jkwiens.com: Classical Mechanics

Answer: Falling Infinite Rope

Friday, 7 of March , 2008 at 8:21 pm

Question: An infinite rope with a linear density of $\lambda$ is placed on a frictionless table. If the end of the rope is placed at the end of the table and starts falling, what is the velocity of the rope as a function of distance?

Answer: According to Newton’s 2nd Law of Motion, we know that

$F_{net} = \frac{\partial p}{\partial t}$

If we create a force diagram, we can easily see that $F_{net} = mg$ .

where $m = \lambda x$ .

Therefore, we can create the equation of motion as follows:

$\frac{\partial p}{\partial t} = \frac{\partial (m \dot{x})}{\partial t} = \dot{x}\dot{m} + m\ddot{x} = mg$
$\dot{x}\frac{\partial (\lambda x)}{\partial t} + \lambda x \ddot{x} = \lambda x g$
$\lambda\dot{x}^2 + \lambda x \ddot{x} = \lambda x g$
$\dot{x}^2 + x \ddot{x} = x g$

In order to solve this differential equation, let $y = \frac{1}{2} \dot{x}^2$ .

This means:

$\dot{y} = \dot{x} \ddot{x} ~\Longrightarrow ~ \ddot{x} = \frac{\dot{y}}{\dot{x}} = \frac{\frac{\partial y}{\partial t}}{\frac{\partial x}{\partial t}} = \frac{\partial y}{\partial x}$

Therefore, the differential equation becomes

$2y + x \frac{dy}{dx} = gx ~\Longrightarrow ~ \frac{dy}{dx} + \frac{2}{x}y = g$

We can solve this using the integrating factor method. According to our differential equation, our integrating factor will be $e^{\int \frac{2 dx}{x}}= x^2$ . If we multiple our integrating factor to our ODE, we get

$x^2 \frac{dy}{dx} + 2xy = g x^2$
$\frac{d(x^2 y)}{dx} = gx^2$
$x^2 y = \int g x^2 dx = \frac{gx^2}{3} + C$
$\therefore y = \frac{g}{3}x + \frac{C}{x^2}$

However, since $y = \frac{1}{2} \dot{x}^2$ , we know that

$\frac{1}{2}\dot{x}^2 = \frac{g}{3}x + \frac{C}{x^2}$
$\dot{x} = \sqrt{\frac{2}{3}gx + \frac{2C}{x^2}}$

If we add the boundary condition, $\dot{x}(0) =0$ , the equation reduces to

$\dot{x} = \sqrt{\frac{2}{3}gx}$

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Category: Answers, Classical Mechanics

Question: Falling Infinite Rope

Sunday, 2 of March , 2008 at 9:42 pm

Question: An infinite rope with a linear density of $\lambda$ is placed on a frictionless table. If the end of the rope is placed at the end of the table and starts falling, what is the velocity of the rope as a function of distance?

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Category: Classical Mechanics, Questions

Answer: A bead sliding on a rotating parabola

Thursday, 13 of December , 2007 at 9:33 pm

After looking at Rod’s Solution on Reasonable Deviations, I have come to the conclusion that my solution was way to complicated. Instead of using high school physics, I decided to solve the problem using Hamiltonian Mechanics.

Question: A bead slides along a smooth wire bent in the shape of a parabola, . The bead rotates in a circle, of radius , when the wire is rotating about its vertical symmetry axis with angular velocity $\omega$ . Find the constant c.

Answer: To solve this problem using Hamiltonian Mechanics, we first need to find the Hamiltonian (which is the Kinetic Energy + Potential) of the system.

The system can easily be shown to have the following form:

$L = K + P = \frac{1}{2} m \arrowvert \vec{v} \arrowvert ^{2} - mgz$

If we use cylindrical coordinates and constrain the path of the bead to the parabola, we can simplify the Hamiltonian to a single degree of freedom. Since the velocity $\vec{v}$ in cylindrical coordinates is $\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k}$ and , the Hamiltonian reduces to:

$L = \frac{1}{2} m (\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k})\cdot(\dot{r}\hat{r} + r \dot{\theta}\hat{\theta} + \dot{z}\hat{k}) - mgcr^2$

and since $\dot{\theta}=\omega$ and $\dot{z}=\frac{d}{dt}(cr^2) = 2 c r \dot{r}$ , the equation becomes:

