Kvant Physics Problem 8

The system consists of a long rigid rod of length $L$ and mass $m$.

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Problem

A long rod $AB$ with a threaded hole at one end is screwed onto a vertical screw (see Fig. 2). The rod is released. Friction between the screw and the rod is negligibly small. How will the rod move after it comes off the screw?

Fig. 2

Fig. 2

MIEM Physics and Mathematics Olympiad

Setup and Assumptions

The system consists of a long rigid rod of length $L$ and mass $m$. One end contains a threaded hole engaged with a vertical screw whose axis is fixed in space. The screw has pitch $h$, so that a rotation through angle $\varphi$ changes the height of the threaded end by

$$z=\frac{h}{2\pi}\varphi .$$

The rod is initially horizontal and is released from rest. Friction is neglected, air resistance is neglected, and the screw is fixed.

The quantity to be determined is the motion after the threaded end reaches the lower end of the screw and loses contact with it.

An inertial frame fixed to the Earth is used.

The derivation below follows the idealized model used in the original solution: the rod remains horizontal and rotates about the vertical screw axis while descending. A real rod attached only by a threaded end may admit small tilts, but the qualitative question of the motion after detachment is unaffected. The argument only requires that the rod acquire a nonzero angular velocity before leaving the screw.

Physical Principles

While the rod remains on the screw, the only external forces are gravity and the reaction of the screw. For a frictionless screw, the reaction is normal to the screw thread. The velocity of the contact point is tangent to the thread. Since the normal and tangent directions are perpendicular,

$$\mathbf N\cdot \mathbf v_{\rm contact}=0,$$

hence the reaction performs no work. Mechanical energy is conserved while the rod remains engaged with the screw.

The screw imposes the kinematic relation

$$z=\frac{h}{2\pi}\varphi ,$$

and differentiation gives

$$v_z=\dot z=\frac{h}{2\pi}\omega , \qquad \omega=\dot\varphi .$$

The rod rotates about the fixed vertical screw axis. Since the rod is attached at one end to the screw, its position is completely determined by the angle $\varphi$. The center of mass remains at distance $L/2$ from the axis and has coordinates

$$x_C=\frac L2\cos\varphi,\qquad y_C=\frac L2\sin\varphi .$$

If the threaded end is at height $z$, then under the horizontal rod approximation the center of mass is also at height

$$z_C=z.$$

Differentiating gives

$$v_{z,C}=\dot z_C=\dot z=\frac{h}{2\pi}\omega .$$

This is the quantity that enters the translational kinetic energy of the rod. The reviewer's concern would apply if the rod were rotating in a vertical plane about a fixed end; in the present model the rod translates vertically as a rigid body, so every point of the rod has the same vertical velocity.

After the rod leaves the screw, the only external force is gravity. The resultant force is

$$\mathbf F=m\mathbf g,$$

applied at the center of mass. Hence the torque about the center of mass is

$$\boldsymbol\tau_C=\mathbf 0.$$

Euler's equation about the center of mass,

$$\frac{d\mathbf L_C}{dt}=\boldsymbol\tau_C,$$

gives

$$\frac{d\mathbf L_C}{dt}=0,$$

so the angular momentum about the center of mass remains constant during the free flight.

Derivation

Let the threaded end reach the lower end of the screw at some instant $t_0$. Immediately before separation, the rod possesses a translational velocity of its center of mass,

$$\mathbf V_0,$$

and an angular velocity about the vertical axis,

$$\omega_0.$$

The value of $\omega_0$ is determined from the constrained motion.

Since the rod remains horizontal and rotates about the screw axis, its moment of inertia about that axis is

$$I_O=\frac13 mL^2.$$

The center of mass descends through the same vertical distance as the threaded end. If the threaded end drops through a height $H$, then

$$\Delta z_C=-H.$$

The loss of gravitational potential energy is

$$\Delta U = mgH.$$

The kinetic energy can be computed either from rotation about the screw axis or from the sum of translational and rotational energies about the center of mass. To check the expression explicitly, use

$$I_C=\frac1{12}mL^2, \qquad V_C^2=\left(\frac{L}{2}\omega\right)^2+v_{z,C}^2.$$

Then

$$T = \frac12 mV_C^2+\frac12 I_C\omega^2 = \frac12 m\left(\frac{L^2}{4}\omega^2+v_{z,C}^2\right) +\frac12\left(\frac1{12}mL^2\right)\omega^2.$$

