Archive for October, 2007


Answer: Terminal Velocity of a Falling Particle

October 31st, 2007

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}

Comments(0) - Answers, Classical Mechanics

Question: Terminal Velocity of a Falling Particle

October 28th, 2007

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.

Comments(0) - Classical Mechanics, Questions

Answer: Electric field of a nonconducting sphere with a spherical cavity

October 24th, 2007

Answer: The electric field inside the cavity is going to be the superposition of the field due to the uncut sphere plus the field due to a sphere the size of the cavity with a uniform charge density of -\rho. The key to solve this problem is to calculate the electric field of each sphere in a different coordinate systems.

First, lets deal with the electric field of the large sphere of charge density \rho. To simplify Gauss’s Law, I am going to use a spherical coordinate system with the origin at the center of the sphere. The coordinate system is going to defined as:

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k} = r \hat{r}

We can now use Gauss’s Law to calculate the electric field of the sphere.

\oint {\vec{E} \cdot dA = } \frac{1}{{\varepsilon _0 }}Q_{inside}

\oint {E dA = } \frac{1}{{\varepsilon _0 }} \frac{4}{3}\pi r^{3}\rho

E 4 \pi r^{2} = \frac{1}{{\varepsilon _0 }} \frac{4}{3}\pi r^{3}\rho

E = \frac{\rho}{{3 \varepsilon _0 }} r

Since the electric field is radial:

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r}

Now that we have calculated the electric field for the big sphere, we can calculate the field of the small sphere of charge density -\rho. We will, again, use a spherical coordinate system with the origin at the center of the sphere. The coordinate system is going to defined as:

\vec{r^{\prime}} = x^{\prime} \hat{i}^{\prime} + y^{\prime} \hat{j}^{\prime} + z^{\prime} \hat{k}^{\prime} = r^{\prime} \hat{r}^{\prime}

When we apply Gauss’s Law to the small sphere, we will get \vec{E^{\prime}} = \frac{-\rho}{{3 \varepsilon _0 }} r^{\prime} \hat{r^{\prime}}

Next we need to take the superposition of both the electric fields. In order to do so, we will need to relate the follow coordinate systems.


Using simple vector addition, we find that \vec{r^{\prime}} + z \hat{j} = \vec{r}. Also because \vec{r} = r \hat{r}, we can reduce this formula to: r^{\prime} \hat{r}^{\prime} = r \hat{r} -z \hat{j}

Summing the electric field of the two spheres we will get:

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r} -\frac{\rho}{{3 \varepsilon _0 }} r^{\prime} \hat{r^{\prime}}

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} r \hat{r} -\frac{\rho}{{3 \varepsilon _0 }} (r \hat{r} -z \hat{j})

\vec{E} = \frac{\rho}{{3 \varepsilon _0 }} z \hat{j}

Comments(2) - Answers, Electricity and Magnetism

Question: Electric field of a nonconducting sphere with a spherical cavity

October 21st, 2007

I got this question a couple times on E&M assignments. The solution to this problem is really elegant.

Question: A sphere of radius a is made of a nonconducting material that has a uniform volume charge density \rho. A spherical cavity of radius b is removed from sphere which is a distance z from the center of the sphere. Assume that a > z + b. What is the electric field in the cavity?

Note: You should only need Gauss’s Law to solve this problem.

Comments(0) - Electricity and Magnetism, Questions

Refactoring jkwiens.com

October 21st, 2007

jkwiens.com has undergone a complete facelift!

Besides the facelift, my vision of jkwiens.com has been completely reconstructed. Instead of meta-blogging, this site will become an outlet for pursuing my interest in programming, physics, and mathematics.

Every week, I plan on posting an interesting problem related to computer science, physics, or mathematics. I will post the solution the following week which will give readers time to solve the problem. If you managed to solve the problem, please email your solution to jkwiens@shaw.ca.

Besides posting weekly problems, I will be creating references which will aid anyone trying to learn programming, physics, and mathematics.

Comments(0) - Announcement

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