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MIT 8 01 - Transcript – Lecture 12

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MIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Please use the following citation format: Walter Lewin, 8.01 Physics I: Classical Mechanics, Fall 1999. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Transcript – Lecture 12 We're going to discuss today resistive forces and drag forces. When you move an object through a medium, whether it's a gas or whether it's a liquid, it experiences a drag force. This drag force depends on the shape of the object, the size of the object, the medium through which you move it and the speed of the object. The medium is immediately obvious. If it's air and you move through air, you feel the wind through your hair-- that's a drag force. If you swim in water, you feel this drag force. In oil, the drag force would be even larger. This drag force, this resistive force is very, very different from the friction that we have discussed earlier when two surfaces move relative to each other. There, the kinetic friction coefficient remains constant independent of the speed. With the drag forces and the resistive forces, they are not at all independent of the speed. In very general terms, the resistive force can be written as k1 times the velocity plus k2 times the velocity squared and always in the opposite direction of the velocity vector. This v here is the speed, so all these signs-- k1, v and k2, and obviously v squared-- they all are positive values. And the k values depend on the shape and the size of the object and on the kind of medium that I have. Today I will restrict myself exclusively to spheres. And when we deal with spheres, we're going to get that the force, the magnitude of the force--so that's this part-- equals C1 times r times the speed plus C2 times r squared times v squared. And again, it's always opposing the velocity vector. C1 in our unit is kilograms per meters per second and C2 has the dimension of density kilogram per cubic meters. We call this the viscous term, and we call this the pressure term. The viscous term has to do with the stickiness of the medium. If you take, for instance, liquids-- water and oil and tar-- there is a huge difference in stickiness. Physicists also refer to that as viscosity. If you have a high viscosity, it's very sticky, then this number, C1, will be very high. So this we call the viscous term, and this we call the pressure term. The C1 is a strong function of temperature. We all know that if you take tar and you heat it that the viscosity goes down. It is way more sticky when it is cold. C2 is not very dependent on the temperature. It's not so easy to see why this pressure term here has a v square. Later in the course when we deal with transfer of momentum, we will understand why there is a v-square term there. But the r square is very easy to see, because if you have a sphere and there is some fluid-- gas or liquid-- streaming onto it, then this has a cross-sectional area which is proportional to r squared, and so it's easy to see that the force that this object experiences-- we call it the pressure term-- is proportional to r square, so that's easy to see. Two liquids with the very same density would have... they could have very different values for C1.They could differ by ten... not ten, by four or five orders of magnitude. But if they have the same density, the liquids, then the C2 is very much the same. C2 is almost the density rho of the liquid--not quite but almost--but there is a very strong correlation between the C2 and the density. If I drop an object and I just let it go, I take an object and I let it fall--we're only dealing with spheres today--then what you will see, I have a mass m, and so there is a force mg--that is gravity. And as it picks up speed, the resistive force will grow, and it will grow and it will grow, and there comes a time... because the speed increases, so the resistive force will grow, and there comes a time that the two are equal. And when the two are equal, then there is no longer acceleration, so the object has a constant speed, and we call that the terminal velocity, and that will be the case when mg equals C1 r v plus C2 r squared v squared. And then we have here terminal velocity. If you know what m is, the mass of an object, the radius, and you know the values for C1 or C2 of that medium in which you move it, then you can calculate what the terminal velocity is. It is a quadratic equation, so you get two solutions of which one of them is nonphysical, so you can reject that one. Very often will we work in a domain, in a regime whereby this viscous term is dominating. I call that regime one, but it also happens--and I will show you examples today--that we're working in a regime where really this force is dominating. I call that regime two. Where one and two are the same--where the force due to the viscous force and the pressure force are the same-- we can make these terms the same, so you get C1 r v equals C2 r square v squared, and that velocity we call the critical velocity, even though there is nothing critical about it. It's not critical at all; it's simply the speed at which the two terms are equal. That's all it means. And that, of course, then equals C1 divided by C2 divided by r. Now we're going to make a clear distinction between the domains one and two, the regimes one and two. Regime one is when the speed is much, much less than the critical velocity. So we then have that mg equals C1 r v terminal, and therefore the terminal velocity equals mg divided by C1 r. If you take objects of the same material-- that means they have the same density, the density of the objects that you drop in the liquid or that you drop in the gas-- so that m equals 4/3 pi rho r cubed-- this is now the rho, the density of the object; it's not the density of the medium-- then you can immediately see, since you get an r cubed here, that this is proportional to the square of the radius if you drop objects in there with the same density. Regime two is the case when v is much larger than v critical, so then mg equals C2 r squared v squared if this is the terminal


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