Classical Mechanics
Classical mechanics is the branch of physics that describes the motion of macroscopic objects under the action of forces. Founded by Newton in the Principia (1687) and refined through the analytical frameworks of Lagrange, Hamilton, and others, it remains the foundation for engineering, orbital mechanics, and everyday physical reasoning. This skill covers kinematics, dynamics, energy, momentum, rotation, gravitation, and oscillations with worked examples and strategy guidance.
Agent affinity: newton (classical mechanics, Opus)
Concept IDs: phys-motion-kinematics, phys-newtons-laws, phys-projectile-motion, phys-circular-motion, phys-gravitation
Classical Mechanics at a Glance
| # | Topic | Key equations | Core idea | |---|---|---|---| | 1 | Kinematics (1D) | x = x_0 + v_0 t + (1/2)a t^2 | Describe motion without asking why | | 2 | Newton's Laws | F_net = ma | Force causes acceleration | | 3 | Free-body diagrams | Sum of forces = ma per axis | Visual decomposition of all forces | | 4 | Friction | f_s <= mu_s N, f_k = mu_k N | Resists relative motion | | 5 | Projectile motion | Independent x and y components | Parabolic trajectory under gravity | | 6 | Circular motion | a_c = v^2/r, F_c = mv^2/r | Centripetal acceleration toward center | | 7 | Rotational mechanics | tau = I alpha, L = I omega | Angular analogs of force and momentum | | 8 | Gravitation | F = GMm/r^2, orbits via Kepler's laws | Universal attraction | | 9 | Work-energy theorem | W_net = Delta KE | Net work equals change in kinetic energy | | 10 | Momentum | p = mv, impulse J = F Delta t | Conserved when no external forces act | | 11 | Collisions | Elastic: KE conserved. Inelastic: KE lost | Momentum always conserved in isolation | | 12 | Oscillations | x(t) = A cos(omega t + phi) | Restoring force proportional to displacement |
Topic 1 — Kinematics in One Dimension
Core idea. Kinematics describes how objects move — position, velocity, acceleration — without explaining why. It is the prerequisite for dynamics.
The five kinematic equations (constant acceleration):
- v = v_0 + at
- x = x_0 + v_0 t + (1/2)a t^2
- v^2 = v_0^2 + 2a(x - x_0)
- x = x_0 + (1/2)(v + v_0)t
- x = x_0 + vt - (1/2)a t^2
Strategy. List knowns (three of five: x, x_0, v, v_0, a, t). Pick the equation that uses your three knowns and one unknown. Solve algebraically before substituting numbers.
Worked example. A car accelerates from rest at 3 m/s^2 for 5 seconds. How far does it travel?
Solution. Known: v_0 = 0, a = 3 m/s^2, t = 5 s. Unknown: x - x_0. Use equation 2: x - x_0 = v_0 t + (1/2)a t^2 = 0 + (1/2)(3)(25) = 37.5 m.
Extension to 2D/3D. Treat each coordinate axis independently. The kinematic equations apply component-wise: x(t), y(t), z(t) each evolve under their own acceleration component.
Topic 2 — Newton's Three Laws
First Law (Inertia). An object at rest stays at rest, and an object in uniform motion stays in uniform motion, unless acted on by a net external force. This defines inertial reference frames.
Second Law (F = ma). The net force on an object equals its mass times its acceleration: F_net = ma. This is the engine of classical mechanics — given forces, compute acceleration; given acceleration, infer forces.
Third Law (Action-Reaction). For every force that object A exerts on object B, object B exerts an equal and opposite force on A. These forces act on different objects and never cancel each other in a free-body diagram.
Worked example. A 5 kg block is pushed across a frictionless surface by a 20 N horizontal force. Find its acceleration.
Solution. F_net = ma. The only horizontal force is 20 N. So a = F/m = 20/5 = 4 m/s^2.
Common mistake. Applying action-reaction pairs to the same object. If you push a wall and the wall pushes back, those forces act on different objects (you and the wall). They do not cancel.
Topic 3 — Free-Body Diagrams
Purpose. A free-body diagram (FBD) isolates a single object and shows every force acting on it as a vector arrow. It is the essential first step for any force problem.
