Kvant Physics Problem 42

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m30s
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 steady uniform motion of the rope. The height of the table is $h$.

Setup and Assumptions

A uniform rope of linear mass density $\lambda$ (kg/m) passes over a frictionless pulley. One part of the rope lies on a horizontal table, and another part hangs vertically down to the floor, which is a height $h$ below the table surface.

After release, the rope reaches a steady regime of uniform motion with constant speed $v$ (m/s). The motion is taken to be one-dimensional along the rope.

The pulley is assumed ideal, with no friction and no rotational inertia. The rope is inextensible and does not slip on the pulley. Air resistance is neglected.

When a segment of rope reaches the floor, it is assumed to come to rest immediately relative to the floor, so its kinetic energy is irreversibly lost at impact. This dissipation is the mechanism that allows a steady state with constant speed.

The goal is to determine the constant speed $v$ in terms of $h$ and the gravitational acceleration $g$ (m/s$^2$).

Physical Principles

The gravitational potential energy change per unit time equals the rate of energy dissipation in the steady regime.

The mass flow rate through any cross section of the rope is $\dot{m} = \lambda v$.

The gravitational potential energy lost per unit mass when descending height $h$ is $gh$.

The kinetic energy per unit mass of the moving rope is $\frac{v^2}{2}$.

Energy conservation in steady motion requires that the power supplied by gravity equals the power dissipated upon inelastic stopping of the rope segments.

Derivation

In the steady regime, a length $v,dt$ of rope passes from the tabletop region through the pulley and ultimately reaches the floor in time $dt$. The corresponding mass is

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

Each element of this mass descends by height $h$, so the gravitational potential energy decrease per unit time is

$$\frac{dE_p}{dt} = \lambda v g h.$$

When this mass segment reaches the floor, it arrives with speed $v$ and loses its kinetic energy upon impact. The kinetic energy carried per unit time is

$$\frac{dE_k}{dt} = \frac{1}{2} \lambda v^3.$$

In steady motion, the rate at which gravitational energy is supplied equals the rate at which mechanical energy is dissipated:

$$\lambda v g h = \frac{1}{2} \lambda v^3.$$

Canceling $\lambda v$ from both sides yields

$$g h = \frac{v^2}{2}.$$

Solving for $v$ gives

$$v^2 = 2gh, \quad v = \sqrt{2gh}.$$

Result

The steady uniform speed is

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

Substituting SI units, $g = 9.81,\text{m/s}^2$ and $h$ in meters,

$$v = \sqrt{2 \cdot 9.81,\text{m/s}^2 \cdot h,\text{m}}.$$

Thus,

$$\boxed{v = \sqrt{19.62,h}\ \text{m/s}}.$$

Sanity Checks

Dimensional consistency follows since $gh$ has units $\text{m}^2/\text{s}^2$, so $\sqrt{2gh}$ has units $\text{m/s}$.

If $h \to 0$, the speed $v \to 0$, consistent with the absence of gravitational driving. If $h$ increases, the speed grows as $\sqrt{h}$, matching the behavior of motion under constant gravitational potential drop per unit length.

The balance equation equates power $\lambda vgh$ with $\frac{1}{2}\lambda v^3$, so any missing factor of $1/2$ in kinetic energy would directly double the predicted speed, making it the most sensitive point for error.