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Question: Brachistochrone Problem

Sunday, 4 of November , 2007 at 10:52 pm

The brachistochrone problem was one of the earliest problems which started the development of calculus of variations. In 1696, Johann Bernoulli posed question to the readers of Acta Eruditorum. Isaac Newton was able to solve the problem the very next day.

Question: A particle starts from rest and slides down a smooth curve under gravity. Find the shape of the curve that will minimize the time taken between two given points.

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Category: Calculus, Classical Mechanics, Questions

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 t = 0, initial conditions are said to be \frac{\partial y}{\partial t}(0) = 0 and y(0) = 0. 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 y(0) = 0, 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}

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Category: Answers, Classical Mechanics

Question: Terminal Velocity of a Falling Particle

Sunday, 28 of October , 2007 at 1:53 pm

Last week I picked up a new Classical Mechanics (Kibble, Berkshire) textbook. I seem to have misplaced my old 3rd year Classical Mechanics textbook, so I wanted a replacement. As a repercussion for the next while I will be solving Classical Mechanics problems. I will save my “fancier” Quantum Mechanics and Relativistic problems for a later date. However, just because General Relativity and Quantum Mechanics have superseded Classical Mechanics, it doesn’t mean that it is irrelevant. Classical Mechanics is accurate over a huge domain. It only seems to break down when dealing with the very fast and the very small. Considering that I’m not THAT small and I’m definitely not THAT fast (and I’m guessing you aren’t either), this area of physics has huge practical implications.

Question: A particle falling under gravity is subjected to a retarding force proportional to its velocity. Find its position as a function of time if it starts from rest and show that it will eventually reach a terminal velocity.

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Category: Classical Mechanics, Questions