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Kinematics

From Wikipedia, the free encyclopedia
Branch of physics describing the motion of objects without considering forces
"Kinematic" redirects here. For the music group, seeKinematic (band).
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

Kinematics is a subfield ofphysics andmathematics, developed inclassical mechanics, that describes themotion of points,bodies (objects), and systems of bodies (groups of objects) without considering theforces that cause them to move.[1][2][3] The study of how forces act on bodies falls withinkinetics ordynamics (includinganalytical dynamics), not kinematics.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of both applied and puremathematics since it can be studied without considering the mass of a body or the forces acting upon it.[4][5][6] A kinematics problem begins by describing the geometry of the system and declaring theinitial conditions of any known values of position, velocity and/or acceleration of points within the system. Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined.

Kinematics is used inastrophysics to describe the motion ofcelestial bodies and collections of such bodies. Inmechanical engineering,robotics, andbiomechanics,[7] kinematics is used to describe the motion of systems composed of joined parts (multi-link systems) such as anengine, arobotic arm or thehuman skeleton.

Geometric transformations, including calledrigid transformations, are used to describe the movement of components in amechanical system, simplifying the derivation of the equations of motion. They are also central todynamic analysis.

Kinematic analysis is the process of measuring the kinematicquantities used to describe motion. In engineering, for instance, kinematic analysis may be used to find the range of movement for a givenmechanism and, working in reverse, usingkinematic synthesis to design a mechanism for a desired range of motion.[8] In addition, kinematics appliesalgebraic geometry to the study of themechanical advantage of amechanical system or mechanism.

Etymology

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The term kinematic is the English version ofA.M. Ampère'scinématique,[9] which he constructed from theGreekκίνημαkinema ("movement, motion"), itself derived fromκινεῖνkinein ("to move").[10][11]

Kinematic and cinématique are related to the French word cinéma, but neither are directly derived from it. However, they do share a root word in common, as cinéma came from the shortened form of cinématographe, "motion picture projector and camera", once again from the Greek word for movement and from the Greekγρᾰ́φωgrapho ("to write").[12]

Kinematics of a particle trajectory in a non-rotating frame of reference

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Kinematic quantities of a classical particle: massm, positionr, velocityv, accelerationa.
Position vectorr, always points radially from the origin.
Velocity vectorv, always tangent to the path of motion.
Acceleration vectora, not parallel to the radial motion but offset by the angular and Coriolis accelerations, nor tangent to the path but offset by the centripetal and radial accelerations.
Kinematic vectors in plane polar coordinates. Notice the setup is not restricted to 2-d space, but a plane in any higher dimension.

Particle kinematics is the study of the trajectory of particles. The position of a particle is defined as the coordinate vector from the origin of a coordinate frame to the particle. For example, consider a tower 50 m south from your home, where the coordinate frame is centered at your home, such that east is in the direction of thex-axis and north is in the direction of they-axis, then the coordinate vector to the base of the tower isr = (0 m, −50 m, 0 m). If the tower is 50 m high, and this height is measured along thez-axis, then the coordinate vector to the top of the tower isr = (0 m, −50 m, 50 m).

In the most general case, a three-dimensional coordinate system is used to define the position of a particle. However, if the particle is constrained to move within a plane, a two-dimensional coordinate system is sufficient. All observations in physics are incomplete without being described with respect to a reference frame.

The position vector of a particle is avector drawn from the origin of thereference 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 vectorr{\displaystyle {\bf {r}}} can be expressed asr=(x,y,z)=xx^+yy^+zz^,{\displaystyle \mathbf {r} =(x,y,z)=x{\hat {\mathbf {x} }}+y{\hat {\mathbf {y} }}+z{\hat {\mathbf {z} }},}wherex{\displaystyle x},y{\displaystyle y}, andz{\displaystyle z} are theCartesian coordinates andx^{\displaystyle {\hat {\mathbf {x} }}},y^{\displaystyle {\hat {\mathbf {y} }}} andz^{\displaystyle {\hat {\mathbf {z} }}} are theunit vectors along thex{\displaystyle x},y{\displaystyle y}, andz{\displaystyle z} coordinate axes, respectively. The magnitude of the position vector|r|{\displaystyle \left|\mathbf {r} \right|} gives the distance between the pointr{\displaystyle \mathbf {r} } and the origin.|r|=x2+y2+z2.{\displaystyle |\mathbf {r} |={\sqrt {x^{2}+y^{2}+z^{2}}}.}Thedirection cosines of the position vector provide a quantitative measure of direction. In general, an object's position vector will depend on the frame of reference; different frames will lead to different values for the position vector.

Thetrajectory of a particle is a vector function of time,r(t){\displaystyle \mathbf {r} (t)}, which defines the curve traced by the moving particle, given byr(t)=x(t)x^+y(t)y^+z(t)z^,{\displaystyle \mathbf {r} (t)=x(t){\hat {\mathbf {x} }}+y(t){\hat {\mathbf {y} }}+z(t){\hat {\mathbf {z} }},}wherex(t){\displaystyle x(t)},y(t){\displaystyle y(t)}, andz(t){\displaystyle z(t)} describe each coordinate of the particle's position as a function of time.

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

Velocity and speed

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Thevelocity of a particle is a vector quantity that describes thedirection as well as the magnitude of motion of the particle. More mathematically, the rate of change of the position vector of a point with respect to time is the velocity of the point. Consider the ratio formed by dividing the difference of two positions of a particle (displacement) by the time interval. This ratio is called theaverage velocity over that time interval and is defined asv¯=ΔrΔt=ΔxΔtx^+ΔyΔty^+ΔzΔtz^=v¯xx^+v¯yy^+v¯zz^{\displaystyle \mathbf {\bar {v}} ={\frac {\Delta \mathbf {r} }{\Delta t}}={\frac {\Delta x}{\Delta t}}{\hat {\mathbf {x} }}+{\frac {\Delta y}{\Delta t}}{\hat {\mathbf {y} }}+{\frac {\Delta z}{\Delta t}}{\hat {\mathbf {z} }}={\bar {v}}_{x}{\hat {\mathbf {x} }}+{\bar {v}}_{y}{\hat {\mathbf {y} }}+{\bar {v}}_{z}{\hat {\mathbf {z} }}\,}whereΔr{\displaystyle \Delta \mathbf {r} } is the displacement vector during the time intervalΔt{\displaystyle \Delta t}. In the limit that the time intervalΔt{\displaystyle \Delta t} approaches zero, the average velocity approaches the instantaneous velocity, defined as the time derivative of the position vector,v=limΔt0ΔrΔt=drdt=vxx^+vyy^+vzz^.{\displaystyle \mathbf {v} =\lim _{\Delta t\to 0}{\frac {\Delta \mathbf {r} }{\Delta t}}={\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}=v_{x}{\hat {\mathbf {x} }}+v_{y}{\hat {\mathbf {y} }}+v_{z}{\hat {\mathbf {z} }}.}Thus, a particle's velocity is the time rate of change of its position. Furthermore, this velocity istangent to the particle's trajectory at every position along its path. In a non-rotating frame of reference, the derivatives of the coordinate directions are not considered as their directions and magnitudes are constants.

