# Rope through a hole

1. Aug 29, 2013

### Saitama

1. The problem statement, all variables and given/known data
A rope of mass M and length $l$ lies on a friction less table, with a short portion, $l_0$ hanging through a hole. Initially the rope is at rest.

a. Find a general solution for x(t), the length of rope through the hole.

(Ans: $x=Ae^{\gamma t}+Be^{-\gamma t}$, where $\gamma^2=g/l$)

b. Evaluate the constants A and B so that the initial conditions are satisfied.

2. Relevant equations

3. The attempt at a solution
The forces acting on the rope are weight and tension (T) due to the part of rope on the table. If x is the length of rope hanging, l-x is the length of rope on the table. Let $\lambda$ be the mass per unit length of rope.
Newton's second law for hanging part,
$$\lambda xg-T=\lambda xa$$
Newton's second law for rope on table,
$$T=\lambda (l-x)a$$
From the two equations,
$$a=\frac{gx}{l+2x}$$
I can substitute a=d^2x/dt^2 but Wolfram Alpha gives no solution for this.

Any help is appreciated. Thanks!

2. Aug 29, 2013

### Staff: Mentor

Hi Pranav-Arora. Your formulation is correct, but check your algebra. It should be la=xg.

3. Aug 29, 2013

### Saitama

Oh yes, sorry about that. Thanks a lot!

At t=0, $x(0)=l_0$, x'(0)=0
$x(0)=A+B=l_0$

Since $x'(t)=A\gamma e^{\gamma t}-B\gamma e^{-\gamma t}\Rightarrow x'(0)=0=A-B$
Solving the two equations, $A=B=l_0/2$.
Hence,
$$x(t)=\frac{1}{2}\left(l_0e^{\gamma t}+l_0e^{-\gamma t}\right)$$
Looks good?

4. Aug 29, 2013

### janhaa

It can be solved this way too, by Newton's second law :

$$F=Ma=\rho g x A$$

$$M\frac{d^2x}{dt^2}=\rho g x A$$

$$M\frac{d^2x}{dt^2}=\frac{M}{l^3} g x l^2$$

$$\frac{d^2x}{dt^2}=\frac{gx}{l}$$

and the solution of this DE is

$$x(t)=x=A\cdot exp(\sqrt{\frac{g}{l}}t)+B\cdot exp(-\sqrt{\frac{g}{l}}t)$$

same as yours...

Last edited: Aug 29, 2013
5. Aug 29, 2013

### Saitama

@janhaa: What are $\rho$ and $A$?

6. Aug 29, 2013

### janhaa

$$\rho$$ is density
and
$$A: area = l^2$$