When energy conservation doesn’t add up (or does it?)

By Marcus Wilson 23/09/2014

In the last few weeks holes have been popping up all over Cambridge. They are being dug by 'ditch-witches'  – pieces of machinery designed for making small-diameter tunnels for cabling – as part of the installation of fibre-optic cables for the much vaunted ultra-fast broadband. A ditch-witch is about the ultimate in machinery-obsessed-toddler heaven. We've been avidly following their movement around the Cambridge streets, or at least the youngest member of our family has. They went down our street about seven weeks ago, and since then have been tracking southwards. I'm tempted to slip a GPS locator beacon on them and then write a ditch-witch locator app to help all those stressed parents cope with constant demands to find them.

So, Sunday saw Benjamin and I get on the bicycle and go on a ditch-witch hunt. (We're going on a ditch-witch hunt… We're going to catch a BIG one…we're not scared…). And, much to my relief, we found them, resting quietly on Thompson Street. 

But this entry isn't about ditch-witches or diggers or cranes or other large pieces of machinery, it's about what we saw on the way. On the front lawn of one house, there was a teenage boy practising 'barrel walking'. He was standing on a barrel, and rolling it forward and backwards around the garden. He was obviously reasonably skilled at this since he had some pretty good control of where he was going. 

An interesting observation is that to get the barrel to roll forwards, the rider has to walk backwards. That must feel a little disconcerting. To get the barrel (and you) moving forward at say 2 km/h, you have to walk backwards at 2 km/h . That's because the bottom of the barrel, in contact with the ground, is instaneously stationary, so if the centre is travelling at 2 km/h forwards, the top of the barrel must be going 4 km/h forwards relative to the ground. In order for you to go at the pace of the centre, 2 km/h forwards (and stay on top), you therefore need to go 2 km/h backwards with respect to the top of the barrel. In terms of mathematics: your speed relative to the ground = 2v – v = v, where the 2v is the speed of the top of the barrel, the  '-v' is the speed of you relative to the barrel, and the 'v' is the speed of the centre. Go it?

That kind of relationship crops up quite a bit in physics. I've talked about a case before – when a satellite in orbit loses energy because it hits air molecules, it speeds up. Uh! How does that work? It's because, as it loses energy, it drops to a lower orbit, one with less potential energy. But lower orbits have higher orbital speeds. It turns out that the loss in potential energy is exactly double the gain in kinetic energy. That is, if the satellite loses 100 J of energy, It's made up of a gain of 100 J of kinetic energy and a loss of 200 J of potential energy. It's another '2 – 1 = 1' sum. 

There's also the neat but confusing case of a parallel plate capacitor at constant voltage. Let's say a capacitor consists of two large flat plates, a distance of 1 cm apart. The plates are maintained at constant voltage of say 12 V by a power-supply (e.g battery). This means that the plates have opposite charge, and so attract each other. (To hold them at constant distance, you have to fix them in place somehow). Now, consider pulling those plates apart. Since they attract each other, it is clear that you have to do work on the system to do this. One might therefore expect that the energy stored in the capacitor has gone up. But no. Do the calculation, and you'll see that the energy goes down. (Energy stored = capacitance times voltage squared, divided by two. The voltage stays the same, and since the capacitance is inversely proportional to plate separation, increasing the separation will decrease the stored energy.) Uh! Where does the energy go then? In this case, you have to consider the power supply. What happens is that you are putting energy back into the battery, by causing a current to flow backwards through it. It turns out in this case, that the work you need to do is exactly half the energy that goes to the power supply. The other half comes from the loss in energy stored in the capacitor. So, if we put in 10 J of energy, we lose 10 J of stored energy in the capacitor, and we gain 20 J of energy in the power supply. So, again, we have the '2 – 1 = 1' sum. 

So, for every kilojoule of energy burned by the ditch-witch, doe the toddler also burns a kilojoule, thus meaning 2 kilojoules of heat end up in the air?  (As neat as it would be if that were true, I don't think the actual figures will come close).