Question

A flywheel is attached to an engine to keep its rotational speed between 100 rad/s and 110 rad/s. If the energy fluctuation in the flywheel between these two speeds is 1.05 kJ then the moment of inertia of the flywheel is kg.$m^2$(round off to 2 decimal places).

• A. 2

Answer 1

The correct option is A 2
Given,
$\Delta E_{max}=1.05kJ,{\omega }_{max}=110\text{rad/s}$;

${\omega }_{min}=100\text{rad/s}$

$\Delta E_{max}=\frac{1}{2}I({\omega }_{max}^2-{\omega }_{min}^2)$

$\therefore 1.05\times {10}^3\times 2=I({110}^2-{100}^2)$

$I=1kg.m^2$