- #1
TheGreatDeadOne
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- 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?
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?
