Physics Classical Mechanics 100% Free Open Access
Chapter 1 • Theory & Derivations

Review of Elementary Principles & D'Alembert's Principle

Mechanics of particle systems, classification of kinematic constraints, generalized coordinates, principle of virtual work, D'Alembert's dynamic principle, derivation of Lagrange's equations, generalized velocity-dependent potentials (Lorentz force), and Rayleigh dissipation functions.

1.1Mechanics of a System of Particles & Conservation Theorems

1. Center of Mass & Total Linear Momentum

Consider an assembly of $N$ discrete particles with masses $m_i$ ($i = 1, 2, \dots, N$) and position vectors $\vec{r}_i$ measured relative to an inertial reference frame. The center of mass position vector $\vec{R}$ is defined by:
$$\vec{R} = \frac{1}{M} \sum_{i=1}^N m_i \vec{r}_i, \quad M = \sum_{i=1}^N m_i$$
Differentiating with respect to time yields the center of mass velocity $\vec{V} = \dot{\vec{R}}$ and the total linear momentum $\vec{P}$:
$$\vec{P} = \sum_{i=1}^N m_i \dot{\vec{r}}_i = M \dot{\vec{R}} = M \vec{V}$$

2. Newton's Second Law for Particle Systems

The net force acting on the $i$-th particle decomposes into an external force $\vec{F}_i^{\text{ext}}$ applied from outside the system and internal pairwise interaction forces $\vec{F}_{ij}$ exerted on particle $i$ by particle $j$:
$$\dot{\vec{p}}_i = m_i \ddot{\vec{r}}_i = \vec{F}_i^{\text{ext}} + \sum_{j \neq i} \vec{F}_{ij}$$
Summing over all $N$ particles gives:
$$\frac{d\vec{P}}{dt} = \sum_{i=1}^N \dot{\vec{p}}_i = \sum_{i=1}^N \vec{F}_i^{\text{ext}} + \sum_{i=1}^N \sum_{j \neq i} \vec{F}_{ij}$$
By Newton's Third Law (weak form), pairwise internal forces are equal and opposite: $\vec{F}_{ij} = -\vec{F}_{ji}$. Therefore, the double summation vanishes identically:
$$\sum_{i=1}^N \sum_{j \neq i} \vec{F}_{ij} = \sum_{i < j} (\vec{F}_{ij} + \vec{F}_{ji}) = 0$$
Hence, the rate of change of total linear momentum equals the net external force:
$$\frac{d\vec{P}}{dt} = M \ddot{\vec{R}} = \vec{F}^{\text{ext}}$$

Conservation Theorem of Linear Momentum: If the total external force vanishes ($\vec{F}^{\text{ext}} = 0$), the total linear momentum is conserved ($\vec{P} = \text{constant}$), and the center of mass moves with constant rectilinear velocity.

3. Angular Momentum & Internal Torques

The total angular momentum about the origin is $\vec{L} = \sum_{i=1}^N \vec{r}_i \times \vec{p}_i$. Its time derivative is:
$$\frac{d\vec{L}}{dt} = \sum_{i=1}^N (\dot{\vec{r}}_i \times \vec{p}_i) + \sum_{i=1}^N (\vec{r}_i \times \dot{\vec{p}}_i) = \sum_{i=1}^N \vec{r}_i \times \vec{F}_i^{\text{ext}} + \sum_{i=1}^N \sum_{j \neq i} \vec{r}_i \times \vec{F}_{ij}$$
Since $\dot{\vec{r}}_i \times m_i \dot{\vec{r}}_i = 0$. Grouping the internal cross products in pairs:
$$\sum_{i < j} (\vec{r}_i \times \vec{F}_{ij} + \vec{r}_j \times \vec{F}_{ji}) = \sum_{i < j} (\vec{r}_i - \vec{r}_j) \times \vec{F}_{ij}$$
By the strong form of Newton's Third Law, internal forces are central—they act along the line connecting the particles, so $\vec{F}_{ij} \parallel (\vec{r}_i - \vec{r}_j)$. Consequently, $(\vec{r}_i - \vec{r}_j) \times \vec{F}_{ij} = 0$, yielding:
$$\frac{d\vec{L}}{dt} = \sum_{i=1}^N \vec{r}_i \times \vec{F}_i^{\text{ext}} = \vec{N}^{\text{ext}}$$

