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

 Page 4 / 12

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

velocity is a physician vector quantity; both magnitude and direction needed to define it. the scalar absolute value ( magnitude) of velocity is called "speed being a coherent derived unite whose quantity is measured in SI ( metric system) as metres per second (m/s) or SI base unit of (m . s^-1).
number of lines passing through area which is placed at some angle. these line are are produced by charge(+ or -).
hstjsbks
define electric flux? find the electric field due to a long strainght line
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 43° with respect to the 50-yard line (the +x-axis) and covers 7.8 m in 1 s. He then runs straight down the field at 90° with respect to the 50-yard line (that is, in the +y-direction) for 17 m, with an elapsed time of 2.9 s. (Express your answers in vector form.) (a) What is Matthews's final displacement (in m) from the start of the play?
What is his average velocity (in m/s)?
Justin
A machine at a post office sends packages out a chute and down a ramp to be loaded into delivery vehicles. (a) Calculate the acceleration of a box heading down a 17.4° slope, assuming the coefficient of friction for a parcel on waxed wood is 0.100. (b) Find the angle of the slope down which this box could move at a constant velocity. You can neglect air resistance in both parts.
what principle is applicable in projectile motion
does rocket and satellite follow the same principle??? which principle do both of these follow???
According to d'Broglie's concept of matter waves matter behaves like wave and the wavelength is h/p. but actually there is not only a wave but a wave packet wich is defined by a wave function and that wave function can defines everything about the particle but restricted by the uncertainty principle
what phenomenon describes Matter behave as a wave???
simple definition of wave
hello
can anyone help me with this problem
Carls
A projectile is shot at a hill, the base of which is 300 m away. The projectile is shot at 60°60° above the horizontal with an initial speed of 75 m/s. The hill can be approximated by a plane sloped at 20°20° to the horizontal. Relative to the coordinate system shown in the following figure, the equation of this straight line is y=(tan20°)x−109.y=(tan20°)x−109. Where on the hill does the projectile land?
Carls
what is velocity
hi, Musa,this moment a lateral
what is moment