Kinematics: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

Encyclopedia

Updated live from Wikipedia, last check: May 21, 2013 08:38 UTC (53 seconds ago)

From Wikipedia, the free encyclopedia

Classical mechanics

$\mathbf{F} = \frac{\mathrm{d}}{\mathrm{d}t}(m \mathbf{v})$
Newton's Second Law

History of ...

Fundamental concepts
Space · Time · Mass · Force Energy · Momentum

Formulations
Newtonian mechanics Lagrangian mechanics Hamiltonian mechanics

Branches
Statics Dynamics Kinematics Applied mechanics Celestial mechanics Continuum mechanics Statistical mechanics

Scientists
Isaac Newton · Jeremiah Horrocks · Leonhard Euler · Jean le Rond d'Alembert · Alexis Clairaut Joseph Louis Lagrange · Pierre-Simon Laplace · William Rowan Hamilton · Siméon-Denis Poisson

Kinematics (from Greek κινεῖν, kinein, to move) is the branch of classical mechanics that describes the motion of objects without consideration of the causes leading to the motion.^[1]^[2]^[3]^[4]

“

It is natural to begin this discussion by considering the various possible types of motion in themselves, leaving out of account for a time the causes to which the initiation of motion may be ascribed; this preliminary enquiry constitutes the science of Kinematics. — ET Whittaker^[4]

”

Kinematics is not to be confused with another branch of classical mechanics: analytical dynamics (the study of the relationship between the motion of objects and its causes), sometimes subdivided into kinetics (the study of the relation between external forces and motion) and statics (the study of the relations in a system at equilibrium). Kinematics also differs from dynamics as used in modern-day physics to describe time-evolution of a system.

The term kinematics is less common today than in the past, but still has a role in physics.^[5] (See analytical dynamics for more detail on usage). The term "kinematics" also finds use in biomechanics and animal locomotion.^[6]

The simplest application of kinematics is for particle motion, translational or rotational. The next level of complexity is introduced by the introduction of rigid bodies, which are collections of particles having time invariant distances amongst themselves. Rigid bodies might undergo translation and rotation or a combination of both. A more complicated case is the kinematics of a system of rigid bodies, possibly linked together by mechanical joints.

Some kinds of computational fluid dynamics use a (complicated) particle-based kinematic model of the fluid, but fluid flow is generally thought of using a continuum mechanics model rather than kinematics.

Linear motion

Linear or translational kinematics^[7]^[8] is the description of the motion in space of a point along a line, also known as trajectory or path.^{[note 1]} This path can be either straight (rectilinear) or curved (curvilinear).

Particle Kinematics

Particle kinematics is the study of the kinematics of a single particle. The results obtained in particle kinematics are used to study the kinematics of collection of particles, dynamics and in many other branches of mechanics.

Position & Reference Frames

The position of a point in space is the most fundamental idea in particle kinematics. To specify the position of a point, one must specify three things: the reference point (often called the origin), distance from the reference point and the direction in space in which the straight line from the reference point to the particle makes. Exclusion of any of these three parameters renders the description of position inconsistent. Consider for example a tower 50 m south from your home. The reference point is home, the distance 50 m and the direction south. If one only says that the tower is 50 m south, the natural question that arises is " from where?". If one says that the tower is southwards from your home, the question that arises is "how far?". If one says the tower is 50 m from your home, the question that arises is "in which direction?". Hence, all these three parameters are crucial to defining uniquely the position of a point in space.

Position is usually described by mathematical quantities that have all these three attributes: the most common are vectors and complex numbers. Usually, only vectors are used. For measurement of distances and directions, usually three dimensional coordinate systems are used the origin coinciding with the reference point. A three dimensional coordinate system (whose origin coincides with the reference point) with some provision for time measurement is called a reference frame or frame of reference or simply frame. All observations in physics are incomplete without the reference frame being specified.

Position Vector

The position vector of a particle is a vector drawn from the origin of the reference frame to the particle. It expresses both the distance of the point from the origin and its direction from the origin. In three-dimensions, the position of point A can be expressed as

$\mathbf{r}_A = (x_A,y_A,z_A),$

where x_A, y_A, and z_A are the Cartesian coordinates of the point. The magnitude of the position vector |r| gives the distance between the point A and the origin.

$|\mathbf{r}| = \sqrt{x_A^{\ 2} + y_A^{\ 2} + z_A^{\ 2}}.$

The direction cosines of the position vector provide a quantitative measure of direction. It is important to note that the position vector of a particle isn't unique. Relative to different frames of reference, the position vector of the same particle can be different.

Rest & Motion

Once the notion of position is firmly established, the ideas of rest and motion naturally follow. If the position vector of the particle (relative to a given reference frame) changes with time, then the particle is said to be in motion with respect to the chosen reference frame. However, if the position vector of the particle (relative to a given reference frame) remains same with time, then the particle is said to be at rest with respect to the chosen frame. Note that rest and motion are relative to the reference frame chosen. It is quite possible that a particle at rest relative to a particular reference frame is in motion relative to the other. Hence, rest and motion aren't absolute terms, rather they are dependent on the reference frame.

Path

The path of locus of endpoints of the reference frame dependent. The path of a particle may be rectilinear (straight line) in one frame, and curved in another.

Displacement

Displacement is a vector describing the difference in position between two points, i.e. it is the change in position the particle undergoes during the time interval. If point A has position r_A = (x_A,y_A,z_A) and point B has position r_B = (x_B,y_B,z_B), the displacement r_AB of B from A is given by

$\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A = (x_B-x_A,y_B-y_A,z_B-z_A).$

Geometrically, displacement is the shortest distance between the points A and B. Displacement, distinct from position vector, is independent of the reference frame. This can be understood as follows: the positions of points is frame dependent, however, the shortest distance between any pair of points is invariant on translation from one frame to another (barring relativistic cases).

The distance traveled is always greater than or equal to the displacement.

Distance

In physics, distance is a distinct quantity from either position or displacement. It is a scalar quantity, describing the length of the path between two points along which the particle has traveled.

When considering the motion of a particle over time, distance is the length of the particle's path as opposed to displacement which is the change from its initial position to its final position. For example, a race car traversing a 10 km closed loop from start to finish travels a distance of 10 km; its displacement, however, is zero because it arrives back at its initial position.

