Question

The maximum fluctuation of energy in a flywheel is equal to

Where: I = Mass moment of inertial of the flywheel

E = Mean kinetic energy of the flywheel

K = Co-efficient of fluctuation of speed

ω = Mean angular speed

\(=\frac{\omega_1+\omega_2}{2}\)

1. Iω(ω₁ - ω₂)
2. Iω²K
3. 2EK
4. All of these

Answer 1

Correct Answer - Option 4 : All of these

Explanation:

Mass moment of inertia of the flywheel = I, Mean speed of the rotation = ω, Kinetic energy = E, Fluctuation of speed = K

Coefficient of fluctuation of energy:

\({C_E} = \frac{{Maximum\;fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)

Coefficient of fluctuation of speed:

\(\begin{array}{l} {K} = \frac{{Maximum\;fluctuation\;of\;speed}}{{Mean\;speed}}\\ {K} = \frac{{{\omega _1} - {\omega _2}}}{\omega } = \frac{{2\left( {{\omega _1} - {\omega _2}} \right)}}{{\left( {{\omega _1} + {\omega _2}} \right)}} \end{array}\)

The maximum fluctuation of energy:

\(\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}I\omega _1^2 - \frac{1}{2}I\omega _2^2 = \frac{1}{2}I\left( {{\omega _1} + {\omega _2}} \right)\left( {{\omega _1} - {\omega _2}} \right)\\ {\rm{\Delta }}E = I\omega \left( {{\omega _1} - {\omega _2}} \right) = I{\omega ^2}\frac{{\left( {{\omega _1} - {\omega _2}} \right)}}{\omega } = I{\omega ^2}{K}\\ E = \frac{1}{2}I{\omega ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{K}}} = \Delta E\;=2KE {{{}}}{{{}}} \end{array}\)

Hence maximum fluctuation energy is the difference between the maximum and minimum energies.