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Here's the problem:

"Suppose that a flexible 4-ft rope starts with 3 ft of its length arranged in a heap right at the edge of a high horizontal table, with the remaining foot hanging (at rest) off the table. At time t = 0 the heap begins to unwind and the rope begins gradually to fall off the table, under the force of gravity pulling on the overhanging part. Under the assumption that frictional forces of all sorts are negligible, how long will it take for all the rope to fall off the table?"

Note that many viewers will be familiar with a problem which is

*similar*to this...but not the same! The rope on the table is arranged in a heap, which means that

*only the rope off the table is accelerating.*The rope on the table remains at rest (it's not being dragged along).

I didn't bring this up because I couldn't solve it--I could, but I got a different answer from the authors in my textbook. They used the definition F = d/dt [ma] as the basis of their model. I used the principle of conservation of energy (the system is, after all, conservative).

It makes sense to set up the potential energy system so the rope on the table has a gravitational potential energy of 0 (and rope hanging off the table has negative gravitational potential). The following is my model:

[tex]Energ{y_{sys}} = C = Energ{y_{gravitational\;potential\;rope}} + Energ{y_{kinetic\;rope}}\;[/tex]

[tex]Energ{y_{grav}} = {m_{off}}gh = (wx)(g)( - x/2) = \frac{{ - wg{x^2}}}{2}[/tex]

[tex]Energ{y_{kinetic}} = \frac{1}{2}{m_{off}}{\left( {\frac{{dx}}{{dt}}} \right)^2} = \frac{1}{2}wx{\left( {\frac{{dx}}{{dt}}} \right)^2}[/tex]

[tex]Energ{y_{grav}} = {m_{off}}gh = (wx)(g)( - x/2) = \frac{{ - wg{x^2}}}{2}[/tex]

[tex]Energ{y_{kinetic}} = \frac{1}{2}{m_{off}}{\left( {\frac{{dx}}{{dt}}} \right)^2} = \frac{1}{2}wx{\left( {\frac{{dx}}{{dt}}} \right)^2}[/tex]

In these equations, m-off means mass off the table. w is the linear density (w is irrelevant, soon canceled in the calculations). The goal now is to find a function t(x), which fits these equations. I have set up the x-axis with 0 at the table and pointing downward. Therefore, positive values of x correspond to negative potential energies.

[tex]{E_{sys}} = {E_{grav}} + {E_{kin}}[/tex]

We can use the initial conditions to find the energy of the system. See below.

[tex] - \frac{{wg}}{2} = - \frac{{wg{x^2}}}{2} + \frac{{wx}}{2}{\left( {\frac{{dx}}{{dt}}} \right)^2}[/tex]

[tex]\sqrt {g\frac{{{x^2} - 1}}{x}} = \frac{{dx}}{{dt}}[/tex]

[tex]\sqrt {g\frac{{{x^2} - 1}}{x}} = \frac{{dx}}{{dt}}[/tex]

And variables are easily separated:

[tex]\frac{{dt}}{{dx}} = \sqrt {\frac{1}{g}} \cdot \sqrt {\frac{x}{{{x^2} - 1}}} [/tex]

[tex]t(4) = \sqrt {\frac{1}{g}} \int_1^4 {\sqrt {\frac{x}{{{x^2} - 1}}} } dx[/tex]

[tex]t(4) = \sqrt {\frac{1}{g}} \int_1^4 {\sqrt {\frac{x}{{{x^2} - 1}}} } dx[/tex]

This integral is improper. A trigonometric substitution does the trick:

[tex]\ell etting\;x = \sec \theta [/tex]

[tex]t(4) = \sqrt {\frac{1}{g}} \int\limits_{{{\cos }^{ - 1}}(1)}^{{{\cos }^{ - 1}}(1/4)} {{{(\sec \theta )}^{3/2}}d\theta } [/tex]

[tex]t(4) \approx .4876(s)\;using\;Simpson's\;Rule\;with\;n\;=\;100[/tex]

[tex]t(4) = \sqrt {\frac{1}{g}} \int\limits_{{{\cos }^{ - 1}}(1)}^{{{\cos }^{ - 1}}(1/4)} {{{(\sec \theta )}^{3/2}}d\theta } [/tex]

