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

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

what is physics
Physics is the tool humans use to understand the properties characteristics and interactions of where they live - the universe. Thus making laws and theories about the universe in a mathematical way derived from empirical results yielded in tons of experiments.
Jomari
This tool, the physics, also enhances their way of thinking. Evolving integrating and enhancing their critical logical rational and philosophical thinking since the greeks fired the first neurons of physics.
Jomari
nice
Satyabrata
Physics is also under the category of Physical Science which deals with the behavior and properties of physical quantities around us.
Angelo
Physical Science is under the category of Physics*... I prefer the most is Theoretical Physics where it deals with the philosophical view of our world.
Jomari
what is unit
Metric unit
Arzoodan
A unit is what comes after a number that gives a precise detail on what the number means. For example, 10 kilograms, 10 is the number while "kilogram" is the unit.
Angelo
there are also different types of units, but metric is the most widely used. It is called the SI system. Please research this on google.
Angelo
Unit? Bahay yon
Jomari
How did you get the value as Dcd=0.2Dab
Why as Dcd=0.2Dab? where are you got this formula?...
Arzoodan
since the distance Dcd=1.2 and the distance Dab=6.0 the ratio 1.2/6.0 gives the equation Dcd=0.2Dab
sunday
Well done.
Arzoodan
how do we add or deduct zero errors from result gotten using vernier calliper?
how can i understand if the function are odd or even or neither odd or even
hamzaani
I don't get... do you mean positive or negative@hamzaani
Aina
Verner calliper is an old calculator
Antonio
Function is even if f(-x) =f(x)
Antonio
Function is odd if f(-x) = - f(x)
Antonio
what physical phenomena is resonance?
is there any resonance in weight?
amrit
Resonance is due to vibrations and waves
Antonio
wait there is a chat here
dare
what is the difference between average velocity and magnitude of displacement
ibrahim
how velocity change with time
ibrahim
average velocity can be zero positive negative but magnitude of displacement is positive
amrit
if there is different displacement in same interval of time
amrit
Displacement can be zero, if you came back
Antonio
Displacement its a [L]
Antonio
Velocity its a vector
Antonio
Speed its the magnitude of velocity
Antonio
[Vt2-Vt1]/[t2-t1] = average velocity,another vector
Antonio
Distance, that and only that can't be negative, and is not a vector
Antonio
Distance its a metrical characteristic of the euclidean space
Antonio
Velocity change in time due a force acting (an acceleration)
Antonio
the change in velocity can be found using conservation of energy if the displacement is known
Jose
BEFORE = AFTER
Jose
kinetic energy + potential energy is equal to the kinetic energy after
Jose
the potential energy can be described as made times displacement times acceleration. I.e the work done on the object
Jose
mass*
Jose
from there make the final velocity the subject and solve
Jose
If its a conservative field
Antonio
So, no frictions in this case
Antonio
right
Jose
and if still conservative but force is in play then simply include work done by friction
Jose
Is not simple, is a very unknown force
Antonio
the vibration of a particle due to vibration of a similar particle close to it.
Aina
No, not so simple
Antonio
Frequency is involved
Antonio
mechanical wave?
Aina
All kind of waves, even in the sea
Antonio
will the LCR circut pure inductive if applied frequency becomes more than the natural frequency of AC circut? if yes , why?
LCR pure inductive? Is an nonsense
Antonio
what is photon
Photon is the effect of the Maxwell equations, it's the graviton of the electromagnetic field
Antonio
a particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.
Areej
Quantum it's not exact, its the elementary particle of electromagnetic field. Its not well clear if quantum theory its so, or if it's classical mechanics improved
Antonio
A photon is first and foremost a particle. And hence obeys Newtonian Mechanics. It is what visible light and other electromagnetic waves is made up of.
eli
No a photon has speed of light, and no mass, so is not Newtonian Mechanics
Antonio
photon is both a particle and a wave (It is the property called particle-wave duality). It is nearly massless, and travels at speed c. It interacts with and carries electromagnetic force.
Angelo
what are free vectors
a vector hows point of action doesn't static . then vector can move bodily from one point to another point located on its original tragectory.
Anuj
A free vector its an element of an Affine Space
Antonio
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90° with respect to the 50-yard line for 12 m, with an elapsed time of 1.2 s. (a) What is Matthews’ final displacement from the start of the play? (b) What is his average velocity?
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45Â° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90Â° with respect to the 50-yard line for 12 m, with an elap
ibrahim
Very easy man
Antonio
how to find time moved by a mass on a spring
Maybe you mean frequency
Antonio
why hot soup is more tastier than cold soup?
energy is involved
michael
hot soup is more energetic and thus enhances the flavor than a cold one.
Angelo
Its not Physics... Firstly, It falls under Anatomy. Your taste buds are the one to be blame not its coldness or hotness. Secondly, it depends on how the soup is done. Different soups possess different flavors and savors. If its on Physics, coldness of the soup will just bore you and if its hot...
Jomari
what is the importance of banking road in the circular path
the coefficient of static friction of the tires and the pavement becomes less important because the angle of the banked curve helps friction to prevent slipping
Jose
an insect is at the end of the ring and the ring is rotating at an angular speed 'w' and it reaches to centre find its angular speed.
Angular speed is the rate at which an object changes its angle (measured) in radians, in a given time period. Angular speed has a magnitude (a value) only.  v represents the linear speed of a rotating object, r its radius, and ω its angular velocity in units of radians per unit of time, then v = rω
Angular speed = (final angle) - (initial angle) / time = change in position/time. ω = θ /t. ω = angular speed in radians/sec.
a boy through a ball with minimum velocity of 60 m/s and the ball reach ground 300 metre from him calculate angle of inclination
what is the fomula for work done
work= force x distance
Guest
force × distance
Akash
Foece and displacement along the same direction as that of the force
nalin
force×displacement×cos∆ where ∆ is the angle between displacement and force.....i.e dot product of force and displacement
Is the angle between direcrion and force...
Arzoodan
Work is F x d = [F] •[d] • cos(a°)
Antonio
Force × distance along the same plane....
Aina