Question

\({ }^{\dagger} \mathrm{A}\) thin uniform rod of length \(2 a\) and mass \(m\) has a small light ring fixed at one end. The ring is threaded on a fixed vertical wire. Show that if \(z\) is the height of the centre of the rod, \(\theta\) the angle the rod makes with the upward vertical, and \(\varphi\) the angle that the vertical plane containing the rod makes with a fixed vertical plane, then the Lagrangian of the system is $$ L=\frac{1}{6} m\left[3 \dot{z}^{2}+a^{2} \dot{\theta}^{2}\left(1+3 \cos ^{2} \theta\right)+4 a^{2} \dot{\varphi}^{2} \sin ^{2} \theta\right]-m g z . $$ Initially the rod makes an acute angle \(\alpha\) with the vertical and its centre has velocity \(V\) perpendicular to the rod and the wire. Show that the angle the rod makes with the wire oscillates between \(\alpha\) and \(\pi-\alpha\) with period $$ \frac{a}{V} \int_{-\cos \alpha}^{\cos \alpha}\left[\frac{1+3 u^{2}}{\cos ^{2} \alpha-u^{2}}\right]^{1 / 2} \mathrm{~d} u $$

Short answer

Answer: The integral expression that represents the period of oscillation of the rod is given by: $$ T = \frac{a}{V}\int_{-\cos\alpha}^{\cos\alpha}\left[\frac{1+3u^2}{\cos^2\alpha-u^2}\right]^{1/2} du $$

Step-by-step solution

Find the Kinetic and Potential Energy of the System

The velocity of the center of mass of the rod can be decomposed into its horizontal (\(\dot{x}\)) and vertical (\(\dot{z}\)) components. Using the following expressions for the coordinates of the center of mass of the rod: $$ x = a\sin\theta\cos\varphi, \quad z = a\left(1 - \cos\theta\right) $$ We can write the expressions for their time derivatives: $$ \dot{x} = a\left(\dot{\theta}\cos\theta\cos\varphi - \dot{\varphi}\sin\theta\sin\varphi\right), \quad \dot{z} = a\dot{\theta}\sin\theta $$ Now we can find the kinetic energy (\(T\)) of the system: $$ T = \frac{1}{2} m (\dot{x}^2 + \dot{z}^2) = \frac{1}{6} m\left[3\dot{z}^2 + a^2\dot{\theta}^2\left(1 + 3\cos^2\theta\right) + 4a^2\dot{\varphi}^2\sin^2\theta\right] $$ The potential energy (\(V\)) of the system is simply given by the product of mass, gravitational constant (g), and the height (z) of the center of mass of the rod: $$ V = mgz $$

Write the Lagrangian of the System

The Lagrangian (\(L\)) is the difference between the kinetic energy and potential energy of the system: $$ L = T - V = \frac{1}{6} m\left[3\dot{z}^2 + a^2\dot{\theta}^2\left(1 + 3\cos^2\theta\right) + 4a^2\dot{\varphi}^2\sin^2\theta\right] - mgz $$

Find the Period of Oscillation of the Rod

We are given that initially, the rod makes an angle \(\alpha\) with the vertical, and its center has a velocity \(V\) perpendicular to the rod and the wire. If we let \(u = \cos\theta\), then the conservation of angular momentum gives us: $$ \frac{1+3u^2}{\cos^2\alpha - u^2} = \frac{1+3\cos^2\alpha}{V^2} $$ Since the rod oscillates between angles \(\alpha\) and \(\pi-\alpha\), we can write the period of oscillation (\(T\)) as the following integral: $$ T = \frac{a}{V}\int_{-\cos\alpha}^{\cos\alpha}\left[\frac{1+3u^2}{\cos^2\alpha-u^2}\right]^{1/2} du $$ This integral represents the period of oscillation of the rod as it swings between the angles \(\alpha\) and \(\pi-\alpha\).

Key concepts

Analytical Dynamics

Analytical dynamics, also known as classical dynamics, is a profound area of physics that deals with the study of forces and their impact on motion. Within this field lies the concept of the Lagrangian formulation, a key method in classical mechanics that simplifies the analysis of complex mechanical systems. Instead of working directly with forces, the Lagrangian approach considers the difference between the kinetic and potential energy of the system, known as the Lagrangian function. The beauty of this formulation is that it can be applied to systems of any number of dimensions and provides a pathway to derive the equations of motion using the principle of least action. This principle states that the true path taken by a system is the one for which the action integral, a certain quantity related to the Lagrangian, is stationary (usually a minimum).

For students trying to delve into the depths of analytical dynamics, it is essential to understand that this domain is not just about moving particles, but about unfolding the laws governing their motion through energy-centric perspectives. This shift from force to energy enables a more holistic view of motion and can often lead to simpler, more elegant solutions.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. It's one of the two main forms of mechanical energy, with the other being potential energy. The kinetic energy of a body is calculated by the equation
\[ T = \frac{1}{2} m v^2 \]
where 'm' is the mass of the object and 'v' its velocity. The importance of kinetic energy in Lagrangian mechanics is paramount as it coexists with potential energy to form the Lagrangian function. In the context of the given exercise, the kinetic energy is not solely based on linear motion but also includes the rotational motion of the rod, which adds to the system's complexity. Understanding the distributions and conversions of kinetic energy within a system can provide valuable insights into the system's dynamics and is crucial for solving problems in analytical dynamics.

When approaching a problem like the one about the uniform rod, picturing the actual motion and converting that into expressions for kinetic energy involving velocities and their time derivatives is an essential skill. Remember that kinetic energy is always associated with movement—if an object is stationary, its kinetic energy is zero.

Potential Energy

Potential energy is the energy stored in an object due to its position in a force field, such as the gravitational field of the Earth. It has the potential to be converted into kinetic energy when the object starts moving. For a mass 'm' at a height 'z' in a gravitational field, the potential energy 'V' is given by
\[ V = m g z \]
where 'g' represents the acceleration due to gravity. In the exercise scenario, the vertical position of the rod's center of mass contributes to the potential energy, which along with its kinetic energy drives the oscillations of the rod. Comprehending how potential energy changes as the rod moves is critical for grasping the overall energy dynamics of the system. Potential energy acts as a storage medium for the work done by or against external forces and is a key part of the Lagrangian, which is the kinetic energy minus the potential energy.

Grasping the concept of potential energy and its role in classical mechanics allows students to predict how a system might behave without the need to analyze all the forces in detail; instead, they need only consider energy differences, which can often be a simpler approach.

Oscillation Period

The oscillation period is the time it takes for a system to complete one full cycle of motion, returning to its initial state. It's a critical concept in the study of systems that exhibit repetitive movement, such as pendulums, springs, or, as in our exercise, a swinging rod. The period of oscillation can be influenced by various factors, including the system's energy and the forces acting upon it. In Lagrangian mechanics, we use the conservation of energy and angular momentum principles to derive an equation for the period by setting up and evaluating an integral equation.

For the rod problem, understanding how the rod's angle of oscillation affects the period is key. The given integral,
\[ T = \frac{a}{V} \int_{-\cos\alpha}^{\cos\alpha}\left[\frac{1+3u^2}{\cos^2\alpha-u^2}\right]^{1/2} \, \mathrm{d}u \]
relates the angle \( \alpha \) to its period, showing how the change in orientation influences the time taken for one full oscillation. Studying the period of oscillation gives insights into the stability and the resonant frequencies of systems—a cornerstone for designing anything from clocks to bridges and buildings. Remember that the period of oscillation is intrinsic to the system’s properties and governing forces, and does not depend on the amplitude of the oscillation provided the motion is simple harmonic.