# Velocity of a rope falling down a table. Three different ways

A system consisting of a rope falling down a table is considered nontrivial by many students due to the differential equation to be solved to obtain the dependence on time of the position and velocity of the rope. Nevertheless,  for those who understand well, and apply correctly the physical concepts involved in this kind of problem, the solution is indeed quite simple from the mathematical point of view.

Let’s consider a homogeneous rope of length $$L$$ and mass $$m$$, released from rest and falling down a frictionless table of height $$h>L$$.  At the initial state, only a portion of the rope of length $$a$$ is hanging from the table and we want to know what is the velocity $$V_d$$ of the rope just when it detaches from the table. A student is likely to tackle the problem by applying the Second Law of Newton, but, despite this is one of the ways to solve the problem, there are more efficient and more physical ways to solve it. The methods we will explain can be used to calculate the velocity and position at any intermediate state between release and detachment, and even friction can be considered. We leave these latter cases as exercises for the student.

## Finding the velocity of a rope falling down a table using conservation of energy

Using the conservations of energy is an efficient way of addressing the problem. Indeed, you just have to write the expressions for the energy at the initial $$E_i$$ and at the detachment instant $$E_d$$, then, conservation of energy $$E_i=E_d$$ will give us the detachment velocity straightforward.

One key idea is to consider the motion of the center of mass of each one of the two sections of the rope, namely, the one laying on the table and the second one hanging at the edge of the table. This allows us to calculate the potential energy and track the motion of the rope. The center of mass of one section is located at the geometric center of that section, and we can consider that the mass of the section is concentrated at its center of mass. The mass of one portion of the rope is just the linear mass density $$\lambda=m/L$$ multiplied by the length of the portion.

The rope is released from rest, therefore, in the beginning, the total energy is only potential, so we need to set the system of coordinates in order to determine the position of the center of mass.

The position of the center of mass of the rope section lying on the table (whose mass is $$m_b$$) with respect to the floor is $$h$$ and is constant, while for the center of mass of the section hanging from the table (whose mass is $$m_a$$) is $$h-\frac{1}{2}a$$. The total initial energy of the system is the sum of the potential energy of the center of mass of both sections as follows:

\begin{align}
E_i & = m_bgh+m_ag\left(h-\frac{1}{2}a\right) \\
E_i & = \lambda (L-a)gh+\lambda ag\left(h-\frac{1}{2} a\right) \\
\end{align}

At the precise moment that the rope is detached from the table, the potential energy is due to the whole string and the new center of mass is at that moment just at the middle of the rope.

The total energy ($$E_d$$ – Energy of detachment) in that moment is just the sum of the potential and kinetic energy of the falling rope.

\begin{align}E_d&=\frac{1}{2}m\overline{V}_d^2+mg\left(h-\frac{L}{2}\right)\\
E_d&=\frac{1}{2} \lambda L\overline{V}_d^2+\lambda Lg\left(h-\frac{L}{2}\right)\end{align}

where $$\overline{V}_d$$ means velocity of the center of mass at the detachment instant. Applying the energy conservation principle, we can find the velocity of detachment of the rope falling down a table with simple mathematics.

E_i=E_d

\lambda (L-a)gh+\lambda ag\left(h-\frac{1}{2} a\right)=\frac{1}{2} \lambda L\overline{V}_d^2+\lambda Lg\left(h-\frac{L}{2}\right)

\boxed{\overline{V}_d=\sqrt{g\left(L-\frac{a^2}{L}\right)}}

The analysis above assumes that $$a$$ is strictly greater than zero at the time of release. There are two limits, when $$a\to0$$,  $$\overline{V}_d\to\sqrt{gL}$$, and in the case $$a=L$$, the detachment velocity is just the initial velocity which is zero.

## Finding the velocity of a rope falling down a table using the Work-Energy Theorem

The velocity of the rope falling down a table can also be found through the work made by the gravity force on the rope. To apply the Work-Energy Theorem, it is important to choose conveniently the direction of the coordinate axis as shown in the figure, taking the positive direction of the $$y$$ axis downwards.

If we decide to locate the origin of the coordinate system on the floor, the calculation turns out more complicated. According to the Work-Energy Theorem, the work done by the total force on the rope equals the change of the kinetic energy of the rope. In this case, it is not necessary to consider the center of mass, the analysis is based on the forces acting on the entire rope. For the horizontal section of the rope, the force of gravity $$\vec{F}=mg\hat{y}$$ is canceled out by the normal force $$\vec{N}$$ perpendicular to the table. Then the gravity force is only working on the hanging section of the rope. The work done by a force is given by the general expression:

W_{AB}=\int_{A}^{B} \vec{F} \cdot \vec{dl}

where $$A$$ and $$B$$ are respectively the initial and final position of the object moved by the force. Since any point of the rope is moving at the same velocity we can consider the motion of the hanging end only.