If the position of the particle is known as a function of time (r = r(t)), the distance s it travels from time t₁ to time t₂ can be found by

$s = \int_{t_1}^{t_2} |d\mathbf{r}| = \int_{t_1}^{t_2} ds =\int_{t_1}^{t_2} \sqrt{dx^2 + dy^2 + dz^2} = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\; dt.$

The formula utilizes the fact that over an infinitesimal time interval, the magnitude of the displacement equals the distance covered in that interval. This is analogous to the geometric fact that infinitesimal arcs on a curved line coincide with the chord drawn between the ends of the arc itself.

Velocity and speed

Average velocity is defined as

$\overline{\mathbf{v}} = \frac {\Delta \mathbf{r}}{\Delta t} \ ,$

where Δr is the change in displacement and Δt is the interval of time over which displacement changes.The direction of v is same as the direction of the displacement Δr as Δt>0.

Velocity is the measure of the rate of change in displacement with respect to time; that is, how the distance of a point changes with each instant of time. Velocity also is a vector. Instantaneous velocity (the velocity at an instant of time) can be defined as the limiting value of average velocity as the time interval Δt becomes smaller and smaller. Both Δr and Δt approach zero but the ratio v approaches a non-zero limit v. That is,

$\mathbf{v} = \lim_{\Delta t\rightarrow0}\frac{\Delta\mathbf{r}}{\Delta t} = \frac {d \mathbf{r}}{d t} \, ,$

where dr is an infinitesimally small displacement and dt is an infinitesimally small length of time.^{[note 2]} As per its definition in the derivative form, velocity can be said to be the time rate of change of displacement. Further, as dr is tangential to the actual path, so is the velocity.

As position vector itself is frame dependent, velocity is also dependent on the reference frame.

The speed of an object is the magnitude |v| of its velocity. It is a scalar quantity:

$|\mathbf{v}| = \left|\frac {d \mathbf{r}}{d t} \right| = \frac {d s}{d t}$

The distance traveled by a particle over time is a non-decreasing quantity. Hence, ds/dt is non-negative, which implies that speed is also non-negative.

Acceleration

Average acceleration (acceleration over a length of time) is defined as:

$\overline{\mathbf{a}} = \frac {\Delta \mathbf{v}}{\Delta t} \ ,$

where Δv is the change in velocity and Δt is the interval of time over which velocity changes.

Acceleration is the vector quantity describing the rate of change with time of velocity. Instantaneous acceleration (the acceleration at an instant of time) is defined as the limiting value of average acceleration as Δt becomes smaller and smaller. Under such a limit, a → a.

$\mathbf{a} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac {d \mathbf{v}}{d t} \, ,$

where dv is an infinitesimally small change in velocity and dt is an infinitesimally small length of time.

Types of motion based on velocity and acceleration

If the acceleration of a particle is zero, then the velocity of the particle is constant over time and the motion is said to be uniform. Otherwise, the motion is non-uniform.

If the acceleration is non-zero but constant, the motion is said to be motion with constant acceleration. On the other hand, if the acceleration is variable, the motion is called motion with variable acceleration. In motion with variable acceleration, the rate of change of acceleration is called the jerk

Integral relations

The above definitions can be inverted by mathematical integration to find:

$\mathbf{v}(t) =\mathbf{v}_0 + \int_{t_0}^t \mathbf{a}(t) \; dt$

$\begin{align} \mathbf{r}(t) &=\mathbf{r}_0 + \int_{t_0}^t \mathbf{v}(t) \; dt \ &= \mathbf{r}_0 + \mathbf{v}_0 t + \int_{t_0}^t \left[\int_{t_0}^{t} \mathbf{a}(t) dt \right]\; dt \ \end{align}$

Kinematics of constant acceleration

Many physical situations can be modeled as constant-acceleration processes, such as projectile motion.

Integrating acceleration a with respect to time t gives the change in velocity. When acceleration is constant both in direction and in magnitude, the point is said to be undergoing uniformly accelerated motion. In this case, the integral relations can be simplified:

$\mathbf{v}(t) = \int_0^{t} \mathbf{a} \; dt' = \mathbf{v}_0 + \mathbf{a}t.$

$\begin{align} \mathbf{r}(t) &= \mathbf{r}_0 + \int_0^t \mathbf{v} \; dt' = \mathbf{r}_0 + \int_0^t (\mathbf{v}_0 + \mathbf{a} t) \; dt' \\ &= \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2} \mathbf{a} t^2. \end{align}$

Additional relations between displacement, velocity, acceleration, and time can be derived. Since a = (v − v₀)/t,

$\mathbf{r}(t) = \mathbf{r}_0 + \left(\frac{\mathbf{v}+ \mathbf{v_0}}{2}\right) t .$

By using the definition of an average, this equation states that when the acceleration is constant average velocity times time equals displacement.

A relationship without explicit time dependence may also be derived for one-dimensional motion. By setting r₀ = 0 and noting that at = v − v₀,

$\mathbf{r} \cdot \mathbf{a} t = \left( \mathbf{v} - \mathbf{v}_0 \right) \cdot \frac{\mathbf{v} + \mathbf{v}_0}{2} t \ ,$

where · denotes the dot product. Dividing the t on both sides and carrying out the dot-products:

$2\mathbf{r} \cdot \mathbf{a} = v^2 - v_0^{\ 2}.$

In the case of straight-line motion, r is parallel to a, and r has magnitude equal to the path length s = r − r₀ at time t. Then

$v^2= v_0^2 + 2 a(r-r_0).$

This relation is useful when time is not known explicitly.

Relative velocity

Main article: Relative velocity

To describe the motion of object A with respect to object B, when we know how each is moving with respect to a reference object O, we can use vector algebra. Choose an origin for reference, and let the positions of objects A, B, and O be denoted by r_A, r_B, and r_O. Then the position of A relative to the reference object O is

$\mathbf{r}_{A/O} = \mathbf{r}_{A} - \mathbf{r}_{O} \,\!$

Consequently, the position of A relative to B is

$\mathbf{r}_{A/B} = \mathbf{r}_A - \mathbf{r}_B = \mathbf{r}_A - \mathbf{r}_O - \left(\mathbf{r}_B-\mathbf{r}_O\right) = \mathbf{r}_{A/O}-\mathbf{r}_{B/O} \ .$

The above relative equation states that the motion of A relative to B is equal to the motion of A relative to O minus the motion of B relative to O. It may be easier to visualize this result if the terms are re-arranged:

$\mathbf{r}_{A/O} = \mathbf{r}_{A/B} + \mathbf{r}_{B/O} \ ,$

or, in words, the motion of A relative to the reference is that of B plus the relative motion of A with respect to B. These relations between displacements become relations between velocities by simple time-differentiation, and a second differentiation makes them apply to accelerations.

