Question

The angular velocity of a process control motor is $\mathrm{\omega }=(20-\frac{1}{2}{\mathrm{t}}^2)\mathrm{rad}/\mathrm{s}$, where t is in seconds.

a. At what time does the motor reverse direction?

b. Through what angle does the motor turn between$t=0s$ and the instant at which it reverses direction?

Short answer

a. $\mathrm{t}=6.32\mathrm{s}$

b.$\mathrm{\theta }=60.32°$

Step-by-step solution

Given Information

We are given with:

$\mathrm{\omega }=(20-\frac{1}{2}{\mathrm{t}}^2)\mathrm{rad}/\mathrm{s}$, where $\mathrm{t}$is in seconds.

b.role="math" localid="1647936746784" $\mathrm{t}=0\mathrm{s}$

Part a) Step 1: Explanation

$\mathrm{\omega }=20-\frac{1}{2}{\mathrm{t}}^2$

Motor reverse it direction when $\mathrm{\omega }=0$

role="math" localid="1647936920144" $$\begin{gathered} 0=20-\frac{1}{2}{\mathrm{t}}^2 \\ 0.5{\mathrm{t}}^2=20 \\ {\mathrm{t}}^2=40 \\ \mathrm{t}=\sqrt{40} \\ \mathrm{t}=6.32\mathrm{s} \end{gathered}$$

Part b) Step 1: Explanation

The angle turns the motor between $\mathrm{t}=0\mathrm{s}$ and the instant at which it reverses direction is

$$\begin{gathered} \frac{\mathrm{d\theta }}{\mathrm{dt}}=\mathrm{\omega } \\ \int \mathrm{d\theta }={\int }_0^{6.3}\mathrm{\omega dt} \\ \int \mathrm{d\theta }={\int }_0^{6.3}(20-\frac{1}{2}{\mathrm{t}}^2)\mathrm{dt} \\ \mathrm{\theta }={(20\mathrm{t}-\frac{0.5}{3}{\mathrm{t}}^3)}_0^{6.3} \\ \mathrm{\theta }={(20\mathrm{t}-0.167\times {\mathrm{t}}^3)}_0^{6.3} \\ \mathrm{\theta }=20\times 6.3-0.167\times 6.3^3 \\ \mathrm{\theta }=60.32° \end{gathered}$$