So for my latest physics homework question, I had to derive an equation for the terminal velocity of a ball falling in some gravitational field assuming that the air resistance force was equal to some constant c multiplied by $v^2.$
So first I started with the differntial equation:
$\frac{dv}{dt}=-mg-cv^2$
Rearranging to get:
$\frac{dv}{dt}=-\left(g+\frac{cv^2}{m}\right)$
From here I tried solving it and ended up with:
$\frac{\sqrt{m}}{\sqrt{c}\sqrt{g}}\arctan \left(\frac{\sqrt{c}v}{\sqrt{g}\sqrt{m}}\right)+C=-t$
I rearranged this to get:
$v\left(t\right)=\left(\frac{\sqrt{g}\sqrt{m}\tan \left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$
In order to calculate the terminal velocity I took the limit as t approaches infinity:
$\lim _{t\to \infty }\left(\frac{\sqrt{g}\sqrt{m}\tan \:\left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$
This reduces to:
$\frac{\sqrt{g}\sqrt{m}\tan \left(\infty \right)}{\sqrt{c}}$
The problem with this is that tan $(\infty)$ is indefinite.
Where did I go wrong? Could someone please help properly solve this equation.
Cheers, Gabriel.
3 Answers
$\begingroup$Write the differential equation as a rate of change of velocity with respect to just aerodynamic drag. Then solve for the time it takes for the drag to equal $mg$.
$$\frac{dV}{dt} = \frac{cv^2}{m}$$ $$\frac{v^{-2}}{c}dV = \frac{dt}{m}$$ $$-\frac{1}{cv} = \frac{t}{m} + C$$ Assuming $t=0, v=0$ then.......
$$v = -\frac{m}{ct}$$ When $cv^2 = -mg, v = -\sqrt{\frac{gm}{c}}$ $$-\sqrt{\frac{gm}{c}} = -\frac{m}{ct}$$ $$t = \frac{m}{c\sqrt{\frac{gm}{c}}}$$ Substituting back.......$$v = \sqrt{\frac{gm}{c}}$$ Does this seem reasonable? Assume $c = .5\cdot C_d\cdot \rho\cdot A = .5\cdot 0.3\cdot 1.225\cdot 0.1 = 0.018$ and $m = 0.5\ kg$
$$v = \sqrt{\frac{9.8\cdot 0.5}{0.018}} = 16.5\ m/s$$
Thinking about this it would have been easier just to set $cv^2 = mg$ to get $$v = \sqrt{\frac{gm}{c}}$$
$\endgroup$ 2 $\begingroup$In order to determine the teriminal velocity, set $m\frac{dv}{dt}=-mg+cv^2=0$, which implies that $v_t=\sqrt{\frac{mg}{c}}$. The differential equation itself can be solved as follows. Since we know $v_t$, we can rewrite the orign differential equation as $\frac{dv}{dt}=g(1-\frac{v^2}{v_t^2})$ with boundary conditions $v(t_0)=v_0$. Then, we can solve this differential equation by integration over both sides. $$ \int_{t_0}^t dt'=\int_{v_0}^{v(t)}\frac{dv'}{g(1-\frac{v'^2}{v_t^2})} $$ Let us write $\tau=\frac{v_t}{g}=\sqrt{\frac{m}{cg}}$. Then $$t-t_0=\tau(\tanh^{-1}\frac{v}{v_t}-\tanh^{-1}\frac{v_0}{v_t})$$ Solving for $v$, we find that $$ v=v_t\tanh(\frac{t-t_0}{\tau}-\tanh^{-1}\frac{v_0}{v_t}) $$
If the body is released at rest at $t_0=0$, $$v=v_t\tanh\frac{t}{\tau}$$
$\endgroup$ $\begingroup$Taking proper sign of air resistance opposing gravity, we have terminal velocity when acceleration vanishes:
$$ \dfrac{dv}{dt}=mg-cv^2 = 0 \rightarrow v= v_{terminal}=\sqrt{\dfrac{mg}{c}}. $$
gets included in the coefficient of tanh function for velocity as an asymptotic value.
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