Since our coordinate origin was located on the table’s surface in the downwards direction, the work done by the gravity force on the rope is:

\begin{align}
W_{aL}&=\int_{a}^{L} \vec{F} \cdot \vec{dy}\\
&=\int_{a}^{L} (mg\hat{y})\cdot (dy\hat{y})\\
&=\int_{a}^{L} g\lambda ydy
\end{align}

And the final expression for the work is:

W_{aL}=\frac{\lambda g}{2} (L^2-a^2)

Now let’s calculate the change of kinetic energy of the falling rope. Since the movement starts from rest, the initial kinetic energy is $$0$$, and at the detachment instant the kinetic energy is

\begin{align}
E_d&=\frac{1}{2}m_LV_{d}^2\\
&=\frac{1}{2} \lambda LV_{d}^2
\end{align}

This energy equals the change in kinetic energy of the rope which must be equal to the work done by the gravity force on the rope. In this way, we can find $$V_{d}$$.

\frac{1}{2} \lambda LV_{d}^2=\frac{\lambda g}{2} \left(L^2-a^2\right)

\boxed{{V}_d=\sqrt{g\left(L-\frac{a^2}{L}\right)}}

## Velocity of detachment of a rope falling down a table – Second Law of Newton solution

The third way of solving this problem is no doubt the more complex one and less intuitive, but it is a great opportunity for reviewing concepts and acquiring tools and training for solving problems.

The Second Law of Newton states that the rate of change of the linear momentum equals the total external force applied to the body. To analyze all the forces acting on the rope let’s take each section separately.

Before writing the equations of the Second Law of Newton, let’s set the system of coordinates. For each section of the rope separately, the movement is in one direction, therefore for the hanging rope, we set just one axis downwards. Assuming the rope is not elastic, all of its points are moving with the same velocity so that we just need to track the movement of the end of the rope which is at a distance $$y$$ at any moment.

For the horizontal section of the rope, we could set the axis wherever we want, but fixing the origin at a distance $$L$$ from the table’s edge makes the analysis and calculation easier. Also, for this section, we only need to track the movement of one point and for the sake of simplicity, the best choice is the left end.

Thanks to the convenient way of choosing the axes, it is clear that $$x$$ and $$y$$ are equal at any time as seen in the above figure. After identifying the forces and setting the axes we now can apply the Second Law of Newton for both sections of the rope.

\frac{d}{dt}(m_xV_x)=T

\frac{d}{dt}(m_yV_y)=m_yg-T

Notice that the mass is not constant in time for each section and it is a function of the distance covered by the ends of the rope, then $$m_x =\lambda (L-x)$$ and $$m_y=\lambda y$$. Since both sections are moving with the same velocity $$V_x=V_y=V$$. Therefore, the resulting equations are:

\frac{d}{dt}(\lambda (L-x)V)=T

\frac{d}{dt}(\lambda yV)=\lambda yg-T

Since we are interested in the velocity at the detachment, let’s replace the tension $$T$$ in equation (15) knowing that $$x=y$$.

\frac{d}{dt}(\lambda yV)=\lambda yg-\frac{d}{dt}(\lambda (L-y)V)

\lambda \frac{d}{dt}[yV+(L-y)V]=\lambda yg

\frac{dV}{dt}=\frac{g}{L}y

In order to find the velocity of detachment, we will apply a method frequently used in this kind of equations. The goal is to transform the equation into a new one that allows us to remove the dependency of time and integrate using the known data. Multiplying both sides of the equations by $$2V$$ and replacing $$V$$ by $$\displaystyle{\frac{dy}{dt}}$$ the solution is straightforward.

$2V\frac{dV}{dt}=2\frac{g}{L}yV$

$\frac{dV^2}{dt}=2\frac{g}{L}y\frac{dy}{dt}$

$dV^2=2\frac{g}{L}ydy$

Taking $$V^2$$ as one variable we can integrate between the initial and final states.

$\int_{0}^{V_d^2}\,dV^2=2\frac{g}{L}\int_{a}^{L} y\,dy$

$V_d^2=\frac{g}{L}y^2 \Biggr|_{a}^{L}$

\boxed{{V}_d=\sqrt{g\left(L-\frac{a^2}{L}\right)}}

After reading this solution many could think: “That solution and its mathematical trick would never come to mind in a hundred years” and that thought is quite normal. These kinds of solutions “don’t come to your mind, you just have to learn them beforehand”, that’s why training in solving many problems and learning solved examples is an essential part for mastering the concepts and getting mental ability in physics.

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