Discussion
A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is
Answer: B
The kinetic energy is about 790 J.
A solid disc has \(I = \tfrac{1}{2}mr^{2}\):
\[ \begin{aligned} I &= \tfrac{1}{2}(20)(0.2)^{2} = 0.4\ \text{kg m}^{2} \\ \omega &= \frac{2\pi(600)}{60} = 62.83\ \text{rad/s} \\ KE &= \tfrac{1}{2}(0.4)(62.83)^{2} = 789.6\ \text{J} \end{aligned} \]
1580 J uses \(I = mr^{2}\), the value for a thin ring, and 3160 J also drops the half in the kinetic energy formula. 395 J is half the correct value. The thickness given is not needed once the mass is known.
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