Unit 1: Classical Mechanics

1. Dynamics of a System of Particles

1.1 Centre of Mass of a Two-Particle System

The centre of mass (CM) of a system of particles is a specific point that behaves as if the total mass of the system were concentrated at that point and all external forces were applied directly to it. For a system of two discrete particles, it provides a simplified point-particle representation of the overall motion.

Definition: For a system of two particles with masses m₁ and m₂ having position vectors r₁ and r₂ respectively relative to a given origin, the position vector of the centre of mass r_cm is defined as:
r_cm = (m₁r₁ + m₂r₂) / (m₁ + m₂) = (1 / M) (m₁r₁ + m₂r₂)
where M = m₁ + m₂ is the total mass of the system.

Cartesian Components of Centre of Mass:

If particle 1 is located at (x₁, y₁, z₁) and particle 2 is located at (x₂, y₂, z₂), the coordinates of the centre of mass are given by:

• x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
• y_cm = (m₁y₁ + m₂y₂) / (m₁ + m₂)
• z_cm = (m₁z₁ + m₂z₂) / (m₁ + m₂)

Physical Interpretation and Observations:

• Internal Position Ratio: The centre of mass lies on the straight line joining the two particles. The distance of the centre of mass from each particle is inversely proportional to their respective masses.
• If m₁ = m₂, then r_cm = (r₁ + r₂) / 2, which is the exact geometric midpoint.
• If m₁ ≫ m₂, then r_cm ≈ r₁, meaning the centre of mass practically coincides with the heavier particle (e.g., Sun-Earth system).

1.2 Velocity, Acceleration, and Linear Momentum of Centre of Mass

To analyze the dynamic behavior of the centre of mass, we differentiate its position vector with respect to time.

1. Velocity of Centre of Mass (v_cm):

Differentiating the position vector r_cm with respect to time t:

v_cm = dr_cm / dt = (m₁ (dr₁ / dt) + m₂ (dr₂ / dt)) / (m₁ + m₂) = (m₁v₁ + m₂v₂) / M

2. Total Linear Momentum (P):

The total linear momentum P of the two-particle system is the vector sum of individual linear momenta p₁ = m₁v₁ and p₂ = m₂v₂:

P = p₁ + p₂ = m₁v₁ + m₂v₂ = M v_cm

Key Insight: The total linear momentum of a system of particles equals the total mass multiplied by the velocity of the centre of mass.

3. Acceleration of Centre of Mass (a_cm):

Differentiating the velocity vector v_cm with respect to time t:

a_cm = dv_cm / dt = (m₁a₁ + m₂a₂) / M
M a_cm = m₁a₁ + m₂a₂ = F₁ + F₂

4. External and Internal Forces:

According to Newton's third law, the internal interaction forces between the two particles are equal and opposite (F₁₂ = -F₂₁), so their vector sum vanishes (F₁₂ + F₂₁ = 0). Thus, internal forces cannot alter the motion of the centre of mass.

F_ext = M a_cm = dP / dt

Law of Conservation of Linear Momentum:

If the net external force acting on the system is zero (F_ext = 0):

• a_cm = 0
• v_cm = constant vector
• P = M v_cm = constant vector

Exam Note / Common Mistake: Internal forces can change individual particle velocities and kinetic energies, but they never change the velocity or momentum of the centre of mass.

1.3 Degrees of Freedom

The degrees of freedom (DOF) of a mechanical system refer to the minimum number of independent coordinates or variables required to completely specify the position and configuration of all parts of the system at any instant of time.

Formula: For a system consisting of N particles subjected to k independent geometric constraint equations, the number of degrees of freedom f is:
f = 3N - k

Detailed Analysis of Cases:

1.4 Constraints and Their Classification

Constraints are geometrical or kinematical restrictions imposed on the motion of a particle or system of particles. They limit the permissible positions and velocities of the particles in the system.

Classification of Constraints:

1. Holonomic vs Non-Holonomic Constraints:

• Holonomic Constraints: Constraints that can be expressed as algebraic equations relating the position coordinates and time in the form:
  f(r₁, r₂, ..., r_N, t) = 0
  Examples: Rigid body constraint (|r_i - r_j| = c_ij); Simple pendulum with a rigid rod of fixed length L (x² + y² + z² - L² = 0).
• Non-Holonomic Constraints: Constraints that cannot be expressed as algebraic equations of coordinates alone. They are expressed as inequalities or as non-integrable differential equations involving velocities:
  f(r₁, ..., r_N, ṙ₁, ..., ṙ_N, t) = 0 or f(r₁, ..., r_N) ≥ 0
  Examples: A sphere rolling without slipping on a rough horizontal plane; Motion of gas molecules inside a spherical container (r ≤ R).

