Question 5.17: Force proportional to velocity. Determine the velocity as a ......
Force proportional to velocity. Determine the velocity as a function of time for an object falling vertically from rest when there is a resistive force linearly proportional to v.
APPROACH This is a derivation and we start with Eq. 5-7, which we rewrite as
\frac{d v}{d t}=g-\frac{b}{m} v \text {. }
m g-b v=m \frac{d v}{d t}, (5-7)
In this equation there are two variables, v and t. We collect variables of the same type on one or the other side of the equation:
\frac{d v}{g-\frac{b}{m} v}=d t \quad \text { or } \quad \frac{d v}{v-\frac{m g}{b}}=-\frac{b}{m} d t \text {. }
Now we can integrate, remembering v=0 at t=0 :
\int_{0}^{v} \frac{d v}{v-\frac{m g}{b}}=-\frac{b}{m} \int_{0}^{t} d t
which gives
\ln \left(v-\frac{m g}{b}\right)-\ln \left(-\frac{m g}{b}\right)=-\frac{b}{m} t
or
\ln \frac{v-m g / b}{-m g / b}=-\frac{b}{m} t
We raise each side to the exponential [note that the natural log and the exponential are inverse operations of each other: e^{\ln x}=x, or \left.\ln \left(e^{x}\right)=x\right] and obtain
v-\frac{m g}{b}=-\frac{m g}{b} e^{-\frac{b}{m} t}
so
v=\frac{m g}{b}\left(1-e^{-\frac{b}{m} t}\right) \text {. }
This relation gives the velocity v as a function of time and corresponds to the graph of Fig. 5-27b. As a check, note that at t=0, and v=0
a(t=0)=\frac{d v}{d t}=\frac{m g}{b} \frac{d}{d t}\left(1-e^{-\frac{b}{m} t}\right)=\frac{m g}{b}\left(\frac{b}{m}\right)=g,
as expected (see also Eq. 5-7). At large t, e^{-\frac{b}{m} t} approaches zero, so v approaches m g / b, which is the terminal velocity, v_{\mathrm{T}}, as we saw earlier. If we set \tau=m / b, then v=v_{\mathrm{T}}\left(1-e^{-t / \tau}\right). So \tau=m / b is the time required for the velocity to reach 63 \% of the terminal velocity (since e^{-1}=0.37 ). Figure 5-27 \mathrm{~b} shows a plot of speed v vs. time t, where the terminal velocity v_{\mathrm{T}}=m g / b.