Thespeed of an object is the magnitude of its velocity. It is a scalar quantity:v=|v|=dsdt,{\displaystyle v=|\mathbf {v} |={\frac {{\text{d}}s}{{\text{d}}t}},}wheres{\displaystyle s} is the arc-length measured along the trajectory of the particle. This arc-length must always increase as the particle moves. Hence,dsdt{\displaystyle {\frac {{\text{d}}s}{{\text{d}}t}}} is non-negative, which implies that speed is also non-negative.

Acceleration

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The velocity vector can change in magnitude and in direction or both at once. Hence, the acceleration accounts for both the rate of change of the magnitude of the velocity vector and the rate of change of direction of that vector. The same reasoning used with respect to the position of a particle to define velocity, can be applied to the velocity to define acceleration. Theacceleration of a particle is the vector defined by the rate of change of the velocity vector. Theaverage acceleration of a particle over a time interval is defined as the ratio.a¯=Δv¯Δt=Δv¯xΔtx^+Δv¯yΔty^+Δv¯zΔtz^=a¯xx^+a¯yy^+a¯zz^{\displaystyle \mathbf {\bar {a}} ={\frac {\Delta \mathbf {\bar {v}} }{\Delta t}}={\frac {\Delta {\bar {v}}_{x}}{\Delta t}}{\hat {\mathbf {x} }}+{\frac {\Delta {\bar {v}}_{y}}{\Delta t}}{\hat {\mathbf {y} }}+{\frac {\Delta {\bar {v}}_{z}}{\Delta t}}{\hat {\mathbf {z} }}={\bar {a}}_{x}{\hat {\mathbf {x} }}+{\bar {a}}_{y}{\hat {\mathbf {y} }}+{\bar {a}}_{z}{\hat {\mathbf {z} }}\,}where Δv is the average velocity and Δt is the time interval.

The acceleration of the particle is the limit of the average acceleration as the time interval approaches zero, which is the time derivative,a=limΔt0ΔvΔt=dvdt=axx^+ayy^+azz^.{\displaystyle \mathbf {a} =\lim _{\Delta t\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {{\text{d}}\mathbf {v} }{{\text{d}}t}}=a_{x}{\hat {\mathbf {x} }}+a_{y}{\hat {\mathbf {y} }}+a_{z}{\hat {\mathbf {z} }}.}

Alternatively,a=lim(Δt)20Δr(Δt)2=d2rdt2=axx^+ayy^+azz^.{\displaystyle \mathbf {a} =\lim _{(\Delta t)^{2}\to 0}{\frac {\Delta \mathbf {r} }{(\Delta t)^{2}}}={\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}=a_{x}{\hat {\mathbf {x} }}+a_{y}{\hat {\mathbf {y} }}+a_{z}{\hat {\mathbf {z} }}.}

Thus, acceleration is the first derivative of the velocity vector and the second derivative of the position vector of that particle. In a non-rotating frame of reference, the derivatives of the coordinate directions are not considered as their directions and magnitudes are constants.

The magnitude of theacceleration of an object is the magnitude |a| of its acceleration vector. It is a scalar quantity:|a|=|v˙|=dvdt.{\displaystyle |\mathbf {a} |=|{\dot {\mathbf {v} }}|={\frac {{\text{d}}v}{{\text{d}}t}}.}

Relative position vector

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A relative position vector is a vector that defines the position of one point relative to another. It is the difference in position of the two points.The position of one pointA relative to another pointB is simply the difference between their positions

rA/B=rArB{\displaystyle \mathbf {r} _{A/B}=\mathbf {r} _{A}-\mathbf {r} _{B}}

which is the difference between the components of their position vectors.

If pointA has position componentsrA=(xA,yA,zA){\displaystyle \mathbf {r} _{A}=\left(x_{A},y_{A},z_{A}\right)}

and pointB has position componentsrB=(xB,yB,zB){\displaystyle \mathbf {r} _{B}=\left(x_{B},y_{B},z_{B}\right)}

then the position of pointA relative to pointB is the difference between their components:rA/B=rArB=(xAxB,yAyB,zAzB){\displaystyle \mathbf {r} _{A/B}=\mathbf {r} _{A}-\mathbf {r} _{B}=\left(x_{A}-x_{B},y_{A}-y_{B},z_{A}-z_{B}\right)}

Relative velocity

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Main article:Relative velocity
Relative velocities between two particles in classical mechanics.

The velocity of one point relative to another is simply the difference between their velocitiesvA/B=vAvB{\displaystyle \mathbf {v} _{A/B}=\mathbf {v} _{A}-\mathbf {v} _{B}}which is the difference between the components of their velocities.

If pointA has velocity componentsvA=(vAx,vAy,vAz){\displaystyle \mathbf {v} _{A}=\left(v_{A_{x}},v_{A_{y}},v_{A_{z}}\right)} and pointB has velocity componentsvB=(vBx,vBy,vBz){\displaystyle \mathbf {v} _{B}=\left(v_{B_{x}},v_{B_{y}},v_{B_{z}}\right)} then the velocity of pointA relative to pointB is the difference between their components:vA/B=vAvB=(vAxvBx,vAyvBy,vAzvBz){\displaystyle \mathbf {v} _{A/B}=\mathbf {v} _{A}-\mathbf {v} _{B}=\left(v_{A_{x}}-v_{B_{x}},v_{A_{y}}-v_{B_{y}},v_{A_{z}}-v_{B_{z}}\right)}

Alternatively, this same result could be obtained by computing the time derivative of the relative position vectorrB/A.

Relative acceleration

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The acceleration of one pointC relative to another pointB is simply the difference between their accelerations.aC/B=aCaB{\displaystyle \mathbf {a} _{C/B}=\mathbf {a} _{C}-\mathbf {a} _{B}}which is the difference between the components of their accelerations.