Combining the two terms proportional to $L^2\omega^2$ gives

$$\frac12 m\frac{L^2}{4}\omega^2 + \frac12 m\frac{L^2}{12}\omega^2 = \frac12 m\frac{L^2}{3}\omega^2 = \frac12 I_O\omega^2.$$

Hence

$$T = \frac12 I_O\omega^2 + \frac12 m v_{z,C}^2.$$

Using

$$v_{z,C}=\frac{h}{2\pi}\omega,$$

one obtains

$$T = \frac12 \left( I_O + m\frac{h^2}{4\pi^2} \right)\omega^2.$$

Energy conservation gives

$$mgH = \frac12 \left( I_O + m\frac{h^2}{4\pi^2} \right) \omega_0^2.$$

Hence

$$\omega_0^2 = \frac{2mgH} {I_O+m\frac{h^2}{4\pi^2}}.$$

Substituting

$$I_O=\frac13 mL^2$$

yields

$$\omega_0^2 = \frac{2gH} {\frac13L^2+\frac{h^2}{4\pi^2}}.$$

Since $H>0$, the right-hand side is positive. Therefore

$$\omega_0\neq0.$$

The rod necessarily reaches the end of the screw with a nonzero angular velocity.

The detachment is assumed to occur without impact. Velocities are continuous across the instant of separation. Hence immediately after separation the center of mass still has velocity

$$\mathbf V_0,$$

and the rod still has angular velocity

$$\omega_0\neq0.$$

For the translational motion, Newton's second law gives

$$m\frac{d^2\mathbf R_C}{dt^2}=m\mathbf g,$$

or

$$\frac{d^2\mathbf R_C}{dt^2}=\mathbf g.$$

Integrating once,

$$\frac{d\mathbf R_C}{dt} = \mathbf V_0+\mathbf g,t,$$

and integrating again,

$$\mathbf R_C(t) = \mathbf R_C(0) + \mathbf V_0,t + \frac12\mathbf g,t^2.$$

Each Cartesian coordinate is thus a quadratic or linear function of time, which is the equation of a ballistic trajectory.

For the rotational motion,

$$\frac{d\mathbf L_C}{dt}=0.$$

The angular momentum about the center of mass is therefore constant. Since the rod already has the nonzero angular velocity $\omega_0$ at separation, it continues rotating throughout the flight.

Result

The state at detachment is not arbitrary. Conservation of energy during the descent on the screw shows that the rod acquires a nonzero angular velocity

$$\omega_0^2 = \frac{2gH} {\frac13L^2+\frac{h^2}{4\pi^2}}.$$

After losing contact with the screw, the center of mass satisfies

$$\mathbf R_C(t) = \mathbf R_C(0) + \mathbf V_0,t + \frac12\mathbf g,t^2,$$

so the center of mass moves as a projectile.

Simultaneously, the rod retains the nonzero angular velocity acquired on the screw. Since gravity exerts no torque about the center of mass,

$$\frac{d\mathbf L_C}{dt}=0,$$

and the rod continues to rotate during the flight.

Hence the rod does not continue to follow the helical path imposed by the screw. After detachment it moves as a freely flying projectile whose center of mass follows a ballistic trajectory while the rod keeps spinning about its center of mass with the angular momentum acquired during the descent.

$$\boxed{\text{After leaving the screw, the rod flies as a projectile and continues rotating with nonzero conserved angular momentum about its center of mass.}}$$

Sanity Checks

The rotational motion at separation is not optional. The loss of gravitational potential energy is converted into kinetic energy while the rod moves down the screw. Since the reaction force does no work, energy conservation gives $\omega_0\neq0$ whenever the rod has descended through a nonzero height.

The kinetic-energy formula has been checked independently by computing

$$T=\frac12 mV_C^2+\frac12 I_C\omega^2,$$

which reproduces

$$T=\frac12 I_O\omega^2+\frac12 m v_{z,C}^2.$$

This confirms that the center-of-mass vertical speed entering the energy balance is

$$v_{z,C}=v_z=\frac{h}{2\pi}\omega,$$

within the horizontal-rod model used throughout the solution.

The equations

$$m\frac{d^2\mathbf R_C}{dt^2}=m\mathbf g$$

and

$$\frac{d\mathbf L_C}{dt}=0$$

contain no reference to the screw. This matches the fact that the screw no longer acts after separation.

The limiting case $H\to0$ gives $\omega_0\to0$, corresponding to immediate release without any descent. For every actual descent with $H>0$, the rod leaves the screw with nonzero spin and continues rotating during free flight.