Classical Mechanics

Position, velocity, acceleration

$\mathbf{r}(t)$

The position vector at time $t$ — three numbers $(x, y, z)$ that locate the particle. Bold letters mean "vector, not a single number."

$\dfrac{d\mathbf{r}}{dt}$

The time derivative of position — how fast each component of $\mathbf{r}$ is changing with time. A vector, called the velocity.

$\dfrac{d^2 \mathbf{r}}{dt^2}$

The second time derivative of position — the derivative of the velocity. Also a vector, called the acceleration.

$t$

Time. A single real number. In classical mechanics, every observer shares the same clock.

Why two derivatives. Velocity tells you where the particle is heading right now. Acceleration tells you how that heading is changing — whether you are speeding up, slowing down, or turning. A car driving in a straight line at 60 mph has velocity but zero acceleration. A car going around a curve at a steady 60 mph still has acceleration, because the direction of $\mathbf{v}$ is changing even though its magnitude is not.

Vector components

$\mathbf{r}$

Position vector — the arrow from the origin to the object's location. Components are coordinates $(r_x, r_y, r_z)$.

$r_x, r_y, r_z$

Signed real numbers, each measured in the same unit as the vector (metres for position, m/s for velocity, m/s² for acceleration).

bold vs. italic

Bold $\mathbf{v}$ is a vector; italic $v$ is its magnitude. The direction is gone. Never substitute one for the other in an equation.

Kinematic vectors. Position $\mathbf{r}$ locates an object. Velocity $\mathbf{v} = d\mathbf{r}/dt$ is how fast and in what direction it moves. Acceleration $\mathbf{a} = d\mathbf{v}/dt$ is how fast and in what direction the velocity is changing. Force $\mathbf{F} = m\mathbf{a}$ has a direction — push left and the object accelerates left. A scalar temperature of 300 K has no direction; it just is. These are fundamentally different objects.

Vector addition

tip-to-tail

A geometric construction: slide $\mathbf{b}$ so its tail sits at the tip of $\mathbf{a}$. The sum is the new arrow from start to finish.

component-wise

The algebraic equivalent: add matching components independently. Both methods give the same answer.

commutativity

$\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$. The order doesn't matter — it makes the same parallelogram.

Kinematics application. Two forces act on a box: $\mathbf{F}_1 = (3, 4)$ N and $\mathbf{F}_2 = (-1, 2)$ N. The net force is $\mathbf{F}_1 + \mathbf{F}_2 = (2, 6)$ N with magnitude $\sqrt{4 + 36} = \sqrt{40} \approx 6.3$ N pointing northeast. If you had just added the magnitudes (5 + 2.2 = 7.2 N) you would have been wrong by 14% and in the wrong direction.

Decomposition into components

$v$

Speed — the magnitude $|\mathbf{v}|$. A non-negative scalar.

$\theta$

The angle the velocity makes with the positive $x$-axis (horizontal), measured counter-clockwise.

$v_x = v\cos\theta$

The horizontal component. Zero when $\theta = 90°$ (straight up); equals $v$ when $\theta = 0°$ (horizontal).

$v_y = v\sin\theta$

The vertical component. Equals $v$ when $\theta = 90°$; zero when $\theta = 0°$.

Why decomposition unlocks kinematics. Once you split a 2D motion into its $x$ and $y$ components, each component evolves completely independently. Projectile motion becomes two separate 1D problems: $x$ moves with constant velocity $v_x$; $y$ accelerates downward at $g$. Without decomposition, the two are hopelessly tangled.

Magnitude and unit vector

$|\mathbf{v}|$

The magnitude — a non-negative scalar. Also written as just $v$ when clear from context.

$\hat{\mathbf{v}}$

The unit vector in the direction of $\mathbf{v}$: $\hat{\mathbf{v}} = \mathbf{v}/|\mathbf{v}|$. Always has magnitude 1. A pure direction, no scale.

$\mathbf{v} = v\,\hat{\mathbf{v}}$

Every vector splits into "how big" ($v$) times "which way" ($\hat{\mathbf{v}}$). This is the cleanest way to read a velocity: 30 m/s to the northeast.

Uniformly accelerated motion

$v_0$, $x_0$

Initial velocity and position (at $t = 0$). These are the constants of integration that the antiderivatives leave behind.

$a$

The (constant) acceleration.

$v(t)$

Velocity at time $t$ — a linear function of time.

$x(t)$

Position at time $t$ — a quadratic function of time. The $\tfrac{1}{2} a t^2$ comes from integrating the linear $v_0 + at$ once more.