If pointC has acceleration componentsaC=(aCx,aCy,aCz){\displaystyle \mathbf {a} _{C}=\left(a_{C_{x}},a_{C_{y}},a_{C_{z}}\right)}and pointB has acceleration componentsaB=(aBx,aBy,aBz){\displaystyle \mathbf {a} _{B}=\left(a_{B_{x}},a_{B_{y}},a_{B_{z}}\right)}then the acceleration of pointC relative to pointB is the difference between their components:aC/B=aCaB=(aCxaBx,aCyaBy,aCzaBz){\displaystyle \mathbf {a} _{C/B}=\mathbf {a} _{C}-\mathbf {a} _{B}=\left(a_{C_{x}}-a_{B_{x}},a_{C_{y}}-a_{B_{y}},a_{C_{z}}-a_{B_{z}}\right)}

Alternatively, this same result could be obtained by computing the second time derivative of the relative position vectorrB/A.[13]

Assuming that the initial conditions of the position,r0{\displaystyle \mathbf {r} _{0}}, and velocityv0{\displaystyle \mathbf {v} _{0}} at timet=0{\displaystyle t=0} are known, the first integration yields the velocity of the particle as a function of time.[a]v(t)=v0+0tadτ=v0+at.{\displaystyle \mathbf {v} (t)=\mathbf {v} _{0}+\int _{0}^{t}\mathbf {a} \,{\text{d}}\tau =\mathbf {v} _{0}+\mathbf {a} t.}

A second integration yields its path (trajectory),r(t)=r0+0tv(τ)dτ=r0+0t(v0+aτ)dτ=r0+v0t+12at2.{\displaystyle \mathbf {r} (t)=\mathbf {r} _{0}+\int _{0}^{t}\mathbf {v} (\tau )\,{\text{d}}\tau =\mathbf {r} _{0}+\int _{0}^{t}\left(\mathbf {v} _{0}+\mathbf {a} \tau \right){\text{d}}\tau =\mathbf {r} _{0}+\mathbf {v} _{0}t+{\tfrac {1}{2}}\mathbf {a} t^{2}.}

Additional relations between displacement, velocity, acceleration, and time can be derived. Since the acceleration is constant,a=ΔvΔt=vv0t{\displaystyle \mathbf {a} ={\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {\mathbf {v} -\mathbf {v} _{0}}{t}}} can be substituted into the above equation to give:r(t)=r0+(v+v02)t.{\displaystyle \mathbf {r} (t)=\mathbf {r} _{0}+\left({\frac {\mathbf {v} +\mathbf {v} _{0}}{2}}\right)t.}

A relationship between velocity, position and acceleration without explicit time dependence can be had by solving the average acceleration for time and substituting and simplifying

t=vv0a{\displaystyle t={\frac {\mathbf {v} -\mathbf {v} _{0}}{\mathbf {a} }}}

(rr0)a=(vv0)v+v02 ,{\displaystyle \left(\mathbf {r} -\mathbf {r} _{0}\right)\cdot \mathbf {a} =\left(\mathbf {v} -\mathbf {v} _{0}\right)\cdot {\frac {\mathbf {v} +\mathbf {v} _{0}}{2}}\ ,}where{\displaystyle \cdot } denotes thedot product, which is appropriate as the products are scalars rather than vectors.2(rr0)a=|v|2|v0|2.{\displaystyle 2\left(\mathbf {r} -\mathbf {r} _{0}\right)\cdot \mathbf {a} =|\mathbf {v} |^{2}-|\mathbf {v} _{0}|^{2}.}

The dot product can be replaced by the cosine of the angleα between the vectors (seeGeometric interpretation of the dot product for more details) and the vectors by their magnitudes, in which case:2|rr0||a|cosα=|v|2|v0|2.{\displaystyle 2\left|\mathbf {r} -\mathbf {r} _{0}\right|\left|\mathbf {a} \right|\cos \alpha =|\mathbf {v} |^{2}-|\mathbf {v} _{0}|^{2}.}

In the case of acceleration always in the direction of the motion and the direction of motion should be in positive or negative, the angle between the vectors (α) is 0, socos0=1{\displaystyle \cos 0=1}, and|v|2=|v0|2+2|a||rr0|.{\displaystyle |\mathbf {v} |^{2}=|\mathbf {v} _{0}|^{2}+2\left|\mathbf {a} \right|\left|\mathbf {r} -\mathbf {r} _{0}\right|.}This can be simplified using the notation for the magnitudes of the vectors|a|=a,|v|=v,|rr0|=Δr{\displaystyle |\mathbf {a} |=a,|\mathbf {v} |=v,|\mathbf {r} -\mathbf {r} _{0}|=\Delta r}[citation needed] whereΔr{\displaystyle \Delta r} can be any curvaceous path taken as the constant tangential acceleration is applied along that path[citation needed], sov2=v02+2aΔr.{\displaystyle v^{2}=v_{0}^{2}+2a\Delta r.}

This reduces the parametric equations of motion of the particle to a Cartesian relationship of speed versus position. This relation is useful when time is unknown. We also know thatΔr=vdt{\textstyle \Delta r=\int v\,{\text{d}}t} orΔr{\displaystyle \Delta r} is the area under a velocity–time graph.[15]

Velocity Time physics graph

We can takeΔr{\displaystyle \Delta r} by adding the top area and the bottom area. The bottom area is a rectangle, and the area of a rectangle is theAB{\displaystyle A\cdot B} whereA{\displaystyle A} is the width andB{\displaystyle B} is the height. In this caseA=t{\displaystyle A=t} andB=v0{\displaystyle B=v_{0}} (theA{\displaystyle A} here is different from the accelerationa{\displaystyle a}). This means that the bottom area istv0{\displaystyle tv_{0}}. Now let's find the top area (a triangle). The area of a triangle is12BH{\textstyle {\frac {1}{2}}BH} whereB{\displaystyle B} is the base andH{\displaystyle H} is the height.[16] In this case,B=t{\displaystyle B=t} andH=at{\displaystyle H=at} orA=12BH=12att=12at2=at22{\textstyle A={\frac {1}{2}}BH={\frac {1}{2}}att={\frac {1}{2}}at^{2}={\frac {at^{2}}{2}}}. Addingv0t{\displaystyle v_{0}t} andat22{\textstyle {\frac {at^{2}}{2}}} results in the equationΔr{\displaystyle \Delta r} results in the equationΔr=v0t+at22{\textstyle \Delta r=v_{0}t+{\frac {at^{2}}{2}}}.[17] This equation is applicable when the final velocityv is unknown.

Figure 2: Velocity and acceleration for nonuniform circular motion: the velocity vector is tangential to the orbit, but the acceleration vector is not radially inward because of its tangential componentaθ that increases the rate of rotation: dω/dt = |aθ|/R.

Particle trajectories in cylindrical-polar coordinates

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See also:Generalized coordinates,Curvilinear coordinates,Orthogonal coordinates, andFrenet-Serret formulas

It is often convenient to formulate the trajectory of a particler(t) = (x(t),y(t),z(t)) using polar coordinates in theXY plane. In this case, its velocity and acceleration take a convenient form.