For example, let Ann move with velocity $\mathbf{V}_{A}$ relative to the reference (we drop the O subscript for convenience) and let Bob move with velocity $\mathbf{V}_{B}$ , each velocity given with respect to the ground (point O). To find how fast Ann is moving relative to Bob (we call this velocity $\mathbf{V}_{A/B}$ ), the equation above gives:

$\mathbf{V}_{A} = \mathbf{V}_{B} + \mathbf{V}_{A/B} \,\! .$

To find $\mathbf{V}_{A/B}$ we simply rearrange this equation to obtain:

$\mathbf{V}_{A/B} = \mathbf{V}_{A} -\mathbf{V}_{B} \,\! .$

At velocities comparable to the speed of light, these equations are not valid. They are replaced by equations derived from Einstein's theory of special relativity.

Example: Rectilinear (1D) motion

Figure A: An object is fired upwards, reaches its apex, and then begins its descent under a constant acceleration.

Consider an object that is fired directly upwards and falls back to the ground so that its trajectory is contained in a straight line. If we adopt the convention that the upward direction is the positive direction, the object experiences a constant acceleration of approximately −9.81 m s⁻². Therefore, its motion can be modeled with the equations governing uniformly accelerated motion.

For the sake of example, assume the object has an initial velocity of +50 m s⁻¹. There are several interesting kinematic questions we can ask about the particle's motion:

How long will it be airborne?

To answer this question, we apply the formula

$x_{\mathrm{f}} - x_{\mathrm{i}} = v_{\mathrm{i}} t + \tfrac{1}{2} at^2.$

Since the question asks for the length of time between the object leaving the ground and hitting the ground on its fall, the displacement is zero.

$0 = v_{\mathrm{i}} t + \tfrac{1}{2} at^2 = t\left(v_{\mathrm{i}} + \tfrac{1}{2} at \right)$

There are two solutions: the first, t = 0, is trivial. The solution of interest is

$t = -\frac{2v_{\mathrm{i}}}{a} = \rm -\frac{2(50\ m\ s^{-1})}{-9.81\ m\ s^{-2}} = 10.2\ s\,.$

What altitude will it reach before it begins to fall?

In this case, we use the fact that the object has a velocity of zero at the apex of its trajectory. Therefore, the applicable equation is:

$v_{\mathrm{f}}^2 = v_{\mathrm{i}}^2 + 2 a (x_{\mathrm{f}} - x_{\mathrm{i}})$

If the origin of our coordinate system is at the ground, then x_i is zero. Then we solve for x_f and substitute known values:

$x_{\mathrm{f}} = \frac{v_{\mathrm{f}}^2 - v_{\mathrm{i}}^2}{2 a} + x_{\mathrm{i}} = \rm \frac{(0\ m\ s^{-1})^2-(50\ m\ s^{-1})^2}{2(-9.81\ m\ s^{-2})} + 0\ m = 127.55\ m\,.$

What will its final velocity be when it reaches the ground?

To answer this question, we use the fact that the object has an initial velocity of zero at the apex before it begins its descent. We can use the same equation we used for the last question, using the value of 127.55 m for x_i.

$v_{\mathrm{f}} = \sqrt{v_{\mathrm{i}}^2 + 2 a (x_{\mathrm{f}} - x_{\mathrm{i}})} = \rm \sqrt{(0\ m\ s^{-1})^2 + 2(-9.81\ m\ s^{-2})(0\ m - 127.55\ m)} = 50\ m\ s^{-1}\,.$

Assuming this experiment were performed in a vacuum (negating drag effects), we find that the final and initial speeds are equal, a result which agrees with conservation of energy.

Example: Projectile (2D) motion

Figure B: An object fired at an angle

θ

from the ground follows a parabolic trajectory.

Suppose that an object is not fired vertically but is fired at an angle θ from the ground. The object will then follow a parabolic trajectory, and its horizontal motion can be modeled independently of its vertical motion. Assume that the object is fired at an initial velocity of 50 m s⁻¹ and 30° from the horizontal.

How far will it travel before hitting the ground?

The object experiences an acceleration of −9.81 m s⁻² in the vertical direction and no acceleration in the horizontal direction. Therefore, the horizontal displacement is

$\Delta x = x_{\mathrm{f}} - x_{\mathrm{i}} = v_{\mathrm{i}} \cos \theta \; t + \tfrac{1}{2} at^2 = v_{\mathrm{i}} \cos \theta \; t$

Solving the equation requires finding t. This can be done by analyzing the motion in the vertical direction. If we impose that the vertical displacement is zero, we can use the same procedure we did for rectilinear motion to find t.

$0 = v_{\mathrm{i}} \sin \theta \; t + \tfrac{1}{2} at^2 = t\left(v_{\mathrm{i}} \sin \theta + \tfrac{1}{2} at \right)$

We now solve for t and substitute this expression into the original expression for horizontal displacement.

$\Delta x = v_{\mathrm{i}} \cos \theta \left(\frac{-2 v_{\mathrm{i}} \sin \theta}{a}\right) = -\frac{v_{\mathrm{i}}^2 \sin 2\theta}{a} = 220.93 \ m\,$

Note the use of the trigonometric identity 2sinθ cosθ = sin 2θ.

Rotational motion

Main article: Circular motion

Figure 1: The angular velocity vector Ω points up for counterclockwise rotation and down for clockwise rotation, as specified by the right-hand rule. Angular position θ(t) changes with time at a rate ω(t) = dθ/dt.

Rotational or angular kinematics is the description of the rotation of an object.^[9] The description of rotation requires some method for describing orientation, for example, the Euler angles. In what follows, attention is restricted to simple rotation about an axis of fixed orientation. The z-axis has been chosen for convenience.

Description of rotation then involves these three quantities:

Angular position: The oriented distance from a selected origin on the rotational axis to a point of an object is a vector r ( t ) locating the point. The vector r(t) has some projection (or, equivalently, some component) r_⊥(t) on a plane perpendicular to the axis of rotation. Then the angular position of that point is the angle θ from a reference axis (typically the positive x-axis) to the vector r_⊥(t) in a known rotation sense (typically given by the right-hand rule).
Angular velocity: The angular velocity ω is the rate at which the angular position θ changes with respect to time t:

$\mathbf{\omega} = \frac {\mathrm{d}\theta}{\mathrm{d}t}$

The angular velocity is represented in Figure 1 by a vector Ω pointing along the axis of rotation with magnitude ω and sense determined by the direction of rotation as given by the right-hand rule.

