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concepts/mechanics/u2/m2-7-drag-terminal.tex

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+\subsection{Drag Forces and Terminal Velocity}
+
+This subsection uses the AP linear-drag model, in which the resistive force depends on the object's instantaneous velocity. The local-first viewpoint is Newton's second law with a velocity-dependent force.
+
+\dfn{Linear drag and terminal velocity}{Let an object move through a fluid with velocity $\vec{v}$ relative to the fluid, and let $b>0$ denote the linear-drag coefficient. In the linear-drag model, the drag force exerted by the fluid on the object is
+\[
+\vec{F}_D=-b\vec{v}.
+\]
+The negative sign means that the drag force always points opposite the velocity.
+
+If an object falls vertically through the fluid and eventually moves with constant velocity, that steady velocity is called the \emph{terminal velocity}. At terminal velocity, the net force is zero, so the acceleration is zero.}
+
+\thm{Vertical-fall ODE and terminal-speed result}{Choose the vertical axis positive downward. Let $v(t)$ denote the downward velocity of an object of mass $m$ at time $t$, let $g$ denote the gravitational field strength, and let $b>0$ denote the linear-drag coefficient. Then the vertical equation of motion is
+\[
+m\frac{dv}{dt}=mg-bv.
+\]
+The terminal speed $v_T$ is the steady-state value obtained by setting the net force equal to zero:
+\[
+mg-bv_T=0,
+\]
+so
+\[
+v_T=\frac{mg}{b}.
+\]
+
+If the initial velocity is $v(0)=v_0$, then
+\[
+v(t)=v_T+(v_0-v_T)e^{-bt/m}.
+\]
+In particular, if the object is released from rest, then
+\[
+v(t)=v_T\left(1-e^{-bt/m}\right).
+\]}
+
+\pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction:
+\[
+m\frac{dv}{dt}=mg-bv.
+\]
+Define the terminal speed by
+\[
+v_T=\frac{mg}{b}.
+\]
+Then the differential equation becomes
+\[
+\frac{dv}{dt}=\frac{b}{m}(v_T-v).
+\]
+Separate variables:
+\[
+\frac{dv}{v_T-v}=\frac{b}{m}\,dt.
+\]
+Integrating gives
+\[
+-\ln|v_T-v|=\frac{b}{m}t+C.
+\]
+Therefore
+\[
+v_T-v=Ce^{-bt/m}
+\]
+for some constant $C$, so
+\[
+v=v_T-Ce^{-bt/m}.
+\]
+Now use the initial condition $v(0)=v_0$:
+\[
+v_0=v_T-C.
+\]
+Thus $C=v_T-v_0$, and
+\[
+v(t)=v_T-(v_T-v_0)e^{-bt/m}=v_T+(v_0-v_T)e^{-bt/m}.
+\]
+If $v_0=0$, this reduces to
+\[
+v(t)=v_T\left(1-e^{-bt/m}\right).
+\]
+As $t\to\infty$, the exponential term approaches $0$, so $v(t)\to v_T$.}
+
+\ex{Illustrative example}{Choose downward as positive. A ball of mass $m=0.20\,\mathrm{kg}$ falls through air with linear drag coefficient $b=0.50\,\mathrm{N\cdot s/m}$. Find the terminal speed and the acceleration when the downward speed is $v=2.0\,\mathrm{m/s}$.
+
+The terminal speed is
+\[
+v_T=\frac{mg}{b}=\frac{(0.20\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{0.50\,\mathrm{N\cdot s/m}}=3.92\,\mathrm{m/s}.
+\]
+When $v=2.0\,\mathrm{m/s}$, Newton's second law gives
+\[
+m\frac{dv}{dt}=mg-bv,
+\]
+so the acceleration is
+\[
+a=g-\frac{b}{m}v=9.8-\frac{0.50}{0.20}(2.0)=4.8\,\mathrm{m/s^2}.
+\]
+Since this value is positive in the downward-positive coordinate system, the ball is still accelerating downward.}
+
+\qs{Worked example}{A small package of mass $m=0.40\,\mathrm{kg}$ is dropped from rest and falls vertically through air. Choose downward as positive. Let $v(t)$ denote the package's downward velocity at time $t$, let $g=9.8\,\mathrm{m/s^2}$, and let the drag force be modeled by
+\[
+\vec{F}_D=-b\vec{v}
+\]
+with $b=0.80\,\mathrm{N\cdot s/m}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the differential equation for $v(t)$.
+\item Find the terminal speed.
+\item Find an explicit formula for $v(t)$.
+\item Find the package's velocity and acceleration at $t=1.0\,\mathrm{s}$.
+\end{enumerate}}
+
+\sol The forces on the package are its weight $\vec{W}$ downward and the drag force $\vec{F}_D$ upward because the package is moving downward. Since downward is chosen as positive, the scalar force equation is
+\[
+mg-bv=m\frac{dv}{dt}.
+\]
+
+For part (a), the differential equation is therefore
+\[
+m\frac{dv}{dt}=mg-bv.
+\]
+Substitute $m=0.40\,\mathrm{kg}$ and $b=0.80\,\mathrm{N\cdot s/m}$:
+\[
+(0.40)\frac{dv}{dt}=(0.40)(9.8)-0.80v.
+\]
+So an equivalent form is
+\[
+\frac{dv}{dt}=9.8-2.0v.
+\]
+
+For part (b), terminal speed occurs when the acceleration is zero, so
+\[
+\frac{dv}{dt}=0.
+\]
+Then
+\[
+mg-bv_T=0,
+\]
+which gives
+\[
+v_T=\frac{mg}{b}=\frac{(0.40)(9.8)}{0.80}=4.9\,\mathrm{m/s}.
+\]
+
+For part (c), because the package is dropped from rest, the initial condition is
+\[
+v(0)=0.
+\]
+Using the linear-drag result for release from rest,
+\[
+v(t)=v_T\left(1-e^{-bt/m}\right).
+\]
+Substitute the values:
+\[
+v(t)=4.9\left(1-e^{-(0.80/0.40)t}\right).
+\]
+Therefore,
+\[
+v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s}.
+\]
+
+For part (d), evaluate this at $t=1.0\,\mathrm{s}$:
+\[
+v(1.0)=4.9\left(1-e^{-2.0}\right)\,\mathrm{m/s}.
+\]
+Since
+\[
+e^{-2.0}\approx 0.135,
+\]
+we get
+\[
+v(1.0)\approx 4.9(0.865)=4.24\,\mathrm{m/s}.
+\]
+So after $1.0\,\mathrm{s}$ the package is moving downward at about
+\[
+4.24\,\mathrm{m/s}.
+\]
+
+Now find the acceleration. From the differential equation,
+\[
+a=\frac{dv}{dt}=9.8-2.0v.
+\]
+At $t=1.0\,\mathrm{s}$,
+\[
+a=9.8-2.0(4.24)=1.32\,\mathrm{m/s^2}.
+\]
+This is positive in the downward-positive coordinate system, so the acceleration is still downward.
+
+Therefore the differential equation is
+\[
+\frac{dv}{dt}=9.8-2.0v,
+\]
+the terminal speed is
+\[
+4.9\,\mathrm{m/s},
+\]
+the velocity function is
+\[
+v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s},
+\]
+and at $t=1.0\,\mathrm{s}$ the package has velocity
+\[
+4.24\,\mathrm{m/s}
+\]
+downward and acceleration
+\[
+1.32\,\mathrm{m/s^2}
+\]
+downward.