Recall that the trajectory of a particleP is defined by its coordinate vectorr measured in a fixed reference frameF. As the particle moves, its coordinate vectorr(t) traces its trajectory, which is a curve in space, given by:r(t)=x(t)x^+y(t)y^+z(t)z^,{\displaystyle \mathbf {r} (t)=x(t){\hat {\mathbf {x} }}+y(t){\hat {\mathbf {y} }}+z(t){\hat {\mathbf {z} }},}where,ŷ, and are theunit vectors along thex,y andz axes of thereference frameF, respectively.

Consider a particleP that moves only on the surface of a circular cylinderr(t) = constant, it is possible to align thez axis of the fixed frameF with the axis of the cylinder. Then, the angleθ around this axis in thexy plane can be used to define the trajectory as,r(t)=rcos(θ(t))x^+rsin(θ(t))y^+z(t)z^,{\displaystyle \mathbf {r} (t)=r\cos(\theta (t)){\hat {\mathbf {x} }}+r\sin(\theta (t)){\hat {\mathbf {y} }}+z(t){\hat {\mathbf {z} }},}where the constant distance from the center is denoted asr, andθ(t) is a function of time.

The cylindrical coordinates forr(t) can be simplified by introducing the radial and tangential unit vectors,r^=cos(θ(t))x^+sin(θ(t))y^,θ^=sin(θ(t))x^+cos(θ(t))y^.{\displaystyle {\hat {\mathbf {r} }}=\cos(\theta (t)){\hat {\mathbf {x} }}+\sin(\theta (t)){\hat {\mathbf {y} }},\quad {\hat {\mathbf {\theta } }}=-\sin(\theta (t)){\hat {\mathbf {x} }}+\cos(\theta (t)){\hat {\mathbf {y} }}.}and their time derivatives from elementary calculus:dr^dt=ωθ^.{\displaystyle {\frac {{\text{d}}{\hat {\mathbf {r} }}}{{\text{d}}t}}=\omega {\hat {\mathbf {\theta } }}.}d2r^dt2=d(ωθ^)dt=αθ^ω2r^.{\displaystyle {\frac {{\text{d}}^{2}{\hat {\mathbf {r} }}}{{\text{d}}t^{2}}}={\frac {{\text{d}}(\omega {\hat {\mathbf {\theta } }})}{{\text{d}}t}}=\alpha {\hat {\mathbf {\theta } }}-\omega ^{2}{\hat {\mathbf {r} }}.}

dθ^dt=ωr^.{\displaystyle {\frac {{\text{d}}{\hat {\mathbf {\theta } }}}{{\text{d}}t}}=-\omega {\hat {\mathbf {r} }}.}d2θ^dt2=d(ωr^)dt=αr^ω2θ^.{\displaystyle {\frac {{\text{d}}^{2}{\hat {\mathbf {\theta } }}}{{\text{d}}t^{2}}}={\frac {{\text{d}}(-\omega {\hat {\mathbf {r} }})}{{\text{d}}t}}=-\alpha {\hat {\mathbf {r} }}-\omega ^{2}{\hat {\mathbf {\theta } }}.}

Using this notation,r(t) takes the form,r(t)=rr^+z(t)z^.{\displaystyle \mathbf {r} (t)=r{\hat {\mathbf {r} }}+z(t){\hat {\mathbf {z} }}.}In general, the trajectoryr(t) is not constrained to lie on a circular cylinder, so the radiusR varies with time and the trajectory of the particle in cylindrical-polar coordinates becomes:r(t)=r(t)r^+z(t)z^.{\displaystyle \mathbf {r} (t)=r(t){\hat {\mathbf {r} }}+z(t){\hat {\mathbf {z} }}.}Wherer,θ, andz might be continuously differentiable functions of time and the function notation is dropped for simplicity. The velocity vectorvP is the time derivative of the trajectoryr(t), which yields:vP=ddt(rr^+zz^)=vr^+rωθ^+vzz^=v(r^+θ^)+vzz^.{\displaystyle \mathbf {v} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }}\right)=v{\hat {\mathbf {r} }}+r\mathbf {\omega } {\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}=v({\hat {\mathbf {r} }}+{\hat {\mathbf {\theta } }})+v_{z}{\hat {\mathbf {z} }}.}

Similarly, the accelerationaP, which is the time derivative of the velocityvP, is given by:aP=ddt(vr^+vθ^+vzz^)=(avω)r^+(a+vω)θ^+azz^.{\displaystyle \mathbf {a} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(v{\hat {\mathbf {r} }}+v{\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}\right)=(a-v\omega ){\hat {\mathbf {r} }}+(a+v\omega ){\hat {\mathbf {\theta } }}+a_{z}{\hat {\mathbf {z} }}.}

The termvωr^{\displaystyle -v\omega {\hat {\mathbf {r} }}} acts toward the center of curvature of the path at that point on the path, is commonly called thecentripetal acceleration. The termvωθ^{\displaystyle v\omega {\hat {\mathbf {\theta } }}} is called theCoriolis acceleration.

Constant radius

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If the trajectory of the particle is constrained to lie on a cylinder, then the radiusr is constant and the velocity and acceleration vectors simplify. The velocity ofvP is the time derivative of the trajectoryr(t),vP=ddt(rr^+zz^)=rωθ^+vzz^=vθ^+vzz^.{\displaystyle \mathbf {v} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }}\right)=r\omega {\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}=v{\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}.}

Planar circular trajectories

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Kinematics of Machinery
Each particle on the wheel travels in a planar circular trajectory (Kinematics of Machinery, 1876).[18]

A special case of a particle trajectory on a circular cylinder occurs when there is no movement along thez axis:r(t)=rr^+zz^,{\displaystyle \mathbf {r} (t)=r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }},}wherer andz0 are constants. In this case, the velocityvP is given by:vP=ddt(rr^+zz^)=rωθ^=vθ^,{\displaystyle \mathbf {v} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }}\right)=r\omega {\hat {\mathbf {\theta } }}=v{\hat {\mathbf {\theta } }},}whereω{\displaystyle \omega } is theangular velocity of the unit vectorθ^ around thez axis of the cylinder.

The accelerationaP of the particleP is now given by:aP=d(vθ^)dt=aθ^vθr^.{\displaystyle \mathbf {a} _{P}={\frac {{\text{d}}(v{\hat {\mathbf {\theta } }})}{{\text{d}}t}}=a{\hat {\mathbf {\theta } }}-v\theta {\hat {\mathbf {r} }}.}

The componentsar=vθ,aθ=a,{\displaystyle a_{r}=-v\theta ,\quad a_{\theta }=a,}are called, respectively, theradial andtangential components of acceleration.