Angular acceleration: The magnitude of the angular acceleration α is the rate at which the angular velocity ω changes with respect to time t:

$\mathbf{\alpha} = \frac {\mathrm{d}\mathbf{\omega}}{\mathrm{d}t}$

The equations of translational kinematics can easily be extended to planar rotational kinematics with simple variable exchanges:

$\omega_{\mathrm{f}} = \omega_{\mathrm{i}} + \alpha t\!$

$\theta_{\mathrm{f}} - \theta_{\mathrm{i}} = \omega_{\mathrm{i}} t + \tfrac{1}{2} \alpha t^2$

$\theta_{\mathrm{f}} - \theta_{\mathrm{i}} = \tfrac{1}{2} (\omega_{\mathrm{f}} + \omega_{\mathrm{i}})t$

$\omega_{\mathrm{f}}^2 = \omega_{\mathrm{i}}^2 + 2 \alpha (\theta_{\mathrm{f}} - \theta_{\mathrm{i}}).$

Here θ_i and θ_f are, respectively, the initial and final angular positions, ω_i and ω_f are, respectively, the initial and final angular velocities, and α is the constant angular acceleration. Although position in space and velocity in space are both true vectors (in terms of their properties under rotation), as is angular velocity, angle itself is not a true vector.

Point object in circular motion

Coordinate systems

In any given situation, the most useful coordinates may be determined by constraints on the motion, or by the geometrical nature of the force causing or affecting the motion. Thus, to describe the motion of a bead constrained to move along a circular hoop, the most useful coordinate may be its angle on the hoop. Similarly, to describe the motion of a particle acted upon by a central force, the most useful coordinates may be polar coordinates. Polar coordinates are extended into three dimensions with either the spherical polar or cylindrical polar coordinate systems. These are most useful in systems exhibiting spherical or cylindrical symmetry respectively.

Fixed rectangular coordinates

In this coordinate system, vectors are expressed as an addition of vectors in the x, y, and z direction from a non-rotating origin. Usually i, j, k are unit vectors in the x-, y-, and z-directions.

The position vector, r, the velocity vector, v, and the acceleration vector, a are expressed using rectangular coordinates in the following way:

$\mathbf{r} = x\, \hat {\mathbf{i}} + y \, \hat {\mathbf{j}} + z \, \hat {\mathbf{ k}} \, \!$

$\mathbf{v} = \dot {\boldsymbol{ r}} = \dot {x} \, \hat {\mathbf{ i}} + \dot {y} \, \hat {\mathbf{ j}} + \dot {z} \, \hat {\mathbf{ k}} \, \!$

$\mathbf{a} = \ddot {\boldsymbol{ r}} = \ddot {x} \, \hat {\mathbf{ i}} + \ddot {y} \, \hat {\mathbf{ j}} + \ddot {z} \, \hat {\mathbf{ k}} \, \!$

Note: $\dot {x} = \frac{\mathrm{d}x}{\mathrm{d}t}$ , $\ddot {x} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$

Two dimensional rotating reference frame

Derivatives of unit vectors

The position, velocity, and acceleration vectors of a given point can be expressed using these coordinate systems, but we have to be a bit more careful than we do with fixed frames of reference. Since the frame of reference is rotating, the unit vectors also rotate, and this rotation must be taken into account when taking the derivative of any of these vectors. If the coordinate frame is rotating at angular rate ω in the counterclockwise direction (that is, Ω = ω k using the right hand rule) then the derivatives of the unit vectors are as follows:

$\dot{\hat {\mathbf{ i}}} = \omega \hat {\mathbf{ k}} \times \hat {\mathbf{ i}} = \omega\hat {\mathbf{ j}}$

$\dot{\hat {\mathbf{ j}}} = \omega \hat {\mathbf{ k}}\times \hat {\mathbf{ j}} = - \omega \hat {\mathbf{ i}}$

Position, velocity, and acceleration

Given these identities, we can now figure out how to represent the position, velocity, and acceleration vectors of a particle using this reference frame.

Position

Position is straightforward:

$\boldsymbol{ r} = x \ \hat {\mathbf{ i}} + y \ \hat {\mathbf{ j}}$

It is just the distance from the origin in the direction of each of the unit vectors.

Velocity

Velocity is the time derivative of position:

$\mathbf{v} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \frac{\mathrm{d} (x \ \hat {\mathbf{ i}})}{\mathrm{d}t} + \frac{\mathrm{d} (y \ \hat {\mathbf{ j}})}{\mathrm{d}t}$

By the product rule, this is:

$\mathbf{v} = \dot x \ \hat {\mathbf{ i}} + x \dot{\ \hat {\mathbf{ i}}} + \dot y \ \hat {\mathbf{ j}} + y \dot{ \ \hat {\mathbf{ j}}}$

Which from the identities above we know to be:

$\mathbf{v} = \dot x \ \hat {\mathbf{ i}} + x \omega \ \hat {\mathbf{ j}} + \dot y \ \hat {\mathbf{ j}} - y \omega \ \hat {\mathbf{ i}} = (\dot x - y \omega) \ \hat {\mathbf{ i}} + (\dot y + x \omega) \ \hat {\mathbf{ j}}$

or equivalently

$\mathbf{v} = (\dot x \ \hat {\mathbf{ i}} + \dot y \ \hat {\mathbf{ j}}) + (y \dot{ \hat {\mathbf{ j}}} + x \dot{\hat {\mathbf{ i}}}) = \mathbf{v}_{rel} + \boldsymbol{\Omega} \times \mathbf{r}$

where v_rel is the velocity of the particle relative to the rotating coordinate system.

Acceleration

First to fourth derivatives of position

Acceleration is the time derivative of velocity.

We know that:

$\boldsymbol{ a} = \frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{ v} = \frac{\mathrm{d} \boldsymbol{ v}_{rel}}{\mathrm{d}t} + \frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{\Omega} \times \boldsymbol{r}$

Consider the $\stackrel{\frac{ \mathrm{d} } { \mathrm{d} t }}{}$ $\boldsymbol{ v}_{rel}$ part. $\boldsymbol{ v}_{rel}$ has two parts we want to find the derivative of: the relative change in velocity ( $\boldsymbol{ a}_{rel}$ ), and the change in the coordinate frame

( $\boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$ ).

