Question

A uniform chain of length $L,$ measured in feet, is held vertically so that the lower end just touches the floor. The chain weighs $2lb/ft.$The upper end that is held is released from rest at$t=0$. and the chain falls straight down. If denotes the length of the chain on the floor at time$t,$ air resistance is ignored, and the positive direction is taken to be downward, then

$(L-x)\frac{d^{2}x}{d{t}^2}-{\left(\frac{dx}{dt}\right)}^2=Lg.$

(a) Solve for in $v$terms of Solve for$x.$ in terms of $t$. Express$v$ in terms of $t$.

(b) Determine how long it takes for the chain to fall completely to the ground.

(c) What velocity does the model in part (a) predict for the upper end of the chain as it hits the ground?

Short answer

$$\begin{gathered} v\left(x\right)=\frac{\sqrt{Lg\left(2Lx-x^2\right)}}{L-x} \\ x\left(t\right)=L-\sqrt{L^2-Lg{t}^2} \\ v\left(t\right)=\frac{\sqrt{L}gt}{\sqrt{L-g{t}^2}} \\ t=\sqrt{L/g}\text{s,}v=\infty \end{gathered}$$

Step-by-step solution

given information

We were given the chain of length $L$, held vertically. The chain weighs 2$\text{lb}/\text{ft}.$ The upper end is released from rest at$t=0.$

Let $x\left(t\right)$denotes the length of the chain on the floor at time $t.$The equation that describes it is

$(L-x)\frac{d^{2}x}{d{t}^2}-{\left(\frac{dx}{dt}\right)}^2=Lg$

solve the various of terms (a)

(a)

Let $v=\frac{dx}{dt}.$Then equation (1) becomes:

$(L-x)\frac{dv}{dt}-v^2=Lg$

Using $\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$and $v=\frac{dx}{dt},$we have

$(L-x)\frac{dv}{dx}·v-v^2=Lg\iff (L-x)\frac{dv}{dx}·v=v^2+Lg$

Let us rewrite it as

$(L-x)\frac{dv}{dx}·v-v^2=Lg\iff (L-x)\frac{dv}{dx}·v=v^2+Lg$

$\frac{v}{v^2+Lg}dv=\frac{dx}{L-x}$

Integrating it, we obtain:

$\frac{1}{2}\ln \left(v^2+Lg\right)=-\ln (L-x)+C$

Let D be constant such that $C=\ln D.$Then the last equation is equivalent to$\ln \sqrt{v^2+Lg}=\ln \frac{D}{L-x}\iff \sqrt{v^2+Lg}=\frac{D}{L-x}$

Since the chain is released from rest, when $x=0,v$is also equal to $0$. Therefore,

$D=L\sqrt{Lg}$

Substitute (3) into (2) and rewrite what you get to find $v\left(x\right).$

$\sqrt{v^2+Lg}=\frac{L\sqrt{Lg}}{L-x}$

$v^2+Lg=\frac{L^{3}g}{{(L-x)}^2}$

$v^2=\frac{L^{3}g-L^{3}g+2L^{2}gx-Lg{x}^2}{{(L-x)}^2}$

$v=\frac{\sqrt{Lg\left(2Lx-x^2\right)}}{L-x}$

Therefore,

$\frac{dx}{dt}=v\left(x\right)=\frac{\sqrt{Lg\left(2Lx-x^2\right)}}{L-x}$

Rewriting (4), we obtain $\frac{L-x}{\sqrt{Lg\left(2Lx-x^2\right)}}dx=dt$

Integrating it (using substitution $u=2Lx-x^2$), we obtain:

$\frac{\sqrt{Lgx(2L-x)}}{Lg}=t+C$

Since $x\left(0\right)=0,C=0$. Therefore,

$\frac{\sqrt{Lgx(2L-x)}}{Lg}=t$

Rewriting this, we obtain quadratic equation:

$Lgx(2L-x)=L^2{g}^2{t}^2$

$2L^{2}gx-Lg{x}^2=L^2{g}^2{t}^2$

$Lg{x}^2-2L^{2}gx+L^2{g}^2{t}^2=0$

Solving this for $x,$we obtain:

$x_{1,2}=\frac{2L^{2}g\pm \sqrt{4L^2{g}^2-4L^3{g}^3{t}^2}}{2Lg}=L\pm \sqrt{L^2-Lg{t}^2}$

Since $x\left(0\right)=0$, the sign before the square root should be. Therefore

$x\left(t\right)=L-\sqrt{L^2-Lg{t}^2}$

To find $v\left(t\right),$ derive the last equation with respect to $t$and use the information that $L>0$and $g>0$:

$x^{\text{'}}\left(t\right)=v\left(t\right)=\frac{\sqrt{L}gt}{\sqrt{L-g{t}^2}}$

Step 3:solve the chain of T

(b)

The chain will fall completely to the ground when $x\left(t\right)=L$. Substitute it into (5) and solve for :

$$\begin{gathered} L=L-\sqrt{L^2-Lg{t}^2} \\ \iff L^2-Lg{t}^2=0 \\ \iff L=g{t}^2 \\ \iff t^2=L/g \end{gathered}$$

Therefore, the chain will completely fall to the ground after$t=\sqrt{L/g}s$

Substitute the upper end of chain

In part (b),

we found how long will it take for chain to fall completely to the ground.

Substitute it into (6) to find what velocity does the model predict for the upper end of the chain as it hits the ground:

$v(L/g)=\frac{\sqrt{L}g·\sqrt{L/g}}{\sqrt{L-g·L/g}}=\frac{L\sqrt{g}}{L-L}=\frac{L\sqrt{g}}{0}$

Therefore, the model predict that the velocity of the upper end of the chain will be infinity as it hits the ground.

Final proof

$$\begin{gathered} v\left(x\right)=\frac{\sqrt{Lg\left(2Lx-x^2\right)}}{L-x} \\ x\left(t\right)=L-\sqrt{L^2-Lg{t}^2} \\ v\left(t\right)=\frac{\sqrt{L}gt}{\sqrt{L-g{t}^2}} \\ t=\sqrt{L/g}\text{s,}v=\infty \end{gathered}$$