The notation for angular velocity andangular acceleration is often defined asω=θ˙,α=θ¨,{\displaystyle \omega ={\dot {\theta }},\quad \alpha ={\ddot {\theta }},}so the radial and tangential acceleration components for circular trajectories are also written asar=rω2,aθ=rα.{\displaystyle a_{r}=-r\omega ^{2},\quad a_{\theta }=r\alpha .}

Point trajectories in a body moving in the plane

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The movement of components of amechanical system are analyzed by attaching areference frame to each part and determining how the various reference frames move relative to each other. If the structural stiffness of the parts are sufficient, then their deformation can be neglected and rigid transformations can be used to define this relative movement. This reduces the description of the motion of the various parts of a complicated mechanical system to a problem of describing the geometry of each part and geometric association of each part relative to other parts.

Geometry is the study of the properties of figures that remain the same while the space is transformed in various ways—more technically, it is the study of invariants under a set of transformations.[19] These transformations can cause the displacement of the triangle in the plane, while leaving the vertex angle and the distances between vertices unchanged. Kinematics is often described as applied geometry, where the movement of a mechanical system is described using the rigid transformations of Euclidean geometry.

The coordinates of points in a plane are two-dimensional vectors inR2 (two dimensional space). Rigid transformations are those that preserve thedistance between any two points. The set of rigid transformations in ann-dimensional space is called the specialEuclidean group onRn, and denotedSE(n).

Displacements and motion

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Boulton & Watt Steam Engine
The movement of each of the components of the Boulton & Watt Steam Engine (1784) is modeled by a continuous set of rigid displacements.

The position of one component of a mechanical system relative to another is defined by introducing a reference frame, sayM, on one that moves relative to a fixed frame,F, on the other. The rigid transformation, or displacement, ofM relative toF defines the relative position of the two components. A displacement consists of the combination of arotation and atranslation.

The set of all displacements ofM relative toF is called theconfiguration space ofM. A smooth curve from one position to another in this configuration space is a continuous set of displacements, called themotion ofM relative toF. The motion of a body consists of a continuous set of rotations and translations.

Matrix representation

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The combination of a rotation and translation in the planeR2 can be represented by a certain type of 3×3 matrix known as a homogeneous transform. The 3×3 homogeneous transform is constructed from a 2×2rotation matrixA(φ) and the 2×1 translation vectord = (dx,dy), as:[T(ϕ,d)]=[A(ϕ)d01]=[cosϕsinϕdxsinϕcosϕdy001].{\displaystyle [T(\phi ,\mathbf {d} )]={\begin{bmatrix}A(\phi )&\mathbf {d} \\\mathbf {0} &1\end{bmatrix}}={\begin{bmatrix}\cos \phi &-\sin \phi &d_{x}\\\sin \phi &\cos \phi &d_{y}\\0&0&1\end{bmatrix}}.}These homogeneous transforms perform rigid transformations on the points in the planez = 1, that is, on points with coordinatesr = (x,y, 1).

In particular, letr define the coordinates of points in a reference frameM coincident with a fixed frameF. Then, when the origin ofM is displaced by the translation vectord relative to the origin ofF and rotated by the angle φ relative to the x-axis ofF, the new coordinates inF of points inM are given by:P=[T(ϕ,d)]r=[cosϕsinϕdxsinϕcosϕdy001][xy1].{\displaystyle \mathbf {P} =[T(\phi ,\mathbf {d} )]\mathbf {r} ={\begin{bmatrix}\cos \phi &-\sin \phi &d_{x}\\\sin \phi &\cos \phi &d_{y}\\0&0&1\end{bmatrix}}{\begin{bmatrix}x\\y\\1\end{bmatrix}}.}

Homogeneous transforms representaffine transformations. This formulation is necessary because atranslation is not alinear transformation ofR2. However, using projective geometry, so thatR2 is considered a subset ofR3, translations become affine linear transformations.[20]

Pure translation

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If a rigid body moves so that itsreference frameM does not rotate (θ = 0) relative to the fixed frameF, the motion is called pure translation. In this case, the trajectory of every point in the body is an offset of the trajectoryd(t) of the origin ofM, that is:r(t)=[T(0,d(t))]p=d(t)+p.{\displaystyle \mathbf {r} (t)=[T(0,\mathbf {d} (t))]\mathbf {p} =\mathbf {d} (t)+\mathbf {p} .}

Thus, for bodies in pure translation, the velocity and acceleration of every pointP in the body are given by:vP=r˙(t)=d˙(t)=vO,aP=r¨(t)=d¨(t)=aO,{\displaystyle \mathbf {v} _{P}={\dot {\mathbf {r} }}(t)={\dot {\mathbf {d} }}(t)=\mathbf {v} _{O},\quad \mathbf {a} _{P}={\ddot {\mathbf {r} }}(t)={\ddot {\mathbf {d} }}(t)=\mathbf {a} _{O},}where the dot denotes the derivative with respect to time andvO andaO are the velocity and acceleration, respectively, of the origin of the moving frameM. Recall the coordinate vectorp inM is constant, so its derivative is zero.

Rotation of a body around a fixed axis

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Main article:Rotation around a fixed axis
Figure 1: The angular velocity vectorΩ points up for counterclockwise rotation and down for clockwise rotation, as specified by theright-hand rule. Angular positionθ(t) changes with time at a rateω(t) = dθ/dt.

Objects like a playgroundmerry-go-round, ventilation fans, or hinged doors can be modeled as rigid bodies rotating about a single fixed axis.[21]: 37  Thez-axis has been chosen by convention.

Position

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This allows the description of a rotation as the angular position of a planar reference frameM relative to a fixedF about this sharedz-axis. Coordinatesp = (x,y) inM are related to coordinatesP = (X, Y) inF by the matrix equation:P(t)=[A(t)]p,{\displaystyle \mathbf {P} (t)=[A(t)]\mathbf {p} ,}

where[A(t)]=[cos(θ(t))sin(θ(t))sin(θ(t))cos(θ(t))],{\displaystyle [A(t)]={\begin{bmatrix}\cos(\theta (t))&-\sin(\theta (t))\\\sin(\theta (t))&\cos(\theta (t))\end{bmatrix}},}is the rotation matrix that defines the angular position ofM relative toF as a function of time.

