Question

The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in $kg{m}^2$ is

• A. 0.10
• B. 0.20
• C. 0.30
• D. 0.40

Answer 1

The correct option is A 0.10

${\omega }_{min}=190$rad/s, ${\omega }_{max}=210$rad/s
$\Delta E=400Nm$

$C_s=\frac{\frac{{\omega }_{max}-{\omega }_{min}}{({\omega }_{max}+{\omega }_{min})}}{2}\frac{\frac{210-190}{(210+190)}}{2}=\frac{20}{200}=0.1$

${\omega }_{mean}=\frac{{\omega }_{min}+{\omega }_{max}}{2}=\frac{190+210}{2}$

$=200\text{rad/s}$

$\Delta E=I{\omega }_{mean}^2{C}_s$

$400=I\times (200{)}^2\times 0.1$

$I=\frac{400}{{200}^2\times 0.1}=0.1kg{m}^2$