Conservation Theorem of Angular Momentum: If the net external torque about a chosen point vanishes ($\vec{N}^{\text{ext}} = 0$), total angular momentum $\vec{L}$ about that point is conserved.

1.2Constraints, Degrees of Freedom & Generalized Coordinates

1. Classification of Mechanical Constraints

In any realistic dynamical setup, particle motions are restricted by geometric or kinematic limitations called constraints. Constraints are rigorously categorized into two primary divisions:
  • Holonomic Constraints: Expressible as algebraic equations involving only coordinates and time:
    $$f_k(\vec{r}_1, \vec{r}_2, \dots, \vec{r}_N, t) = 0, \quad k = 1, 2, \dots, m$$
    Examples include a rigid rod connecting two masses ($|\vec{r}_1 - \vec{r}_2|^2 - L^2 = 0$) or a particle sliding on a spherical surface ($x^2 + y^2 + z^2 - R^2 = 0$).
  • Non-Holonomic Constraints: Cannot be integrated into coordinate-only relations. They occur as non-integrable differentials of velocities:
    $$\sum_{i=1}^{3N} a_{ki} dq_i + a_{kt} dt = 0$$
    or inequalities ($r^2 - R^2 \ge 0$, e.g., a gas in a container or a bead rolling off a sphere). A classic non-holonomic example is a rolling disc without slipping.

2. Temporal Dependence: Scleronomic vs. Rheonomic

  • Scleronomic: Constraint equations do not depend explicitly on time $t$: $f_k(\vec{r}_i) = 0$ (e.g., rigid pendulum with fixed pivot).
  • Rheonomic: Constraint equations depend explicitly on time $t$: $f_k(\vec{r}_i, t) = 0$ (e.g., pendulum whose support oscillates vertically $z_0 = A \cos \omega t$).

3. Degrees of Freedom & Generalized Coordinates

For a system of $N$ particles subjected to $m$ independent holonomic constraints, the number of independent degrees of freedom $n$ is:
$$n = 3N - m$$
Instead of managing $3N$ constrained Cartesian coordinates and calculating unknown constraint forces, we introduce a set of $n$ independent variables called generalized coordinates:
$$q_1, q_2, \dots, q_n$$
The Cartesian position of every particle is a function of the generalized coordinates and possibly time:
$$\vec{r}_i = \vec{r}_i(q_1, q_2, \dots, q_n, t), \quad i = 1, 2, \dots, N$$
The velocities are obtained via the multivariate chain rule:
$$\vec{v}_i = \dot{\vec{r}}_i = \sum_{j=1}^n \frac{\partial \vec{r}_i}{\partial q_j} \dot{q}_j + \frac{\partial \vec{r}_i}{\partial t}$$

1.3Principle of Virtual Work & Virtual Displacements

1. Definition of Virtual Displacements

A virtual displacement $\delta \vec{r}_i$ is defined as an infinitesimal, arbitrary change in the coordinates of the system that is:
  • Purely geometric and instantaneous: It takes place at a fixed instant of time ($\delta t = 0$).
  • Consistent with all instantaneous kinematic constraints: For holonomic constraints $f_k(\vec{r}_i, t) = 0$, the virtual variations satisfy:
    $$\sum_{i=1}^N \nabla_i f_k \cdot \delta \vec{r}_i = 0$$

In contrast, a real displacement $d\vec{r}_i = \vec{v}_i dt$ occurs over a time interval $dt$ during which constraints may change explicitly with time.