Velocity

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If the pointp does not move inM, its velocity inF is given byvP=P˙=[A˙(t)]p.{\displaystyle \mathbf {v} _{P}={\dot {\mathbf {P} }}=[{\dot {A}}(t)]\mathbf {p} .}It is convenient to eliminate the coordinatesp and write this as an operation on the trajectoryP(t),vP=[A˙(t)][A(t)1]P=[Ω]P,{\displaystyle \mathbf {v} _{P}=[{\dot {A}}(t)][A(t)^{-1}]\mathbf {P} =[\Omega ]\mathbf {P} ,}where the matrix[Ω]=[0ωω0],{\displaystyle [\Omega ]={\begin{bmatrix}0&-\omega \\\omega &0\end{bmatrix}},}is known as the angular velocity matrix ofM relative toF. The parameterω is the time derivative of the angleθ, that is:ω=dθdt.{\displaystyle \omega ={\frac {{\text{d}}\theta }{{\text{d}}t}}.}

Acceleration

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The acceleration ofP(t) inF is obtained as the time derivative of the velocity,AP=P¨(t)=[Ω˙]P+[Ω]P˙,{\displaystyle \mathbf {A} _{P}={\ddot {P}}(t)=[{\dot {\Omega }}]\mathbf {P} +[\Omega ]{\dot {\mathbf {P} }},}which becomesAP=[Ω˙]P+[Ω][Ω]P,{\displaystyle \mathbf {A} _{P}=[{\dot {\Omega }}]\mathbf {P} +[\Omega ][\Omega ]\mathbf {P} ,}where[Ω˙]=[0αα0],{\displaystyle [{\dot {\Omega }}]={\begin{bmatrix}0&-\alpha \\\alpha &0\end{bmatrix}},}is the angular acceleration matrix ofM onF, andα=d2θdt2.{\displaystyle \alpha ={\frac {{\text{d}}^{2}\theta }{{\text{d}}t^{2}}}.}

The 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 vectorr(t) locating the point. The vectorr(t) has some projection (or, equivalently, some component)r(t) on a plane perpendicular to the axis of rotation. Then theangular position of that point is the angleθ from a reference axis (typically the positivex-axis) to the vectorr(t) in a known rotation sense (typically given by theright-hand rule).
  • Angular velocity: the angular velocityω is the rate at which the angular positionθ changes with respect to timet:ω=dθdt{\displaystyle \omega ={\frac {{\text{d}}\theta }{{\text{d}}t}}} The angular velocity is represented in Figure 1 by a vectorΩ pointing along the axis of rotation with magnitudeω and sense determined by the direction of rotation as given by theright-hand rule.
  • Angular acceleration: the magnitude of the angular accelerationα is the rate at which the angular velocityω changes with respect to timet:α=dωdt{\displaystyle \alpha ={\frac {{\text{d}}\omega }{{\text{d}}t}}}

The equations of translational kinematics can easily be extended to planar rotational kinematics for constant angular acceleration with simple variable exchanges:ωf=ωi+αt{\displaystyle \omega _{\mathrm {f} }=\omega _{\mathrm {i} }+\alpha t\!}θfθi=ωit+12αt2{\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }=\omega _{\mathrm {i} }t+{\tfrac {1}{2}}\alpha t^{2}}θfθi=12(ωf+ωi)t{\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }={\tfrac {1}{2}}(\omega _{\mathrm {f} }+\omega _{\mathrm {i} })t}ωf2=ωi2+2α(θfθi).{\displaystyle \omega _{\mathrm {f} }^{2}=\omega _{\mathrm {i} }^{2}+2\alpha (\theta _{\mathrm {f} }-\theta _{\mathrm {i} }).}

Hereθi andθf are, respectively, the initial and final angular positions,ωi andωf are, respectively, the initial and final angular velocities, andα 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 trajectories in body moving in three dimensions

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Important formulas in kinematics define thevelocity and acceleration of points in a moving body as they trace trajectories in three-dimensional space. This is particularly important for the center of mass of a body, which is used to derive equations of motion using eitherNewton's second law orLagrange's equations.

Position

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In order to define these formulas, the movement of a componentB of a mechanical system is defined by the set of rotations [A(t)] and translationsd(t) assembled into the homogeneous transformation [T(t)]=[A(t),d(t)]. Ifp is the coordinates of a pointP inB measured in the movingreference frameM, then the trajectory of this point traced inF is given by:P(t)=[T(t)]p=[P1]=[A(t)d(t)01][p1].{\displaystyle \mathbf {P} (t)=[T(t)]\mathbf {p} ={\begin{bmatrix}\mathbf {P} \\1\end{bmatrix}}={\begin{bmatrix}A(t)&\mathbf {d} (t)\\0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}.}This notation does not distinguish betweenP = (X, Y, Z, 1), andP = (X, Y, Z), which is hopefully clear in context.

This equation for the trajectory ofP can be inverted to compute the coordinate vectorp inM as:p=[T(t)]1P(t)=[p1]=[A(t)TA(t)Td(t)01][P(t)1].{\displaystyle \mathbf {p} =[T(t)]^{-1}\mathbf {P} (t)={\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}={\begin{bmatrix}A(t)^{\text{T}}&-A(t)^{\text{T}}\mathbf {d} (t)\\0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}.}This expression uses the fact that the transpose of a rotation matrix is also its inverse, that is:[A(t)]T[A(t)]=I.{\displaystyle [A(t)]^{\text{T}}[A(t)]=I.\!}

Velocity

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The velocity of the pointP along its trajectoryP(t) is obtained as the time derivative of this position vector,vP=[T˙(t)]p=[vP0]=(ddt[A(t)d(t)01])[p1]=[A˙(t)d˙(t)00][p1].{\displaystyle \mathbf {v} _{P}=[{\dot {T}}(t)]\mathbf {p} ={\begin{bmatrix}\mathbf {v} _{P}\\0\end{bmatrix}}=\left({\frac {d}{dt}}{\begin{bmatrix}A(t)&\mathbf {d} (t)\\0&1\end{bmatrix}}\right){\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}={\begin{bmatrix}{\dot {A}}(t)&{\dot {\mathbf {d} }}(t)\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}.}The dot denotes the derivative with respect to time; becausep is constant, its derivative is zero.

This formula can be modified to obtain the velocity ofP by operating on its trajectoryP(t) measured in the fixed frameF. Substituting theinverse transform forp into the velocity equation yields:vP=[T˙(t)][T(t)]1P(t)=[vP0]=[A˙d˙00][Ad01]1[P(t)1]=[A˙d˙00]A1[1d0A][P(t)1]=[A˙A1A˙A1d+d˙00][P(t)1]=[A˙ATA˙ATd+d˙00][P(t)1]vP=[S]P.{\displaystyle {\begin{aligned}\mathbf {v} _{P}&=[{\dot {T}}(t)][T(t)]^{-1}\mathbf {P} (t)\\[4pt]&={\begin{bmatrix}\mathbf {v} _{P}\\0\end{bmatrix}}={\begin{bmatrix}{\dot {A}}&{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}A&\mathbf {d} \\0&1\end{bmatrix}}^{-1}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}&{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}A^{-1}{\begin{bmatrix}1&-\mathbf {d} \\0&A\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}A^{-1}&-{\dot {A}}A^{-1}\mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}A^{\text{T}}&-{\dot {A}}A^{\text{T}}\mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[6pt]\mathbf {v} _{P}&=[S]\mathbf {P} .\end{aligned}}}The matrix [S] is given by:[S]=[ΩΩd+d˙00]{\displaystyle [S]={\begin{bmatrix}\Omega &-\Omega \mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}}where[Ω]=A˙AT,{\displaystyle [\Omega ]={\dot {A}}A^{\text{T}},}is the angular velocity matrix.

