Kvant Physics Problem 23

Verified: yes
Verdicts: PASS + PASS
Solve time: 13m51s
Source on kvant.digital

Problem

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

III All-Union Physics Olympiad (1969)

Setup and Assumptions

Two identical thin-walled tubes, each of mass $m$ and radius $R$, lie on a horizontal plane. One tube is initially at rest. The other moves toward it while rolling without slipping. Its center-of-mass velocity is $v_0$, and its angular velocity is

$$\omega_0=\frac{v_0}{R}.$$

The tubes are thin-walled rings, so the moment of inertia of each tube about its axis is

$$I=mR^2.$$

The collision between the tubes is perfectly elastic. Friction during the very short impact between the tubes is negligible, so the impulse acts only along the line joining their centers. The axes of the tubes remain parallel throughout the motion.

The coefficient of kinetic friction between each tube and the horizontal plane is $k$. Air resistance is neglected. The gravitational acceleration is $g$.

The positive direction is chosen as the initial direction of motion of the rolling tube.

The task is to determine the translational velocities $v_1(t)$ and $v_2(t)$ and the angular velocities $\omega_1(t)$ and $\omega_2(t)$ of the two tubes, and to describe their graphs.

Physical Principles

The collision is perfectly elastic and the masses are equal. Since the impact force acts through the centers of the tubes and friction between the tubes is negligible, the angular velocities do not change during the collision.

For a one-dimensional elastic collision of equal masses, the translational velocities are exchanged.

During subsequent sliding on the horizontal plane, the kinetic friction force has magnitude

$$F_f=kmg.$$

The translational motion obeys

$$m\frac{dv}{dt}=F_f .$$

The rotational motion obeys

$$I\frac{d\omega}{dt}=M_f,$$

where $M_f$ is the torque of friction about the tube axis.

For a thin-walled tube,

$$I=mR^2.$$

Rolling without slipping corresponds to

$$v=R\omega .$$

Derivation

Before the collision, tube 1 rolls without slipping:

$$v_1=v_0, \qquad \omega_1=\frac{v_0}{R}.$$

Tube 2 is at rest:

$$v_2=0, \qquad \omega_2=0.$$

Because the impact is elastic and the masses are equal, the translational velocities are exchanged:

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

The impact force passes through the centers, so no angular impulse is produced. Hence

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

Immediately after the collision, tube 1 has pure rotation and tube 2 has pure translation.

Consider tube 1 first.

Its bottom point moves relative to the ground to the left with speed $R\omega_1$, so kinetic friction acts to the right. The translational equation is

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

hence

$$\frac{dv_1}{dt}=kg.$$

The friction torque opposes the rotation:

$$mR^2\frac{d\omega_1}{dt}=-kmgR,$$

which gives

$$\frac{d\omega_1}{dt} = -\frac{kg}{R}.$$

Integrating with the initial conditions,

$$v_1(t)=kg,t,$$

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

Sliding ends when rolling without slipping is established:

$$v_1=R\omega_1.$$

Substituting the expressions above,

$$kg,t = v_0-kg,t.$$

Thus

$$t_*=\frac{v_0}{2kg}.$$

At this instant,

$$v_1(t_*) = \frac{v_0}{2},$$

$$\omega_1(t_*) = \frac{v_0}{2R}.$$

After $t=t_*$, friction vanishes and both quantities remain constant.

Now consider tube 2.

Immediately after impact it translates without rotating. The bottom point slides to the right, so friction acts to the left.

The translational equation is

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

hence

$$\frac{dv_2}{dt}=-kg.$$

The friction torque produces clockwise rotation:

$$mR^2\frac{d\omega_2}{dt} = kmgR,$$

so

$$\frac{d\omega_2}{dt} = \frac{kg}{R}.$$

Integrating,

$$v_2(t)=v_0-kg,t,$$

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

Rolling begins when

$$v_2=R\omega_2.$$

Substitution gives

$$v_0-kg,t = kg,t,$$

hence again

$$t_*=\frac{v_0}{2kg}.$$

At that instant,

$$v_2(t_*) = \frac{v_0}{2},$$

$$\omega_2(t_*) = \frac{v_0}{2R}.$$

For $t>t_*$, friction disappears and the motion remains uniform:

$$v_2=\frac{v_0}{2}, \qquad \omega_2=\frac{v_0}{2R}.$$

The complete time dependence is therefore

$$v_1(t)= \begin{cases} kg,t, & 0\le t\le t_,\[4pt] \dfrac{v_0}{2}, & t\ge t_, \end{cases}$$

$$\omega_1(t)= \begin{cases} \dfrac{v_0}{R}-\dfrac{kg}{R}t, & 0\le t\le t_,\[8pt] \dfrac{v_0}{2R}, & t\ge t_, \end{cases}$$

$$v_2(t)= \begin{cases} v_0-kg,t, & 0\le t\le t_,\[4pt] \dfrac{v_0}{2}, & t\ge t_, \end{cases}$$

$$\omega_2(t)= \begin{cases} \dfrac{kg}{R}t, & 0\le t\le t_,\[8pt] \dfrac{v_0}{2R}, & t\ge t_. \end{cases}$$

Result

The duration of sliding after the collision is

$$\boxed{ t_*=\frac{v_0}{2kg} }.$$

For $0\le t\le t_*$,

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

and

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

For $t\ge t_*$,

$$\boxed{ v_1=v_2=\frac{v_0}{2} }$$

and

$$\boxed{ \omega_1=\omega_2=\frac{v_0}{2R} }.$$

Since no numerical values are given, the answer remains symbolic. The graphs consist of straight-line segments. The velocity $v_1$ rises linearly from $0$ to $v_0/2$, while $v_2$ decreases linearly from $v_0$ to $v_0/2$. The angular velocity $\omega_1$ decreases linearly from $v_0/R$ to $v_0/(2R)$, while $\omega_2$ increases linearly from $0$ to $v_0/(2R)$. All four curves become horizontal at $t=t_*$.

Sanity Checks

The dimensions are correct. The quantity $kg$ has dimensions of acceleration, so $v_0/(kg)$ has dimensions of time. The angular accelerations have magnitude $kg/R$, which has dimensions of $\mathrm{s}^{-2}$.

In the limit $k\to0$, the frictional acceleration vanishes and

$$t_*\to\infty.$$

The tubes never reach a common rolling state, which matches the absence of friction.

In the limit of very large $k$, the time $t_*$ becomes very short, corresponding to rapid establishment of rolling without slipping.

The total linear momentum immediately after the collision is

$$mv_0,$$

and after the sliding phase it is

$$m\frac{v_0}{2}+m\frac{v_0}{2} = mv_0,$$

showing that the horizontal friction forces from the ground cancel when both tubes are considered together.

The easiest place for a sign error is the determination of the friction direction for each tube. Reversing either friction force changes the signs of the translational and angular accelerations and leads to a violation of the rolling condition at the end of the sliding interval.