Classical mechanics

In short

Classical mechanics studies the motion of particles, systems of particles and rigid bodies under forces, using Newton's laws and the equivalent Lagrangian and Hamiltonian formulations. It supplies the conservation laws (momentum, angular momentum, energy) and the mathematical language — generalised coordinates, constraints, variational principles — on which every later physics course is built. In B.Sc (CBCS core paper "Mechanics") it is the largest single block of marks, and it is examined heavily in GATE Physics, JAM, JEST, NET-JRF and in state and central university semester papers. Employers test it in analyst and engineering screening (ISRO, DRDO, BARC/OCES, HAL, L&T, Tata Steel, CAE and simulation firms) and in teaching recruitment such as TGT/PGT and NET.

Explained

Newton's laws, frames and force laws

• First law: a body continues in uniform motion unless acted on by a net external force; this *defines* an inertial frame — a frame in which the law holds. All frames moving at constant velocity relative to an inertial frame are inertial.
• Second law: F = dp/dt. For constant mass this reduces to F = ma; for variable mass (rockets, conveyor belts, falling chains) you must keep the dp/dt form and account for the momentum carried by the mass entering or leaving.
• Third law: forces occur in equal, opposite pairs on *different* bodies along the line joining them (for central forces) — this is what makes total momentum of an isolated system conserved.
• Galilean transformation: r′ = r − Vt, t′ = t; acceleration and hence F = ma are invariant. Classical mechanics is valid for v ≪ c and for actions ≫ ħ.
• Force laws met in the syllabus: gravitation, spring (Hooke, F = −kx), friction (f ≤ μN, static vs kinetic), viscous drag (F = −bv at low Reynolds number, ∝ v² at high), tension and normal reaction as constraint forces.
• Non-inertial frames: in a frame accelerating with A, add pseudo-force −mA. In a frame rotating with ω, the effective force picks up centrifugal (−mω×(ω×r)), Coriolis (−2mω×v′) and Euler (−mω̇×r) terms.
• Earth as a rotating frame: g varies with latitude (≈ 9.78 m/s² at the equator, ≈ 9.83 m/s² at the poles); Coriolis force deflects winds and projectiles right in the Northern hemisphere, explains cyclonic circulation and the Foucault pendulum (precession rate ω sin λ).

Work, energy and conservation laws

• Work W = ∫F·dr; Work–energy theorem: net work = change in kinetic energy, valid for any force, conservative or not.
• A force is conservative if the work round any closed path is zero, equivalently ∇×F = 0, equivalently F = −∇U. Then mechanical energy E = T + U is conserved. Friction and drag are non-conservative (dissipative, path dependent).
• Potential energy diagrams: equilibrium where dU/dx = 0; stable if d²U/dx² > 0, unstable if < 0, neutral if zero. Turning points where U(x) = E; motion is bounded between consecutive turning points.
• Conservation theorems and their symmetry origin (Noether): homogeneity of space → linear momentum; isotropy of space → angular momentum; homogeneity of time → energy.
• Centre of mass: R = Σmᵢrᵢ/Σmᵢ; the CM moves as if the total external force acted on the total mass. Kinetic energy splits as T = ½MV_cm² + T′ (König's theorem), and angular momentum as L = R×MV + L′ (spin about the CM).
• Impulse J = ∫F dt = Δp — used for collisions, bat-and-ball and jet problems.

Collisions, variable mass and the two-body problem

• Elastic collision: momentum and kinetic energy both conserved. In one dimension with equal masses, velocities are exchanged; a light body hitting a very heavy one rebounds with nearly the same speed.
• Inelastic collision: momentum conserved, KE not. Coefficient of restitution e = (relative speed of separation)/(relative speed of approach), 0 ≤ e ≤ 1; e = 1 elastic, e = 0 perfectly inelastic. Maximum KE loss occurs in the perfectly inelastic case and equals ½μ(u₁−u₂)² with μ the reduced mass.
• Centre-of-mass (zero-momentum) frame simplifies collisions: in it, total momentum is zero and for an elastic collision each speed is unchanged, only the direction turns through the scattering angle.
• Two-body problem → one-body problem: with r = r₁ − r₂ and reduced mass μ = m₁m₂/(m₁+m₂), the relative motion obeys μ r̈ = F(r); the CM moves uniformly.
• Variable mass: rocket equation from momentum conservation; a rocket works in vacuum because thrust is the reaction to ejected mass, not a push against air. Multi-staging raises the achievable Δv because dead mass is shed (the basis of PSLV/GSLV/LVM3 staging).

