$t=5s$ is$15\text{\hspace{0.17em}rad}/{\text{s}}^2$

We know that

$\theta =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^3}{6}$

Since in one revolution disk move $2\pi$so, in 2 rev $\theta =4\pi$

$\begin{array}{l}4\pi =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^3}{6}\\ t=3.98\text{\hspace{0.17em}\hspace{0.17em}s}\end{array}$

Now finding the angular velocity substituting $t=3.98\text{\hspace{0.17em}\hspace{0.17em}s}$in the equation of angular velocity.

Angular velocity

$$\begin{gathered} \omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{t^2}{2} \\ \begin{array}{l}\omega =\left(\frac{6}{5}\text{\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^3\right)\frac{{\left(3.98\text{\hspace{0.17em}\hspace{0.17em}s}\right)}^2}{2}\\ \omega =9.5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^2\end{array} \end{gathered}$$

Hence the angular velocity when the disk turns 2 revs is $9.5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}rad}/{\text{s}}^2$