Kvant Physics Problem 42

Verified: yes
Verdicts: PASS + PASS
Solve time: 10m01s
Source on kvant.digital

Problem

A rope is passed over a pulley, with part of the rope lying on a table and part on the floor. After the rope is released, it begins to move. Find the speed of the rope in its steady uniform motion. The height of the table is $h$.

Setup and Assumptions

A flexible rope passes over an ideal pulley. One part of the rope lies on a horizontal table whose surface is at height $h$ above the floor. The remaining part hangs vertically from the table edge to the floor. The rope is released and begins to move. After some time it reaches a regime of steady uniform motion, in which the speed $v$ of every element of rope is constant. The rope has uniform linear mass density $\lambda$. The pulley is assumed massless and frictionless. The rope is inextensible and perfectly flexible. Air resistance is neglected. Friction between the rope and the table is neglected. The motion is observed in an inertial reference frame fixed to the Earth.

In the steady regime the shape of the rope does not change, and the linear speed of all rope elements is constant. The mass flow rate of rope passing over the table edge is uniform and given by

$$\dot m=\lambda v.$$

The problem is to find the steady speed $v$ of the rope.

Physical Principles

The solution is based on the principle of conservation of mechanical energy for an open system. Choose as the system all rope that is moving above the floor, namely the portion on the table together with the hanging vertical portion. Rope continuously leaves this system at the floor and joins the stationary pile there.

The kinetic energy of the chosen system is

$$K=\frac12\int \lambda v^2,dl.$$

In the steady regime, the speed $v$ is constant and the shape and length of the moving rope are unchanged. Hence $K$ is constant and

$$\frac{dK}{dt}=0.$$

The only external force doing work on the moving rope is gravity. The table exerts only vertical reactions on rope elements moving horizontally, so its power is zero. The pulley is ideal and does no net work.

During a time interval $dt$, a mass

$$dm=\dot m,dt$$

moves from the table edge to the floor. Its gravitational potential energy decreases by

$$dE_g=dm,gh.$$

Thus the power supplied by gravity is

$$P_g=\dot m,gh.$$

When the same mass element reaches the floor, it is brought from speed $v$ to rest and becomes part of the stationary pile. The kinetic energy removed from the moving rope is

$$dE_{\rm diss}=\frac12,dm,v^2,$$

so the corresponding dissipation rate is

$$\dot E_{\rm diss}=\frac12,\dot m,v^2.$$

For the open system consisting of the moving rope, the energy balance is

$$\frac{dK}{dt}=P_g-\dot E_{\rm diss}.$$

Substituting the steady-state condition $\frac{dK}{dt}=0$ gives

$$P_g=\dot E_{\rm diss}.$$

Hence

$$\dot m,gh=\frac12,\dot m,v^2.$$

This establishes the required power balance from the open-system energy equation rather than assuming it.

Derivation

During a small interval of time $dt$, a length

$$dl=v,dt$$

of rope passes over the table edge and descends to the floor. The corresponding mass is

$$dm=\lambda,dl=\lambda v,dt.$$

The mass flow rate is

$$\dot m=\frac{dm}{dt}=\lambda v.$$

The gravitational potential energy lost by this mass while descending through height $h$ is

$$dE_g=dm,gh =\lambda vgh,dt.$$

Hence the gravitational power is

$$P_g=\frac{dE_g}{dt} =\lambda vgh =\dot m,gh.$$

The kinetic energy carried by the same mass element immediately before it reaches the floor is

$$dE_k=\frac12,dm,v^2 =\frac12,\lambda v^3,dt.$$

Since the element comes to rest on the floor, this kinetic energy is dissipated. The dissipation rate is

$$\dot E_{\rm diss} =\frac{dE_k}{dt} =\frac12,\dot m,v^2.$$

Using the steady-state energy balance

$$0=\frac{dK}{dt} =P_g-\dot E_{\rm diss},$$

gives

$$\dot m,gh = \frac12,\dot m,v^2.$$

Cancelling the nonzero factor $\dot m$ yields

$$gh=\frac12,v^2.$$

Solving for $v$,

$$v=\sqrt{2gh}.$$

Result

The steady speed of the rope is

$$\boxed{v=\sqrt{2gh}}.$$

Substituting $g=9.81,\mathrm{m/s^2}$ gives

$$v=\sqrt{19.62,h},\mathrm{m/s},$$

with $h$ measured in metres. For example, if $h=1.00,\mathrm{m}$,

$$v=\sqrt{19.62}\approx 4.43,\mathrm{m/s}.$$

Sanity Checks

The quantity $gh$ has dimensions

$$[gh] = \mathrm{m/s^2}\cdot \mathrm{m} = \mathrm{m^2/s^2},$$

$$\sqrt{gh}$$

has dimensions of velocity.

If $h\to0$, the gravitational power input tends to zero, and the formula gives $v\to0$.

The dependence

$$v\propto \sqrt{h}$$

matches the fact that each unit mass loses gravitational energy $gh$ while descending from the table to the floor.

The result coincides with the free-fall speed from height $h$,

$$v=\sqrt{2gh},$$

which is consistent with the steady-state energy balance: each unit mass releases potential energy $gh$ and ultimately dissipates kinetic energy $\frac12 v^2$ when it joins the pile on the floor.