Kvant Physics Problem 8
A long rigid rod $AB$ is connected at end $A$ to a vertical screw by a threaded hole.
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Solve time: 1m08s
Source on kvant.digital
Problem
A long rod $AB$ with a threaded hole at its end is screwed onto a vertical screw (see Fig. 2). The rod is released. Friction between the screw and the rod is negligible. How will the rod move after it comes off the screw?
Fig. 2
MIEM Physico-Mathematical Olympiad
Setup and Assumptions
A long rigid rod $AB$ is connected at end $A$ to a vertical screw by a threaded hole. The screw axis is vertical and fixed in space. The rod is initially engaged with the thread and then released from external constraints except for the contact interaction with the screw while the thread engagement persists.
The motion occurs under gravity, but friction between the rod and the screw thread is neglected. The screw is assumed to impose a holonomic geometric constraint corresponding to a helix of pitch $h$ per full revolution. The rod is treated as a rigid body of mass $m$, with its center of mass located on the screw axis while engaged.
At the instant the rod loses contact with the screw, internal constraint forces vanish and the rod becomes a free rigid body. The unknown is the subsequent translational and rotational motion of the rod after detachment.
The reference frame is an inertial frame fixed to the Earth, with the vertical axis aligned with the screw axis.
Physical Principles
The motion while the rod is engaged with the screw is governed by rigid body kinematics under a holonomic constraint relating vertical displacement $z$ and angular coordinate $\varphi$ through the screw pitch,
$$z = \frac{h}{2\pi}\varphi.$$
Differentiation gives the kinematic constraint between vertical velocity $v_z$ and angular velocity $\omega$,
$$v_z = \frac{h}{2\pi}\omega.$$
After detachment, the rod becomes a free rigid body. In the absence of external torque about its center of mass, the angular momentum $\mathbf{L}$ about the center of mass is conserved,
$$\frac{d\mathbf{L}}{dt} = 0.$$
In the absence of external force other than gravity acting uniformly on all mass elements, the center of mass moves as a particle under gravity,
$$m\frac{d\mathbf{V}}{dt} = m\mathbf{g}.$$
Since gravity acts through the center of mass, it produces no torque about the center of mass,
$$\boldsymbol{\tau}_{\text{ext, CM}} = 0,$$
so the angular velocity vector remains constant in direction and magnitude after detachment if no additional constraints act.
Derivation
While the rod is still threaded, its center of mass lies on the screw axis, so its velocity is purely vertical. Let the angular velocity of the rod about the screw axis at the moment of release be $\omega_0$. The kinematic constraint of the thread gives the vertical velocity of the center of mass at detachment,
$$V_{z0} = \frac{h}{2\pi}\omega_0.$$
At the instant the rod loses contact with the screw, no impulsive force is assumed. Therefore both the linear momentum of the center of mass and the angular momentum of the rod are continuous through detachment.
Immediately after detachment, the center of mass has velocity $\mathbf{V}0 = V{z0},\hat{\mathbf{z}}$. The rod also has angular velocity $\boldsymbol{\omega}_0 = \omega_0 \hat{\mathbf{z}}$ about the vertical axis.
After detachment, the equation of motion of the center of mass reduces to
$$m\frac{dV_z}{dt} = -mg,$$
so the vertical velocity evolves as
$$V_z(t) = V_{z0} - gt.$$
Integrating gives the trajectory of the center of mass,
$$z(t) = z_0 + V_{z0}t - \frac{gt^2}{2}.$$
The horizontal coordinates of the center of mass remain constant because there is no horizontal force,
$$x(t) = x_0, \quad y(t) = y_0.$$
The torque about the center of mass is zero, so the angular velocity remains constant,
$$\boldsymbol{\omega}(t) = \omega_0 \hat{\mathbf{z}}.$$
Since both the direction and magnitude of $\boldsymbol{\omega}$ remain fixed, the rod continues to rotate uniformly about a fixed vertical axis while simultaneously translating along that same axis.
Result
After leaving the screw, the rod performs superposition of uniform rotation about a fixed vertical axis and translational motion of its center of mass along a vertical trajectory under gravity.
The initial center of mass velocity is
$$V_{z0} = \frac{h}{2\pi}\omega_0.$$
The subsequent motion is
$$z(t) = z_0 + \frac{h}{2\pi}\omega_0 t - \frac{gt^2}{2}, \quad x(t) = x_0, \quad y(t) = y_0,$$
with constant angular velocity
$$\omega(t) = \omega_0.$$
The rod therefore rises or falls along a vertical straight line while spinning uniformly about the same vertical axis.
Sanity Checks
The expression $V_{z0} = \frac{h}{2\pi}\omega_0$ has dimensions
$$\frac{\text{length}}{1}\cdot \frac{1}{\text{time}} = \frac{\text{m}}{\text{rad}} \cdot \frac{\text{rad}}{\text{s}} = \frac{\text{m}}{\text{s}},$$
which matches velocity units.
The center of mass motion is purely vertical because no horizontal external force acts after detachment, consistent with Newton’s second law.
The angular velocity remaining constant follows from zero external torque about the center of mass, and gravity produces no torque because it acts through the center of mass.
If $\omega_0 \to 0$, the vertical velocity tends to zero and the rod simply undergoes free fall, which is consistent with $z(t) = z_0 - \frac{gt^2}{2}$. If $g \to 0$, the motion reduces to uniform straight-line motion with constant vertical speed $\frac{h}{2\pi}\omega_0$ while rotation remains uniform.