Central forces, gravitation and orbits

• A central force F = f(r)r̂ gives torque zero about the centre, so angular momentum is conserved: motion is planar and the areal velocity dA/dt = L/2m is constant (Kepler's second law).
• Effective potential U_eff(r) = U(r) + L²/2mr² reduces the problem to one-dimensional radial motion; minima of U_eff give circular orbits, and their stability decides whether orbits are closed.
• Inverse-square force gives conic-section orbits: r = (L²/mk)/(1 + e cos θ). Eccentricity e < 1 ellipse (E < 0), e = 0 circle, e = 1 parabola (E = 0, escape), e > 1 hyperbola (E > 0).
• Kepler's laws: (1) orbits are ellipses with the Sun at a focus; (2) equal areas in equal times; (3) T² ∝ a³. The third law follows from the inverse-square law and is used for satellite periods.
• Gravitational field and potential: g = −GM r̂/r²; V = −GM/r. A uniform spherical shell exerts no field inside and behaves as a point mass outside (shell theorem). Inside a uniform solid sphere, g ∝ r.
• Escape velocity v_e = √(2GM/R) ≈ 11.2 km/s for Earth; orbital velocity at the surface ≈ 7.9 km/s; v_e = √2 v_orbital.
• Satellites: a geostationary orbit needs a period of one sidereal day (≈ 23 h 56 min), radius ≈ 42,164 km, i.e. about 35,786 km altitude, equatorial and prograde — the orbit used by India's INSAT and GSAT satellites; IRS and Cartosat remote-sensing craft use low, near-polar sun-synchronous orbits. NavIC uses a mix of geostationary and inclined geosynchronous orbits.
• Only the inverse-square and the linear (Hooke) force give closed, non-precessing orbits (Bertrand's theorem); the extra precession of Mercury's perihelion is a general-relativity correction.

Rigid body dynamics

• A rigid body has six degrees of freedom — three translational, three rotational. Rotation about a fixed axis: τ = Iα, L = Iω, T = ½Iω².
• Moment of inertia I = Σmᵢrᵢ² = ∫r²dm; depends on the axis. Radius of gyration k = √(I/M).
• Parallel axis theorem: I = I_cm + Md². Perpendicular axis theorem (plane laminae only): I_z = I_x + I_y.
• Rolling without slipping: v_cm = ωR, and friction does no work; T = ½Mv²(1 + k²/R²). On an incline, a = g sinθ/(1 + k²/R²), so a ring rolls down slower than a disc, which is slower than a sphere — independent of mass and radius.
• Angular momentum conservation: a skater pulling in arms, or a diver tucking, spins faster because I falls and Iω is constant; T rises because internal work is done.
• Inertia tensor I_ij; principal axes diagonalise it. For a symmetric top, free precession of the symmetry axis occurs at Ω = (I₃−I₁)ω₃/I₁.
• Gyroscope / top: a torque applied to a spinning body causes precession at Ω ≈ τ/(Iω) about an axis perpendicular to both; small superposed oscillation is nutation. Used in gyrocompasses and inertial navigation.
• Compound pendulum: T = 2π√(I/Mgd); equivalent simple pendulum length L_eq = I/Md — used in the bar-pendulum g measurement, a standard B.Sc lab experiment.

Lagrangian and Hamiltonian formulations

• Constraints: holonomic (expressible as f(r,t) = 0, e.g. rigid rod, bead on a wire) versus non-holonomic (inequalities or velocity constraints, e.g. a rolling sphere on a plane, gas in a box). Degrees of freedom = 3N − (number of holonomic constraints).
• Generalised coordinates q₁…q_n remove constraint forces from the equations — the main practical advantage over Newtonian force analysis.
• D'Alembert's principle → Lagrange's equations: d/dt(∂L/∂q̇ⱼ) − ∂L/∂qⱼ = 0 with L = T − U. For dissipative or applied non-conservative forces, add the generalised force Qⱼ on the right.
• Hamilton's principle: the actual path makes the action S = ∫L dt stationary (δS = 0) — a variational statement equivalent to Newton's laws.
• Cyclic (ignorable) coordinate: if L does not contain qⱼ, the conjugate momentum pⱼ = ∂L/∂q̇ⱼ is conserved; this is how angular momentum conservation drops out of a rotationally symmetric L.
• Hamiltonian: H = Σp q̇ − L; when the constraints are time-independent and U is velocity-independent, H = T + U = E. Hamilton's equations: q̇ = ∂H/∂p, ṗ = −∂H/∂q.
• Advantages: scalar functions instead of vectors, any convenient coordinates, constraint forces eliminated, symmetries and conserved quantities read off directly, and a direct bridge to quantum mechanics (H → Ĥ) and statistical mechanics (phase space, Liouville's theorem).
• Limitations: Lagrangian form does not directly give constraint reaction forces (need Lagrange multipliers); awkward for strong friction; fails at relativistic speeds without modification and at atomic scales entirely.
• Phase space, Poisson brackets {A,B} and canonical transformations are included in some CBCS programmes as an introduction to advanced mechanics.

