Q3-3.4-16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the equation for the angular velocity $\mathit{\omega }$  in Problem15, assuming that the retarding torque is proportional to   $\sqrt{\mathbf{\omega }}$

Step-by-Step Solution

Answer

The equation of angular velocity is  $(\mathrm{K}\sqrt{\omega }-\mathrm{T} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\omega }\right|)=(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{T} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|)-\frac{{\mathrm{K}}^2\mathrm{t}}{2\mathrm{I}}$

Step1: Find the equation for the angular velocity

Here the notations are T= torque for motor,  $\omega$= angular velocity, I = moment of inertia and  ${\omega }_0$= initial angular velocity.

 

According To the question retarding torque due to friction is proportional to the angular velocity so,  ${\mathrm{T}}_1=-\mathrm{K}\sqrt{\omega }$  (K is proportionality constant)

 

Now moment of inertia × angular velocity = sum of the torques

$$\begin{gathered} \mathrm{I}\frac{\mathrm{d}\omega }{\mathrm{dt}}=\mathrm{T}-\mathrm{K}\sqrt{\omega } \\ \frac{\mathrm{I} \mathrm{d\omega }}{\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}}=\mathrm{dt} \left(\text{Variable separating}\right) \\ \frac{-2\mathrm{I}\sqrt{\mathrm{\omega }}}{\mathrm{K}}-\frac{2\mathrm{IT} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{{\mathrm{K}}^2}=\mathrm{t}+\mathrm{C} \left(\text{Integrating on both sides}\right) \\ \frac{2\mathrm{I}}{\mathrm{K}}(\sqrt{\mathrm{\omega }}-\frac{\mathrm{IT} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{\mathrm{K}})=\mathrm{t}+\mathrm{C} \\ (\sqrt{\mathrm{\omega }}-\frac{\mathrm{IT} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{\mathrm{K}})=-\frac{\mathrm{Kt}}{2\mathrm{I}}+{\mathrm{C}}_1 \\ (\mathrm{K}\sqrt{\mathrm{\omega }}-\mathrm{T} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|)=\frac{-{\mathrm{K}}^2\mathrm{t}}{2\mathrm{I}}+\mathrm{A} \end{gathered}$$

Step 2: Find the value of A

Put $\omega \left(0\right)={\omega }_0$ then value of A.

$$\begin{gathered} \mathrm{A}=(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{T} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|) \\ (\mathrm{K}\sqrt{\omega }-\mathrm{T} \ln \left|\mathrm{T}-\mathrm{K}\sqrt{\omega }\right|)=(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{Tln}\left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|)-\frac{{\mathrm{K}}^2\mathrm{t}}{2\mathrm{I}} \end{gathered}$$

Hence, the equation of angular velocity is  $(\mathrm{K}\sqrt{\omega }-\mathrm{T} ln\left|\mathrm{T}-\mathrm{K}\sqrt{\omega }\right|)=(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{T} ln\left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|)-\frac{{\mathrm{K}}^2\mathrm{t}}{2\mathrm{I}}$.