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
${f}_{R}=\text{\u2212}bv,$
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
$\frac{dv}{g-(b\text{/}m)v}=dt.$
Assuming that
$v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
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
Since
$\text{ln}A-\text{ln}B=\text{ln}(A\text{/}B),$ and
$\text{ln}(A\text{/}B)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
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
${f}_{R}=\text{\u2212}bv.$ (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
With the motor stopped, the only horizontal force on the boat is
${f}_{R}=\text{\u2212}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{\u2212}bv,$
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
As time increases,
${e}^{\text{\u2212}bt\text{/}m}\to 0,$ 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
Therefore, the boat’s velocity and position have essentially reached their final values.
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
$1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}){e}^{\text{\u2212}(b\text{/}m)(10\phantom{\rule{0.2em}{0ex}}\text{s})},$ so
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imam
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