Technopedia

The technologist’s encyclopedia

Dynamics

Dynamics relates forces to the acceleration they produce. It is the causal layer of mechanics: given the state (position, velocity) and the forces, it determines the future trajectory. The special case of zero acceleration (static equilibrium) and the Lagrangian/Hamiltonian reformulations are treated here, as they are the same physics expressed with different tools.

The kinematic quantities (position, velocity, acceleration) are defined in Kinematics. The force models that serve as input to the equations of motion (gravity, electromagnetism, contact forces) are catalogued in Interactions.

Authoritative sources

Newton’s laws

F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a}
Details

The second law is the fundamental equation. All other equations of motion (Lagrangian, Hamiltonian) are equivalent reformulations. The first law defines inertial frames; the third law constrains the force model (internal forces cancel in a closed system).

For a system of NN particles:

mia⃗i=F⃗iext+∑j≠iF⃗ijm_i \vec{a}_i = \vec{F}_i^{\text{ext}} + \sum_{j \neq i} \vec{F}_{ij}

Summing over all particles, internal forces cancel (third law), giving:

Ma⃗CM=F⃗netextM\vec{a}_{\text{CM}} = \vec{F}_{\text{net}}^{\text{ext}}

where MM is total mass and a⃗CM\vec{a}_{\text{CM}} is the center-of-mass acceleration.

Momentum and impulse

p⃗=mv⃗,F⃗=dp⃗dt\vec{p} = m\vec{v}, \quad \vec{F} = \frac{d\vec{p}}{dt}

Impulse: J⃗=∫F⃗ dt=Δp⃗\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}.

Details

The momentum form is more general than F=maF = ma (it handles variable-mass systems like rockets). Conservation of total momentum follows when F⃗netext=0\vec{F}_{\text{net}}^{\text{ext}} = 0.

Angular momentum

For a point particle at position r⃗\vec{r} with momentum p⃗\vec{p}:

L⃗=r⃗×p⃗,τ⃗=dL⃗dt\vec{L} = \vec{r} \times \vec{p}, \quad \vec{\tau} = \frac{d\vec{L}}{dt}

where τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F} is the torque about the origin. When τ⃗=0\vec{\tau} = 0, angular momentum is conserved.

Details

The angular momentum L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p} is a pseudovector perpendicular to the plane of motion. Its magnitude is L=mvrsin⁡αL = mvr\sin\alpha where α\alpha is the angle between r⃗\vec{r} and p⃗\vec{p}.

The torque equation τ⃗=dL⃗/dt\vec{\tau} = d\vec{L}/dt is the rotational analogue of F⃗=dp⃗/dt\vec{F} = d\vec{p}/dt. For a system of particles, the total angular momentum about the center of mass is conserved when the net external torque vanishes.

In the Lagrangian formulation, conservation of L⃗\vec{L} follows from rotational symmetry via Noether’s theorem (see below). The rigid-body extension L⃗=I ω⃗\vec{L} = \mathbf{I}\,\vec{\omega} (inertia tensor times angular velocity) is treated in Rigid-Body Mechanics.

Example: A figure skater pulling in their arms reduces the moment of inertia II, so ω\omega increases to keep L=IωL = I\omega constant.

Work and energy

W=∫F⃗⋅dr⃗,K=12mv2W = \int \vec{F} \cdot d\vec{r}, \quad K = \tfrac{1}{2}mv^2

Work–energy theorem: Wnet=ΔKW_{\text{net}} = \Delta K.

For conservative forces: W=−ΔUW = -\Delta U, and total mechanical energy E=K+UE = K + U is conserved.

Details

A force is conservative if ∇×F⃗=0\nabla \times \vec{F} = 0 (in a simply connected region), which guarantees the existence of a potential UU with F⃗=−∇U\vec{F} = -\nabla U. Non-conservative forces (friction, drag) dissipate mechanical energy into heat; see Thermodynamics.

Static equilibrium

When a⃗=0\vec{a} = 0 for all parts of a body:

∑F⃗=0,∑τ⃗=0\sum \vec{F} = 0, \quad \sum \vec{\tau} = 0
Details

Static equilibrium is the a=0a = 0 special case of dynamics. It applies to structures (beams, trusses, frames) where all accelerations vanish. The two conditions (force balance and moment balance) give up to 6 scalar equations in 3-D (3 force + 3 moment), which determine unknown reaction forces.

For a 2-D problem: 3 equations (∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0, ∑τO=0\sum \tau_O = 0).

Example: A uniform beam of mass mm and length LL rests on two supports at its ends. Each support carries mg/2mg/2.

Central forces and the two-body problem

A central force acts along the line joining two bodies: F⃗=f(r) r^\vec{F} = f(r)\,\hat{r}. The two-body problem (two masses m1m_1, m2m_2 interacting via a central force) reduces to a one-body problem with reduced mass:

μ=m1m2m1+m2\mu = \frac{m_1 m_2}{m_1 + m_2}

The relative coordinate r⃗=r⃗1−r⃗2\vec{r} = \vec{r}_1 - \vec{r}_2 obeys μ r⃗¨=F⃗(r)\mu\,\ddot{\vec{r}} = \vec{F}(r).

Details

Reduction: The center of mass moves uniformly (no external force). In the center-of-mass frame, the problem is equivalent to a single particle of mass μ\mu in the potential V(r)V(r).