2. Scleronomic vs Rheonomic Constraints:

• Scleronomic Constraints: Constraints that do not depend explicitly on time (time-independent).
  f(r₁, r₂, ..., r_N) = 0
  Example: Simple pendulum with a fixed support and fixed length string.
• Rheonomic Constraints: Constraints that depend explicitly on time (time-dependent).
  f(r₁, r₂, ..., r_N, t) = 0
  Example: Simple pendulum whose point of suspension moves with a given function of time t.

Summary Table of Constraint Classifications:

2. Motion Under Central Force

2.1 Characteristics of Motion Under Central Force

A central force is a force acting on a particle whose magnitude depends only on the distance r between the particle and a fixed point (force centre), and whose direction is always directed along the line joining the particle to that fixed point.

Mathematical Definition:
F(r) = f(r) r̂ = f(r) (r / r)
where f(r) > 0 represents an attractive central force (if opposite to r̂, f(r) < 0), and r̂ is the unit vector along the position vector r.

Fundamental Characteristics:

1. Central Force is Conservative:

Since F(r) = f(r) r̂, calculating the curl yields ∇ × F = 0. Therefore, central forces are conservative, meaning the force can be derived from a scalar potential energy function V(r):

F(r) = - dV(r)/dr r̂ ⇒ V(r) = - ∫ f(r) dr

2. Conservation of Angular Momentum:

The torque N acting on a particle about the force centre is given by:

N = r × F(r) = r × [f(r) r̂] = 0 (since r is parallel to r̂)

Since N = dL/dt = 0, the total orbital angular momentum L is a constant vector in both magnitude and direction:

L = r × p = m (r × v) = constant vector

3. Motion is Planar:

Because L = r × p is a fixed constant vector in space, the position vector r must remain perpendicular to L at all times (r · L = 0). Consequently, the entire trajectory of the particle lies in a fixed two-dimensional plane perpendicular to L.

4. Constancy of Areal Velocity:

The position vector r sweeps out an infinitesimal area dA in time dt given by dA = (1/2) |r × dr| = (1/2) |r × v| dt.

Areal Velocity = dA / dt = |r × v| / 2 = L / (2m) = constant

This provides a direct theoretical proof for Kepler's Second Law.

2.2 Reduction of Two-Body Problem to an Equivalent One-Body Problem

Consider an isolated system of two particles of masses m₁ and m₂ interacting with each other via a mutual central force dependent only on their relative distance r = |r₁ - r₂|.

Step 1: Equations of Motion in Inertial Frame

• m₁ (d²r₁ / dt²) = F₁₂ = f(r) r̂
• m₂ (d²r₂ / dt²) = F₂₁ = -f(r) r̂

Step 2: Introduction of Relative and Centre of Mass Coordinates

Define relative position vector r and centre of mass vector R_cm:

• Relative position: r = r₁ - r₂
• Centre of mass: R_cm = (m₁r₁ + m₂r₂) / (m₁ + m₂)

Step 3: Derivation of Reduced Mass Equation

Divide the force equations by m₁ and m₂ respectively and subtract:

d²r₁ / dt² - d²r₂ / dt² = (1/m₁ + 1/m₂) F₁₂

d²r / dt² = ((m₁ + m₂) / (m₁ m₂)) f(r) r̂

Reduced Mass (μ) Definition:
1 / μ = 1 / m₁ + 1 / m₂ ⇒ μ = (m₁ m₂) / (m₁ + m₂)

Step 4: Separated Equations of Motion

1. Centre of Mass Motion: M (d²R_cm / dt²) = 0 ⇒ R_cm moves with constant velocity (uniform translational motion).
2. Relative Motion (Equivalent One-Body Problem):

   μ (d²r / dt²) = f(r) r̂

Conclusion: The complex two-body interaction is mathematically reduced to the motion of a single fictitious particle of mass μ (reduced mass) moving in a central force field f(r) r̂ centered at the origin.

2.3 Central Force and Motion in a Plane

Since the motion is confined to a single plane, we introduce polar coordinates (r, θ) in that plane.