$\frac{\mathrm{d} \boldsymbol{ v}_{rel}}{\mathrm{d}t} = \boldsymbol{ a}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

Next, consider $\stackrel{\frac{\mathrm{d}}{\mathrm{d}t}}{}$ $(\boldsymbol{\Omega} \times\boldsymbol{ r})$ . Using the chain rule:

$\frac{\mathrm{d} (\boldsymbol{\Omega} \times \boldsymbol{ r})}{\mathrm{d}t} = \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times \dot{\boldsymbol{ r}}$

$\dot{\boldsymbol{ r}}=\boldsymbol{ v}=\boldsymbol{ v}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ r}$ from above:

$\frac{\mathrm{d} (\boldsymbol{\Omega} \times \boldsymbol{ r})}{\mathrm{d}t} = \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times\boldsymbol{ r}) + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

So all together:

$\boldsymbol{ a} = \boldsymbol{ a}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel} + \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{ r}) + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

And collecting terms:^[10]

$\boldsymbol{ a} = \boldsymbol{ a}_{rel} + 2(\boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}) + \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{ r})\ .$

Kinematic constraints

A kinematic constraint is any condition relating properties of a dynamic system that must hold true at all times. Below are some common examples:

Rolling without slipping

An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass,

$\boldsymbol{ v}_G(t) = \boldsymbol{\Omega} \times \boldsymbol{ r}_{G/O}$ .

For the case of an object that does not tip or turn, this reduces to v = R ω.

Inextensible cord

This is the case where bodies are connected by some cord that remains in tension and cannot change length. The constraint is that the sum of all components of the cord is the total length, and accordingly the time derivative of this sum is zero. See Kelvin and Tait^[11]^[12] and Fogiel.^[13] A dynamic problem of this type is the pendulum. Another example is a drum turned by the pull of gravity upon a falling weight attached to the rim by the inextensible cord.^[14] An equilibrium problem (not kinematic) of this type is the catenary.^[15]

Notes

^ In mathematics, a line refers to a straight trajectory, and a curve to a trajectory which may have curvature. In mechanics and kinematics, "line' and "curve" both refer to any trajectory, in particular a line may be a complex curve in space. Any position along a specified trajectory can be described by a single coordinate, the distance traversed along the path, or arc length. The motion of a particle along a trajectory can be described by specifying the time dependence of its position, for example by specification of the arc length locating the particle at each time t. The following words refer to curves and lines:
- "linear" (= along a straight or curved line;
- "rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread),
- "curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread).
^ Because magnitude of dr is necessarily the distance between two infinitesimally spaced points along the trajectory of the point, it is the same as an increment in arc length along the path of the point, customarily denoted ds.

References

Moon, Francis C. (2007). The Machines of Leonardo Da Vinci and Franz Reuleaux, Kinematics of Machines from the Renaissance to the 20th Century. Springer. ISBN 9781402055980.

^ Joseph Stiles Beggs (1983). Kinematics. Taylor & Francis. p. 1. ISBN 0891163557. http://books.google.com/books?id=y6iJ1NIYSmgC&printsec=frontcover&dq=kinematics&lr=&as_brr=0&sig=brRJKOjqGTavFsydCzhiB3u_8MA#PPA1,M1.
^ O. Bottema & B. Roth (1990). Theoretical Kinematics. Dover Publications. reface. ISBN 0486663469. http://books.google.com/books?id=f8I4yGVi9ocC&printsec=frontcover&dq=kinematics&lr=&as_brr=0&sig=YfoHn9ImufIzAEp5Kl7rEmtYBKc#PPR7,M1.
^ Thomas Wallace Wright (1896). Elements of Mechanics Including Kinematics, Kinetics and Statics. E and FN Spon. Chapter 1. http://books.google.com/books?id=-LwLAAAAYAAJ&printsec=frontcover&dq=mechanics+kinetics&lr=&as_brr=0#PPA6,M1.
^ ^a ^b Edmund Taylor Whittaker & William McCrea (1988). A Treatise on the Analytical Dynamics of Particles and Rigid Bodies. Cambridge University Press. Chapter 1. ISBN 0521358833. http://books.google.com/books?id=epH1hCB7N2MC&printsec=frontcover&dq=inauthor:%22E+T+Whittaker%22&lr=&as_brr=0&sig=SN7_oYmNYM4QRSgjULXBU5jeQrA&source=gbs_book_other_versions_r&cad=0_2#PPA1,M1.
^ See, for example: Russell C. Hibbeler (2009). "Kinematics and kinetics of a particle". Engineering Mechanics: Dynamics (12th ed.). Prentice Hall. p. 298. ISBN 0136077919. http://books.google.com/books?id=tOFRjXB-XvMC&pg=PA298. , Ahmed A. Shabana (2003). "Reference kinematics". Dynamics of Multibody Systems (2nd ed.). Cambridge University Press. ISBN 0521544114. http://books.google.com/books?id=zxuG-l7J5rgC&pg=PA28. , P. P. Teodorescu (2007). "Kinematics". Mechanical Systems, Classical Models: Particle Mechanics. Springer. p. 287. ISBN 1402054416. http://books.google.com/books?id=k4H2AjWh9qQC&pg=PA287.
^ A. Biewener (2003). Animal Locomotion. Oxford University Press. ISBN 19850022X. http://books.google.com/books?id=yMaN9pk8QJAC.
^ James R. Ogden & Max Fogiel (1980). The Mechanics Problem Solver. Research and Education Association. p. 184. ISBN 0878915192. http://books.google.com/books?id=XVyD9pJpW-cC&pg=PA184&dq=%22curvilinear+kinematics%22&lr=&as_brr=0&sig=WW7us4UJzSWOA19pfdAbwTJvPR4.
^ R. Douglas Gregory (2006). Classical Mechanics: An Undergraduate Text. Cambridge UK: Cambridge University Press. Chapter 2. ISBN 0521826780. http://books.google.com/books?id=uAfUQmQbzOkC&printsec=frontcover&dq=%22rigid+body+kinematics%22&lr=&as_brr=0#PRA1-PA25,M1.
^ R. Douglas Gregory (2006). Chapter 16. Cambridge: Cambridge University. ISBN 0521826780. http://books.google.com/books?id=uAfUQmQbzOkC&printsec=frontcover&dq=%22rigid+body+kinematics%22&lr=&as_brr=0#PRA1-PA457,M1.
^ R. Douglas Gregory (2006). pp. 475-476. Cambridge: Cambridge University. ISBN 0521826780. http://books.google.com/books?id=uAfUQmQbzOkC&printsec=frontcover&dq=%22rigid+body+kinematics%22&lr=&as_brr=0#PRA1-PA475,M1.
^ William Thomson Kelvin & Peter Guthrie Tait (1894). Elements of Natural Philosophy. Cambridge University Press. p. 4. ISBN 1573929840. http://books.google.com/books?id=dHASAAAAIAAJ&pg=PA4&dq=%22inextensible+cord%22&lr=&as_brr=0&as_pt=ALLTYPES.
^ William Thomson Kelvin & Peter Guthrie Tait (1894). op. cit.. p. 296. http://books.google.com/books?id=ahtWAAAAMAAJ&pg=PA296&dq=%22inextensible+cord%22&lr=&as_brr=0&as_pt=ALLTYPES#PPA296,M1.
^ M. Fogiel (1980). "Problem 17-11". The Mechanics Problem Solver. Research & Education Assoc.. p. 613. ISBN 0878915192. http://books.google.com/books?id=XVyD9pJpW-cC&pg=PA613&dq=%22inextensible+cord%22&lr=&as_brr=0&as_pt=ALLTYPES.
^ Irving Porter Church (1908). Mechanics of Engineering. Wiley. p. 111. ISBN 1110365276. http://books.google.com/books?id=7-40AAAAMAAJ&pg=PA111&dq=%22inextensible+cord%22&lr=&as_brr=0&as_pt=ALLTYPES.
^ Morris Kline (1990). Mathematical Thought from Ancient to Modern Times. Oxford University Press. p. 472. ISBN 0195061365. http://books.google.com/books?id=aO-v3gvY-I8C&pg=PA472&dq=%22inextensible+cord%22&lr=&as_brr=0&as_pt=ALLTYPES.