2. Virtual Work & Ideal Constraints

Let the total force on particle $i$ be decomposed into applied external force $\vec{F}_i$ and constraint force $\vec{f}_i$:
$$\vec{F}_i^{\text{total}} = \vec{F}_i + \vec{f}_i$$
The virtual work $\delta W$ done by all forces in an arbitrary virtual displacement is:
$$\delta W = \sum_{i=1}^N \vec{F}_i^{\text{total}} \cdot \delta \vec{r}_i = \sum_{i=1}^N \vec{F}_i \cdot \delta \vec{r}_i + \sum_{i=1}^N \vec{f}_i \cdot \delta \vec{r}_i$$

Postulate of Ideal Constraints: In standard classical mechanics, the net virtual work done by constraint forces vanishes identically for any virtual displacement consistent with constraints:

$$\sum_{i=1}^N \vec{f}_i \cdot \delta \vec{r}_i = 0$$

Examples of ideal constraints include rigid interatomic bonds, frictionless surfaces (where normal force $\vec{N} \perp \delta \vec{r}$), and rolling without slipping (where instantaneous point of contact has zero velocity).

3. The Principle of Virtual Work for Static Equilibrium

For a system in static equilibrium, $\vec{F}_i^{\text{total}} = 0$. Incorporating the ideal constraint postulate, the condition for equilibrium reduces purely to the applied forces:
$$\delta W = \sum_{i=1}^N \vec{F}_i \cdot \delta \vec{r}_i = 0$$

This principle enables solving equilibrium problems without determining internal constraint forces.

1.4D'Alembert's Principle & Dynamic Generalization

1. Dynamic Inertial Forces

Jean le Rond d'Alembert (1743) converted dynamical problems into equivalent static problems by rewriting Newton's equation $\vec{F}_i^{\text{total}} = \dot{\vec{p}}_i$ as:
$$\vec{F}_i + \vec{f}_i - \dot{\vec{p}}_i = 0$$
where $-\dot{\vec{p}}_i = -m_i \ddot{\vec{r}}_i$ is the reversed effective force (inertial force).

2. Statement of D'Alembert's Principle

Taking the dot product with an arbitrary virtual displacement $\delta \vec{r}_i$ and summing over all $N$ particles:
$$\sum_{i=1}^N (\vec{F}_i - \dot{\vec{p}}_i) \cdot \delta \vec{r}_i + \sum_{i=1}^N \vec{f}_i \cdot \delta \vec{r}_i = 0$$
Invoking the postulate of ideal constraints ($\sum \vec{f}_i \cdot \delta \vec{r}_i = 0$), we obtain D'Alembert's Principle:
$$\sum_{i=1}^N (\vec{F}_i - \dot{\vec{p}}_i) \cdot \delta \vec{r}_i = 0$$

This principle is the cornerstone of analytical mechanics: it governs dynamics without requiring explicit knowledge of constraint forces.

1.5Derivation of Lagrange's Equations from D'Alembert's Principle

1. Transformation to Generalized Coordinates

Since $\vec{r}_i = \vec{r}_i(q_1, \dots, q_n, t)$, the virtual displacement at fixed $t$ is:
$$\delta \vec{r}_i = \sum_{j=1}^n \frac{\partial \vec{r}_i}{\partial q_j} \delta q_j$$
Substituting into D'Alembert's principle:
$$\sum_{j=1}^n \left[ \sum_{i=1}^N \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} - \sum_{i=1}^N m_i \ddot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} \right] \delta q_j = 0$$

2. Generalized Force Definition

We define the generalized force $Q_j$ associated with coordinate $q_j$ as:
$$Q_j = \sum_{i=1}^N \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j}$$

