The rotating blade of a blender turns with constant angular acceleration$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ . (a) How much time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

The angular acceleration is${\alpha }_z=1.50rad/s^2$ .

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

The final angular speed is${\omega }_z=36rad/s$ .

The straight-line motion is particularly simple when the acceleration is constant. This is also true of rotational motion about a fixed axis.

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

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

Here, 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,

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

Here, is Angular position of body at time 0.

Substitute ,${\omega }_{0_z}=0rad./s,{\omega }_z=36rad/s,and{\alpha }_z=1.5rad/s^2$in equation (1).

$$\begin{gathered} 36rev/s=0rev/s+\left(1.5rad/s^2\right)t \\ t=\frac{36rev/s}{1.5rad/s^2} \\ =24s \end{gathered}$$

Therefore, the required time is 24s .

Substitute${\theta }_0=0,{\omega }_z=0rad/s,t=24s,and{\alpha }_z=1.5rad/s^2$ in equation (2).

$$\begin{gathered} \theta =0+\left(0\right)\left(24s\right)+\frac{1}{2}\left(1.5rad/s^2\right){\left(24s\right)}^2 \\ =432rad \\ =\left(432rad\right)\left(\frac{1ev}{2\mathrm{\pi }\mathrm{rad}}\right) \\ =68.8\mathrm{rev} \end{gathered}$$

Therefore, the number of revolutions are$68.8rev$ .