Kvant Physics Problem 171
The system consists of two identical point masses of mass $m$ connected by a rigid, massless rod of length $l$, forming a dumbbell initially oriented vertically.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m17s
Source on kvant.digital
Problem
A dumbbell consisting of a weightless rod with two identical small balls at its ends is placed vertically on a smooth horizontal table (see Fig. 3). The upper ball is given a horizontal velocity $\overrightarrow{v}$ by an impulse. For what maximum length of the dumbbell $l$ will the lower ball immediately lose contact with the table?
Figure 3
A. R. Zilberman
Setup and Assumptions
The system consists of two identical point masses of mass $m$ connected by a rigid, massless rod of length $l$, forming a dumbbell initially oriented vertically. The lower mass is in contact with a smooth horizontal table, and the upper mass is located directly above it along the vertical direction.
At the initial moment, a horizontal impulse is applied only to the upper mass, giving it velocity $v$ in a direction parallel to the table. The table is smooth, so no frictional force acts in the horizontal direction. Gravity acts uniformly with acceleration $g$ downward.
The rod is perfectly rigid and massless, so both masses share the same rigid-body translational and rotational motion immediately after the impulse. The contact force from the table acts only on the lower mass and is denoted by $N$. The unknown is the maximum rod length $l_{\max}$ for which the lower mass loses contact with the table immediately after the impulse.
The analysis is carried out in the inertial reference frame of the table.
Physical Principles
The motion of the system immediately after the impulse is determined by conservation of linear momentum and conservation of angular momentum about the center of mass, since the impulse is external and instantaneous.
The center of mass velocity is given by
$$\mathbf{V}_{\mathrm{cm}} = \frac{\mathbf{P}}{2m},$$
where $\mathbf{P}$ is the total linear momentum imparted by the impulse.
The angular momentum about the center of mass is
$$\mathbf{L}_{\mathrm{cm}} = I \boldsymbol{\omega},$$
where $I$ is the moment of inertia of the system about the center of mass and $\boldsymbol{\omega}$ is the angular velocity.
The rigid-body acceleration of a point at position $\mathbf{r}$ relative to the center of mass is
$$\mathbf{a} = \mathbf{a}_{\mathrm{cm}} + \boldsymbol{\alpha} \times \mathbf{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}).$$
Newton’s second law applied to the lower mass in the vertical direction gives
$$N - mg = m a_y,$$
where upward is taken as positive.
Loss of contact corresponds to $N = 0$.
Derivation
The impulse $J$ applied to the upper mass satisfies $J = m v$, since it produces the velocity $v$ of that mass immediately after impact. The total momentum of the system is therefore
$$\mathbf{P} = m v ,\mathbf{i}.$$
The center of mass velocity follows as
$$\mathbf{V}_{\mathrm{cm}} = \frac{m v}{2m} \mathbf{i} = \frac{v}{2}\mathbf{i}.$$
The center of mass is located at the midpoint of the rod. Taking the origin at the center of mass, the position vectors of the upper and lower masses are
$$\mathbf{r}{u} = \frac{l}{2}\mathbf{k}, \quad \mathbf{r}{\ell} = -\frac{l}{2}\mathbf{k}.$$
The angular momentum about the center of mass is produced only by the impulse on the upper mass,
$$\mathbf{L}{\mathrm{cm}} = \mathbf{r}{u} \times (m v \mathbf{i}) = \frac{l}{2} m v (\mathbf{k} \times \mathbf{i}) = \frac{l m v}{2}\mathbf{j}.$$
The moment of inertia of two point masses about the center of mass is
$$I = 2m\left(\frac{l}{2}\right)^2 = \frac{m l^2}{2}.$$
The angular velocity follows from $\mathbf{L}_{\mathrm{cm}} = I \boldsymbol{\omega}$,
$$\boldsymbol{\omega} = \frac{\mathbf{L}_{\mathrm{cm}}}{I} = \frac{\frac{l m v}{2}}{\frac{m l^2}{2}} \mathbf{j} = \frac{v}{l}\mathbf{j}.$$
The angular acceleration is zero at the initial instant because gravity exerts no torque about the center of mass, since both weights act along the rod line passing through the center of mass.
The vertical acceleration of the lower mass is obtained from rigid-body kinematics,
$$\mathbf{a}{\ell} = \mathbf{a}{\mathrm{cm}} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{\ell}).$$
The center of mass has no initial vertical acceleration, so $\mathbf{a}_{\mathrm{cm},y} = 0$. The centripetal term is evaluated step by step. First,
$$\boldsymbol{\omega} \times \mathbf{r}_{\ell} = \frac{v}{l}\mathbf{j} \times \left(-\frac{l}{2}\mathbf{k}\right) = -\frac{v}{2}(\mathbf{j} \times \mathbf{k}) = -\frac{v}{2}\mathbf{i}.$$
Then,
$$\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{\ell}) = \frac{v}{l}\mathbf{j} \times \left(-\frac{v}{2}\mathbf{i}\right) = -\frac{v^2}{2l}(\mathbf{j} \times \mathbf{i}) = \frac{v^2}{2l}\mathbf{k}.$$
Thus the lower mass has upward acceleration
$$a_y = \frac{v^2}{2l}.$$
Applying Newton’s second law in the vertical direction,
$$N - mg = m\frac{v^2}{2l}.$$
Solving for the normal force,
$$N = mg + m\frac{v^2}{2l}.$$
The condition for immediate loss of contact is $N \le 0$. The limiting case corresponds to $N = 0$, giving
$$0 = mg + m\frac{v^2}{2l}.$$
Rearranging for $l$ yields the critical condition
$$l = \frac{v^2}{2g}.$$
Result
The maximum rod length for which the lower mass can lose contact immediately is
$$l_{\max} = \frac{v^2}{2g}.$$
Substituting units, $v^2$ has units $\mathrm{m^2/s^2}$ and $g$ has units $\mathrm{m/s^2}$, so $l$ has units $\mathrm{m}$.
$$\boxed{l_{\max} = \frac{v^2}{2g}\ \mathrm{m}}$$
Sanity Checks
Dimensional consistency follows from $[v^2/g] = (\mathrm{m^2/s^2})/(\mathrm{m/s^2}) = \mathrm{m}$.
In the limit $v \to 0$, the critical length tends to zero, consistent with the fact that without horizontal impulse no rotational effects arise and no contact loss is possible.
In the limit of large $g$, the critical length decreases, reflecting that stronger gravity suppresses separation from the table.
The expression depends inversely on $l$ through the induced angular velocity magnitude $\omega = v/l$, so any sign error in the cross products would reverse the direction of the centripetal acceleration term and incorrectly change the contact condition.