3. Mathematical Identities for the Inertial Term

Consider the term $\sum_i m_i \ddot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j}$:
$$\ddot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} = \frac{d}{dt}\left( \dot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} \right) - \dot{\vec{r}}_i \cdot \frac{d}{dt}\left(\frac{\partial \vec{r}_i}{\partial q_j}\right)$$
We apply two fundamental kinematic lemmas:
  1. Cancellation of Dots: Since $\vec{v}_i = \sum_k \frac{\partial \vec{r}_i}{\partial q_k}\dot{q}_k + \frac{\partial \vec{r}_i}{\partial t}$, differentiating with respect to $\dot{q}_j$ gives:
    $$\frac{\partial \vec{v}_i}{\partial \dot{q}_j} = \frac{\partial \vec{r}_i}{\partial q_j}$$
  2. Interchange of Time and Partial Derivatives:
    $$\frac{d}{dt}\left(\frac{\partial \vec{r}_i}{\partial q_j}\right) = \sum_k \frac{\partial^2 \vec{r}_i}{\partial q_k \partial q_j} \dot{q}_k + \frac{\partial^2 \vec{r}_i}{\partial t \partial q_j} = \frac{\partial}{\partial q_j}\left( \sum_k \frac{\partial \vec{r}_i}{\partial q_k} \dot{q}_k + \frac{\partial \vec{r}_i}{\partial t} \right) = \frac{\partial \vec{v}_i}{\partial q_j}$$
Substituting both lemmas:
$$\sum_{i=1}^N m_i \ddot{\vec{r}}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j} = \sum_{i=1}^N m_i \left[ \frac{d}{dt}\left(\vec{v}_i \cdot \frac{\partial \vec{v}_i}{\partial \dot{q}_j}\right) - \vec{v}_i \cdot \frac{\partial \vec{v}_i}{\partial q_j} \right] = \frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j}$$
where $T = \frac{1}{2} \sum_{i=1}^N m_i v_i^2$ is the total kinetic energy.

4. Final Form of Lagrange's Equations

D'Alembert's equation becomes:
$$\sum_{j=1}^n \left[ \frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} - Q_j \right] \delta q_j = 0$$
Since the generalized coordinates $q_j$ are completely independent, each coefficient must vanish identically:
$$\frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} = Q_j, \quad j = 1, 2, \dots, n$$
If forces are conservative, $\vec{F}_i = -\nabla_i V(\vec{r})$, then $Q_j = -\frac{\partial V}{\partial q_j}$. Since $V$ is independent of generalized velocities $\dot{q}_j$, defining the Lagrangian $L = T - V$ yields:
$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0, \quad j = 1, 2, \dots, n$$

1.6Velocity-Dependent Potentials & Dissipation Functions

1. Generalized Potentials for Velocity-Dependent Forces

If generalized forces can be expressed as:
$$Q_j = -\frac{\partial U}{\partial q_j} + \frac{d}{dt}\left( \frac{\partial U}{\partial \dot{q}_j} \right)$$
then the Euler-Lagrange equations retain their standard canonical form $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} = 0$ with $L = T - U$.

2. The Electromagnetic Lorentz Force as a Generalized Potential

For a charged particle of charge $q$ moving with velocity $\vec{v}$ in an electromagnetic field described by scalar potential $\Phi(\vec{r}, t)$ and vector potential $\vec{A}(\vec{r}, t)$, the Lorentz force is:
$$\vec{F} = q \left( \vec{E} + \vec{v} \times \vec{B} \right) = q \left( -\nabla \Phi - \frac{\partial \vec{A}}{\partial t} + \vec{v} \times (\nabla \times \vec{A}) \right)$$
Using the vector identity $\vec{v} \times (\nabla \times \vec{A}) = \nabla(\vec{v} \cdot \vec{A}) - (\vec{v} \cdot \nabla)\vec{A}$:
$$\vec{F} = -\nabla \left( q \Phi - q \vec{v} \cdot \vec{A} \right) - q \left[ \frac{\partial \vec{A}}{\partial t} + (\vec{v} \cdot \nabla)\vec{A} \right] = -\nabla U - q \frac{d\vec{A}}{dt}$$
Notice that $\frac{\partial U}{\partial \vec{v}} = -q \vec{A}$. Hence:
$$\frac{d}{dt}\left( \frac{\partial U}{\partial \vec{v}} \right) = -q \frac{d\vec{A}}{dt}$$
Therefore, the electromagnetic force is derived from the generalized velocity-dependent potential:
$$U(\vec{r}, \vec{v}, t) = q \Phi(\vec{r}, t) - q \vec{v} \cdot \vec{A}(\vec{r}, t)$$
The corresponding Lagrangian for a non-relativistic charged particle is:
$$L = \frac{1}{2} m v^2 - q \Phi + q \vec{v} \cdot \vec{A}$$
The canonical momentum $\vec{p}$ conjugate to $\vec{r}$ is:
$$\vec{p} = \frac{\partial L}{\partial \vec{v}} = m \vec{v} + q \vec{A}$$