[tex]t(4) \approx .4876(s)\;using\;Simpson's\;Rule\;with\;n\;=\;100[/tex]

This is my answer. As you can see, I got it using a pretty straight-forward application of conservation of energy. If anyone doesn't understand the model or work, just ask. I'm pretty sure it's all right. Now, my textbook does NOT begin with conservation of energy. Instead, it begins with an application of F = d/dt [ma]. Here's the model:

[tex]F = \frac{{dP}}{{dt}} = [mv];\quad m = wx;\quad F = mg = wxg[/tex]

[tex]wxg = \frac{d}{{dt}}[wxv] = w\left( {x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}} \right)[/tex]

[tex]xg = x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}[/tex]

[tex]wxg = \frac{d}{{dt}}[wxv] = w\left( {x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}} \right)[/tex]

[tex]xg = x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}[/tex]

Here, the author cleverly uses the chain rule to eliminate t and reduce the equation to two variables: v and x, where v is a function of the independent variable x.

[tex]xg = x\frac{{dv}}{{dx}}\frac{{dx}}{{dt}} + v \cdot v[/tex]

[tex]xg = xv\frac{{dv}}{{dx}} + {v^2}[/tex]

[tex]xg = xv\frac{{dv}}{{dx}} + {v^2}[/tex]

Now, the author sets up the equation in differential form to be solved as an exact differential equation:

[tex]0 = \left( {xv} \right)\,dv\, + ({v^2}\, - xg)\,dx[/tex]

Multiplying both sides by the simple integrating factor x yields an exact equation:

[tex]0 = ({x^2}v)\,dv + ({v^2}x - {x^2}g)dx = dF[/tex]

Given this differential form of F, dF, we can easily find F to be:

[tex]F(x,v) = \frac{1}{2}{x^2}{v^2} - \frac{g}{3}{x^3} = C[/tex]

It's easy to check that

[tex]\frac{{\delta F}}{{\delta v}} = {x^2}v\,;\,\frac{{\delta F}}{{\delta x}} = {v^2}x - {x^2}g[/tex]

And that F is a general solution to the differential equation. Now, we use the initial condition to find that

[tex]C = - \frac{g}{3}[/tex]

We then replace v with dx/dt, then we separate variables and integrate.

[tex]{v^2} = \frac{{2g}}{3}\left( {\frac{{{x^3} - 1}}{{{x^2}}}} \right)[/tex]

[tex]\frac{{dx}}{{dt}} = \sqrt {\frac{{2g}}{3}} \cdot \sqrt {\frac{{{x^3} - 1}}{{{x^2}}}} [/tex]

[tex]t = \sqrt {\frac{3}{{2g}}} \int {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]

[tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_1^4 {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]

[tex]\frac{{dx}}{{dt}} = \sqrt {\frac{{2g}}{3}} \cdot \sqrt {\frac{{{x^3} - 1}}{{{x^2}}}} [/tex]

[tex]t = \sqrt {\frac{3}{{2g}}} \int {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]

[tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_1^4 {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]

Which, like in the first example, is improper. Trig sub!

[tex]\ell etting\;{x^3} = {\sec ^2}\theta [/tex]

[tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_0^{{{\cos }^{ - 1}}(1/8)} {{{(\sec \theta )}^{4/3}}d\theta } [/tex]

[tex]t(4) \approx 0.5411(s)[/tex]

[tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_0^{{{\cos }^{ - 1}}(1/8)} {{{(\sec \theta )}^{4/3}}d\theta } [/tex]

[tex]t(4) \approx 0.5411(s)[/tex]

Now, here's an easy exercise. Show that this model does NOT obey conservation of energy. How is this possible? In the first model, using conservation of energy from the start, we get that the rope falls off in 0.4876 seconds. In the second model, using F = dP/dt, we get that the rope falls off in 0.5411 seconds. What's going on here?

*The calculations are right, so is there a problem with one of the models? Obviously, there must be.*Thanks in advance for any thoughtful responses.

--Mazerakham.

All credit for this problem and the second solution go to the authors C.H. Edwards, Jr. and David E. Penney of the book Elementary Differential Equations With Boundary Value Problems, 2nd Edition.