Question

A fly wheel of gasoline engine is required to give up $300J$ of kinetic energy while its angular velocity decreases from $600rev/min$ to $540rev/min$. What moment of inertia is required?

Answer 1

Given that,

Kinetic energy $K.E=300J$

Angular velocity ${\omega }_1=600rev/min=62.83rad/s$

Angular velocity ${\omega }_2=540rev/min=56.55rad/s$

We know that,

$K.E=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)$

$300J=\frac{1}{2}I(62.83-56.55)$

$I=\frac{600}{6.28}$

$I=95.5kg{m}^2$

Hence, the moment of inertia is $95.5kg{m}^2$