Question 10.43: A 5 kW induction motor running at 750 rpm operates a rivetti...

\text { Mean speed, } \omega_{m}=2 \pi \times \frac{750}{60}=78.54 rad / s .

\text { Energy supplied by the motor in one hour }=5 \times 10^{3} \times 3600=18 \times 10^{6} N m .

\text { Energy required for one riveting operation }=10 \times 10^{3} \times 1=10^{4} Nm .

\text { Number of rivets closed per hour }=\frac{18 \times 10^{6}}{10^{4}}=1800 .

\text { Energy supplied by the motor in } 1 s=5 \times 10^{3} \times 1=5000 Nm .

Energy to be supplied by the flywheel=10000 – 5000=5000 N m.

E_{f}=E_{\max }-E_{\min }=0.5 I\left(\omega_{\max }^{2}-\omega_{\min }^{2}\right) .

\text { Now } \omega_{\max }=\omega_{m} .

5000=0.5 \times 80 \times(0.45)^{2}\left[\left(78.540^{2}-\omega_{\min }^{2}\right]\right. .

\omega_{\min }=74.5 rad / s .

or N_{\min }=711.5 rpm .

Fall in speed  =750-711.5=38.5 rpm .