Question

When $W$ joule of work is done on a flywheel, its frequency of rotation increases from ${\nu }_1\text{Hz}$ to ${\nu }_2\text{Hz}$. The moment of inertia of the flywheel about its axis of rotation is given by

• A. $\frac{W}{2{\pi }^2({\nu }_2^2-{\nu }_1^2)}$
• B. $\frac{W}{2{\pi }^2({\nu }_2^2+{\nu }_1^2)}$
• C. $\frac{W}{4{\pi }^2({\nu }_2^2-{\nu }_1^2)}$
• D. $\frac{W}{4{\pi }^2({\nu }_2^2+{\nu }_1^2)}$

Answer 1

The correct option is A $\frac{W}{2{\pi }^2({\nu }_2^2-{\nu }_1^2)}$

According to work energy theorem

Work done $=$ increase in kinetic energy

$W=\frac{1}{2}I{\omega }_2^2-\frac{1}{2}I{\omega }_1^2$

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

$W=2{\pi }^{2}I({\nu }_2^2-{\nu }_1^2)(\because \omega =2\pi \nu )$

$\therefore I=\frac{W}{2{\pi }^2({\nu }_2^2-{\nu }_1^2)}$

Hence, the correct choice is $\left(\text{A}\right)$.