jkwiens.com: Answer: Terminal Velocity of a Falling Particle

Answer: Terminal Velocity of a Falling Particle

Wednesday, 31 of October , 2007 at 9:48 pm

According to Newton’s Second Law of Motion,

$\vec{F_t} = \sum{\vec{F_i}} = m \vec{a}$

Since the following forces we are dealing with is gravity and drag force, we will get the following equation:

$\vec{F_t} = m \vec{a} = (mg - kv)\hat{j}$
$mg - k \frac{\partial y}{\partial t} = m \frac{\partial^{2} y}{\partial^{2} t}$
$\frac{\partial^{2} y}{\partial^{2} t} = g - \frac{k}{m} \frac{\partial y}{\partial t}$

If we integrate both sides, we can reduce this equation to:

$\int\frac{\partial^{2} y}{\partial^{2} t}dt = \int(g - \frac{k}{m} \frac{\partial y}{\partial t})dt$
$\frac{\partial y}{\partial t} = \int(g - \frac{k}{m} \frac{\partial y}{\partial t})dt = gt - \frac{k}{m} y + C$

At , initial conditions are said to be $\frac{\partial y}{\partial t}(0) = 0$ and . Because of these initial conditions, we can reduce the equation to a first order non-homogeneous differential equation.

$\frac{\partial y}{\partial t} + \frac{k}{m} y = gt$

To solve this differential equation, I will be using the integrating factor method. According to our differential equation, our integrating factor will be: $e^{\int \frac{k}{m}dt} = e^{\frac{k}{m}t}$ .

If we multiple our integrating factor to our differential equation, we will get the following:

$e^{\frac{k}{m}t} \frac{\partial y}{\partial t} + e^{\frac{k}{m}t} \frac{k}{m} y = e^{\frac{k}{m}t} gt$
$\frac{\partial}{\partial t}(y e^{\frac{k}{m}t}) = e^{\frac{k}{m}t} gt$
$y e^{\frac{k}{m}t} = \int (e^{\frac{k}{m}t} gt) dt$
$y e^{\frac{k}{m}t} = \frac{m g t}{k}e^{\frac{k}{m}t} - \frac{m^{2}g}{k^{2}}e^{\frac{k}{m}t}+ C$
$y = \frac{m g t}{k} - \frac{m^{2}g}{k^{2}} + Ce^{\frac{-k}{m}t}$

If we use the initial condition , the constant C can be calculated to be:

$y(0) = \frac{m g 0}{k} - \frac{m^{2}g}{k^{2}} + Ce^{\frac{-k}{m} 0} = 0$
$y(0) = \frac{-m^{2}g}{k^{2}} + C = 0$
$C = \frac{m^{2}g}{k^{2}}$

Therefore,

$y(t) = \frac{m g t}{k} - \frac{m^{2}g}{k^{2}} + \frac{m^{2}g}{k^{2}}e^{\frac{-k}{m}t}$
and
$\frac{\partial }{\partial t}y(t) = \frac{m g}{k} - \frac{mg}{k}e^{\frac{-k}{m}t}$

In order to find the terminal velocity of the particle, we need to take the limit of the velocity as t goes to zero.

$\lim_{t \rightarrow \infty} \frac{\partial }{\partial t}y(t) = \lim_{t \rightarrow \infty}( \frac{m g}{k} - \frac{mg}{k}e^{\frac{-k}{m}t})$
$\lim_{t \rightarrow \infty} \frac{\partial }{\partial t}y(t) = \frac{m g}{k}$

Category: Answers, Classical Mechanics

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