Multiplying by the operator [S], the formula for the velocityvP takes the form:vP=[Ω](Pd)+d˙=ω×RP/O+vO,{\displaystyle \mathbf {v} _{P}=[\Omega ](\mathbf {P} -\mathbf {d} )+{\dot {\mathbf {d} }}=\omega \times \mathbf {R} _{P/O}+\mathbf {v} _{O},}where the vectorω is the angular velocity vector obtained from the components of the matrix [Ω]; the vectorRP/O=Pd,{\displaystyle \mathbf {R} _{P/O}=\mathbf {P} -\mathbf {d} ,}is the position ofP relative to the originO of the moving frameM; andvO=d˙,{\displaystyle \mathbf {v} _{O}={\dot {\mathbf {d} }},}is the velocity of the originO.

Acceleration

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The acceleration of a pointP in a moving bodyB is obtained as the time derivative of its velocity vector:AP=ddtvP=ddt([S]P)=[S˙]P+[S]P˙=[S˙]P+[S][S]P.{\displaystyle \mathbf {A} _{P}={\frac {d}{dt}}\mathbf {v} _{P}={\frac {d}{dt}}\left([S]\mathbf {P} \right)=[{\dot {S}}]\mathbf {P} +[S]{\dot {\mathbf {P} }}=[{\dot {S}}]\mathbf {P} +[S][S]\mathbf {P} .}

This equation can be expanded firstly by computing[S˙]=[Ω˙Ω˙dΩd˙+d¨00]=[Ω˙Ω˙dΩvO+AO00]{\displaystyle [{\dot {S}}]={\begin{bmatrix}{\dot {\Omega }}&-{\dot {\Omega }}\mathbf {d} -\Omega {\dot {\mathbf {d} }}+{\ddot {\mathbf {d} }}\\0&0\end{bmatrix}}={\begin{bmatrix}{\dot {\Omega }}&-{\dot {\Omega }}\mathbf {d} -\Omega \mathbf {v} _{O}+\mathbf {A} _{O}\\0&0\end{bmatrix}}}and[S]2=[ΩΩd+vO00]2=[Ω2Ω2d+ΩvO00].{\displaystyle [S]^{2}={\begin{bmatrix}\Omega &-\Omega \mathbf {d} +\mathbf {v} _{O}\\0&0\end{bmatrix}}^{2}={\begin{bmatrix}\Omega ^{2}&-\Omega ^{2}\mathbf {d} +\Omega \mathbf {v} _{O}\\0&0\end{bmatrix}}.}

The formula for the accelerationAP can now be obtained as:AP=Ω˙(Pd)+AO+Ω2(Pd),{\displaystyle \mathbf {A} _{P}={\dot {\Omega }}(\mathbf {P} -\mathbf {d} )+\mathbf {A} _{O}+\Omega ^{2}(\mathbf {P} -\mathbf {d} ),}orAP=α×RP/O+ω×ω×RP/O+AO,{\displaystyle \mathbf {A} _{P}=\alpha \times \mathbf {R} _{P/O}+\omega \times \omega \times \mathbf {R} _{P/O}+\mathbf {A} _{O},}whereα is the angular acceleration vector obtained from the derivative of the angular velocity vector;RP/O=Pd,{\displaystyle \mathbf {R} _{P/O}=\mathbf {P} -\mathbf {d} ,}is the relative position vector (the position ofP relative to the originO of the moving frameM); andAO=d¨{\displaystyle \mathbf {A} _{O}={\ddot {\mathbf {d} }}}is the acceleration of the origin of the moving frameM.

Kinematic constraints

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Kinematic constraints are constraints on the movement of components of a mechanical system. Kinematic constraints can be considered to have two basic forms, (i) constraints that arise from hinges, sliders and cam joints that define the construction of the system, calledholonomic constraints, and (ii) constraints imposed on the velocity of the system such as the knife-edge constraint of ice-skates on a flat plane, or rolling without slipping of a disc or sphere in contact with a plane, which are callednon-holonomic constraints. The following are some common examples.

Kinematic coupling

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Akinematic coupling exactly constrains all 6 degrees of freedom.

Rolling without slipping

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An object that rolls against asurface without slipping obeys the condition that thevelocity of itscenter of mass is equal to thecross product of its angular velocity with a vector from the point of contact to the center of mass:vG(t)=Ω×rG/O.{\displaystyle {\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 tov=rω{\displaystyle v=r\omega }.

Inextensible cord

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This is the case where bodies are connected by an idealized cord that remains in tension and cannot change length. The constraint is that the sum of lengths of all segments of the cord is the total length, and accordingly the time derivative of this sum is zero.[22][23][24] A dynamic problem of this type is thependulum. Another example is a drum turned by the pull of gravity upon a falling weight attached to the rim by the inextensible cord.[25] Anequilibrium problem (i.e. not kinematic) of this type is thecatenary.[26]

Kinematic pairs

[edit]
Main article:Kinematic pair

Reuleaux called the ideal connections between components that form a machinekinematic pairs. He distinguished between higher pairs which were said to have line contact between the two links and lower pairs that have area contact between the links. J. Phillips shows that there are many ways to construct pairs that do not fit this simple classification.[27]

Lower pair

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A lower pair is an ideal joint, or holonomic constraint, that maintains contact between a point, line or plane in a moving solid (three-dimensional) body to a corresponding point line or plane in the fixed solid body. There are the following cases:

  • A revolute pair, or hinged joint, requires a line, or axis, in the moving body to remain co-linear with a line in the fixed body, and a plane perpendicular to this line in the moving body maintain contact with a similar perpendicular plane in the fixed body. This imposes five constraints on the relative movement of the links, which therefore has one degree of freedom, which is pure rotation about the axis of the hinge.
  • Aprismatic joint, or slider, requires that a line, or axis, in the moving body remain co-linear with a line in the fixed body, and a plane parallel to this line in the moving body maintain contact with a similar parallel plane in the fixed body. This imposes five constraints on the relative movement of the links, which therefore has one degree of freedom. This degree of freedom is the distance of the slide along the line.
  • A cylindrical joint requires that a line, or axis, in the moving body remain co-linear with a line in the fixed body. It is a combination of a revolute joint and a sliding joint. This joint has two degrees of freedom. The position of the moving body is defined by both the rotation about and slide along the axis.
  • A spherical joint, or ball joint, requires that a point in the moving body maintain contact with a point in the fixed body. This joint has three degrees of freedom.
  • A planar joint requires that a plane in the moving body maintain contact with a plane in fixed body. This joint has three degrees of freedom.

