CLASS6 - 1 CHAPTER 6 MOTION IN A RESISTING MEDIUM 1....

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1 CHAPTER 6 MOTION IN A RESISTING MEDIUM 1. Introduction In studying the motion of a body in a resisting medium, we assume that the resistive force on a body, and hence its deceleration, is some function of its speed. Such resistive forces are not generally conservative, and kinetic energy is usually dissipated as heat. For simple theoretical studies one can assume a simple force law, such as the resistive force is proportional to the speed, or to the square of the speed, or to some function that we can conveniently handle mathematically. For slow, laminar, nonturbulent motion through a viscous fluid, the resistance is indeed simply proportional to the speed, as can be shown at least by dimensional arguments. One thinks, for example, of Stokes's Law for the motion of a sphere through a viscous fluid. For faster motion, when laminar flow breaks up and the flow becomes turbulent, a resistive force that is proportional to the square of the speed may represent the actual physical situation better. 2 . Uniformly Accelerated Motion. Before studying motion in a resisting medium, a brief review of uniformly accelerating motion might be in order. That is, motion in which the resistance is zero. Any formulas that we develop for motion in a resisting medium must go to the formulas for uniformly accelerated motion as the resistance approaches zero. One may imagine a situation in which a body starts with speed v 0 and then accelerates at a rate a . One may ask three questions: How fast is it moving after time t ? How far has it moved in time t ? How fast is it moving after it has covered a distance x ? The answers to these questions are well known to any student of physics: v v = + 0 at , 6.2.1 xt a t =+ v 0 1 2 2 , 6.2.2 v v 2 0 2 2 ax . 6.2.3 Since the acceleration is uniform, there is no need to use calculus to derive these. The first follows immediately from the meaning of acceleration. Distance travelled is the area under a speed : time graph. Figure VI.1 shows a speed : time graph for constant acceleration, and equation 6.2.2 is obvious from a glance at the graph. Equation 6.2.3 can be obtained by elimination of t between equations 6.2.1 and 6.2.2. (It can also be deduced from energy considerations, though that is rather putting the cart before the horse.)
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2 Nevertheless, although calculus is not necessary, it is instructive to see how calculus can be used to analyse uniformly accelerated motion, since calculus will be necessary in less simple situations. We shall be using calculus to answer the three questions posed earlier in the section. For uniformly accelerated motion, the equation of motion is . a x = & & 6 . 2 . 4 To answer the first question, we write , / as dt d x v & & and then the integral (with initial condition x =0 when t =0) is v v = + 0 at . 6.2.5 This is the first time integral.
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This note was uploaded on 10/25/2010 for the course MECHANIC Mechanic taught by Professor Monfered during the Fall '10 term at MIT.

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CLASS6 - 1 CHAPTER 6 MOTION IN A RESISTING MEDIUM 1....

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