When I challenged my students to explain why the bike was stable, I got a lot of "because of the weight underneath," but not much concrete, convincing explanation to justify that intuition. OK; let's back it up a bit. Why - in terms of energy - is a regular bike, when upright and motionless (for simplicity), unstable?
They connected stability (since we had said the word about ten times at this point) to the potential energy graph, and then just needed to do the trig to determine the gravitational potential energy of the Earth/bike/person system and graph it.
The diagram:
The gravitational potential energy (taking h=0 to be the vertical position):
$$U(\theta) = mgh = -mg\dfrac{l}{2}(1-\cos\theta)$$
The graph:
Why is it unstable? The force exerted on the bike that will act to change the angle is given by $F = -\dfrac{dU}{ds}$, which is another way of saying that the direction of the force is the opposite of the slope of the U graph or... that the system will evolve in the same way that a ball would, if it were rolling on a hill of the same shape as the U graph. Dome shaped? It'll roll downhill, away from the equilibrium point, so the equilibrium is unstable.
OK, let's add the mass underneath. I arbitrarily decided that it was on a massless pole of the same length as the bike's height, and that its mass was greater than the bike/person mass. This made qualitative analysis easier at the end, but they see how the parameters could be modified once the analysis is done.
The diagram:
$$U(\theta) = mgh = -mg\dfrac{l}{2}(1-\cos\theta) + Mgl(1-\cos\theta)$$
The big deal here conceptually is that, when the bike/person goes down, the mass goes up, and it gains more potential energy than the bike/person lost, meaning that we've turned a dome into a bowl, so that we now have a stable equilibrium.
Two interesting asides: if $M = \dfrac{m}{2}$, then $U=0$ for all angles, and the equilibrium is neutral, so the rider could stably sit at whatever angle. Not a super-fun idea, so it's a good time to talk about engineering and designing around such possibilities.
Also, how do we make the ride more stable? What does that mean graphically? It'd mean making the $U$ graph steeper, which we could do by increasing $M$. Note that the masses are intimately related to the "heights" of the domes/bowls:
The connection between forces and potential energy is often a topic that gets short shrift in AP Physics - seen as a small tidbit or something mathematical to be explored, but it's actually a very deep and applicable concept. I've also incorporated some programming exercises on this for students as well. Its use to justify the ball-and-spring model of matter (by approximating the Lenard-Jones potential as a parabola near the equilibrium point) is one of the lynchpins of my want for it in a physics course that uses Matter and Interactions.
No comments:
Post a Comment