W8.2 Green buses.

You are working in a research group investigating more energy efficient city buses. The design strategy is to store energy in the rotation of a flywheel when the bus is brought to a stop and then use it to subsequently accelerate the bus. The flywheel under consideration is a disk of uniform construction except that it has a massive, thin rim on its edge, comprising half of the flywheel’s mass (which is 100 kg in total).

You’ve been asked to advise various engineering teams:

  1. The people building a prototype of the bus want you to estimate the radius of the flywheel that is needed. [The design brief says that the bus will be brought to rest from its cruising speed of 30 mph by accelerating the flywheel from 2 revolutions per second to 20 revolutions per second].
  2. The engineers building the flywheel need to know if the join between the different materials of the disc and the rim of the flywheel can withstand rapid acceleration when the bus needs to come to an emergency stop. They think that they can design a join to withstand (tangential) accelerations of 12g for a short period. Is this going to be safe? [The brief says that the bus must be capable of stopping from 30 mph in 1 second]
  3. A third team want advice on how they should mount the flywheel in the bus. What can you tell them?
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Hint

Hint: you can estimate the mass of a fully-laden bus. The expressions for moments of inertia for a disc and a hoop of mass m are (respectively) $\frac{mr^2}{2}$ and mr2
 

Solution

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Solution

Let’s start with some (rather idealistic) assumptions

  • All the kinetic energy of the bus can be transferred without loss to the flywheel, during ’braking’.
  • The mass of a laden bus is 104kg.
  • The acceleration of the flywheel is uniform

So our strategy and solution for each of the three questions is as follows:

  1. Conservation of energy; translational ke of bus and rotational ke of flywheel

    $$1/2 M (v^2-v_0^2) = 1/2 I (\omega^2 - \omega_0^2)$$
    where

    • M is the mass of the bus (104kg)
    • $(v^2-v_0^2) = \Delta v^2$ is the change in speed of the bus (30mph=13.4ms-1)
    • $(\omega^2 - \omega_0^2) = \Delta \omega^2$ is the change in angular speed of the flywheel ([(20-2)×2π]2)rad2s-2
    • I is the total moment of inertia of the flywheel, whose radius is r and mass is m=100kg. ($I = I_d + I_h = \frac{mr^2}{4} + \frac{mr^2}{2} = \frac{3mr^2}{4})$

    Rearranging,

    $$ r^2 = {\frac{4 M \Delta v^2}{3 \Delta\omega^2 m} } \Rightarrow r \simeq 1.2 m$$

  2. Calculate tangential acceleration of a point on the join as the flywheel is accelerated up to full speed.

    We have assumed constant acceleration of the flywheel, so we can use ω=ω0+αt. We find α=36πrads-2.

    A point on the join (let’s assume the heavy rim is very thin, so the join is effectively located at the edge of the flywheel) is thus accelerating with a tangential acceleration given by

    a=rα=150ms-2=14gms-2

    It looks like the engineers will need to find better glue.....

  3. There’s the added complication here that a rotating flywheel acts as a gyroscope ( see About gyroscopes), that will resist changes in its orientation. This will be a problem if the bus tries to turn (or tilt) when the flywheel is spinning fast. So it would need to be mounted on gimbals to maintain a fixed orientation relative to the bus on which it is mounted. If you haven’t yet seen this demonstrated in lectures (not with a bus, obviously........) you will soon.