Effective potential: Because angular momentum L=μr2θ˙L = \mu r^2 \dot{\theta} is conserved (central force → zero torque), the radial equation becomes:

μr¨=−dVdr+L2μr3\mu \ddot{r} = -\frac{dV}{dr} + \frac{L^2}{\mu r^3}

The term L2/(2μr2)L^2/(2\mu r^2) acts as a centrifugal barrier. The effective potential is:

Veff(r)=V(r)+L22μr2V_{\text{eff}}(r) = V(r) + \frac{L^2}{2\mu r^2}

Kepler’s laws follow for V(r)=−GMμ/rV(r) = -GM\mu/r:

  1. Orbits are conic sections (ellipse, parabola, hyperbola).
  2. Equal areas are swept in equal times (angular momentum conservation).
  3. T2∝a3T^2 \propto a^3 (period squared proportional to semi-major axis cubed).

The force law itself (Newton’s law of gravitation) is treated in Gravity; this section covers the orbital mechanics that follows from it.

Variational principle (Hamilton’s principle)

The actual path taken by a mechanical system between two fixed endpoints makes the action stationary:

S[q⃗(t)]=∫t1t2L(q⃗(t), q⃗˙(t), t) dtS[\vec{q}(t)] = \int_{t_1}^{t_2} L\bigl(\vec{q}(t),\, \dot{\vec{q}}(t),\, t\bigr)\, dt

That is, δS=0\delta S = 0 for all virtual variations δqi(t)\delta q_i(t) that vanish at the endpoints.

Details

Hamilton’s principle is the variational foundation of all of analytical mechanics. The Euler–Lagrange equations are not independent postulates; they are the Euler–Lagrange equations of calculus of variations applied to the functional SS.

Derivation sketch: Let qi(t)→qi(t)+ϵ ηi(t)q_i(t) \to q_i(t) + \epsilon\,\eta_i(t) with ηi(t1)=ηi(t2)=0\eta_i(t_1) = \eta_i(t_2) = 0. Requiring dS/dϵ∣ϵ=0=0dS/d\epsilon|_{\epsilon=0} = 0:

0=∫t1t2(∂L∂qi ηi+∂L∂q˙i η˙i)dt0 = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial q_i}\,\eta_i + \frac{\partial L}{\partial \dot{q}_i}\,\dot{\eta}_i\right) dt

Integrating the second term by parts and discarding the boundary term (since ηi\eta_i vanishes at endpoints):

0=∫t1t2(∂L∂qi−ddt∂L∂q˙i)ηi dt0 = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}\right)\eta_i\, dt

Since ηi\eta_i is arbitrary, the integrand must vanish, giving the Euler–Lagrange equations.

This variational viewpoint explains why the equations work: the true trajectory is the one that extremizes the action, just as a light ray extremizes optical path length (Fermat’s principle). It also generalizes naturally to fields (classical field theory) and to quantum mechanics (path integrals).

Lagrangian mechanics

For a system with generalized coordinates qiq_i and nn degrees of freedom:

L=T−V,ddt∂L∂q˙i−∂L∂qi=0L = T - V, \quad \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0
Details

The Lagrangian L=T−VL = T - V encodes the dynamics in a scalar. The Euler–Lagrange equations are equivalent to Newton’s second law but work naturally with constraints (e.g., a pendulum’s fixed length) and in any coordinate system.

Connection to Newton’s second law: For a single particle in Cartesian coordinates, T=12m(x˙2+y˙2+z˙2)T = \tfrac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2) and V=V(x,y,z)V = V(x,y,z). The Euler–Lagrange equation for coordinate xx:

ddt∂L∂x˙−∂L∂x=ddt(mx˙)−∂V∂x=mx¨+Fx=0\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = \frac{d}{dt}(m\dot{x}) - \frac{\partial V}{\partial x} = m\ddot{x} + F_x = 0

which is exactly mx¨=−∂V/∂x=Fxm\ddot{x} = -\partial V/\partial x = F_x. The same holds for yy and zz, recovering F⃗=ma⃗\vec{F} = m\vec{a}. The two formulations are the same physics; the Lagrangian form simply packages the three component equations into a coordinate-free variational statement.

Noether’s theorem: Every continuous symmetry of LL yields a conserved quantity.

This is the deep reason conservation laws exist: they are consequences of the symmetries of space and time, not independent postulates.

Hamiltonian mechanics and phase space

pi=∂L∂q˙i,H=∑piq˙i−Lp_i = \frac{\partial L}{\partial \dot{q}_i}, \quad H = \sum p_i \dot{q}_i - L

Hamilton’s equations:

q˙i=∂H∂pi,p˙i=−∂H∂qi\dot{q}_i = \frac{\partial H}{\partial p_i}, \quad \dot{p}_i = -\frac{\partial H}{\partial q_i}
Details

Phase space is the 2n2n-dimensional space of (q1,…,qn,p1,…,pn)(q_1, \ldots, q_n, p_1, \ldots, p_n). Hamilton’s equations describe a flow in this space. Liouville’s theorem states that the phase-space volume is preserved under this flow, which is the mechanical foundation of Statistical Mechanics.

The Hamiltonian HH equals the total energy T+VT + V when the potential does not depend on velocities and the coordinate transformation is time-independent.

Interactive simulation

Example problem

A block of mass mm slides down a frictionless incline of angle θ\theta. Find its acceleration and speed after traveling distance ss from rest.

Solution

Forces along the incline: mgsin⁡θmg\sin\theta. Newton’s second law: ma=mgsin⁡θma = mg\sin\theta, so a=gsin⁡θa = g\sin\theta.

Speed after distance ss: v2=2as=2gssin⁡θv^2 = 2as = 2gs\sin\theta, so v=2gssin⁡θv = \sqrt{2gs\sin\theta}.

Energy check: mgh=12mv2mgh = \tfrac{1}{2}mv^2 where h=ssin⁡θh = s\sin\theta. Same result.