Discussion
The change in kinetic energy \(\Delta E\) of an engine is 300 J, and minimum and maximum shaft speeds are \(\omega_{min} = 220\) rad/s and \(\omega_{max} = 280\) rad/s, respectively. Assume that the torque-time function is purely harmonic. To achieve a coefficient of fluctuation of 0.05, the moment of inertia (in kg.m2) of the flywheel to be mounted on the engine shaft is
Answer: B
The moment of inertia is 0.096 kg.m2.
The flywheel energy relation links the fluctuation of energy to the coefficient of fluctuation of speed \(C_{s}\) and the mean speed:
\[ \begin{aligned} \omega_{mean} &= \frac{220 + 280}{2} = 250\ \text{rad/s} \\ \Delta E &= I\,\omega_{mean}^{2}\,C_{s} \\ I &= \frac{300}{250^{2} \times 0.05} = \frac{300}{3125} = 0.096 \end{aligned} \]
The stated speeds are used only for the mean speed; the target coefficient 0.05 is the design value the flywheel must achieve. Using \(\omega_{max}\) or \(\omega_{min}\) in place of the mean speed gives about 0.077 or 0.124, neither of which is offered; 0.113, 0.071 and 0.053 do not follow from the data.
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