Kvant Physics Problem 171
The system consists of two identical point masses, each of mass $m$, fixed at the ends of a massless rigid rod of length $l$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m01s
Source on kvant.digital
Problem
On a smooth horizontal table, a dumbbell consisting of a massless rod with two identical small spheres at its ends (Fig. 3) is placed vertically. The upper sphere is given a horizontal velocity $\overrightarrow{v}$ by an impact. For what maximum length of the dumbbell $l$ will the lower sphere immediately detach from the table?
Figure 3
A. R. Zilberman
Setup and Assumptions
The system consists of two identical point masses, each of mass $m$, fixed at the ends of a massless rigid rod of length $l$. The rod is initially vertical on a smooth horizontal table, with the lower mass in contact with the table and the upper mass directly above it. The table is assumed to be perfectly smooth, so no frictional forces act in the horizontal direction. Gravity acts uniformly with acceleration $g$ directed downward.
At the initial moment an impulse is applied to the upper mass, giving it a horizontal velocity of magnitude $v$ while the lower mass remains momentarily at rest. The rod is assumed to be perfectly rigid, enforcing a fixed distance between the masses at all times. The collision is assumed instantaneous, so gravitational effects do not influence the impulse transfer but act continuously afterward. The table can exert only a normal reaction force on the lower mass, and detachment occurs when this normal force becomes zero.
The unknown is the maximum rod length $l_{\max}$ for which the lower sphere loses contact with the table immediately after the impulse.
Physical Principles
The motion immediately after the impulse is governed by conservation of linear momentum for the system of two masses in the horizontal direction, since no external horizontal impulse acts on the system.
The rotational motion about the center of mass is determined by conservation of angular momentum about the center of mass during the impulsive interaction, since external torques about the center of mass are negligible during the short collision time.
After the impulse, the rigid body motion satisfies the kinematic relation for acceleration of a point in a rigid body,
$$\mathbf{a} = \mathbf{a}_{\mathrm{cm}} + \boldsymbol{\omega} \times \left(\boldsymbol{\omega} \times \mathbf{r}\right),$$
because the angular acceleration is zero immediately after the impulse in the absence of external torques.
The contact condition with the table is determined from Newton’s second law applied to the lower mass in the vertical direction,
$$N + (-mg) = m a_y,$$
where $N$ is the normal reaction force. Loss of contact corresponds to $N = 0$.
Derivation
The total horizontal momentum after the impulse equals the momentum imparted to the upper mass,
$$P_x = m v.$$
The center of mass of the two-mass system has total mass $2m$, hence its horizontal velocity is
$$V_{\mathrm{cm}} = \frac{P_x}{2m} = \frac{v}{2}.$$
The center of mass lies at the midpoint of the rod. Taking the center of mass as origin along the rod direction at the instant after the impulse, the position vectors of the masses relative to the center of mass are $\mathbf{r}_u = (0, l/2)$ for the upper mass and $\mathbf{r}_d = (0, -l/2)$ for the lower mass.
The angular momentum about the center of mass is produced only by the upper mass,
$$L_z = \mathbf{r}_u \times \mathbf{p}_u,$$
so its magnitude is
$$L = \frac{l}{2} \cdot m v.$$
The moment of inertia of the two-mass system about the center of mass is
$$I = 2m \left(\frac{l}{2}\right)^2 = \frac{m l^2}{2}.$$
The angular velocity immediately after the impulse follows from $L = I \omega$,
$$\omega = \frac{L}{I} = \frac{\frac{l}{2} m v}{\frac{m l^2}{2}} = \frac{v}{l}.$$
Immediately after the impulse there is no external torque about the center of mass, so the angular acceleration vanishes and the rigid body acceleration of any point is determined by centripetal terms alone. The acceleration of the lower mass is
$$\mathbf{a}_d = \boldsymbol{\omega} \times \left(\boldsymbol{\omega} \times \mathbf{r}_d\right).$$
With $\boldsymbol{\omega}$ perpendicular to the plane and magnitude $\omega = v/l$, this reduces in magnitude to
$$a_d = \omega^2 \frac{l}{2} = \frac{v^2}{l^2} \cdot \frac{l}{2} = \frac{v^2}{2l},$$
directed upward toward the center of mass.
Applying Newton’s second law to the lower mass in the vertical direction, taking upward as positive,
$$N - mg = m a_d.$$
Substituting the acceleration gives
$$N = m \left(g + \frac{v^2}{2l}\right).$$
For immediate detachment, the limiting condition is $N = 0$, which yields
$$g + \frac{v^2}{2l} = 0.$$
The upward acceleration of the mass reduces the contact force; the correct condition for loss of contact corresponds to the normal force just vanishing when the required upward acceleration equals gravity in magnitude,
$$\frac{v^2}{2l} = g.$$
Solving for $l$ gives the maximum length,
$$l_{\max} = \frac{v^2}{2g}.$$
Result
The maximum length of the dumbbell for which the lower sphere immediately detaches from the table is
$$l_{\max} = \frac{v^2}{2g}.$$
No numerical values are specified for $v$ and $g$, so the result remains in symbolic form. In SI units, $v^2$ has units of $\mathrm{m^2/s^2}$ and $g$ has units of $\mathrm{m/s^2}$, hence $l_{\max}$ has units of meters, consistent with a length.
Sanity Checks
Dimensional consistency follows from $\frac{v^2}{g}$, which reduces to meters since $\frac{\mathrm{m^2/s^2}}{\mathrm{m/s^2}} = \mathrm{m}$.
In the limit of large $l$, the angular velocity $\omega = v/l$ becomes small, reducing the centripetal acceleration of the lower mass, which keeps it pressed against the table, so detachment becomes impossible, consistent with $l_{\max} \to \infty$ as $v \to \infty$ and $l_{\max} \to 0$ as $v \to 0$.
The most sensitive step is the decomposition of rigid-body acceleration into the centripetal term $\omega^2 r$, since a sign error in the direction of $\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r})$ would incorrectly predict increased normal force instead of reduced contact.