91Ó°ÊÓ

The equation of angular displacement is as follows.

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

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

$\mathit{Ó\neg }\mathbf{=}\frac{\mathbf{d}\mathbf{θ}}{\mathbf{d}\mathbf{t}}$

The angular acceleration is given by,

$\mathit{Î\pm }\mathbf{=}\frac{\mathbf{dÓ\neg }}{\mathbf{dt}}$

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

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

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.

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

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

Find the expression for angular acceleration as follows.

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

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

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

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

Substitute t = 4.23 s in equation .

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

Therefore, the required number of revolution is 93.3 rev.

Substitute t = 0 in equation (1).

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

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

Find the angular displacement at t = 0 .

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

Find the angular displacement at t = 4.23 s .

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

Find the average angular velocity as follows.

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

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