The one near the Earth's surface. Freely falling objects near the Earth have $a = g \approx 9.81\,\text{m/s}^2$ pointing downward. Drop a rock from rest ($v_0 = 0$) and after $t$ seconds it has fallen $\tfrac{1}{2} g t^2$ meters with speed $g t$. You can check the first second: you fall about 4.9 meters and are moving at about 9.8 m/s by the end. Galileo verified this rolling balls down inclined planes in the early 1600s, long before anyone had a stopwatch accurate to tenths of a second.

Projectile trajectory

$\theta$

Launch angle above the horizontal, in radians (or degrees if you convert).

$v_0 \cos\theta$, $v_0 \sin\theta$

Horizontal and vertical components of the initial velocity, via elementary trigonometry.

$g$

Gravitational acceleration near Earth's surface, $9.81\,\text{m/s}^2$.

$-\tfrac{1}{2} g t^2$

Vertical drop due to gravity. Negative because up is positive and gravity pulls down.

The shape. Eliminating $t$ between the two equations gives $y$ as a quadratic function of $x$ — a parabola. The projectile rises, peaks, and falls symmetrically. Maximum range (on flat ground) occurs at $\theta = 45^\circ$, where horizontal and vertical components are balanced. You will verify this in the interactive figure below.

Newton's second law

$\mathbf{F}$

The net force — the vector sum of all forces acting on the body.

$\mathbf{p} = m\mathbf{v}$

Linear momentum. A vector quantity, pointing in the direction of motion.

$\dfrac{d\mathbf{p}}{dt}$

Time derivative of momentum. If mass is constant, this is $m \, d\mathbf{v}/dt = m\mathbf{a}$.

$m\mathbf{a}$

Mass times acceleration. The familiar high-school form, valid only when $m$ is constant.

When mass isn't constant. The full form with $d\mathbf{p}/dt$ matters when you have systems whose mass is changing — rockets ejecting fuel, conveyor belts catching falling sand, accreting black holes. For those, use the momentum form. For everyday engineering where mass stays put, $\mathbf{F} = m\mathbf{a}$ is fine.

Atwood machine force balance

$T$

Tension in the string. Because the string is massless and the pulley frictionless, the tension is the same everywhere along the string.

$m_i g$

Weight of mass $i$ — the downward gravitational force, with $g = 9.81\,\text{m/s}^2$.

$a$

The (signed) acceleration of $m_1$; the acceleration of $m_2$ is $-a$ because the string constrains them.

The trick. Two unknowns ($T$ and $a$), two equations. Add the two equations to eliminate $T$, or subtract to eliminate $a$. This "draw a free-body diagram, write $\mathbf{F} = m\mathbf{a}$ for each body, then solve the system" procedure is how essentially every classical mechanics problem starts.

Atwood acceleration

$m_2 - m_1$

Difference in masses. When the masses are equal this is zero, and so is the acceleration — the system is in equilibrium.

$m_1 + m_2$

Total mass the force has to accelerate.

$g$

The gravitational acceleration that ultimately drives everything.

Why Atwood liked this. By choosing $m_1$ and $m_2$ close in value, you can make $a$ arbitrarily smaller than $g$. Atwood used this to slow down free fall enough to time it with 1780s-era clocks and confirm Galileo's $x = \tfrac{1}{2} g t^2$. It was an early triumph of experimental precision.

Incline acceleration

$\theta$

Incline angle. $\theta = 0$ is flat ground; $\theta = 90^\circ$ is a vertical drop.

$g \sin\theta$

The component of gravitational acceleration along the slope. At $\theta = 0$ this is zero (nothing happens); at $\theta = 90^\circ$ it is $g$ (free fall).

$m g \cos\theta$

The perpendicular component — squashed into the ramp and cancelled by the ramp's normal force.

Why it does not depend on mass. The same logic as Galileo's leaning tower: doubling the mass doubles both the gravitational force and the inertia, and they cancel in $F = ma$. Heavy and light blocks on a frictionless slope slide side by side.

Work done by a force

$W$

Work — a scalar (single number), with SI unit joule, $1\,\text{J} = 1\,\text{N}\cdot\text{m}$.

$\mathbf{F} \cdot d\mathbf{r}$

The dot product of the force with a tiny displacement along the path. The dot product extracts the component of force that is parallel to the motion.

$\int_{\mathbf{r}_1}^{\mathbf{r}_2}$

A line integral along the actual path from position $\mathbf{r}_1$ to position $\mathbf{r}_2$.