Position, Velocity, and Acceleration in Polar Coordinates:

• Position vector: r = r r̂
• Velocity vector: v = dr/dt = ṙ r̂ + r θ̇ θ̂
• Acceleration vector: a = dv/dt = (r̈ - r θ̇²) r̂ + (2 ṙ θ̇ + r θ̈) θ̂

Equations of Motion:

Applying Newton's second law μ a = F(r) = f(r) r̂ gives two component equations:

1. Radial Component Equation:

   μ (r̈ - r θ̇²) = f(r)

2. Transverse Component Equation:

   μ (2 ṙ θ̇ + r θ̈) = (1 / r) d/dt (μ r² θ̇) = 0

Angular Momentum Conservation:

Integrating the transverse equation yields:

L = μ r² θ̇ = constant ⇒ θ̇ = L / (μ r²)

2.4 Equations of Motion and Differential Equation of Orbit

To find the trajectory r(θ) (the orbit) of the particle, we eliminate time t from the radial equation using the substitution u = 1/r.

Step-by-Step Derivation:

1. Express θ̇ in terms of u:

θ̇ = L / (μ r²) = (L / μ) u²

2. Transform radial velocity ṙ:

ṙ = dr / dt = (dr / dθ) (dθ / dt) = (d(1/u) / dθ) θ̇ = (-1/u²) (du / dθ) ((L / μ) u²) = - (L / μ) (du / dθ)

3. Transform radial acceleration r̈:

r̈ = d(ṙ) / dt = (d/dθ [- (L / μ) (du / dθ)]) θ̇ = - (L / μ) (d²u / dθ²) ((L / μ) u²) = - (L² u² / μ²) (d²u / dθ²)

4. Substitute r̈ and θ̇ into the Radial Equation:

μ [ - (L² u² / μ²) (d²u / dθ²) - (1/u) ((L u² / μ)²) ] = f(1/u)

- (L² u² / μ) [ d²u / dθ² + u ] = f(1/u)

Differential Equation of Orbit:
d²u / dθ² + u = - (μ / (L² u²)) f(1/u)

2.5 Kepler’s Laws of Motion and Their Deductions

For planetary motion governed by Newton's law of universal gravitation, the attractive central force is an inverse-square force:

f(r) = - k / r² = - k u² (where k = G M m)

Substituting f(1/u) = - k u² into the differential equation of orbit:

d²u / dθ² + u = - (μ / (L² u²)) (- k u²) = μ k / L²

This is a linear second-order non-homogeneous differential equation with constant solution:

u = 1 / r = (μ k / L²) [ 1 + ε cos(θ - θ₀) ]

Choosing θ₀ = 0, the equation of orbit becomes:

1 / r = (1 / p) [ 1 + ε cos θ ]
where semi-latus rectum p = L² / (μ k), and eccentricity ε = √(1 + (2 E L²) / (μ k²)), with total mechanical energy E.

Deductions of Kepler’s Three Laws:

1. Kepler’s First Law (Law of Orbits):

Statement: All planets move in elliptical orbits around the Sun, with the Sun situated at one focus of the ellipse.

Deduction: For bound planetary motion, total energy E < 0. Consequently, eccentricity ε satisfies 0 ≤ ε < 1. The polar equation 1/r = (1/p)(1 + ε cos θ) is precisely the equation of an ellipse in polar coordinates with the focus at the origin.

2. Kepler’s Second Law (Law of Areas):

Statement: A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Deduction: As derived earlier, areal velocity is proportional to orbital angular momentum:

dA / dt = L / (2 μ)

Since the central force produces zero torque, L is strictly constant in time, rendering dA/dt constant.

3. Kepler’s Third Law (Law of Periods):

Statement: The square of the orbital period T of a planet is directly proportional to the cube of the semi-major axis a of its elliptical orbit (T² ∝ a³).

Deduction: Integrating areal velocity over one complete orbital period T gives the total area A of the ellipse:

A = ∫₀^T (dA / dt) dt = (L / (2 μ)) T

The total area of an ellipse with semi-major axis a and semi-minor axis b is A = π a b. Thus:

π a b = (L / (2 μ)) T ⇒ T = (2 μ π a b) / L

Squaring both sides:

T² = (4 π² μ² a² b²) / L²

Using geometrical relationships for an ellipse: b = a √(1 - ε²) and semi-latus rectum p = b² / a = L² / (μ k), we get b² = a L² / (μ k).

Substitute b² into T²:

T² = (4 π² μ² a² / L²) × (a L² / (μ k)) = (4 π² μ / k) a³

T² = (4 π² μ / k) a³ ⇒ T² ∝ a³