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45

Kinematics

Field: Newtonian Mechanics

The branch of mechanics that describes the motion of points, bodies (objects), and systems of bodies without considering the forces that cause them to move.

Sequence of Expressions

Definition

Displacement

Let r:[t1,t2]→R3\mathbf{r}: [t_1, t_2] \to \mathbb{R}^3 be a continuous and differentiable position vector function describing the trajectory of a particle in Cartesian coordinates. The displacement vector Δr\Delta \mathbf{r} between time t1t_1 and time t2t_2 is formally defined as the difference between the final position vector r(t2)\mathbf{r}(t_2) and the initial position vector r(t1)\mathbf{r}(t_1): \n\nΔr=r(t2)−r(t1)\Delta \mathbf{r} = \mathbf{r}(t_2) - \mathbf{r}(t_1) \n\nIf r(t)=⟨x(t),y(t),z(t)⟩\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle, then the components of the displacement are given by:\n\nΔr=⟨x(t2)−x(t1),y(t2)−y(t1),z(t2)−z(t1)⟩\Delta \mathbf{r} = \langle x(t_2) - x(t_1), y(t_2) - y(t_1), z(t_2) - z(t_1) \rangle
Definition

Velocity

Let r:I→Rn\mathbf{r}: I \to \mathbb{R}^n be the position vector of a particle, where I=[t0,tf]⊂RI = [t_0, t_f] \subset \mathbb{R} is the time interval. The velocity vector v(t)\mathbf{v}(t) is defined as the first-order time derivative of the position vector: v(t)=drdt=lim⁡Δt→0r(t+Δt)−r(t)Δt.\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}. If r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)), then the components are given by: v(t)=(dxdt,dydt,dzdt)=(x˙(t),y˙(t),z˙(t)).\mathbf{v}(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = \left( \dot{x}(t), \dot{y}(t), \dot{z}(t) \right). The magnitude (speed) is then the Euclidean norm: ∣v(t)∣=(dxdt)2+(dydt)2+(dzdt)2.|\mathbf{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}.
Definition

Acceleration

Let x:I→R3\mathbf{x}: I \to \mathbb{R}^3 be a differentiable position vector, where I=[t0,tf]⊂RI = [t_0, t_f] \subset \mathbb{R} is the time interval. Define the velocity vector v(t)\mathbf{v}(t) as the first derivative of x(t)\mathbf{x}(t) with respect to time tt: v(t)=dxdt\mathbf{v}(t) = \frac{d\mathbf{x}}{dt}. The acceleration vector a(t)\mathbf{a}(t) is then defined as the time derivative of the velocity vector: a(t)=dvdt=d2xdt2.\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{x}}{dt^2}. Furthermore, if a\mathbf{a} is expressed in Cartesian coordinates, a(t)=(ax(t),ay(t),az(t))\mathbf{a}(t) = (a_x(t), a_y(t), a_z(t)), where ai(t)=d2xidt2a_i(t) = \frac{d^2x_i}{dt^2} for i∈{x,y,z}i \in \{x, y, z\}.
Definition

Trajectory

Let I=[t0,tf]⊂RI = [t_0, t_f] \subset \mathbb{R} be the time interval. Define the position vector r:I→R3\mathbf{r}: I \to \mathbb{R}^3 such that r(t)=⟨x(t),y(t),z(t)⟩\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle, where x(t),y(t),z(t)x(t), y(t), z(t) are continuously differentiable functions of tt. The trajectory T\mathcal{T} is the image set of this mapping: T=r(I)={r(t)∣t∈I}.\mathcal{T} = \mathbf{r}(I) = \{\mathbf{r}(t) \mid t \in I\}. Furthermore, the velocity vector v(t)\mathbf{v}(t) and acceleration vector a(t)\mathbf{a}(t) are defined by the derivatives: v(t)=drdt=r′(t)\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \mathbf{r}'(t) and a(t)=d2rdt2=r′′(t).\mathbf{a}(t) = \frac{d^2\mathbf{r}}{dt^2} = \mathbf{r}''(t).
Definition

