Question

In Problem 16, let I = 50 kg-m² and the retarding torque be N-mIf the motor is turned off with the angular velocity at 225 rad/sec, determine how long it will take for the flywheel to come to rest.

Short answer

The flywheel takes to come to rest in300 sec.

Step-by-step solution

Find the value of time

The equation is

$$\begin{gathered} \mathrm{I}\frac{\mathrm{d}\omega }{\mathrm{dt}}=-{\mathrm{T}}_1 \\ 50\frac{\mathrm{d}\omega }{\mathrm{dt}}=-5\sqrt{\omega } \\ -10\int \frac{\mathrm{d}\omega }{\sqrt{\omega }}=\int \mathrm{dt}\backslash \mathrm{begingathered}-20\sqrt{\mathrm{\omega }}=\mathrm{t}+\mathrm{c} \\ \mathrm{\omega }\left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+\mathrm{c}{)}^2 \\ \backslash \mathrm{endgathered} \\ -20\sqrt{\mathrm{\omega }}=\mathrm{t}+\mathrm{c} \\ \mathrm{\omega }\left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+\mathrm{c}{)}^2 \end{gathered}$$

Apply the given conditions

When ${\omega }_0$=225 , c = 15

$$\begin{gathered} \omega \left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+15{)}^2 \\ \mathrm{t}=20(15-\sqrt{\omega \left(\mathrm{t}\right)}) \end{gathered}$$

at the moment when flywheel stopes rotating (t)=0,

so,t = 300 sec

Hence, the flywheel takes to come to rest in 300 sec.