Key points

• Inertial frame - frame where Newton's first law holds; pseudo-forces appear only in non-inertial frames.
• Second law in general form - F = dp/dt; essential for rocket and variable-mass problems.
• Work–energy theorem - total work equals ΔKE always; energy conservation only when all forces are conservative.
• Conservative force test - ∇×F = 0 ⇔ F = −∇U ⇔ zero work round a closed loop.
• Noether link - space homogeneity → p, isotropy → L, time homogeneity → E.
• Centre of mass - external forces govern CM motion; internal forces cancel in pairs.
• Reduced mass - μ = m₁m₂/(m₁+m₂) converts the two-body problem into one-body motion in a central field.
• Coefficient of restitution - e = separation speed / approach speed; e = 1 elastic, e = 0 perfectly inelastic.
• Central force - torque-free about the centre, so L is conserved and the orbit is planar with constant areal velocity.
• Effective potential - U + L²/2mr²; its minimum fixes the circular orbit radius.
• Orbit type by energy - E < 0 ellipse, E = 0 parabola, E > 0 hyperbola.
• Kepler's third law - T² = 4π²a³/GM; gives satellite periods and planetary masses.
• Escape and orbital speed - v_e = √2 v_orb; 11.2 km/s and 7.9 km/s for Earth.
• Moment of inertia theorems - parallel axis I = I_cm + Md²; perpendicular axis only for laminae.
• Rolling - a = g sinθ/(1 + k²/R²); shape decides the ranking, not mass or radius.
• Precession - Ω = τ/Iω for a fast spinning top; the axis moves perpendicular to the applied torque.
• Degrees of freedom - 3N minus holonomic constraints; a rigid body has 6.
• Cyclic coordinate - coordinate absent from L ⇒ its conjugate momentum is conserved.

Formulas and examples

Example 1 — Geostationary orbit radius. Find the radius of a circular equatorial orbit with period T = 86,164 s (one sidereal day). GM = 3.986×10¹⁴ m³/s².

• From T² = 4π²r³/GM, r³ = GM T²/4π² = (3.986×10¹⁴ × 7.4242×10⁹)/39.478.
• Numerator = 2.9594×10²⁴; divide by 39.478 → r³ = 7.496×10²² m³.
• r = (7.496×10²²)^1/3 ≈ 4.216×10⁷ m = 42,164 km, so altitude = 42,164 − 6,378 ≈ 35,786 km, and v = 2πr/T ≈ 3.07 km/s. This is the orbit of GSAT communication satellites.

Example 2 — Rolling bodies on an incline. A ring, a solid disc and a solid sphere roll without slipping down a 30° incline of length 4 m from rest. Compare final speeds, using a = g sinθ/(1 + k²/R²), g = 9.8 m/s².

• The sphere wins because the smallest fraction of its energy goes into rotation; mass and radius do not enter.

Example 3 — Inelastic collision and energy loss. A 2 kg block moving at 6 m/s strikes a stationary 4 kg block; e = 0.5.

• Momentum: 2×6 = 2v₁ + 4v₂ → 2v₁ + 4v₂ = 12.
• Restitution: v₂ − v₁ = 0.5 × 6 = 3.
• Substituting v₂ = v₁ + 3: 2v₁ + 4v₁ + 12 = 12 → v₁ = 0 m/s, v₂ = 3 m/s.
• KE before = ½×2×36 = 36 J; after = ½×4×9 = 18 J; loss = 18 J.
• Check with the formula: μ = (2×4)/6 = 1.333 kg; loss = ½×1.333×36×(1−0.25) = 24×0.75 = 18 J. ✔

Example 4 — Rocket (Tsiolkovsky). A stage of initial mass 50,000 kg burns to a final mass 10,000 kg with exhaust speed 2,700 m/s, in free space.

• Δv = 2,700 × ln(5) = 2,700 × 1.6094 ≈ 4,345 m/s.
• If the propellant burns in 120 s, mean mass flow = 40,000/120 = 333.3 kg/s and thrust = 2,700 × 333.3 ≈ 900 kN.
• Burning vertically near Earth would cost gravity loss ≈ gt = 9.8×120 ≈ 1,176 m/s, leaving about 3,169 m/s — why launchers pitch over early and use multiple stages.

Example 5 — Lagrangian of a simple pendulum. Mass m, massless rod of length l, angle θ from the vertical (one degree of freedom).

• T = ½ml²θ̇², U = −mgl cosθ, so L = ½ml²θ̇² + mgl cosθ.
• ∂L/∂θ̇ = ml²θ̇ ⇒ d/dt = ml²θ̈; ∂L/∂θ = −mgl sinθ.
• Lagrange's equation gives ml²θ̈ + mgl sinθ = 0, i.e. θ̈ = −(g/l) sinθ; for small θ, θ̈ = −(g/l)θ with T = 2π√(l/g).
• For l = 1 m, g = 9.8 m/s²: T = 2π√(1/9.8) = 2π × 0.3194 ≈ 2.01 s.

