Question

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

Short answer

The puck moves in a downward spiral, combining constant angular momentum and gravitational acceleration.

Step-by-step solution

Define the system and coordinates

The puck is a frictionless point mass constrained to move between two concentric cylinders. It uses cylindrical coordinates (\(\rho, \phi, z\)) with a fixed radial distance \(\rho = R\). The other coordinates, \(\phi\) and \(z\), can vary. The system is under the influence of gravity acting in the \(-z\) direction.

Apply Newton's 2nd Law for radial motion

Since the puck's radial position \(\rho = R\) is constant due to the constraints, there is no radial acceleration or force component in the radial direction. Thus, we consider only tangential and vertical components.

Apply Newton's 2nd Law for tangential motion

For tangential motion (variation in \(\phi\)), consider Newton's second law, \(m\rho\dot{\phi}^2 = 0\). Since there is no tangential force acting on the puck, this implies the angular momentum is constant: \ mR^2\dot{\phi} = L \ where \(L\) is the constant angular momentum.

Apply Newton's 2nd Law for vertical motion

In the vertical \(z\)-direction, gravity acts on the puck. Using Newton's second law, we have: \ m\ddot{z} = -mg \ By integrating this equation, we find that \ \ddot{z} = -g, \ so \(\dot{z} = -gt + C_1\), and \(z = -\frac{1}{2}gt^2 + C_1t + C_2\). Here, \(C_1\) and \(C_2\) are constants determined by initial conditions.

Combine the results

The solution shows that the puck has a constant angular velocity \(\dot{\phi}\) around the vertical axis, due to the conservation of angular momentum. Vertically, it undergoes constant acceleration due to gravity with an initial velocity component depending on its initial condition. Thus, it follows a helical path as it spirals downward with acceleration.

Key concepts

Cylindrical Coordinates

In physics and engineering, cylindrical coordinates are used to describe positions in a three-dimensional space using a combination of linear and angular measurements. The system comprises three components:

• Radial distance (\(\rho\)) from a central axis.
• Angular position (\(\phi\)) around the central axis.
• Height (\(z\)) along the axis.

For the exercise involving concentric cylinders, the puck is constrained to move at a fixed radial distance, meaning its \(\rho\) value is constant. This significantly simplifies the problem, as motion only varies angularly and vertically. By isolating changes to \(\phi\) and \(z\), and assuming no radial motion, we can effectively apply Newton's second law to assess the puck's behavior.

Angular Momentum

Angular momentum is a measure of an object's rotational motion. It is essentially the rotational equivalent of linear momentum. For a point mass in cylindrical coordinates, its angular momentum \(L\) is given by:\[L = mR^2\dot{\phi}\]where \(m\) is the mass of the puck, \(R\) is the fixed radial distance, and \(\dot{\phi}\) is the angular velocity.
According to the conservation of angular momentum, if no external torque acts on the object, the angular momentum remains constant. This principle is pivotal in our exercise, as it maintains the puck's rotational speed constant around the cylinder's axis. This means that while the puck might be accelerating vertically due to gravity, its rate of rotation around the \(z\)-axis does not change, leading to a spiraling motion.

Tangential Motion

Tangential motion refers to movement along the circumference of a circle at a distance from the center. In the exercise, the change in the angular position \(\phi\) of the puck epitomizes this motion. Newton's second law dictates: \[m\rho\dot{\phi}^2 = 0\] This indicates that in the absence of tangential forces, there's no acceleration along the angle \(\phi\), corroborating that angular momentum stays constant.
Thus, the tangential motion aspect of the puck's path does not induce any force, ensuring the puck retains a steady angular speed. This contributes to the helical path by maintaining a consistent rate of rotation supported by conservation principles. It highlights the delicate balance between forces in motion.

Vertical Motion

Vertical motion involves movement in the vertical direction, typically influenced by forces like gravity. In this problem, gravity acting on the puck in the direction of \(-z\) must be considered carefully. By applying Newton's second law in the vertical direction, we derive the differential equation:\[m\ddot{z} = -mg\]which simplifies to acceleration:\[\ddot{z} = -g\]Integration gives velocity and position equations as functions of time. The position equation is:\[z = -\frac{1}{2}gt^2 + C_1t + C_2\]Here, \(C_1\) and \(C_2\) are constants reflecting initial velocity and initial position respectively.
This reveals that the puck accelerates downward continuously due to gravity. Combined with constant angular speed, this vertical descent creates a helical path. Understanding vertical motion here underscores Newton's second law in describing how gravitational force dictates the puck's downward trajectory.