3. Rayleigh's Dissipation Function

When frictional or viscous forces are proportional to velocity, $\vec{F}_{f, i} = -k_i \vec{v}_i$, Lord Rayleigh introduced the dissipation function $\mathcal{F}$:
$$\mathcal{F} = \frac{1}{2} \sum_{i=1}^N (k_{ix} v_{ix}^2 + k_{iy} v_{iy}^2 + k_{iz} v_{iz}^2)$$
In generalized coordinates, the non-conservative generalized dissipative force is $Q_{j}^{\text{diss}} = -\frac{\partial \mathcal{F}}{\partial \dot{q}_j}$. The extended Lagrange equations become:
$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} + \frac{\partial \mathcal{F}}{\partial \dot{q}_j} = 0$$
The rate of energy dissipation from the system satisfies $\frac{dE}{dt} = -2 \mathcal{F} \le 0$.

Standard University Exam Solved Problems

SOLVED PROBLEM 1.1

Double Incline with Connected Masses via D'Alembert's Principle

Two masses $m_1$ and $m_2$ rest on frictionless planes inclined at angles $\alpha$ and $\beta$ respectively. They are connected by an inextensible string passing over a frictionless massless pulley. Use D'Alembert's principle to determine the acceleration of the system and the tension in the string.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Formulate the Holonomic Constraint & Virtual Displacements
$$s_1 + s_2 = L \implies \delta s_2 = -\delta s_1$$

Let $s_1$ be the displacement of mass $m_1$ down plane $\alpha$, and $s_2$ the position of $m_2$ from the pulley along plane $\beta$. Inextensibility requires $s_1 + s_2 = L$, so any virtual displacement satisfies $\delta s_2 = -\delta s_1$.

Step 2: Apply D'Alembert's Equation
$$\sum_{i=1}^2 (F_i - m_i a_i) \delta s_i = (m_1 g \sin \alpha - m_1 a)\delta s_1 + (m_2 g \sin \beta - m_2 (-a))(-\delta s_1) = 0$$

Since $a_1 = a$ and $a_2 = -a$, substituting applied forces along the planes and collecting terms in $\delta s_1$ gives $(m_1 g \sin \alpha - m_2 g \sin \beta - (m_1 + m_2)a)\delta s_1 = 0$.

Step 3: Solve for System Acceleration
$$a = g \frac{m_1 \sin \alpha - m_2 \sin \beta}{m_1 + m_2}$$

Because $\delta s_1$ is arbitrary, the coefficient must vanish. String tension is then found from single particle dynamics $T = m_1(g \sin \alpha - a) = \frac{m_1 m_2 g(\sin \alpha + \sin \beta)}{m_1 + m_2}$.

Final Answer & Physical Insight

System acceleration: a = g (m1 sin α - m2 sin β)/(m1 + m2); Tension: T = m1 m2 g (sin α + sin β)/(m1 + m2).