Procedure:
- Identify the object of interest.
- Draw it as a dot or simple shape.
- Draw all forces acting ON that object: weight (mg, downward), normal force (perpendicular to surface), friction (parallel to surface, opposing motion), tension, applied forces, spring forces.
- Choose a coordinate system (often tilted for inclined planes).
- Decompose forces into components.
- Apply Newton's second law per axis.
Worked example. A 10 kg block rests on a 30-degree incline. Find the normal force and the friction force required to keep it stationary.
Solution. Tilt coordinates: x along the incline (positive up-slope), y perpendicular to incline.
Forces: Weight W = mg = 98 N (vertically down). Decompose: W_x = mg sin(30) = 49 N (down-slope), W_y = mg cos(30) = 84.9 N (into surface).
y-axis: N - W_y = 0, so N = 84.9 N. x-axis: f_s - W_x = 0, so f_s = 49 N (up-slope).
Topic 4 — Friction
Static friction: f_s <= mu_s N. The force adjusts up to a maximum before sliding begins.
Kinetic friction: f_k = mu_k N. Constant magnitude once sliding, always opposing relative motion.
Key insight. Static friction is not always at its maximum. It is whatever value is needed to prevent motion, up to the limit mu_s N. Only at the threshold of motion does f_s = mu_s N.
Worked example. A 20 kg crate on a flat floor has mu_s = 0.4 and mu_k = 0.3. You push horizontally with 100 N. Does it slide? If so, what is its acceleration?
Solution. Normal force: N = mg = 196 N. Maximum static friction: f_s,max = 0.4 * 196 = 78.4 N. Applied force (100 N) > f_s,max (78.4 N), so yes, it slides.
Once sliding: f_k = 0.3 * 196 = 58.8 N. Net force = 100 - 58.8 = 41.2 N. Acceleration a = 41.2/20 = 2.06 m/s^2.
Topic 5 — Projectile Motion
Setup. An object launched with initial velocity v_0 at angle theta above horizontal, subject only to gravity (no air resistance).
Decomposition: v_0x = v_0 cos(theta), v_0y = v_0 sin(theta). Horizontal: a_x = 0. Vertical: a_y = -g.
Key results:
- Time of flight: T = 2 v_0 sin(theta) / g
- Maximum height: H = v_0^2 sin^2(theta) / (2g)
- Range: R = v_0^2 sin(2 theta) / g
- The range is maximized at theta = 45 degrees.
- Complementary angles (e.g., 30 and 60 degrees) give the same range.
Worked example. A ball is launched at 20 m/s at 60 degrees above horizontal. Find the range and maximum height. Use g = 9.8 m/s^2.
Solution. R = (20^2)(sin 120)/9.8 = (400)(0.866)/9.8 = 35.3 m. H = (20^2)(sin^2 60)/(2 * 9.8) = (400)(0.75)/19.6 = 15.3 m.
When NOT to use these formulas. When the launch and landing heights differ, or when air resistance matters. In those cases, return to the component equations and solve directly.
Topic 6 — Circular Motion and Centripetal Force
Uniform circular motion. An object moving at constant speed v in a circle of radius r has centripetal acceleration a_c = v^2/r directed toward the center. The centripetal force is F_c = mv^2/r.
Critical point. Centripetal force is not a new kind of force. It is the net radial component of whatever forces are present — gravity, tension, normal force, friction.
Worked example. A 1500 kg car rounds a flat curve of radius 50 m at 15 m/s. What friction force is required?
Solution. The friction force provides the centripetal force: f = mv^2/r = (1500)(225)/50 = 6750 N.
Check: Is this achievable? f <= mu_s N = mu_s mg. So mu_s >= 6750/(1500 * 9.8) = 0.459. Dry rubber on asphalt has mu_s around 0.7, so yes.
Non-uniform circular motion. When speed changes, there is also a tangential acceleration a_t = dv/dt. The total acceleration is the vector sum of radial (centripetal) and tangential components.
Topic 7 — Rotational Mechanics
Angular analogs. Every translational quantity has a rotational counterpart:
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