## Differential Equations and Mechanics: Drag Force on an Inclined Plane

One of my regrets on how I spent my time as an undergraduate is that I didn’t study more physics. I am turned off by the nonrigorous use of mathematical formalism because I cannot accept a professor’s declaration that “X is true” followed by hand waving of the details. Also, a number of physics classes have lab components, which did not seem like the most enjoyable way to spend several hours every week or so.

Anyway, recently I have been trying to learn more about the Lagrangian and Hamiltonian formalisms, in part by using MIT OCW. In this process, I have been brushing up on my Newtonian mechanics. I would like to share with you the following problem that is an assignment exercise for MIT Course 8.09.

Suppose a particle of mass $m$ travels down a frictionless plane with incline $\theta\in(0,\frac{\pi}{2}]$ under the influence of gravity. Suppose that a force of magnitude $f=kmv^{2}$, where $v$ is the speed of the particle, acts anti-parallel to the direction of the motion of the particle. Then the time required for the particle to move a distance $d$ after starting from rest is

$\displaystyle t=\dfrac{\cosh^{-1}(e^{kd})}{\sqrt{kg\sin\theta}}$

Proof. Using elementary trigonometry, we see that the component of the gravity vector which is parallel to the inclined plane has magnitude $mg\sin\theta$. Hence, the net force in this direction is given by

$\displaystyle m\dot{v}=f_{\text{net}}=mg\sin\theta - kmv^{2}=m(g\sin\theta - kv^{2})$

Dividing both sides by $m$, we obtain the differential equation

$\displaystyle\dot{v}=g\sin\theta-kv^{2}\Longleftrightarrow\dfrac{\dot{v}}{g\sin\theta-kv^{2}}=1$,

where we divide both sides by $g\sin-kv^{2}$. The diligent reader will observe that we must restrict the domain of $v$ to $[0,\sqrt{k^{-1}g\sin\theta})$ so as not to divide by zero. We use partial fractions expansion to write

$\displaystyle1=\dfrac{\dot{v}}{g\sin\theta-kv^{2}}=\dfrac{1}{2\sqrt{kg\sin\theta}}\left[\dfrac{\dot{v}}{\sqrt{k^{-1}g\sin\theta}-v}+\dfrac{\dot{v}}{\sqrt{k^{-1}g\sin\theta}+v}\right]$

$\displaystyle t+c=\dfrac{1}{2\sqrt{kg\sin\theta}}\cdot\ln\dfrac{\sqrt{k^{-1}g\sin\theta}+v}{\sqrt{k^{-1}g\sin\theta}-v}$

Since the initial velocity is zero, we see that

$\displaystyle e^{2\sqrt{kg\sin\theta}t}=\dfrac{\sqrt{k^{-1}g\sin\theta}+v}{\sqrt{k^{-1}g\sin\theta}-v}$

$\displaystyle v=\sqrt{k^{-1}g\sin\theta}\cdot\dfrac{e^{2\sqrt{kg\sin\theta}t}-1}{e^{2\sqrt{kg\sin\theta}t}+1}=\sqrt{k^{-1}g\sin\theta}\cdot\left[\dfrac{2e^{2\sqrt{kg\sin\theta}t}}{1+e^{2\sqrt{kg\sin\theta}t}}-1\right]$

Integrating both sides again with respect to $t$, we obtain the equation

$\displaystyle x=\dfrac{1}{k}\ln\left(1+e^{2\sqrt{kg\sin\theta}t}\right)-\sqrt{k^{-1}g\sin\theta}t+c$

Since $x(0)=0$, we see that $c=-k^{-1}\ln 2$.

$\displaystyle d=\dfrac{1}{k}\ln\dfrac{1+e^{2\sqrt{kg\sin\theta}t}}{2}-\sqrt{k^{-1}g\sin\theta}t=\dfrac{1}{k}\ln\dfrac{1+e^{2\sqrt{kg\sin\theta}t}}{2e^{\sqrt{kg\sin\theta}t}}=\dfrac{1}{k}\ln\cosh\left(e^{\sqrt{kg\sin\theta}t}\right)$

Rearranging, we see that

$\displaystyle\cosh\left(\sqrt{kg\sin\theta}t\right)=e^{kd}\Longrightarrow t=\dfrac{\cosh^{-1}(e^{kd})}{\sqrt{kg\sin\theta}}$

$\Box$