Question

The torque provided by an engine is given by $T\left(\theta \right)=12000+2500sin\left(2\theta \right)$N.m, where $\theta$ is the angle turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a machine that provides a constant resisting torque. If variation of the speed form the mean speed is not to exceed $\pm 0.5\%$ the minimum mass momnet of inertia of the flywheel should bekg.m${}^2$ (round off to the nearest integer).

Answer 1

$T_{mean}=12000$Nm
$\omega =\frac{\pi \times 200}{30}=20.9439$ rad/s
$\Delta E={\int }_0^{\frac{\pi }{2}}(T-T_{mean})d\theta =2500{\int }_0^{\frac{\pi }{2}}sin2\theta d\theta$
$=2500\times 1=2500$ J
$\Delta E=I{\omega }^2{C}_s$
$=2500=I\times {20.9439}^2\times 1$
$\Rightarrow I=569.934kg{m}^2\underset{–}{\sim }570kg.m^2$