Question

Falling Chain A portion of a uniform chain of length 8ft is loosely coiled around a peg at the edge of a high horizontal platform, and the remaining portion of the chain hangs at rest over the edge of the platform. See Figure 2.4.2. Suppose that the length of the overhanging chain is 3ft, that the chain weighs 2lb/ft, and that the positive direction is downward. Starting at t=0 seconds, the weight of the overhanging portion causes the chain on the table to uncoil smoothly and to fall to the floor. If x(t) denotes the length of the chain overhanging the table at time t>0, then v=dx/dt is its velocity. When all resistive forces are ignored, it can be shown that a mathematical model relating v to x is given by $\mathit{x}\mathit{v}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathbf{32}\mathit{x}$

(a) Rewrite this model in differential form. Proceed as in Problems 31-36 and solve the DE for v in terms of x by finding an appropriate integrating factor. Find an explicit solution v(x).

(b) Determine the velocity with which the chain leaves the platform.

Short answer

The explicit solution is $v=8\sqrt{\frac{x}{3}-\frac{9}{x^2}}$and v=12.7

Step-by-step solution

The Problem equation (a)

The given equation is $xv\frac{dv}{dt}+v^2=32$.

Which can rewritten as,

$\begin{array}{c}xv\times dv=\left(32x-v^2\right)\times dx\\ xv\times dv+\left(v^2-32x\right)\times dx=0\end{array}$

Exact equation form is given as,

$\frac{dM}{dv}=2v$and $\frac{dN}{dx}=v$.

So$\frac{M_v-N_x}{N}=\frac{1}{x}$

It is not exact, but it can be by find the integrating factor for it. Which can be found as below:

$\begin{array}{c}\mu \left(x\right)=e^{\int \frac{M_v-N_x}{N}dx}\\ =e^{\int \frac{1}{x}dx}\\ =x\end{array}$

Multiply the integrating factor with the given equation.

$x^{2}v\times dv+\left(v^{2}x-32x^2\right)\times dx=0$

Find the Integral of N

Get the integral of N

$\begin{array}{c}f(x,v)=x^2\int v\times dv\\ =\frac{x^2{v}^2}{2}+g\left(x\right)\end{array}$

Find the derivative of the obtained function with respect to x.

$\frac{dF}{dx}=x{v}^2+g^{\text{'}}\left(x\right)$

Here, M is equal to the derivative with respect to x of the function. Then we have,

$\begin{array}{c}x{v}^2-32x^2=x{v}^2+g^{\text{'}}\left(x\right)\\ \Rightarrow g^{\text{'}}\left(x\right)=-32x^2\end{array}$

Integrate $g^{\text{'}}\left(x\right)$to get $g\left(x\right)$.

Which implies that $g\left(x\right)=\frac{-32x^3}{3}$.

Then the solution becomes,

$\frac{-32x^3}{3}+\frac{x^2{v}^2}{2}=c$

Substitute the initial value $v\left(0\right)=3$into the solution so you can get the value of your constant.

$\begin{array}{c}\frac{-32{\left(3\right)}^3}{3}+\frac{{\left(3\right)}^2{\left(0\right)}^2}{2}=c\\ -288=c\end{array}$

Hence, the solution is $v=8\sqrt{\frac{x}{3}-\frac{9}{x^2}}$.

To solve for V (b)

Now solve for v by using x=8.

$\begin{array}{c}v=8\sqrt{\frac{\left(8\right)}{3}-\frac{9}{{\left(8\right)}^2}}\\ \approx 12.7\end{array}$

Hence, the required value is $v=12.7$.