jkwiens.com: Answer: Falling Infinite Rope

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

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

Category: Answers, Classical Mechanics

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