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A photograph of geese flying in a V formation.
Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and also allows them a better way to communicate. (credit: “Julo”/Wikimedia Commons)

In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.

The calculus of velocity-dependent frictional forces

When a body slides across a surface, the frictional force on it is approximately constant and given by μ k N . Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by

f R = b v ,

where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.

Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in [link] . Newton’s second law in the vertical direction gives the differential equation

m g b v = m d v d t ,

where we have written the acceleration as d v / d t . As v increases, the frictional force – bv increases until it matches mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity v T . From the previous equation,

m g b v T = 0 ,

so

v T = m g b .
The free body diagram shows forces m times vector g pointing vertically down and b times vector v pointing vertically up. The velocity, vector v, is vertically down. The positive y direction is also vertically down.
Free-body diagram of an object falling through a resistive medium.

We can find the object’s velocity by integrating the differential equation for v . First, we rearrange terms in this equation to obtain

d v g ( b / m ) v = d t .

Assuming that v = 0 at t = 0 , integration of this equation yields

0 v d v g ( b / m ) v = 0 t d t ,

or

m b ln ( g b m v ) | 0 v = t | 0 t ,

where v ' and t ' are dummy variables of integration. With the limits given, we find

m b [ ln ( g b m v ) ln g ] = t .

Since ln A ln B = ln ( A / B ) , and ln ( A / B ) = x implies e x = A / B , we obtain

g ( b v / m ) g = e b t / m ,

and

v = m g b ( 1 e b t / m ) .

Notice that as t , v m g / b = v T , which is the terminal velocity.

The position at any time may be found by integrating the equation for v . With v = d y / d t ,

d y = m g b ( 1 e b t / m ) d t .

Assuming y = 0 when t = 0 ,

0 y d y = m g b 0 t ( 1 e b t ' / m ) d t ,

which integrates to

y = m g b t + m 2 g b 2 ( e b t / m 1 ) .

Effect of the resistive force on a motorboat

A motorboat is moving across a lake at a speed v 0 when its motor suddenly freezes up and stops. The boat then slows down under the frictional force f R = b v . (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?

Solution

  1. With the motor stopped, the only horizontal force on the boat is f R = b v , so from Newton’s second law,
    m d v d t = b v ,

    which we can write as
    d v v = b m d t .

    Integrating this equation between the time zero when the velocity is v 0 and the time t when the velocity is v , we have
    0 v d v v = b m 0 t d t .

    Thus,
    ln v v 0 = b m t ,

    which, since ln A = x implies e x = A , we can write this as
    v = v 0 e b t / m .

    Now from the definition of velocity,
    d x d t = v 0 e b t / m ,

    so we have
    d x = v 0 e b t / m d t .

    With the initial position zero, we have
    0 x d x ' = v 0 0 t e b t ' / m d t ' ,

    and
    x = m v 0 b e b t ' / m | 0 t = m v 0 b ( 1 e b t / m ) .

    As time increases, e b t / m 0 , and the position of the boat approaches a limiting value
    x max = m v 0 b .

    Although this tells us that the boat takes an infinite amount of time to reach x max , the boat effectively stops after a reasonable time. For example, at t = 10 m / b , we have
    v = v 0 e −10 4.5 × 10 −5 v 0 ,

    whereas we also have
    x = x max ( 1 e −10 ) 0.99995 x max .

    Therefore, the boat’s velocity and position have essentially reached their final values.
  2. With v 0 = 4.0 m/s and v = 1.0 m/s, we have 1.0 m/s = ( 4.0 m/s ) e ( b / m ) ( 10 s ) , so
    ln 0.25 = ln 4.0 = b m ( 10 s ) ,

    and
    b m = 1 10 ln 4.0 s -1 = 0.14 s -1 .

    Now the boat’s limiting position is
    x max = m v 0 b = 4.0 m/s 0.14 s −1 = 29 m .

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Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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