Why the dot product. If you push sideways on a refrigerator you are sliding straight ahead, you do no work. The force component perpendicular to the motion accomplishes nothing. Only the component in the direction of travel transfers energy, and the dot product $\mathbf{F} \cdot d\mathbf{r} = |\mathbf{F}||d\mathbf{r}|\cos\phi$ pulls out exactly that component.

Kinetic energy

$T$

Kinetic energy — a scalar with units of joules. Some textbooks call it $K$ or $E_k$; we use $T$ because it is the Lagrangian-mechanics convention.

$m$

Mass of the particle.

$v^2$

Speed squared — the dot product $\mathbf{v} \cdot \mathbf{v}$. Always nonnegative.

Why squared. Kinetic energy grows with the square of speed, not linearly. Double the speed of a car and it has four times the kinetic energy to dissipate on a crash. This is why highway crashes at 100 km/h are not just "twice as bad" as 50 km/h crashes — they're four times as bad. The $v^2$ is also why braking distance grows quadratically with speed.

Work-energy theorem

$W_\text{net}$

The total work done by the net force along the path.

$\Delta T$

The change in kinetic energy between the start and end of the path.

Where it comes from. Take Newton's second law $\mathbf{F} = m\,d\mathbf{v}/dt$, dot both sides with $d\mathbf{r} = \mathbf{v}\,dt$, integrate along the path, and the right side is $\int m \mathbf{v}\cdot d\mathbf{v} = \tfrac{1}{2} m v^2$ evaluated between the endpoints. It is Newton's second law rewritten so only endpoints matter.

Force from a potential

$V(\mathbf{r})$

Potential energy as a function of position. A scalar field — one number per point in space.

$\nabla V$

The gradient of $V$ — a vector pointing in the direction of steepest increase of $V$.

$-\nabla V$

The force — pointing in the direction of steepest decrease. Objects "roll downhill" in the potential.

The hill picture. Imagine $V$ as the height of a landscape. The force on a marble placed on that landscape points downhill, in the direction where the height drops fastest. Gravity near Earth gives $V = m g h$ (a tilted plane); a spring gives $V = \tfrac{1}{2} k x^2$ (a parabolic bowl); a planet orbiting the sun sits in a $V \propto -1/r$ well. All three have their forces given by $-\nabla V$.

Conservation of mechanical energy

$E$

Total mechanical energy — kinetic plus potential. In joules.

$T = \tfrac{1}{2} m v^2$

Kinetic energy: energy of motion.

$V(\mathbf{r})$

Potential energy: energy of position, stored in whatever field the force comes from.

The trade. As a ball falls, $V$ decreases and $T$ increases by the same amount. At the bottom of a swing a pendulum is moving fastest and has minimum potential; at the top of the swing it has maximum potential and zero speed. The sum stays put. This is the single most useful equation in elementary mechanics — it lets you find speeds at specific positions without ever integrating a differential equation.

Impulse

$\mathbf{J}$

The impulse delivered by a force over a time interval — a vector with units of kg·m/s (or equivalently N·s).

$\Delta \mathbf{p}$

The change in momentum of the object.

$\int \mathbf{F}\,dt$

The time integral of the force. If the force is constant, this is just $\mathbf{F}\Delta t$; if not, you integrate whatever time profile you have.

Why engineers love impulse. For short collisions (airbags, bumpers, catching a ball with bent knees), the change in momentum is fixed — it equals the final momentum minus the initial momentum. But the peak force depends on how long the collision lasts. Spread the same impulse over a longer time and the peak force drops. That is what crumple zones, bike helmets, and trampolines all exploit.

1D elastic collision (target at rest)

$v_1, v_1'$

Speed of the projectile mass before and after the collision.

$v_2'$

Speed of the target mass after the collision. It was at rest before.

$(m_1 - m_2) / (m_1 + m_2)$

A dimensionless ratio between $-1$ and $+1$. Controls how much of the projectile's velocity is retained vs transferred.

Three limiting cases. Equal masses ($m_1 = m_2$): $v_1' = 0$ and $v_2' = v_1$ — the projectile stops dead and the target takes off with the original speed. This is Newton's cradle. Projectile much heavier ($m_1 \gg m_2$): $v_1' \approx v_1$ and $v_2' \approx 2 v_1$ — the target is kicked ahead at twice the incoming speed while the projectile barely slows. Projectile much lighter ($m_1 \ll m_2$): $v_1' \approx -v_1$ and $v_2' \approx 0$ — the projectile bounces back and the target is unmoved. This is a ball bouncing off a wall.