SOLVED PROBLEM 1.2

Lagrangian of a Bead Sliding on a Uniformly Rotating Wire Hoop

A bead of mass $m$ slides without friction on a circular wire hoop of radius $R$ in a vertical plane. The hoop rotates with constant angular velocity $\omega$ about its vertical diameter. Set up the Lagrangian and derive the equation of motion for the angular position $\theta$ of the bead.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Define Coordinates in Terms of Generalized Coordinate θ
$$x = R \sin \theta \cos(\omega t), \quad y = R \sin \theta \sin(\omega t), \quad z = -R \cos \theta$$

Here $\theta$ is the angle of the bead measured from the lowest point of the hoop. The vertical axis is $z$, and the hoop rotates at azimuth $\phi = \omega t$.

Step 2: Calculate Velocities and Kinetic & Potential Energies
$$v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2 = R^2 \dot{\theta}^2 + R^2 \omega^2 \sin^2 \theta$$

The kinetic energy is $T = \frac{1}{2}m R^2(\dot{\theta}^2 + \omega^2 \sin^2 \theta)$. Taking $z=0$ at the center, potential energy is $V = -m g R \cos \theta$.

Step 3: Form the Lagrangian and Euler-Lagrange Equation
$$L = \frac{1}{2} m R^2 \dot{\theta}^2 + \frac{1}{2} m R^2 \omega^2 \sin^2 \theta + m g R \cos \theta$$

Evaluating $\frac{\partial L}{\partial \dot{\theta}} = m R^2 \dot{\theta}$ and $\frac{\partial L}{\partial \theta} = m R^2 \omega^2 \sin \theta \cos \theta - m g R \sin \theta$, we obtain $\ddot{\theta} - \left(\omega^2 \cos \theta - \frac{g}{R}\right)\sin \theta = 0$.

Final Answer & Physical Insight

Equation of motion: θ̈ = (ω² cos θ - g/R) sin θ. A supercritical pitchfork bifurcation occurs at critical rotation speed ω_c = √(g/R).

SOLVED PROBLEM 1.3

Canonical Momentum & Lagrangian for a Charged Particle in Crossed E and B Fields

Find the Lagrangian, generalized momentum, and equations of motion for a particle of charge $q$ and mass $m$ moving in a uniform magnetic field $\vec{B} = B_0 \hat{z}$ and uniform electric field $\vec{E} = E_0 \hat{y}$ using the Landau gauge $\vec{A} = -B_0 y \hat{x}$, $\Phi = -E_0 y$.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Construct the Velocity-Dependent Lagrangian
$$L = \frac{1}{2}m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - q(-E_0 y) + q(-B_0 y \dot{x})$$

Substituting vector potential components $A_x = -B_0 y, A_y = A_z = 0$ and scalar potential $\Phi = -E_0 y$ into $L = T - q\Phi + q \vec{v}\cdot\vec{A}$.

Step 2: Identify Cyclic Coordinates and Canonical Momenta
$$p_x = \frac{\partial L}{\partial \dot{x}} = m \dot{x} - q B_0 y = \text{const}, \quad p_z = m \dot{z} = \text{const}$$

Coordinates $x$ and $z$ are cyclic since they do not appear explicitly in $L$. Their canonical momenta $p_x$ and $p_z$ are strict constants of motion.

Step 3: Derive Equation of Motion for the Non-Cyclic Coordinate y
$$\frac{d}{dt}(m \dot{y}) - \left( q E_0 - q B_0 \dot{x} \right) = 0 \implies m \ddot{y} = q E_0 - q B_0 \dot{x}$$

Substituting $\dot{x} = \frac{p_x + q B_0 y}{m}$ yields $m \ddot{y} + \omega_c^2 y = q E_0 - \omega_c p_x$, showing harmonic cycloidal drift at cyclotron frequency $\omega_c = q B_0 / m$.

Final Answer & Physical Insight

px = m ẋ - q B0 y = const; pz = m ż = const; y-motion undergoes cycloidal drift with drift velocity v_d = E0/B0.