At t = 0 the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by $\mathit{\theta }\left(t\right)\mathbf{=}\left(250rad/s\right)\mathit{t}\mathbf{-}\left(20.0rad/s^2\right){\mathit{t}}^{\mathbf{2}}\mathbf{-}\left(1.50rad/s^3\right){\mathit{t}}^{\mathbf{3}}$. (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at t = 0 , when the current was reversed? (e) Calculate the average angular velocity for the time period from t = 0 to the time calculated in part (a).

The equation of angular displacement is as follows.

$\mathrm{\theta }\left(\mathrm{t}\right)=(250\mathrm{rad}/\mathrm{s})\mathrm{t}-(20.0\mathrm{rad}/{\mathrm{s}}^2){\mathrm{t}}^2-(1.50\mathrm{rad}/{\mathrm{s}}^3){\mathrm{t}}^3$

The angular velocity of the body in the time dt is the ratio of the angular displacement$\mathit{d}\mathit{\theta }$ to dt.

$\mathit{\omega }\mathbf{=}\frac{\mathbf{d}\mathbf{\theta }}{\mathbf{d}\mathbf{t}}$

The angular acceleration is given by,

$\mathit{\alpha }\mathbf{=}\frac{\mathbf{d\omega }}{\mathbf{dt}}$

Substitute$\mathrm{\theta }\left(\mathrm{t}\right)=(250\mathrm{rad}/\mathrm{s})\mathrm{t}-(20.0\mathrm{rad}/{\mathrm{s}}^2){\mathrm{t}}^2-(1.50\mathrm{rad}/{\mathrm{s}}^3){\mathrm{t}}^3$ in the angular velocity equation.

$$\begin{gathered} \omega \left(t\right)=\frac{d}{dt}\left[\left(250rad/s\right)t-\left(20.0rad/s^2\right)t^2-\left(1.50rad/s^3\right)t^3\right] \\ =250rad/s-40t-4.50t^2.........\left(1\right) \end{gathered}$$

Calculate the time, when the angular velocity of shaft is zero.

$$\begin{gathered} 250-40t-4.50t^2=0 \\ 4.50t^2+40t-250=0 \end{gathered}$$

Simplify the quadratic equation.

$$\begin{gathered} t=\frac{-40\pm \sqrt{{40}^2+4\times 250\times 4.5}}{2\times 4.5} \\ =\frac{-40\pm \sqrt{6100}}{9} \\ =4.23s \end{gathered}$$

Therefore, the angular velocity of the motor shaft will be zero at t = 4.23 s.

Find the expression for angular acceleration as follows.

$$\begin{gathered} \alpha \left(t\right)=\frac{d}{dt}\left[\omega t\right] \\ =\frac{d}{dt}\left[250rad/s-40t-4.50t^2\right] \\ =-40-9t..........\left(2\right) \end{gathered}$$

Substitute t = 4.23 s in equation (2) to find initial value of angular acceleration.

$$\begin{gathered} \mathrm{\alpha }{\left(\mathrm{t}\right)}_{\mathrm{t}=4.23\mathrm{s}}=-40-9(4.23\mathrm{s}) \\ =-78.07\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the required angular acceleration is$-78.07\mathrm{rad}/{\mathrm{s}}^2$ .

Substitute t = 4.23 s in equation .

$$\begin{gathered} \mathrm{\theta }\left(\mathrm{t}\right)=\left(250\mathrm{rad}/\mathrm{s}\right)\mathrm{t}-\left(20.0\mathrm{rad}/{\mathrm{s}}^2\right){\mathrm{t}}^2-\left(1.50\mathrm{rad}/{\mathrm{s}}^3\right){\mathrm{t}}^3 \\ \mathrm{\theta }\left(\mathrm{t}\right)=\left(250\mathrm{rad}/\mathrm{s}\right)\left(4.23\mathrm{s}\right)-\left(20.0\mathrm{rad}/{\mathrm{s}}^2\right){(4.23\mathrm{s})}^2-\left(1.50\mathrm{rad}/\mathrm{s}3\right){\left(4.23\mathrm{s}\right)}^3\approx 586\mathrm{rad} \\ =\left(586\mathrm{rad}\right)\left(\frac{1\mathrm{rev}}{2\mathrm{\pi }\mathrm{rad}}\right) \\ =93.3\mathrm{rev} \end{gathered}$$

Therefore, the required number of revolution is 93.3 rev.

Substitute t = 0 in equation (1).

$$\begin{gathered} \mathrm{\omega }\left(\mathrm{t}\right){|}_{\mathrm{t}=0}=250\mathrm{rad}/\mathrm{s}-40\left(0\right)-4.50{\left(0\right)}^2 \\ =250\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the required speed of the shaft is 250 rad/s.

Find the angular displacement at t = 0 .

$$\begin{gathered} {\mathrm{\theta }}_1=\left(250\mathrm{rad}/\mathrm{s}\right)\left(0\right)-\left(20.0\mathrm{rad}/{\mathrm{s}}^2\right){\left(0\right)}^2-\left(1.50\mathrm{rad}/{\mathrm{s}}^3\right){\left(0\right)}^3 \\ =0 \end{gathered}$$

Find the angular displacement at t = 4.23 s .

$$\begin{gathered} {\mathrm{\theta }}_1=\left(250\mathrm{rad}/\mathrm{s}\right)(4.23\mathrm{s})\left(0\right)-\left(20.0\mathrm{rad}/{\mathrm{s}}^2\right)(4.23\mathrm{s})-\left(1.50\mathrm{rad}/{\mathrm{s}}^3\right)(4.23\mathrm{s}) \\ =586.11\mathrm{rad} \end{gathered}$$

Find the average angular velocity as follows.

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{av}-\mathrm{z}}=\frac{{\mathrm{\theta }}_2-{\mathrm{\theta }}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1} \\ =\frac{586.11\mathrm{rad}-0\mathrm{rad}}{4.23\mathrm{s}-0\mathrm{s}} \\ =138.56\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the average angular velocity is 138.56 rad/s.