Kvant Physics Problem 23

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Verdicts: SKIP + SKIP
Solve time: 1m15s
Source on kvant.digital

Problem

On a horizontal plane there are two identical thin-walled tubes, each of mass $m$. Their axes are parallel, and their radii are equal to $R$. Initially, one of the tubes is at rest, and the other rolls without slipping toward the first until collision. The translational speed of the tube is $v_0$. How do the translational and angular velocities of the tubes depend on time (draw graphs)? The coefficient of sliding friction between the tubes and the horizontal surface is $k$; friction between the tubes during the collision is negligible, and the collision is perfectly elastic.

III All-Union Physics Olympiad (1969)

Setup and Assumptions

Two identical thin-walled cylindrical tubes of mass $m$ and radius $R$ move on a horizontal rough plane. The coefficient of kinetic friction between each tube and the plane is $k$, and the normal force equals $mg$, so the maximal friction force magnitude is $kmg$. Each tube has moment of inertia $I = mR^2$ about its symmetry axis.

At the initial moment tube 1 moves to the right with translational speed $v_0$ while rolling without slipping, so its angular velocity satisfies $\omega_0 = -\frac{v_0}{R}$ when counterclockwise is taken as positive. Tube 2 is at rest.

The collision between tubes is central, occurs along the line of centers in the horizontal direction, is perfectly elastic, and produces no tangential impulse and no frictional torque between tubes. The collision duration is negligible compared with the time scale of frictional evolution, so velocities change instantaneously only at the collision moment.

The unknowns are the translational velocities $v_1(t), v_2(t)$ and angular velocities $\omega_1(t), \omega_2(t)$ as functions of time, together with their qualitative time graphs.

Physical Principles

Newton’s second law for translation gives $m a = F_{\text{ext}}$, where $F_{\text{ext}}$ is kinetic friction when sliding occurs.

Newton’s second law for rotation about the center of mass gives $I \alpha = \tau$, where $\tau$ is the torque of the friction force acting at the contact point.

For kinetic friction, the magnitude is $F_f = kmg$, directed opposite to the velocity of the contact point relative to the ground.

For a perfectly elastic collision of identical masses in one dimension, translational velocities are exchanged while angular velocities remain unchanged because the impulsive force acts through the line of centers and produces no torque about each tube’s center.

The rolling without slipping condition is $v + \omega R = 0$, which may be temporarily violated after the collision, leading to sliding and kinetic friction.

Derivation

Before the collision, tube 1 rolls without slipping. Static friction adjusts to enforce $v_1 + \omega_1 R = 0$, and does no work, so translational motion is uniform. Thus

$v_1(t) = v_0, \qquad \omega_1(t) = -\frac{v_0}{R},$

while tube 2 remains at rest,

$v_2(t) = 0, \qquad \omega_2(t) = 0.$

Let the collision occur at time $t = t_c$. For identical masses and a perfectly elastic central collision,

$v_1(t_c^+) = 0, \qquad v_2(t_c^+) = v_0,$

while angular velocities remain unchanged,

$\omega_1(t_c^+) = -\frac{v_0}{R}, \qquad \omega_2(t_c^+) = 0.$

Immediately after collision, tube 1 has zero translational velocity but nonzero rotation, so the contact point moves relative to the ground with speed $v_{\text{rel},1} = \omega_1 R = -v_0$, implying slip. The friction force acts to the right with magnitude $kmg$. Hence for tube 1,

$m \frac{dv_1}{dt} = kmg,$

giving

$v_1(t) = kgt.$

The torque about the center is positive since friction at the bottom produces counterclockwise rotation,

$I \frac{d\omega_1}{dt} = kmgR,$

so with $I = mR^2$,

$\frac{d\omega_1}{dt} = \frac{kg}{R},$

and therefore

$\omega_1(t) = -\frac{v_0}{R} + \frac{kg}{R}t.$

For tube 2 immediately after collision, the bottom point moves right with speed $v_0$, so friction acts to the left with magnitude $kmg$. Thus,

$m \frac{dv_2}{dt} = -kmg,$

which yields

$v_2(t) = v_0 - kgt.$

The torque is negative,

$I \frac{d\omega_2}{dt} = -kmgR,$

$\omega_2(t) = -\frac{kg}{R}t.$

For tube 1 the slip velocity of the contact point is

$u_1(t) = v_1(t) + \omega_1(t)R = kgt - v_0 + kgt = 2kgt - v_0,$

which changes sign at

$t_1 = \frac{v_0}{2kg}.$

For tube 2 the slip velocity is

$u_2(t) = v_2(t) + \omega_2(t)R = v_0 - kgt - kgt = v_0 - 2kgt,$

which also changes sign at

$t_2 = \frac{v_0}{2kg}.$

At $t = t_1 = t_2$, friction reverses direction for both tubes. After this moment, the same equations hold with opposite sign of friction, leading to symmetric deceleration and spin adjustment, but the key qualitative behavior remains that both tubes undergo piecewise linear evolution of $v(t)$ and $\omega(t)$ with a single switching time determined by $v_0/(2kg)$.

Result

Before collision at $t < t_c$,

$v_1(t) = v_0, \qquad \omega_1(t) = -\frac{v_0}{R}, \qquad v_2(t) = 0, \qquad \omega_2(t) = 0.$

At collision $t = t_c$,

$v_1(t_c^+) = 0, \qquad \omega_1(t_c^+) = -\frac{v_0}{R}, \qquad v_2(t_c^+) = v_0, \qquad \omega_2(t_c^+) = 0.$

For $t_c < t < t_c + \frac{v_0}{2kg}$,

$v_1(t) = kg(t - t_c), \qquad \omega_1(t) = -\frac{v_0}{R} + \frac{kg}{R}(t - t_c),$

$v_2(t) = v_0 - kg(t - t_c), \qquad \omega_2(t) = -\frac{kg}{R}(t - t_c).$

The key switching time for friction reversal is

$\Delta t = \frac{v_0}{2kg}.$

Sanity Checks

Dimensional consistency holds since $kg$ has units of acceleration, so $kg,t$ has units of velocity and $\frac{kg}{R}$ has units of angular acceleration.

The collision conserves momentum and kinetic energy in translation for identical masses with an elastic central impact, consistent with velocity exchange.

The friction-induced acceleration magnitude $kg$ is independent of mass, matching kinetic friction behavior on a horizontal plane.

The angular acceleration scale $\frac{kg}{R}$ correctly scales inversely with radius, so larger tubes rotate more slowly under the same friction force.

The slip velocity $u(t)$ evolves linearly in time and changes sign at a finite time, which identifies the instant when friction direction reverses; any sign error in torque direction would eliminate this root or shift it incorrectly, making it the most sensitive step in the derivation.