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GATE 2024 ME – Question 23

Theory of Machines · Flywheels and governors · 1 mark · Multiple choice

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.m$^2$) of the flywheel to be mounted on the engine shaft is

  1. 0.113
  2. 0.096
  3. 0.071
  4. 0.053

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Correct answer: (B) 0.096

Explanation

The mean speed is $\omega = \frac{220 + 280}{2} = 250$ rad/s. The energy fluctuation is $\Delta E = I\omega^2C_s$, so $I = \frac{300}{250^2 \times 0.05} = \frac{300}{3125} = 0.096$ kg·m$^2$.