External links

Java applet of 1D kinematics
Flash animated tutorial for 1D kinematics
Physclips: Mechanics with animations and video clips from the University of New South Wales
KINEMATICS FOR HIGH SCHOOL AND IIT JEE LEVEL
Kinematic Models for Design Digital Library (KMODDL)
Movies and photos of hundreds of working mechanical-systems models at Cornell University. Also includes an e-book library of classic texts on mechanical design and engineering.

Kinematics

← Integrate … Differentiate → Displacement (Distance) \| Velocity (Speed) \| Acceleration \| Jerk \| Jounce \| Crackle and pop

Categories: Classical mechanics | Kinematics

Study guide

Up to date as of January 14, 2010

From Wikiversity

1 Introduction
2 Concepts and Definitions
3 Strains
4 Relation between axial vector and displacement
5 Relation between axial vector and strain
6 Related Content

Introduction

Kinematics
- Displacements, strains and the relations between displacements and strains.
- We view the theory of infinitesimal linear deformations as a first-order approximation of the theory of finite deformations.
- Note that the linear theory can be derived independently of the finite theory and is completely self-consistent on its own.

Concepts and Definitions

The following concepts and definitions are based on Gurtin (1972) and Truesdell and Noll (1992). These definitions are useful both for the linear and the nonlinear theory of elasticity.

Body

We usually denote a body by the symbol $\textstyle B$ . A body is essentially a set of points in Euclidean space. For mathematical definition see Truesdell and Noll (1992)

Configuration

A configuration of a body is denoted by the symbol $\textstyle \boldsymbol{\chi}$ . A configuration of a body is just what the name suggests. Sometimes a configuration is also referred to as a placement.

Mathematically, we can think of a configuration as a smooth one-to-one mapping of a body into a region of three-dimensional Euclidean space.

Thus, we can have a reference configuration $\textstyle \boldsymbol{\chi}(\mathbf{X})$ and a current configuration $\textstyle \boldsymbol{\chi}(\mathbf{x})$ .

A one-to-one mapping is also called a homeomorphism.

Deformation

A deformation is the relationship between two configurations and is usually denoted by $\textstyle \boldsymbol{\varphi}$ . Deformations include both volume and shape changes and rigid body motions.

For a continuous body, a deformation can be thought of as a smooth mapping from one configuration ( $\textstyle \mathbf{X}$ ) to another ( $\textstyle \mathbf{x}$ ). The inverse mapping should be possible.

This means that

$\begin{align} \mathbf{x} &\equiv \{x_1,x_2,x_3\} = \boldsymbol{\varphi}(\mathbf{X}) \\ \mathbf{X} &\equiv \{X_1,X_2,X_3\} = \boldsymbol{\varphi}^{-1}(\mathbf{x}) \end{align}$

For the inverse mapping to exist, we require that the Jacobian of the deformation is positive, i.e., $\textstyle J = \det(\boldsymbol{\nabla}{\boldsymbol{\varphi}}) > 0$ .

Deformation Gradient

The deformation gradient is usually denoted by $\textstyle \boldsymbol{F}$ and is defined as

$\boldsymbol{F} := \boldsymbol{\nabla}{\boldsymbol{\varphi}}$

In index notation

$F_{ij} = \frac{\partial x_i }{\partial X_j}$

For a deformation to be allowable, we must be able to invert $\textstyle \boldsymbol{F}$ . That is why we require that $\textstyle J = \det\boldsymbol{F} > 0$ . Otherwise, the body may undergo deformations that are unphysical.

Displacement

The displacement is usually denoted by the symbol $\textstyle \mathbf{u}$ .

The displacement is defined as a vector from the location of a material point in one configuration to the location of the same material point in another configuration.

The definition is

$\mathbf{u}(\mathbf{X}) := \boldsymbol{\varphi}(\mathbf{X}) - \mathbf{X} = \mathbf{x} - \mathbf{X} \,$

In index notation

$u_i = x_i - X_i \,$

Displacement Gradient

The gradient of the displacement is denoted by $\textstyle \boldsymbol{\nabla}\mathbf{u}$ .

The displacement gradient is given by

$\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\nabla}{(\mathbf{x} - \mathbf{X})} = \boldsymbol{\nabla}{\mathbf{x}} - \boldsymbol{\nabla}{\mathbf{X}} = \boldsymbol{\nabla}{(\boldsymbol{\varphi}(\mathbf{X}) - \mathbf{X})} = \boldsymbol{F} - \boldsymbol{\it{1}}$

In index notation,

$\frac{\partial u_i }{\partial X_j} = \frac{\partial x_i }{\partial X_j} - \frac{\partial X_i }{\partial X_j} = F_{ij} - \delta_{ij}$

Strains

Finite Strain Tensor

The finite strain tensor ( $\textstyle \boldsymbol{E}$ ) is also called the Green-St. Venant Strain Tensor or the Lagrangian Strain Tensor.