Time

Let MM be the state space of the physical system, and let x:I→M\boldsymbol{x}: I \to M be the trajectory, where I=[t0,tf]⊂RI = [t_0, t_f] \subset \mathbb{R} is the time interval. Define the time parameter tt such that it induces a differentiable structure on the system's evolution. The state x(t)∈Rn\boldsymbol{x}(t) \in \boldsymbol{R}^n is governed by the system of first-order ordinary differential equations (ODEs): \begin{equation} \frac{d\boldsymbol{x}}{dt} = \boldsymbol{f}(\boldsymbol{x}, t) \end{equation} where f:M×I→Rn\boldsymbol{f}: M \times I \to \mathbb{R}^n is the vector field representing the instantaneous rate of change of the state. The time tt is the independent variable parameterizing the flow ρt(x0)\boldsymbol{\rho}_t(\boldsymbol{x}_0) generated by the vector field f\boldsymbol{f}, satisfying the initial value problem x(t0)=x0\boldsymbol{x}(t_0) = \boldsymbol{x}_0. Furthermore, the time elapsed Δt\Delta t between two states x(t1)\boldsymbol{x}(t_1) and x(t2)\boldsymbol{x}(t_2) is defined by the integral of the unit time measure: \begin{equation} \Delta t = \int_{t_1}^{t_2} dt \end{equation} such that t2>t1t_2 > t_1 implies Δt>0\Delta t > 0.
Let SIS_I be the inertial reference frame defined by the basis vectors eI=(eIx,eIy,eIz)\mathbf{e}_I = (\mathbf{e}_{Ix}, \mathbf{e}_{Iy}, \mathbf{e}_{Iz}) and the origin RI(t)\mathbf{R}_I(t). Let SS be the moving reference frame, whose origin R(t)\mathbf{R}(t) and basis vectors e=(ex,ey,ez)\mathbf{e} = (\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z) are time-dependent. The position vector r\mathbf{r} of a point PP in SS is related to the position vector R\mathbf{R} of the origin of SS in SIS_I by the transformation: r(t)=R(t)+ρ(t)\mathbf{r}(t) = \mathbf{R}(t) + \boldsymbol{\rho}(t), where ρ(t)\boldsymbol{\rho}(t) is the vector from the origin of SS to PP, expressed in SS's coordinates. The transformation between the coordinate systems is given by the rotation matrix RSI(t)∈SO(3)\mathbf{R}_{SI}(t) \in SO(3), such that e=RSI(t)eI\mathbf{e} = \mathbf{R}_{SI}(t) \mathbf{e}_I. The velocity v\mathbf{v} of PP in SIS_I is then derived using the transport theorem (or relative velocity formula): v=drdt=dRdt+vrel+∇R⋅vrot\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d\mathbf{R}}{dt} + \mathbf{v}_{rel} + \boldsymbol{\boldsymbol{\nabla}}\mathbf{R} \cdot \mathbf{v}_{rot} where vrel\mathbf{v}_{rel} is the velocity of PP relative to SS, and ∇R⋅vrot\boldsymbol{\boldsymbol{\nabla}}\mathbf{R} \cdot \mathbf{v}_{rot} represents the contribution from the rotation of the frame SS itself, defined by the angular velocity ω=12(ex×deydt+ey×dezdt+ez×dexdt)\boldsymbol{\omega} = \frac{1}{2} \left(\mathbf{e}_x \times \frac{d\mathbf{e}_y}{dt} + \mathbf{e}_y \times \frac{d\mathbf{e}_z}{dt} + \mathbf{e}_z \times \frac{d\mathbf{e}_x}{dt}\right).
Let x(t)∈Rn\mathbf{x}(t) \in \mathbb{R}^n be the position vector of a particle, v(t)=dxdt\mathbf{v}(t) = \frac{d\mathbf{x}}{dt} its velocity vector, and a(t)=dvdt\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} its acceleration vector. Assume the acceleration is constant, i.e., a(t)=a0∈Rn\mathbf{a}(t) = \mathbf{a}_0 \in \mathbb{R}^n. Then, the following relationships hold for the initial conditions x(0)=x0\mathbf{x}(0) = \mathbf{x}_0 and v(0)=v0\mathbf{v}(0) = \mathbf{v}_0: \begin{enumerate} \item \textbf{Velocity:} v(t)=v0+∫0ta0τdτ=v0+a0t\mathbf{v}(t) = \mathbf{v}_0 + \int_{0}^{t} \mathbf{a}_0 \tau d\tau = \mathbf{v}_0 + \mathbf{a}_0 t. \item \textbf{Position:} x(t)=x0+∫0tv(τ)dτ=x0+v0t+12a0t2\mathbf{x}(t) = \mathbf{x}_0 + \int_{0}^{t} \mathbf{v}(\tau) d\tau = \mathbf{x}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a}_0 t^2. \item \textbf{Displacement (Time-independent):} x(t)−x0=x0+v0t+12a0t2−x0= v0t+12a0t2\mathbf{x}(t) - \mathbf{x}_0 = \mathbf{x}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a}_0 t^2 - \mathbf{x}_0 = \ \mathbf{v}_0 t + \frac{1}{2} \mathbf{a}_0 t^2. (This is derived by eliminating tt between the first two equations, yielding the scalar form vf2=vi2+2a(xf−xi)v_f^2 = v_i^2 + 2a(x_f - x_i) when considering the magnitude squared, ∣v(t)−v0∣2=∣a0t∣2|\mathbf{v}(t) - \mathbf{v}_0|^2 = |\mathbf{a}_0 t|^2).
Theorem

