Q3-3.4-15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A rotating flywheel is being turned by a motor that exerts a constant torque T (see Figure 3.10). A retarding torque due to friction is proportional to the angular velocity v. If the moment of inertia of the flywheel, is I and its initial angular velocity is , find the equation for the angular velocity v as a function of time. [Hint: Use Newton’s second law for rotational motion, that is, moment of inertia * angular acceleration = sum of the torques.]

Step-by-Step Solution

Answer

The equation of angular velocity is $\omega \left(\mathrm{t}\right)=\frac{\mathrm{T}}{\mathrm{K}}+({\omega }_{\mathrm{o}}-\frac{\mathrm{T}}{\mathrm{K}}){\mathrm{e}}^{\frac{-\mathrm{KT}}{\mathrm{I}}}$.

Step1: Find the equation for the angular velocity

Here the notations are T= torque for motor, = angular velocity, I = moment of inertia and 

${\omega }_{\mathbf{o}}$= initial angular velocity.

 

Acc. 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

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

Step 2: Find the value of C

Put  $\omega \left(0\right)={\omega }_0$then $\mathrm{C}=\mathrm{T}-\mathrm{K}{\omega }_{\mathrm{o}}$

 $\omega \left(\mathrm{t}\right)=\frac{\mathrm{T}}{\mathrm{K}}+({\omega }_{\mathrm{o}}-\frac{\mathrm{T}}{\mathrm{K}}){\mathrm{e}}^{\frac{-\mathrm{KT}}{\mathrm{I}}}$

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