This strain tensor is defined as

$\boldsymbol{E} := \cfrac{1}{2}(\boldsymbol{F}^T\bullet\boldsymbol{F} - \boldsymbol{1})$

In index notation,

$E_{ij} = \cfrac{1}{2}(F_{ki}F_{kj} - \delta_{ij})$

Infinitesimal Strain Tensor

In the limit of small strains, the Lagrangian finite strain tensor reduces to the infinitesimal strain tensor ( $\textstyle \boldsymbol{\epsilon}$ ).

This strain tensor is defined as

$\boldsymbol{\varepsilon} := \cfrac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T)$

In index notation,

$\varepsilon_{ij} = \cfrac{1}{2}\left(\cfrac{\partial u_i}{\partial x_j}+ \cfrac{\partial u_j}{\partial x_i}\right)$

Therefore we can see that the finite strain tensor and the infinitesimal strain tensor are related by

$\boldsymbol{E} = \boldsymbol{\varepsilon} + \cfrac{1}{2}\boldsymbol{\nabla}\mathbf{u}^T\bullet\boldsymbol{\nabla}\mathbf{u}$

If $\textstyle \epsilon = |\boldsymbol{\nabla}\mathbf{u}|$ , then

$\boldsymbol{E} = \boldsymbol{\varepsilon} + O(\epsilon^2)$

For small strains, $\textstyle \epsilon^2 \rightarrow 0$ and

$\boldsymbol{E} \approx \boldsymbol{\varepsilon}$ .

Infinitesimal Rotation Tensor

For small deformation problems, in addition to small strains we can also have small rotations ( $\textstyle \boldsymbol{W}$ ). The infinitesimal rotation tensor is defined as

$\boldsymbol{W} := \cfrac{1}{2}(\boldsymbol{\nabla}\mathbf{u} - \boldsymbol{\nabla}\mathbf{u}^T)$

In index notation,

$W_{ij} = \cfrac{1}{2}\left(\frac{\partial u_i}{\partial x_j}- \frac{\partial u_j}{\partial x_i}\right)$

If $\boldsymbol{A}$ is a skew-symmetric tensor, then for any vector $\textstyle \mathbf{a}$ we have

$\boldsymbol{A}\bullet\mathbf{a} = \boldsymbol{\omega}\times{\mathbf{a}}~.$

The vector $\boldsymbol{\omega}$ is called the axial vector of the skew-symmetric tensor.

In our case, $\boldsymbol{W}$ is the skew-symmetric infinitesimal rotation tensor. The corresponding axial vector is the rotation vector $\boldsymbol{\theta}$ defined as

$\boldsymbol{W}\bullet\mathbf{a} = \boldsymbol{\theta}\times{\mathbf{a}}~.$

where

$\boldsymbol{\theta}=\cfrac{1}{2}\boldsymbol{\nabla}\times \boldsymbol{u}$

Volume Change Due To Finite Deformation

The change in volume ( $\textstyle \delta V$ ) during a finite deformation is given by

$\delta V = \int_{B} \det (\boldsymbol{F} -\boldsymbol{1}) dv$

Volume Change Due To Infinitesimal Deformation

The volume change during an infinitesimal deformation ( $\textstyle \delta V$ ) is given by

$\delta V = \int_{B} \boldsymbol{\nabla}\bullet\mathbf{u} dv = \int_{\partial B} \mathbf{u}\bullet\mathbf{n} ds$

because

$\det\boldsymbol{F}=\det(\boldsymbol{1}+\boldsymbol{\nabla}\mathbf{u})=1+\text{Tr}\boldsymbol{\nabla}\mathbf{u} + O(\epsilon^2) = 1+\boldsymbol{\nabla}\bullet\mathbf{u}+O(\epsilon^2)$

The quantity $\textstyle \boldsymbol{\nabla}\bullet\mathbf{u} = \text{Tr}(\boldsymbol{\varepsilon})$ is called the dilatation.

A volume change is isochoric (volume preserving) if

$\textstyle \text{Tr}(\boldsymbol{\varepsilon})=0$ .

Relation between axial vector and displacement

Let $\mathbf{u}$ be a displacement field. The displacement gradient tensor is given by $\boldsymbol{\nabla}\mathbf{u}$ . Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be

$\boldsymbol{\omega} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} - \boldsymbol{\nabla}\mathbf{u}^T) ~.$

Let $\boldsymbol{\theta}$ be the axial vector associated with the skew symmetric tensor $\boldsymbol{\omega}$ . Show that

$\boldsymbol{\theta} = \frac{1}{2}~\boldsymbol{\nabla} \times \mathbf{u} ~.$

Proof:

The axial vector $\mathbf{w}$ of a skew-symmetric tensor $\boldsymbol{W}$ satisfies the condition

$\boldsymbol{W}\cdot\mathbf{a} = \mathbf{w}\times\mathbf{a}$

for all vectors $\mathbf{a}$ . In index notation (with respect to a Cartesian basis), we have

$W_{ip}~a_p = e_{ijk}~w_j~a_k$

Since $e i j k = - e i k j$ , we can write

$W_{ip}~a_p = -e_{ikj}~w_j~a_k \equiv -e_{ipq}~w_q~a_p$

or,

$W_{ip} = -e_{ipq}~w_q ~.$

Therefore, the relation between the components of $\boldsymbol{\omega}$ and $\boldsymbol{\theta}$ is

$\omega_{ij} = -e_{ijk}~\theta_k ~.$

Multiplying both sides by $e p i j$ , we get

$e_{pij}~\omega_{ij} = -e_{pij}~e_{ijk}~\theta_k = -e_{pij}~e_{kij}~\theta_k~.$

Recall the identity

$e_{ijk}~e_{pqk} = \delta_{ip}~\delta_{jq} - \delta_{iq}~\delta_{jp}~.$

Therefore,

$e_{ijk}~e_{pjk} = \delta_{ip}~\delta_{jj} - \delta_{ij}~\delta_{jp} = 3\delta_{ip} - \delta_{ip} = 2\delta_{ip}$

Using the above identity, we get

$e_{pij}~\omega_{ij} = -2\delta_{pk}~\theta_k = -2\theta_p~.$

Rearranging,

$\theta_p = -\frac{1}{2}~e_{pij}~\omega_{ij}$

Now, the components of the tensor $\boldsymbol{\omega}$ with respect to a Cartesian basis are given by

$\omega_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i}\right)$

Therefore, we may write

$\theta_p = -\cfrac{1}{4}~e_{pij} \left(\frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i}\right)$

Since the curl of a vector $\mathbf{v}$ can be written in index notation as

$\boldsymbol{\nabla} \times \mathbf{v} = e_{ijk}~\frac{\partial u_k}{\partial x_j}~\mathbf{e}_i$

we have

$e_{pij}~\frac{\partial u_j}{\partial x_i}~ = [\boldsymbol{\nabla} \times \mathbf{u}]_p \qquad \text{and} \qquad e_{pij}~\frac{\partial u_i}{\partial x_j}~ = - e_{pji}\frac{\partial u_i}{\partial x_j} = - [\boldsymbol{\nabla} \times \mathbf{u}]_p$

where $[~]_p$ indicates the $p$ -th component of the vector inside the square brackets.

