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This is in reference to the following problem:

$\textbf{Problem}$: (Coil of Rope) A uniform rope of mass density $\lambda$ per unit length is coiled on a smooth horizontal table. One end is pulled straight up with constant speed $v_0$.

(a) Find the force exerted on the end of the rope as a function of the height $y$.

(b) Compare the power delivered to the rope with the rate of change of the rope's total mechanical energy.

Now according to this paper https://bshotwell.physics.ucsd.edu/Rope-Pull.pdf, the answers to (a) and (b) resp. are : $\lambda yg+\lambda v_0^2$ and that energy is not conserved.

$\textbf{My Solution}$: Since the rope is continuous, the transition part of the rope which is between the 'stationary' part at rest and the moving part must move with velocity $v/2$. Thus using the Meshcherskii Equation, $$ \Sigma F_{ext}=m dv/dt - v_{rel}dm/dt$$

I get $F=\lambda yg+\lambda v_0^2/2$ which agrees with energy conservation.

I think my answer is more appropriate since the rope is continuous, and inelastic collisions between the parts of the rope do not occur( as in the case of , say a chain).

I would like to know which answer is correct, and why. Also I am open to further suggestions/conceptual clarifications to my solution.

Qmechanic
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tensorman666
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1 Answers1

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I've never heard of the Meshcherskii Equation and I do not understand what you're using for $dv/dt$ so I'll just give a really straightforward calculation.

The length $l$ of rope in motion after time $t$ is $l = v_0 t$. The momentum of that length of rope is $$ p = \text{mass}\times \text{velocity} = \underbrace{l \lambda}_\text{mass} \times v_0 = v_0^2 \lambda t \, .$$ Newton's law is that force equals time rate of change of momentum, so $$F = \frac{d}{dt} \left(v_0^2 \lambda t \right) = v_0^2 \lambda \, .$$ Part of that force is gravity $$F_\text{gravity} = -\underbrace{l \lambda}_\text{mass} g = -v_0 t \lambda g$$ so \begin{align} F_\text{pull} + F_\text{gravity} &= v_0^2 \lambda \\ F_\text{pull} - v_0 t \lambda g &= v_0^2 \lambda \\ F_\text{pull} &= \lambda v_0 \left( v_0 + t g\right) \, . \end{align}

DanielSank
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