First Day Back!

Today was my first day of Upper Sixth, and I had my first Physics lesson of the year. I always come away from Physics lessons feeling a little bit inspired, which was very useful after six weeks to myself.

We started by quickly covering some momentum concepts that we squeezed in last year. Specifically, we were asked to come up with an experiment involving a toy helicopter, a wind speed measure, and a metre rule that we could use to measure the density of air. We also had a mass balance, but that list was getting a little long.

So, my first instinct was to calculate the terminal velocity of the helicopter (which I've done before, and I know involves air density) and then to drop it. This is quite a fun calculation to do:

Firstly, picture the helicopter falling through the air. The average air velocity is 0, and the helicopter's downwards velocity is v. This isn't a very helpful frame of reference, so we're going to change coordinate system to one in which the helicopter is stationary - that's the one which is going downwards at speed v. In this frame, the average is velocity is not 0, but v upwards, towards the helicopter. We now have a stationary helicopter, and a wind with speed v blowing on it. We want to find the force due to the wind.

The wind is made of particles with average momentum p, that hit the helicopter and are scattered. Here we are going to make the assumption that after hitting the helicopter, the momentum is randomised to some momentum in the range [-p,p] so that the average particle loses pNs of momentum when hitting the copter. Now, we find the total change in momentum after t seconds: in t seconds, the helicopter sweeps out a volume of (vt)(A), where vt is the length of a prism of cross section A. The momentum of that volume of air is p=mv=dVv=dvtAv=dAv^2t.

Force is rate of change of momentum, so to find the average force exerted, we simply divide by time - F=dAv^2. To find d (the density of air), we would set F=mg by dropping the helicopter and waiting until it reached terminal velocity. My teacher, however, disagreed about this: he pointed out that we hadn't a stopwatch to time the falls, or any other equipment for accurately measuring velocity.

Crestfallen, I did the task seriously: the obvious strategy is to find the thrust from the helicopter, hope it involves density, then have the helicopter hover so that the thrust equals the weight, and then solve for density. Luckily, my refresher course on air resistance had helped me considerably: the thrust from the helicopter could be assumed to be from pushing air down with the rotors. And if the air was being pushed down at a speed v from a rotor of area A, then it just so happens (not actually a coincedence, try to prove the following) that the force exerted is F=dAv^2 - exactly the same as for air resistance. So now you have the following:

dAv^2 = mg

d=mg(A^-1)(v^-2)

Where A is the rotor area, v is the air speed out of the rotors, and mg is the weight.

As it happens, there's another much quicker way to get at this formula: guess. The dimensions of density are [m * L^-3], so you immediately know d = km, as nothing else other than the helicopter's mass has dimensions of mass. The L^-3 takes some more work, and there are several combinations that work but you have some help: A has dimensions [L^2], whereas both g and v have time in their dimensions somewhere - but density doesn't. This puts a limit on what g and v can be, because their 'time parts' have to cancel out. Once you've done this, you can either guess that it should be F=kmg, and then you can solve for the correct equation, or you can motivate it in a few other ways as well.

A word of warning though: the dimensional analysis (advanced guessing) does not guarentee the right answer. For instance, if you used the diameter of the rotors, 'D' in your calculations, rather than the area, then the 'correct' density expression is:

d=(4/pi)mg(D^-2)(v^-2)

but the dimensional analysis will lose the factor (4/pi) in front. In this case, using the area instead makes the guess exact, but dimensional analysis can always be off by a constant factor.