Question

If $C_f$ is the coefficient of speed fluctuation of a flywheel then the ratio of ${\omega }_{max}/{\omega }_{min}$ will be

• A. $\frac{1-2C_f}{1+2C_f}$
• B. $\frac{2-C_f}{2+C_f}$
• C. $\frac{1+2C_f}{1-2C_f}$
• D. $\frac{2+C_f}{2-C_f}$

Answer 1

The correct option is D $\frac{2+C_f}{2-C_f}$

$C_f=\frac{{\omega }_{max}-{\omega }_{min}}{({\omega }_{max}+{\omega }_{min})/2}=\frac{2({\omega }_{max}-{\omega }_{min})}{{\omega }_{max}+{\omega }_{min}}$

$\frac{C_f}{2}=\frac{{\omega }_{max}-{\omega }_{min}}{{\omega }_{max}+{\omega }_{min}}$

$C_f{\omega }_{max}+C_f{\omega }_{min}=2{\omega }_{max}-2{\omega }_{min}$

$C_f{\omega }_{min}+2{\omega }_{min}=2{\omega }_{max}-C_f{\omega }_{max}$

$(C_f+2){\omega }_{min}={\omega }_{max}(2-C_f)$

$\frac{C_f+2}{2-C_f}=\frac{{\omega }_{max}}{{\omega }_{min}}$

$\frac{{\omega }_{max}}{{\omega }_{min}}=\frac{2+C_f}{2-C_f}$