Higher pairs

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Generally speaking, a higher pair is a constraint that requires a curve or surface in the moving body to maintain contact with a curve or surface in the fixed body. For example, the contact between a cam and its follower is a higher pair called acam joint. Similarly, the contact between the involute curves that form the meshing teeth of two gears are cam joints.

Kinematic chains

[edit]
Illustration of a Four-bar linkage from Kinematics of Machinery, 1876
Illustration of a four-bar linkage fromKinematics of Machinery, 1876

Rigid bodies ("links") connected bykinematic pairs ("joints") are known askinematic chains.Mechanisms and robots are examples of kinematic chains. Thedegree of freedom of a kinematic chain is computed from the number of links and the number and type of joints using themobility formula. This formula can also be used to enumerate thetopologies of kinematic chains that have a given degree of freedom, which is known astype synthesis in machine design.

Examples

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The planar one degree-of-freedomlinkages assembled fromN links andj hinges or sliding joints are:

  • N = 2,j = 1 : a two-bar linkage that is the lever;
  • N = 4,j = 4 : thefour-bar linkage;
  • N = 6,j = 7 : asix-bar linkage. This must have two links ("ternary links") that support three joints. There are two distinct topologies that depend on how the two ternary linkages are connected. In theWatt topology, the two ternary links have a common joint; in theStephenson topology, the two ternary links do not have a common joint and are connected by binary links.[28]
  • N = 8,j = 10 : eight-bar linkage with 16 different topologies;
  • N = 10,j = 13 : ten-bar linkage with 230 different topologies;
  • N = 12,j = 16 : twelve-bar linkage with 6,856 topologies.

For larger chains and their linkage topologies, see R. P. Sunkari andL. C. Schmidt, "Structural synthesis of planar kinematic chains by adapting a Mckay-type algorithm",Mechanism and Machine Theory #41, pp. 1021–1030 (2006).

See also

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References

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  1. ^Edmund Taylor Whittaker (1904).A Treatise on the Analytical Dynamics of Particles and Rigid Bodies. Cambridge University Press. Chapter 1.ISBN 0-521-35883-3.
  2. ^Joseph Stiles Beggs (1983).Kinematics. Taylor & Francis. p. 1.ISBN 0-89116-355-7.
  3. ^Thomas Wallace Wright (1896).Elements of Mechanics Including Kinematics, Kinetics and Statics. E and FN Spon. Chapter 1.
  4. ^Russell C. Hibbeler (2009)."Kinematics and kinetics of a particle".Engineering Mechanics: Dynamics (12th ed.). Prentice Hall. p. 298.ISBN 978-0-13-607791-6.
  5. ^Ahmed A. Shabana (2003)."Reference kinematics".Dynamics of Multibody Systems (2nd ed.). Cambridge University Press.ISBN 978-0-521-54411-5.
  6. ^P. P. Teodorescu (2007)."Kinematics".Mechanical Systems, Classical Models: Particle Mechanics. Springer. p. 287.ISBN 978-1-4020-5441-9..
  7. ^A. Biewener (2003).Animal Locomotion. Oxford University Press.ISBN 019850022X.
  8. ^J. M. McCarthy and G. S. Soh, 2010,Geometric Design of Linkages, Springer, New York.
  9. ^Ampère, André-Marie (1834).Essai sur la Philosophie des Sciences. Chez Bachelier.
  10. ^Merz, John (1903).A History of European Thought in the Nineteenth Century. Blackwood, London. pp. 5.
  11. ^O. Bottema & B. Roth (1990).Theoretical Kinematics. Dover Publications. preface, p. 5.ISBN 0-486-66346-9.
  12. ^Harper, Douglas."cinema".Online Etymology Dictionary.
  13. ^Crash course physics
  14. ^2.4 Integration, MIT, 2 June 2017,archived from the original on 2021-11-13, retrieved2021-07-04
  15. ^https://www.youtube.com/watch?v=jLJLXka2wEM Crash course physics integrals
  16. ^https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html Area of Triangles Without Right Angles
  17. ^kinematics.gif (508×368) (Image). Retrieved3 November 2023.
  18. ^Reuleaux, F.; Kennedy, Alex B. W. (1876),The Kinematics of Machinery: Outlines of a Theory of Machines, London: Macmillan
  19. ^Geometry: the study of properties of given elements that remain invariant under specified transformations."Definition of geometry". Merriam-Webster on-line dictionary. 31 May 2023.
  20. ^Paul, Richard (1981).Robot manipulators: mathematics, programming, and control : the computer control of robot manipulators. MIT Press, Cambridge, MA.ISBN 978-0-262-16082-7.
  21. ^Gregory, R. Douglas (2006-04-13).Classical Mechanics (1 ed.). Cambridge University Press.doi:10.1017/cbo9780511803789.ISBN 978-0-521-82678-5.
  22. ^William Thomson Kelvin & Peter Guthrie Tait (1894).Elements of Natural Philosophy. Cambridge University Press. p. 4.ISBN 1-57392-984-0.
  23. ^William Thomson Kelvin & Peter Guthrie Tait (1894).Elements of Natural Philosophy. p. 296.
  24. ^M. Fogiel (1980)."Problem 17-11".The Mechanics Problem Solver. Research & Education Association. p. 613.ISBN 0-87891-519-2.
  25. ^Irving Porter Church (1908).Mechanics of Engineering. Wiley. p. 111.ISBN 1-110-36527-6.
  26. ^Morris Kline (1990).Mathematical Thought from Ancient to Modern Times. Oxford University Press. p. 472.ISBN 0-19-506136-5.
  27. ^Phillips, Jack (2007).Freedom in Machinery, Volumes 1–2 (reprint ed.). Cambridge University Press.ISBN 978-0-521-67331-0.
  28. ^Tsai, Lung-Wen (2001).Mechanism design:enumeration of kinematic structures according to function (illustrated ed.). CRC Press. p. 121.ISBN 978-0-8493-0901-4.
  1. ^Whileτ is used as the variable of integration, some authors may uset′ as the variable of integration, although that can be confused with Lagrange's notation for derivatives[14]

Further reading

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External links

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