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{\bf Question}

A particle of mass $m$ falls from rest due to gravity in a fluid
which offers resistance force of size $k|v|^\beta$, where $k$ and
$\beta$ are positive constants. Write down Newton's second law for
the motion. Without solving the equation determine the limiting
speed of the object in terms of the constants $k$ and $\beta$.  [
Hint: if a finite limiting velocity exits then at that speed the
resistance balances the gravitational force.] (note: beware of the
direction of the forces)

For the particular case $\beta=2$ solve the equations to find the
relation between the velocity and the time.


\vspace{0.25in}

{\bf Answer}

Falling particle $\ds \Rightarrow v<0$ and the air resistance acts
upwards.

$\ds m\frac{d^2x}{dt^2}=m\frac{dv}{dt}=-mg+k|v|^\beta$

The limiting velocity is the same as the equilibrium point, it is
when $\ds -mg+k|v|^\beta=0$

$\ds |v|^\beta=\frac{k}{mg} \Rightarrow
|v|=\left(\frac{k}{mg}\right)^\frac{1}{\beta}$ = speed.

for $\ds \beta=2, \,\,\,\, m\frac{dv}{dt}=-mg+kv^2$ this is
separable and bernoulli.

$\ds \frac{dv}{dt}=-g+\frac{k}{m}v^2$

$\ds \int\frac{dv}{\frac{k}{m}v^2-g}=\int dt$, now use partial
fractions, to get

$\ds \int\frac{\frac{1}{2\sqrt g}}{\left(\sqrt{\frac{k}{m}}v-\sqrt
g\right)}+\frac{-\frac{1}{2\sqrt
g}}{\left(\sqrt{\frac{k}{m}}v+\sqrt g\right)}dv=t+A$

$\ds \frac{1}{2}\sqrt{\frac{m}{gk}}\ln\left(\frac
{v-\sqrt{\frac{gm}{k}}}{v+\sqrt{\frac{gm}{k}}}\right)=t+A$

$\ds
v(t)=\left(\frac{1+ce^{2\sqrt{\frac{gk}{m}}t}}{1-ce^{2\sqrt{\frac{gk}{m}}t}}\right)
\sqrt{\frac{gm}{k}}$


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