Torque and angular momentum

$\boldsymbol{\tau}$

Torque. A vector, computed as the cross product of the position vector (from the pivot) with the applied force. SI units: N·m.

$\mathbf{r}$

Position vector from the pivot point to where the force is applied.

$\mathbf{F}$

Applied force.

$\mathbf{L}$

Angular momentum — the rotational analog of linear momentum.

$\times$

Vector cross product. The result is perpendicular to both inputs, with magnitude $|\mathbf{r}||\mathbf{F}|\sin\phi$ where $\phi$ is the angle between them.

Why the lever arm. Pushing a door at the hinge does nothing; pushing at the handle swings it easily. The torque depends on the distance from the pivot ($|\mathbf{r}|$), the force magnitude, and the sine of the angle between them — pushing along the door does nothing, pushing perpendicular to it is maximally effective. Cross products were invented exactly to bookkeep this geometry.

Rotation about a fixed axis

$I$

Moment of inertia — the rotational analog of mass. For a point mass $m$ at distance $r$ from the axis, $I = m r^2$. For extended bodies, integrate $\int r^2\,dm$ over the mass distribution.

$\alpha$

Angular acceleration — the time derivative of $\omega$, in rad/s².

$\omega$

Angular velocity — how fast the body is rotating, in rad/s.

$T_\text{rot}$

Rotational kinetic energy — the energy stored in rotation.

Why moment of inertia depends on shape. Two objects with the same mass can have very different moments of inertia if the mass is distributed differently. A solid disk has $I = \tfrac{1}{2} M R^2$; a hoop of the same mass and radius has $I = M R^2$, twice as large. This is why a hoop rolls down a ramp slower than a disk — more of its energy goes into rotation, less into translation.

The Lagrangian

$L$

The Lagrangian — a single scalar function that encodes the entire dynamics. Not to be confused with angular momentum, which is also often written $L$.

$T$

Kinetic energy of the system, expressed in the chosen coordinates.

$V$

Potential energy, also expressed in the chosen coordinates.

$q_i$

Generalized coordinates — any convenient set of independent variables that describe the configuration of the system. For a pendulum, the angle $\theta$ is a natural choice; for a bead on a wire, distance along the wire.

$\dot{q}_i$

Time derivative of $q_i$. A dot means "derivative with respect to time," Newton's notation for rates.

The minus sign. The Lagrangian is $T - V$, not $T + V$. This looks strange — it is not the total energy. It is a mathematical auxiliary whose stationary points pick out the true trajectory, via the principle of least action discussed below.

Euler-Lagrange equations

$\dfrac{\partial L}{\partial \dot q_i}$

Partial derivative of the Lagrangian with respect to the velocity $\dot q_i$, treating all other variables as constant. Called the "generalized momentum" conjugate to $q_i$.

$\dfrac{d}{dt}$

Total time derivative of whatever follows.

$\dfrac{\partial L}{\partial q_i}$

Partial derivative of $L$ with respect to the coordinate $q_i$. Called the "generalized force."

$= 0$

One equation per coordinate. If you have $n$ coordinates, you get $n$ equations.

Why this gives the right answer. Writing $T = \tfrac{1}{2} m \dot x^2$ and $V = V(x)$, the Euler-Lagrange equation in one dimension gives $m\ddot x + V'(x) = 0$, which is exactly $F = -dV/dx = m a$. So Lagrangian mechanics reproduces Newton's second law in the simple cases. Its power appears in constrained systems: for a pendulum, write $T$ and $V$ as functions of the angle $\theta$ (not $x$ and $y$), plug into the Euler-Lagrange equation, and out pops the pendulum equation of motion with the tension in the string automatically eliminated. You never draw a free-body diagram.

The action

$S[q]$

The action — a single number assigned to each possible trajectory $q(t)$. The square brackets signal that $S$ is a function of a whole path, not of a point.

$\int_{t_1}^{t_2}$

Time integral from the start to the end of the motion.

$L\,dt$

The Lagrangian evaluated along the path, times an infinitesimal time element. Each instant contributes a little to the total action.

The principle. Among all conceivable paths connecting fixed endpoints at $t_1$ and $t_2$, the true trajectory is the one for which $S$ is stationary — a minimum, maximum, or saddle point with respect to small perturbations of the path. This is Hamilton's principle, also called the principle of least action. The Euler-Lagrange equations are exactly the condition for $S$ to be stationary. It is a kind of optimization problem: nature picks the path whose Lagrangian integral has extremal value.