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Q3-3.4-16E

Expert-verified
Found in: Page 116

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Find the equation for the angular velocity ${\mathbit{\omega }}$ in Problem15, assuming that the retarding torque is proportional to role="math" localid="1663966970646" $\sqrt{\mathbf{\omega }}$

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

See the step by step solution

## 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

$\mathrm{I}\frac{\mathrm{d}\omega }{\mathrm{dt}}=\mathrm{T}-\mathrm{K}\sqrt{\omega }\phantom{\rule{0ex}{0ex}}\frac{\mathrm{I}\mathrm{d\omega }}{\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}}=\mathrm{dt}\left(\text{Variable separating}\right)\phantom{\rule{0ex}{0ex}}\frac{-2\mathrm{I}\sqrt{\mathrm{\omega }}}{\mathrm{K}}-\frac{2\mathrm{IT}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{{\mathrm{K}}^{2}}=\mathrm{t}+\mathrm{C}\left(\text{Integrating on both sides}\right)\phantom{\rule{0ex}{0ex}}\frac{2\mathrm{I}}{\mathrm{K}}\left(\sqrt{\mathrm{\omega }}-\frac{\mathrm{IT}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{\mathrm{K}}\right)=\mathrm{t}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\left(\sqrt{\mathrm{\omega }}-\frac{\mathrm{IT}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|}{\mathrm{K}}\right)=-\frac{\mathrm{Kt}}{2\mathrm{I}}+{\mathrm{C}}_{1}\phantom{\rule{0ex}{0ex}}\left(\mathrm{K}\sqrt{\mathrm{\omega }}-\mathrm{T}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{\mathrm{\omega }}\right|\right)=\frac{-{\mathrm{K}}^{2}\mathrm{t}}{2\mathrm{I}}+\mathrm{A}\phantom{\rule{0ex}{0ex}}$

## Step 2: Find the value of A

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

$\mathrm{A}=\left(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{T}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{K}\sqrt{\omega }-\mathrm{T}\mathrm{ln}\left|\mathrm{T}-\mathrm{K}\sqrt{\omega }\right|\right)=\left(\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}-\mathrm{Tln}\left|\mathrm{T}-\mathrm{K}\sqrt{{\omega }_{\mathrm{o}}}\right|\right)-\frac{{\mathrm{K}}^{2}\mathrm{t}}{2\mathrm{I}}$

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