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

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

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

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,

so

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

Assuming that $v=0\text{at}t=0,$ integration of this equation yields

or

where $v\text{'}\text{and}t\text{'}$ are dummy variables of integration. With the limits given, we find

Since $\text{ln}A-\text{ln}B=\text{ln}(A\text{/}B),$ and $\text{ln}(A\text{/}B)=x\text{implies}{e}^x=A\text{/}B,$ we obtain

and

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 $v=dy\text{/}dt,$

Assuming $y=0\text{when}t=0,$

which integrates to

Effect of the resistive force on a motorboat

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
   
   ${\displaystyle {\int }_0^v\frac{d{v}^{\prime }}{v^{\prime }}}=-\frac{b}{m}{\displaystyle {\int }_0^{t}d{t}^{\prime }.}$
   
   Thus,
   
   $\text{ln}\frac{v}{v_0}=-\frac{b}{m}t,$
   
   which, since $\text{ln}A=x\text{implies}{e}^x=A,$ 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
   
   ${\displaystyle {\int }_0^{x}dx\text{'}=v_0{\displaystyle {\int }_0^t{e}^{\text{−}bt\text{'}\text{/}m}}}dt\text{'},$
   
   and
   
   ${x=-\frac{m{v}_0}{b}{e}^{\text{−}bt\text{'}\text{/}m}|}_0^t=\frac{m{v}_0}{b}(1-e^{\text{−}bt\text{/}m}).$
   
   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
   
   $v=v_0{e}^{\mathrm{-10}}\simeq 4.5\times {10}^{\mathrm{-5}}{v}_0,$
   
   whereas we also have
   
   $x=x_{\text{max}}(1-e^{\mathrm{-10}})\simeq 0.99995x_{\text{max}}.$
   
   Therefore, the boat’s velocity and position have essentially reached their final values.
2. With $v_0=4.0\text{m/s}$ and $v=1.0\text{m/s,}$ we have $1.0\text{m/s}=(4.0\text{m/s})e^{\text{−}(b\text{/}m)(10\text{s})},$ so
   
   $\text{ln}0.25=\text{−}\text{ln}4.0=-\frac{b}{m}(10\text{s}),$
   
   and
   
   $\frac{b}{m}=\frac{1}{10}\text{ln}4.0{\text{s}}^{\text{-1}}=0.14{\text{s}}^{\text{-1}}\text{.}$
   
   Now the boat’s limiting position is
   
   $x_{\text{max}}=\frac{m{v}_0}{b}=\frac{4.0\text{m/s}}{0.14{\text{s}}^{\text{−1}}}=29\text{m}\text{.}$