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201907181434 Homework 1 (Q3)

Obtain the equation of motion for a particle falling vertically under the influence of gravity when the frictional forces obtainable from a dissipation function kv^2/2 are present. Integrate the equation to obtain the velocity as a function of time and show that maximum possible velocity for a fall from rest is v=mg/k.

Solution.

Write the Lagrangian \mathcal{L}=T-V by noting

T=\displaystyle{\frac{1}{2}m\dot{\mathbf{y}}^2} and V=-mg|\mathbf{y}|,

where the upward direction is taken to be positive. The frictional force is

\mathcal{F}=\displaystyle{\frac{k\dot{\mathbf{y}}^2}{2}}.

I wish to obtain the Euler-Lagrange equation, by computing the derivatives below:

\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial y}} & = mg \\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}} & =m\dot{y} \\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) & = m\ddot{y} \\ \displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & = k\dot{y}\\ \end{aligned}

Hence I obtain the E-L equation (with dissipation):

\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) -\displaystyle{\frac{\partial \mathcal{L}}{\partial y}}+\displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & =0\\ m\ddot{y}-mg+k\dot{y} & =0\\ \ddot{y}+\displaystyle{\frac{k}{m}}\dot{y} & =g\\ \end{aligned}

Treating u=\dot{y} as variable, I may obtain a first-order differential equation:

\dot{u}+\displaystyle{\frac{k}{m}u}-g=0

Solving it,

\begin{aligned} v(t) & =e^{\int \frac{k}{m}\mathrm{d}t}=e^{kt/m}\\ e^{kt/m}\dot{u}+\frac{k}{m}ue^{kt/m} & =ge^{kt/m}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(ue^{kt/m}) & = ge^{kt/m}\\ ue^{kt/m} & = \displaystyle{\frac{mg}{k}}e^{kt/m}+\textrm{constant }C\\ u(t)& =\displaystyle{\frac{mg}{k}}+Ce^{-kt/m}\\ v_{\textrm{max.}}=\dot{y} & =\displaystyle{\frac{mg}{k}}\qquad\qquad (e^{-\frac{kt}{m}}\rightarrow 0\enspace \textrm{as}\enspace t\rightarrow \infty )\\ \end{aligned}

In conclusion, it is proven that the maximum possible speed for a fall from rest is v=mg/k.

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