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

In summary: In that case the reaction force should be large.It is interesting to calculate the reaction force from the nail to the rope. The loop of the rope can detach off the nail and jump to the ground. In that case the reaction force should be large.
  • #1
TheGreatDeadOne
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0
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
.
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?
 
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  • #2
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.
 
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  • #3
haruspex said:
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?
 
  • #4
TheGreatDeadOne said:
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).
 
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  • #5
haruspex said:
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!
 
  • #6
Not sure that you got either of the points I was making.
TheGreatDeadOne said:
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?
TheGreatDeadOne said:
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).
 
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  • #7
It is interesting to calculate the reaction force from the nail to the rope. The loop of the rope can detach off the nail and jump
 
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What is the speed of a hanging rope sliding on a nail?

The speed of a hanging rope sliding on a nail can be determined using the principle of energy conservation. This means that the initial potential energy of the rope is converted into kinetic energy as it slides down the nail. The speed can be calculated using the equation: v = √(2gh), where v is the speed, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the rope above the nail.

How does the height of the rope affect its speed?

The height of the rope above the nail directly affects its speed. The higher the rope is, the greater the potential energy it has, and therefore the higher the speed it will have as it slides down the nail. This relationship is described by the equation: v = √(2gh), where h is the height of the rope.

What is the role of energy conservation in determining the speed of the rope?

Energy conservation is a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred from one form to another. In the case of a hanging rope sliding on a nail, the potential energy of the rope is converted into kinetic energy as it slides down. By applying the principle of energy conservation, we can determine the speed of the rope using the equation: v = √(2gh).

Does the mass of the rope affect its speed?

The mass of the rope does not directly affect its speed. According to the principle of energy conservation, the speed of the rope is determined by its initial potential energy and the height of the rope above the nail. However, if the mass of the rope is significantly different, it may affect the friction between the rope and the nail, which can impact the speed of the rope.

What are the assumptions made when calculating the speed of the rope using energy conservation?

When using energy conservation to calculate the speed of a hanging rope sliding on a nail, there are a few assumptions that are made. These include assuming that there is no air resistance or friction between the rope and the nail, and that the rope is a perfect, massless string. In reality, these assumptions may not hold true, but they provide a simplified model for calculating the speed of the rope.

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