$L = \frac{1}{2} m (\dot{r}^2 + r^2\omega^2 + 4c^2r^2\dot{r}^2) - mgcr^2$

Now that we have the Hamiltonian reduced to its generalized coordinates, we can apply it to the Euler-Lagrange equation. Since our Hamiltonian has only one degree of freedom, the Euler-Lagrange equation will have the following form:

$\frac{\partial L}{\partial r} - \frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{r}}) = 0$

First, lets take the derivatives needed for the differential equation. They can be shown to be:

$\frac{\partial L}{\partial r} = m ( 4c^2r\dot{r}^2 + r \omega ^2) - 2mgcr$

$\frac{\partial L}{\partial \dot{r}} = m (\dot{r} + 4 c^2 r^2 \dot{r})$

$\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{r}}) = m (\ddot{r} + 4 c^2 r^2 \ddot{r} + 8 c^2 r \dot{r}^2)$

When we put these values into the differential equation, we will get:

$m(4c^2r\dot{r}^2 + r \omega^2) - 2mgcr - m(\ddot{r} + 4c^2r^2\ddot{r} + 8c^2r\dot{r}^2) = 0$

$m(1 + 4 c^2 r^2)\ddot{r} + m(4 c^2 r)\dot{r}^2 + m(2gc - \omega ^2)r = 0$

Since the position of the bead is confined to a circle of radius R, we can simplify the differential equation by using the following conditions: $\ddot{r} = 0$ , $\dot{r} = 0$ , and . The differential equation will then reduce to:

$m(2gc - \omega ^2)R = 0$

Therefore, the value of c must be:

$c = \frac{\omega ^2}{2g}$

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Category: Answers, Classical Mechanics

Question: A bead sliding on a rotating parabola

Sunday, 9 of December , 2007 at 9:24 pm

Question: A bead slides along a smooth wire bent in the shape of a parabola, . The bead rotates in a circle, of radius , when the wire is rotating about its vertical symmetry axis with angular velocity $\omega$ . Find the constant c.

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Category: Classical Mechanics, Questions

Answer: Brachistochrone Problem

Saturday, 10 of November , 2007 at 5:13 pm

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.

Answer: First, we need to find the distance dl of two neighboring points on the curve f(x).

$dl = \sqrt{dx^{2} + dy^{2}} = \sqrt{dx^{2} (1 + \frac{dy^{2}}{dx^{2}})} = \sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx$

We can then use the relationship $dt = \frac{dl}{v}$ to determine the time required to travel the distance dl. If we integrate dt over the entire path of f(x), we will obtain the function we want to minimize.

$t = \oint \frac{dl}{v} = \int \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx}{v}$
In order to obtain the value v, we can use the fact that energy is conserved. Therefore the following relation will hold:
$\frac{1}{2}mv^{2} = mgy \rightarrow v = \sqrt{2gy}$
We can now insert this relationship into our integral to get the following:
$t = \int \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}dx}{\sqrt{2gy}}$
In order to minimize t, we will need to minimize the following function
$\Psi(x,y) = \frac{\sqrt{1 + \frac{dy^{2}}{dx^{2}}}}{\sqrt{2gy}}$
This function, according to Calculus of Variations, will be an extrema when the Euler-Lagrange differential equation is satisfied.
$\frac{\partial\Psi}{\partial y} - \frac{\partial}{\partial x}\partial\frac{\Psi}{\partial \dot{y}} = 0$
If we insert $\Psi(x,y)$ into the Euler-Lagrange differential equation, we obtain the following differential equation (with a little bit of algebra).
$2\ddot{y}y + \dot{y}^{2} + 1 = 0$
With a little more manipulation, we can reduce this differential equation into something more manageable.
$2\ddot{y}y + \dot{y}^{2} + 1 = 0$
$2\ddot{y}y + \dot{y}^{3} + \dot{y} = 0$
$\frac{\partial}{\partial x}(y\dot{y}^{2}) + \frac{\partial y}{\partial x} = 0$
$\frac{\partial}{\partial x}(y\dot{y}^{2} + y) = 0$
If we integrate this differential equation, we get the following:
$y\dot{y}^{2} + y = C$
$\dot{y} = \sqrt{\frac{C - y}{y}}$
This equation is solved by the parametric equations:
$x = \frac{C}{2}(\theta - \sin{\theta})$
$y = \frac{C}{2}(1 - \cos{\theta})$
which is a cycloid.

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Category: Answers, Calculus, Classical Mechanics