Hence,

$\theta_p = -\cfrac{1}{4}~\left(-[\boldsymbol{\nabla} \times \mathbf{u}]_p - [\boldsymbol{\nabla} \times \mathbf{u}]_p\right) = \frac{1}{2}~[\boldsymbol{\nabla} \times \mathbf{u}]_p ~.$

Therefore,

${ \boldsymbol{\theta} = \frac{1}{2} \boldsymbol{\nabla} \times \mathbf{u} \qquad \square }$

Relation between axial vector and strain

Let $\mathbf{u}$ be a displacement field. Let $\boldsymbol{\varepsilon}$ be the strain field (infinitesimal) corresponding to the displacement field and let $\boldsymbol{\theta}$ be the corresponding infinitesimal rotation vector. Show that

$\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} ~.$

Proof:

The infinitesimal strain tensor is given by

$\boldsymbol{\varepsilon} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) ~.$

Therefore,

$\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) + \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T)] ~.$

Recall that

$\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) = \mathbf{0}\qquad\text{and}\qquad \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u}) ~.$

Hence,

$\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u})] ~.$

Also recall that

$\boldsymbol{\theta} = \frac{1}{2} \boldsymbol{\nabla} \times \mathbf{u}~.$

Therefore,

${ \boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} \qquad \square }$

1911 encyclopedia

Up to date as of January 14, 2010

(Redirected to Database error article)

From LoveToKnow 1911

(There is currently no text in this page)

Wikibooks

Up to date as of January 23, 2010

From Wikibooks, the open-content textbooks collection

Current revision (unreviewed)

Preface

Introduction

Kinematics is the branch of physics dealing with the motion of particles or bodies. It defines movement at the level of position, velocity and acceleration, without incorporating masses and forces.

Prerequisites

A proper study of the subject requires an understanding of 2D and 3D vectors. Of course you also need to review basic algebra, trignometry and calculus. A proper understanding of kinematics require an understanding of vector calculus too.

Chapter 0: A Mathematics Review

Algebra Review
Geometry and Trig Review
Calculus Review
Vector Review

Chapter 1: Plane Kinematics

Chapter 2: Kinematics in Three Space

Two Dimensional Kinematics

Displacement, Velocity, and Acceleration

Displacement is the change in position with respect to the frame of reference. In the case of a car moving on a road, the displacement is the change in position with respect to the road. This is important because the Earth itself is rotating and orbiting at considerable speed, which you do not notice due to inertia. Since omnipotence is not possible, all observations are made from a specific frame of reference. Displacement is usually expressed as a length measurement. Velocity is the change in displacement with respect to a change in time. The velocity of an object is again relative to the frame of reference. Acceleration is the change in velocity with respect to a change in time. Therefore, we know that, for functions for displacement, velocity, and acceleration, respectively d, v, t :

$d(t)=\int_0^t v(x)dx$

$v(t)=\frac{dd}{dt}=\int_0^t a(x)dx$

$a(t)=\frac{dv}{dt}$

Kinematics Formulas

Using the above can create the basic kinematics formulas(subject to acceleration being constant):

$x=x_0+v_0t+\frac{1}{2}at^2$

v = v 0 + a t

$v^2=v_0^2+2ax$

Applications of Kinematics

Motion

Motion is by far the simplest application of Kinematics, and uses the formulas with no adaptation.

Example 1

Gravity and Falling Objects

Gravity affects most all earthly motion, and therefore is discussed first. Gravity pulls all objects the same distance from the earth with equal acceleration, regardless of mass, as discovered by Galileo in his famous test on the Leaning Tower of Pisa. In this section we will assume that friction and air resistance are negligible. The Gravitational constant, 'g', is the is equal to 9.8ms^-2, and is the acceleration done by the earth's gravity at sea level. Therefore, a falling object fall accelerating at rate g. When using this with the Kinematics formulas, g replaces a.

Example 1

Show that the force (F) of a body is directly proportional to its momentum

Example 2

Subjects: Kinematics | Mechanics

Hidden categories: Alphabetical/K | Dewey/500 | Library of Congress/QC

The Full Wiki

Search:

More info on Kinematics

Related topics

Kinematics: Wikis

Related top topics

Encyclopedia

From Wikipedia, the free encyclopedia

Contents

Linear motion

Particle Kinematics

Position & Reference Frames

Position Vector

Rest & Motion

Path

Displacement

Distance

Velocity and speed

Acceleration

Types of motion based on velocity and acceleration

Integral relations

Kinematics of constant acceleration

Relative velocity

Rotational motion

Point object in circular motion

Coordinate systems

Fixed rectangular coordinates

Two dimensional rotating reference frame

Derivatives of unit vectors

Position, velocity, and acceleration

Position

Velocity

Acceleration

Kinematic constraints

Rolling without slipping

Inextensible cord

See also

Notes

References

External links

Study guide

From Wikiversity

Contents

Introduction

Concepts and Definitions

Body

Configuration

Deformation

Deformation Gradient

Displacement

Displacement Gradient

Strains

Finite Strain Tensor

Infinitesimal Strain Tensor

Infinitesimal Rotation Tensor

Volume Change Due To Finite Deformation

Volume Change Due To Infinitesimal Deformation

Relation between axial vector and displacement

Relation between axial vector and strain

Related Content

1911 encyclopedia

From LoveToKnow 1911

Wikibooks

From Wikibooks, the open-content textbooks collection

Preface

Chapter 0: A Mathematics Review

Chapter 1: Plane Kinematics

Chapter 2: Kinematics in Three Space

Two Dimensional Kinematics

Displacement, Velocity, and Acceleration

Kinematics Formulas

Applications of Kinematics

Motion

Example 1

Gravity and Falling Objects

Example 1

Example 2

Related links

Related topics