Discussion
Consider a flywheel of mass 10 kg, uniform thickness 15 mm, and radius 100 mm. If it rotates at 300 RPM, then the KE of the flywheel (in Joules) is:
Answer: A
The flywheel is a uniform solid disc, so its moment of inertia about its axis is \(I = \tfrac{1}{2}mr^2 = \tfrac{1}{2}(10)(0.1)^2 = 0.05\ \text{kg m}^2\). The speed is 300 rpm, so \(\omega = 2\pi \times 300/60 = 31.42\) rad/s. The rotational kinetic energy is
\[ \begin{aligned} KE = \tfrac{1}{2}I\omega^2 \\ &= \tfrac{1}{2}(0.05)(31.42)^2 \approx 24.7\ \text{J} \end{aligned} \]The nearest option is 24.65 J. The thickness of 15 mm is not needed since the mass is given. The larger options would come from using the wrong moment of inertia, such as that of a ring (\(mr^2\)), which would double the answer.
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