# 6.4 Drag force and terminal speed  (Page 4/12)

 Page 4 / 12 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 ${\mu }_{\text{k}}N.$ 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

${f}_{R}=\text{−}bv,$

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

$mg-bv=m\frac{dv}{dt},$

where we have written the acceleration as $dv\text{/}dt.$ 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 ${v}_{\text{T}}.$ From the previous equation,

$mg-b{v}_{\text{T}}=0,$

so

${v}_{\text{T}}=\frac{mg}{b}.$

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

$\frac{dv}{g-\left(b\text{/}m\right)v}=dt.$

Assuming that $v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields

${\int }_{0}^{v}\frac{d{v}^{\prime }}{g-\left(b\text{/}m\right){v}^{\prime }}={\int }_{0}^{t}d{t}^{\prime },$

or

${-\frac{m}{b}\text{ln}\left(g-\frac{b}{m}{v}^{\prime }\right)|}_{0}^{v}={{t}^{\prime }|}_{0}^{t},$

where $v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find

$-\frac{m}{b}\left[\text{ln}\left(g-\frac{b}{m}v\right)-\text{ln}g\right]=t.$

Since $\text{ln}A-\text{ln}B=\text{ln}\left(A\text{/}B\right),$ and $\text{ln}\left(A\text{/}B\right)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain

$\frac{g-\left(bv\text{/}m\right)}{g}={e}^{\text{−}bt\text{/}m},$

and

$v=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$

Notice that as $t\to \infty ,v\to mg\text{/}b={v}_{\text{T}},$ which is the terminal velocity.

The position at any time may be found by integrating the equation for v . With $v=dy\text{/}dt,$

$dy=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right)dt.$

Assuming $y=0\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=0,$

${\int }_{0}^{y}d{y}^{\prime }=\frac{mg}{b}{\int }_{0}^{t}\left(1-{e}^{\text{−}bt\text{'}\text{/}m}\right)d{t}^{\prime },$

which integrates to

$y=\frac{mg}{b}t+\frac{{m}^{2}g}{{b}^{2}}\left({e}^{\text{−}bt\text{/}m}-1\right).$

## 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 ${f}_{R}=\text{−}bv.$ (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 ${f}_{R}=\text{−}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{−}bv,$

which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$

Integrating this equation between the time zero when the velocity is ${v}_{0}$ and the time t when the velocity is $v$ , we have
${\int }_{0}^{v}\frac{d{v}^{\prime }}{{v}^{\prime }}=-\frac{b}{m}{\int }_{0}^{t}d{t}^{\prime }.$

Thus,
$\text{ln}\frac{v}{{v}_{0}}=-\frac{b}{m}t,$

which, since $\text{ln}A=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A,$ we can write this as
$v={v}_{0}{e}^{\text{−}bt\text{/}m}.$

Now from the definition of velocity,
$\frac{dx}{dt}={v}_{0}{e}^{\text{−}bt\text{/}m},$

so we have
$dx={v}_{0}{e}^{\text{−}bt\text{/}m}dt.$

With the initial position zero, we have
${\int }_{0}^{x}dx\text{'}={v}_{0}{\int }_{0}^{t}{e}^{\text{−}bt\text{'}\text{/}m}dt\text{'},$

and
${x=-\frac{m{v}_{0}}{b}{e}^{\text{−}bt\text{'}\text{/}m}|}_{0}^{t}=\frac{m{v}_{0}}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$

As time increases, ${e}^{\text{−}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$

Although this tells us that the boat takes an infinite amount of time to reach ${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at $t=10m\text{/}b,$ we have
$v={v}_{0}{e}^{-10}\simeq 4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}{v}_{0},$

whereas we also have
$x={x}_{\text{max}}\left(1-{e}^{-10}\right)\simeq 0.99995{x}_{\text{max}}.$

Therefore, the boat’s velocity and position have essentially reached their final values.
2. With ${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and $v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have $1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=\left(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\right){e}^{\text{−}\left(b\text{/}m\right)\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right)},$ so
$\text{ln}\phantom{\rule{0.2em}{0ex}}0.25=\text{−}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0=-\frac{b}{m}\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right),$

and
$\frac{b}{m}=\frac{1}{10}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}=0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}\text{.}$

Now the boat’s limiting position is
${x}_{\text{max}}=\frac{m{v}_{0}}{b}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}}{0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{−1}}}=29\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

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