Question 6.11: Falling body with air resistance. A particle of mass m, a ra...

The solution may be read from the foregoing results in which the drag force is  modeled by Stokes’s law (6.29) so that   \mathbf{F}_D=-R(v) \mathbf{t}=-c v \mathbf{t}.   Hence, use of  R(v)=c v  in (6.31) gives

\mathbf{F}_D=-c \mathbf{v}                       (6.29)

\dot{v}=g-\frac{R(v)}{m} \equiv F(v)                  (6.31)

\frac{d v}{d t}=g  –  v v \equiv F(v) \quad \text { with } \quad v \equiv \frac{c}{m} \text {. }                    (6.34a)

For the initial condition v(O) = 0, we find by (6.32)

t=\int \frac{d v}{F(v)}  +  c_0             (6.32)

t=\int_0^v \frac{d v}{g  –  v v}=-\frac{1}{v} \log \left(1  –  \frac{v v}{g}\right) \text {; }

so, the rectilinear speed of the particle in its fall from rest is

v(t)=v_{\infty}\left(1  –  e^{-v t}\right) \quad \text { with } \quad v_{\infty}\equiv \frac{g}{v}.          (6.34b)

In consequence, as  t \rightarrow \infty,  the particle speed v approaches a constant value  v_{\infty} \equiv g / \nu=W / c,  named the terminal speed. When the particle achieves its terminal speed, its  weight is balanced by the drag force so that  c v_{\infty}=W,  and the particle continues to fall without further acceleration.

These facts are  illustrated in Fig. 6.9. Equation (6.34b) shows that the rate at which v(t)  changes is governed by the coefficient of dynamic viscosity v =c / m , which has the physical dimensions  [v]=[F / M V]=\left[T^{-1}\right].  Thus, at the in stant  t=v^{-1},  by (6.34b),  v\left(v^{-1}\right)=v_{\infty}\left(1  –  e^{-1}\right) \approx 0.632 v_{\infty}.  Therefore, the speed reaches 63.2% of the terminal speed in the time  t=v^{-1},  called the retardation time. The straight line of slope 1 in Fig. 6.9 shows that this also is the time at which the speed would reach the terminal value if it had continued to change at its initial constant rate a(0) = g, without air resistance. As the particle’s speed approaches the terminal speed of its ultimate uniform motion shown by the  horizontal asymptote, the weight  \mathbf{W}=W \mathbf{t}  is balanced by the drag force  \mathbf{F}_D \rightarrow \mathbf{F}_{\infty}=-c v_{\infty} \mathbf{t} .

Finally, with  v=d s / d t  and the initial condition  s(0)=0,  (6.34b) yields the distance through which the particle falls in time t :

s(t)=v_{\infty} \int_0^t\left(1  –  e^{-v t}\right) d t=v_{\infty} t  –  \frac{v_{\infty}}{v}\left(1  –  e^{-v t}\right) .                    (6.34c)

Hence, the distance traveled in the retardation time interval is  s(1 / \nu)=v_{\infty} /(v e) \approx 0.368 v_{\infty} / v.  The result (6.34c) also may be read from (6.33).

s=\int \frac{v d v}{F(v)}  +  c_1               (6.33)

The reader may verify that in the absence of air resistance when  v \rightarrow 0  the limit solutions of (6.34b) and (6.34c) are the elementary solutions (6.24). Now consider the case when the viscosity v is  small. First, recall the power series expansion of  e^z  about z = 0:

\mathbf{x}(P, t)=\frac{1}{2} \mathbf{g} t^2, \quad \mathbf{v}(P, t)=\mathbf{g} t, \quad \mathbf{a}(P, t)=\mathbf{g}               (6.24)

e^z=1+z+\frac{z^2}{2 !}+\frac{z^3}{3 !}+\cdots .                (6.34d)

Then use of (6.34d) in (6.34b) and (6.34c) yields, to the first order in v, an approximate solution for the case of small air resistance:

v(t)=g t\left(1  –  \frac{v}{2} t\right), \quad s(t)=\frac{g}{2} t^2\left(1  –  \frac{v}{3} t\right) .                 (6.34e)

When  v \rightarrow 0,  we again recover (6.24) for which air is  absent.