Teaching Physics with the Physics Suite Edward F. Redish |
We know that as an object passes through the air, the air exerts a resistive force on it. We can get some idea as to the form of this force by dimensional analysis. Suppose we have a spherical object of radius R and mass m. What might the force plausibly depend on?
(b) Explain why it is plausible that the force the air exerts depends on the objects speed through it, v, but not on its position, x, or acceleration, a.
(c) Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.
Solution
(a) Let's consider trying to hold a sphere fixed against a wind. Would the force we have to use depend on the mass of the sphere? We can see that this is the same question by imagining sitting on an open rail car. If we didn't look at the scenery, we couldn't tell whether the car was stopped and the wind was blowing with a speed v or whether the air was still and the car was moving at a speed v. But it's easier to think about the force we need to exert to hold the object fixed if we imagine ourselves fixed and the wind blowing. Suppose we have two styrofoam balls of the same size, one filled and one hollow. The wind is pushing on both and exerts a force on both. Since the wind is only interacting with the surface of the two balls, the molecules of air are bouncing off in the same way. They have no way of knowing whether the ball is solid or hollow. The amount of momentum they transfer to the balls through their collisions will be the same in the same time interval. By the impulse-momentum theorem, the forces they exert on the two balls would have to be the same.
In plain language, the wind is exerting its force on the surface of the ball. Whether the ball is hollow or solid doesn't matter. (Of course the responses of the two balls will be different through a = F^{net}/m. But we are only asking about the forces.) Therefore the mass shouldn't matter. But the radius determines how much air the ball is blocking. The more air that is pushing on the ball, the more we expect it to feel. Therefore, the force should depend on the radius.
(b) The force should not depend on the value of the coordinate since the coordinate origin can be chosen anywhere and that change of how we describe the system shouldn't change the force the object feels. (Of course the density of the air can depend on your location and that will affect the force.) The force therefore shouldn't depend on x. The force should definitely depend on the velocity since if a strong wind is blowing we know there is a strong force on an object. Only the relative velocity of the object and air matters, as described above. There is no reason we know of that the force should not depend on the acceleration or on the higher derivatives of the position. But if it did depend on acceleration, then there would be a term in the force that looked like "constant times a" that we could move to the other side and combine with "ma". This force would then make it appear that the mass of the object changed when it was moving through the air. We hope this doesn't happen and will assume that it does not until we are forced to change our assumption. (Something like this actually does happen to an electron in a material or a quark inside a proton.)
(c) Now we can try to build a force out of the dimensioned quantities we have to work with: [R] = L, [v] = L/T, and [r] = M/L^{3}. Force has dimensions [F] = ML/T^{2}. We need something with an M. The only thing we have to work with is the density. So our expression giving the force must have one power of r in order to get the M right. We need 1/T^{2} in our expression for force and the only quantity we have to work with is the velocity. It has a 1/T, so to get 1/T^{2} we need two powers of the velocity. So far we have F = (something depending on R) times rv^{2}. We'll choose the number of powers of R we use to make the L's come out right. In our force expression, we need one power of L. So far (from r v^{2}) we have (1/L^{3})L^{2} = 1/L. To get L, we have to multiply by something that has dimensions of L^{2}. So we choose to multiply by R^{2}. We therefore get the expression for the drag force:
This is known as the Newton drag law. C is a dimensionless constant that we cannot determine by dimensional analysis.
Page last modified September 6, 2004: D28