Free Fall

Let r⃗(t)=⟨x(t),y(t),z(t)⟩∈R3\vec{r}(t) = \langle x(t), y(t), z(t) \rangle \in \mathbb{R}^3 be the position vector of a particle of mass mm at time tt. The gravitational force F⃗g\vec{F}_g is defined by F⃗g=mg⃗\vec{F}_g = m\vec{g}, where g⃗=⟨0,0,−g⟩\vec{g} = \langle 0, 0, -g \rangle and g=GMR2g = \frac{GM}{R^2} (assuming Earth-like gravity). By Newton's Second Law, the equation of motion is:\d2r⃗dt2=F⃗gm=g⃗\frac{d^2\vec{r}}{dt^2} = \frac{\vec{F}_g}{m} = \vec{g}. Integrating this constant acceleration yields the velocity v⃗(t)=dr⃗dt=v⃗0+g⃗t\vec{v}(t) = \frac{d\vec{r}}{dt} = \vec{v}_0 + \vec{g}t, and the position r⃗(t)=r⃗0+v⃗0t+12g⃗t2\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{g}t^2. Specifically, for initial position r⃗0=⟨x0,y0,z0⟩\vec{r}_0 = \langle x_0, y_0, z_0 \rangle and initial velocity v⃗0=⟨v0x,v0y,v0z⟩\vec{v}_0 = \langle v_{0x}, v_{0y}, v_{0z} \rangle, the position is:\r⃗(t)=⟨x0+v0xt,y0+v0yt,z0+v0zt−12gt2⟩.\vec{r}(t) = \langle x_0 + v_{0x}t, y_0 + v_{0y}t, z_0 + v_{0z}t - \frac{1}{2}gt^2 \rangle.
Let r:I→Rn\mathbf{r}: I \to \mathbb{R}^n be the position vector of a particle, where I=[t0,tf]⊂RI = [t_0, t_f] \subset \mathbb{R} is the time interval. The condition for Uniform Motion is defined by the constancy of the velocity vector v(t)=drdt\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}. Mathematically, this requires that the acceleration vector a(t)\mathbf{a}(t) vanishes identically: a(t)=dvdt=d2rdt2=0\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} = \mathbf{0}. Integrating this second-order ordinary differential equation (ODE) yields the general solution for the position r(t)\mathbf{r}(t): \n\nr(t)=r0+v0(t−t0)\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 (t - t_0) \n\nwhere r0=r(t0)\mathbf{r}_0 = \mathbf{r}(t_0) is the initial position vector, and v0=v(t0)\mathbf{v}_0 = \mathbf{v}(t_0) is the constant initial velocity vector. This solution defines a trajectory that is a straight line parameterized linearly by time.
Let r(t)∈R3\mathbf{r}(t) \in \mathbb{R}^3 be the position vector of a particle at time t∈Rt \in \mathbb{R}. Define the acceleration vector a(t)\mathbf{a}(t) such that a(t)=a0\mathbf{a}(t) = \mathbf{a}_0, where a0∈R3\mathbf{a}_0 \in \mathbb{R}^3 is a constant vector. The velocity vector v(t)\mathbf{v}(t) and the position vector r(t)\mathbf{r}(t) are then determined by the following differential equations and initial conditions:\\begin{align*} \frac{d^2\mathbf{r}}{dt^2} &= \mathbf{a}_0 \\ \frac{d\mathbf{r}}{dt} &= \mathbf{v}(t) \\ \mathbf{r}(0) &= \mathbf{r}_0 \\ \mathbf{v}(0) &= \mathbf{v}_0 \end{align*}\newline\text{The unique solution to this system is given by the explicit kinematic equations:}\newline\mathbf{r}(t) &= \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a}_0 t^2 \\ \mathbf{v}(t) &= \mathbf{v}_0 + \mathbf{a}_0 t