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

 

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