# Speed of a hanging rope sliding on a nail (using energy conservation)

Homework Statement:
A rope of full length 2l hangs balanced on a smooth nail, length l on each side. A small impulse causes the rope starts to slide over the nail. Get the string speed module at the moment when it hangs with length x on one side and 2l - x on the other. Disregard the nail dimensions and assume x> l.
Relevant Equations:
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I solved this problem easily using Newton's second law, but I had problems trying to use mechanical energy conservation to solve it.
How I solved using Newton's second law:
##\text{(part of the rope that is on the left)}\, m_1=x\rho g,\, \text{(part of the rope that is on the right)}\, m_2=(2l-x)\rho ##
$$F=x\rho g - (2l-x)\rho g=2(x-l)g\rho$$
$$\Rightarrow \frac{dF}{dt}=m\frac{dv}{dt} =2l\rho\frac{dv}{dt}$$
Thus,
$$\frac{dv}{dt}=\frac{x-l}{l}g, \quad \mbox{by the chain rule:} \quad \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$
Replacing and integrating:
$$\int s\frac{dv}{dx}=\int \frac{x-l}{l}g \, dx \Rightarrow\frac{g}{l}\frac{(x-l)^2}{2}=\frac{v^2}{2}+C$$

For ##x=l\, \rightarrow v=0##, then C=0, so:

$$\boxed {v=(x-l)\sqrt{\frac{g}{l}} }$$

Now for conservation of energy I had trouble writing the relations. What I've tried to do so far:
(Assuming the nail is at a distance h from the ground, and h>l)

For the left side \begin{align}
E_{iL}& = l\rho g (h-l) \\
E_{fL}&=(x)\rho g (h-l-x) +(x)\rho \frac{v^2}{2}
\end{align}

For the right side \begin{align}
E_{iR}& = l\rho g (h-l) \\
E_{fR}&=(2l-x)\rho g (h-2l+x) +(2l-x)\rho \frac{v^2}{2}
\end{align}

And as I wrote above, it is very wrong. The problems I'm having are as follows:

1) Should I consider when the rope reaches its length and falls to the ground?
2) Should I consider potential gravitational energy when the string is in balance or not?

Homework Helper
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You are making extra work for yourself by worrying about where the ground is. Much simpler to take the height of the nail as the zero height.

• Lnewqban
You are making extra work for yourself by worrying about where the ground is. Much simpler to take the height of the nail as the zero height.
You're right! It's as simple as that. But, the logic used to express the initial and final energies do you think is right?

Homework Helper
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You're right! It's as simple as that. But, the logic used to express the initial and final energies do you think is right?
We can take the equations you wrote and substitute h=0.
In (1) and (3), what is the initial height of the mass centre of each side?
In (2), you have a factor x(l-x) in the GPE term, but it is (l-x)2 in (4). Do you see the asymmetry? Note that substituting x=l and v=0 in (2) and (4) should produce (1) and (3).

• We can take the equations you wrote and substitute h=0.
In (1) and (3), what is the initial height of the mass centre of each side?
In (2), you have a factor x(l-x) in the GPE term, but it is (l-x)2 in (4). Do you see the asymmetry? Note that substituting x=l and v=0 in (2) and (4) should produce (1) and (3).
"In (1) and (3), what is the initial height of the mass center of each side?"

I was trying to calculate from the end of the string, but really from the center of mass it seems a lot simpler.

"In (2), you have a factor x (l-x) in the GPE term, but it is (l-x)^2 in (4). Do you see the asymmetry? Note that substituting x = l and v = 0 in (2) and (4) should produce (1) and (3) "

I saw this, I was trying to describe that when one side of the rope slips, the length of the other side decreases and the mass of rope on that side decreases too.
Thank you again!

Homework Helper
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Not sure that you got either of the points I was making.
I was trying to calculate from the end of the string, but really from the center of mass it seems a lot simpler.
If we take the nail as height zero, what is the initial height of the mass centres of the two halves of the string? So what are their initial GPEs?
I was trying to describe that when one side of the rope slips, the length of the other side decreases and the mass of rope on that side decreases too.
Your two expressions for the GPEs when one string has length x are inconsistent. Either (2) or (4) must be wrong.
After you have got equations (1) and (3) right, check (2) and (4) by plugging in x=l and v=0. If (2) and (4) are right then the resulting equations should be (1) and (3).

• • 