Where it is used

• Space and defence: ISRO orbit mechanics and launch-vehicle staging (PSLV, LVM3), Chandrayaan trajectory design, DRDO ballistics and missile guidance; inertial navigation using gyroscopes.
• Mechanical and automotive design: flywheels, gearboxes, balancing of rotating machinery, vibration isolation, crash and impulse analysis at Maruti, Tata Motors, Mahindra and ARAI.
• Civil and structural practice: dynamic loads, seismic analysis to IS 1893 (Part 1): 2016 and wind loads to IS 875 (Part 3): 2015 build directly on rigid-body and oscillatory dynamics.
• Simulation and software roles: multibody dynamics and FEA/CFD engineers using ANSYS, Abaqus, ADAMS and open-source solvers; physics engines in game and animation studios.
• Research and teaching: B.Sc/M.Sc laboratories (bar pendulum for g, Kater's pendulum, flywheel moment of inertia, Searle's and torsion apparatus); NET/GATE-based teaching and research careers at IITs, IISERs, BARC and TIFR.
• Sports science, robotics and drones: gait and projectile analysis, manipulator dynamics solved in Lagrangian form, quadcopter attitude control.

Interview questions

• Q: Why is Newton's first law not just a special case of the second? A: The first law defines the class of frames (inertial) in which the second law can be written; without it F = ma has no reference frame to be true in.
• Q: A rocket accelerates in vacuum — what does it push against? A: Nothing external; thrust is the reaction to the momentum of the ejected exhaust, so F = v_ex |dm/dt| needs no surrounding medium.
• Q: Is the work–energy theorem valid when friction acts? A: Yes — total work by all forces equals ΔKE; what fails is conservation of mechanical energy, because friction's work is negative and dissipated as heat.
• Q: Why is angular momentum conserved for any central force, whatever its r-dependence? A: The force is along r, so τ = r×F = 0, hence dL/dt = 0; this makes the orbit planar with constant areal velocity.
• Q: Which is faster down an incline, a hollow cylinder or a solid sphere, and why? A: The sphere, because its k²/R² = 2/5 is smaller than the cylinder's 1, so less of the energy goes into rotation; mass and radius are irrelevant.
• Q: What does a cyclic coordinate tell you? A: That the Lagrangian does not depend on it, so its conjugate momentum is conserved — for example, ignorable φ in a central-force problem gives conservation of angular momentum.
• Q: When is the Hamiltonian equal to the total energy? A: When the transformation to generalised coordinates is time-independent and the potential is velocity-independent; otherwise H is still conserved if ∂L/∂t = 0, but it need not equal T + U.
• Q: Is the Coriolis force real? A: It is a pseudo-force with no physical agent, appearing only because the frame rotates; in an inertial frame the same motion is explained entirely by real forces.
• Q: Why must a geostationary satellite be equatorial? A: Only an equatorial prograde orbit keeps the satellite above a fixed ground point; an inclined geosynchronous orbit traces a figure-of-eight instead.
• Q: Why does a spinning top not fall? A: Gravity's torque is perpendicular to its large angular momentum, so it changes the direction of L rather than its magnitude — producing precession at Ω ≈ τ/Iω.
• Q: Give one holonomic and one non-holonomic constraint. A: A bead threaded on a rigid wire (x²+y² = a², holonomic); a sphere rolling without slipping on a plane, whose velocity constraint cannot be integrated into a coordinate relation.
• Q: Why does the skater spin faster with arms pulled in, and where does the extra kinetic energy come from? A: I falls while L = Iω is conserved, so ω rises; T = L²/2I increases, and the energy comes from the muscular work done pulling the arms inward.

Common mistakes

• Using F = ma for variable-mass problems instead of F = dp/dt, and then mis-deriving the rocket equation.
• Treating centrifugal or Coriolis forces as real forces in an inertial frame, or forgetting them in a rotating frame.
• Applying the perpendicular axis theorem to three-dimensional bodies — it holds only for plane laminae.
• Assuming momentum conservation in collisions where an external impulse acts (a ball bouncing off a wall, or a block on rough ground over a long time).
• Writing e as a ratio of kinetic energies, or forgetting that momentum is still conserved when e < 1.
• Taking L = T + U (or H = T − U) by slip, and forgetting the generalised force term when non-conservative forces act.
• Mixing radius and diameter in I, using degrees in place of radians for ω and α, or quoting kinetic energy of a rolling body as ½mv² only.

Related topics

• Oscillations and waves · Properties of matter · Special relativity · Fluid mechanics · Mathematical methods: vector calculus and differential equations · Electrodynamics and the Lagrangian of charged particles