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

  • #1
TheGreatDeadOne
11
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?
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,216
8,528
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.
 
  • #3
TheGreatDeadOne
11
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,216
8,528
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).
 
  • Like
Likes TheGreatDeadOne
  • #5
TheGreatDeadOne
11
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,216
8,528
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).
 
  • Like
Likes TheGreatDeadOne
  • #7
wrobel
Science Advisor
Insights Author
997
859
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
 
  • Like
Likes TheGreatDeadOne

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

Replies
6
Views
335
Replies
7
Views
496
Replies
19
Views
519
Replies
3
Views
248
Replies
1
Views
665
Replies
4
Views
398
  • Last Post
Replies
11
Views
431
Top