Viscosity-and-Brownian

# Viscosity-and-Brownian - Why does a constant force on an...

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Why does a constant force on an object in a viscous medium produce a constant (terminal) velocity ? You know that the viscous drag force F drag is proportional to the velocity v (and in the opposite direction); we’ll use the symbol f to represent the coefficient of proportionality: F drag = – fv When an object reaches terminal velocity, the drag force is equal and opposite to the constant external force, so the velocity is constant: F ext = – F drag so v = F ext f We seek a microscopic model for viscosity that can explain this observation. As the object is being pulled through the fluid by a constant force, the object is also subjected to numerous microscopic collisions with the molecules of the fluid. Let’s assume that these collisions are periodic: there is one collision per time ! t . Let’s also assume that as a result of each collision, our object’s velocity is “reset” to some random value v o . Then, before the next collision, the object moves freely, influenced only by the constant external force. The constant force produces a constant acceleration a = F ext / m , so the object’s velocity is given by the usual kinematic equation for constant acceleration: v = v o + F ext m " # \$ % & ( t For a constant “downward” force (for instance, a gravitational force) the velocity of such a particle might look like this:

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Take a close look at this graph. Each black circle shows an initial (random) velocity that follows a collision. On average, those velocities are zero (i.e. the random velocity is equally likely to be positive or negative). However, during the period between collisions, the object is subjected to a constant downward acceleration, so its velocity decreases linearly between each collision in a constant, predictable way. Thus, the average of all the velocities on the graph is a negative number: on average, the object is falling “down” with a constant velocity. If we integrate the velocity we get the position as a function of time. The kinematic expression is: x = x o + v o " t + 1 2 F ext m # \$ % & ( " t ( ) 2 and a graph of position versus time (using the velocities shown on the previous graph) is: Each dot on the above graph shows one collision. Can you see that the object is in parabolic free-fall between each collision? The dashed line represents the average downward trajectory of the object. We see that, although the instantaneous motion of the object is a constant-acceleration trajectory (as expected for a constant force), the average motion is a constant-velocity trajectory (as observed in a viscous medium).
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## This note was uploaded on 03/14/2012 for the course PHYS SCI PS2 taught by Professor Loganmccarty during the Fall '10 term at Harvard.

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Viscosity-and-Brownian - Why does a constant force on an...

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