Flywheel

Classical Mechanics Level 3

A bus sometimes has a flywheel. When the bus brakes, translational kinetic energy from the bus is transferred to this flywheel. When the bus starts up again, energy from the flywheel is then transferred back to the wheels of the bus.

Suppose a $15\:000\:\text{kg}$ bus is initially traveling at a speed of $20\:\text{m/s}$ . During the braking process, 20% of the kinetic energy is transferred to a flywheel. This is a uniform solid disc that can rotate about its central axis; it has a radius of $1.50\:\text{m}$ and a mass of $300\:\text{kg}$ . What will be the approximate rotational period of the flywheel?

1 s 0.001 s 10 s 0.01 s 0.1 s

1 solution

Arjen Vreugdenhil
Feb 18, 2016

Let's solve the problem generally for a bus of mass $M$ at speed $v$ , with a flywheel of mass $m$ , radius $r$ , and constant of gyration $\gamma$ . (That is, $I = \gamma m r^2$ ; for a solid disc, $\gamma = \tfrac12$ .) The efficiency of the process is $\epsilon$ . (In this case, $\epsilon = 0.2$ .)

initial kinetic energy of the bus: $K = \tfrac12 M v^2$
energy transferred to the flywheel: $K' = \epsilon K = \frac \epsilon 2 M v^2$
angular speed of the flywheel: $K' = \tfrac12 I \omega^2 = \frac\gamma 2 m r^2 \omega^2\\ \omega = \sqrt{\frac{2K}{\gamma m r^2}} = \sqrt{\frac{\epsilon M v^2}{\gamma m r^2}}$
period of the flywheel: $T = \frac{2\pi}\omega = 2\pi \sqrt{\frac{\gamma m r^2}{\epsilon M v^2}} = 2\pi \sqrt{\frac{\gamma m}{\epsilon M}} \frac r v.$

Plug in the given values: $T = 2\pi \sqrt{\frac{0.5\cdot 300}{0.2\cdot 15\:000}}\frac{1.50}{20} = 0.11\:\text{s}.$ Thus we select the option $T \approx 0.1\:\text{s}$ .

Flywheel

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