Kinematics of Particle

Chapter 1 : Introduction to DynamicsMechanics

Particle/Rigid-body Deformable-body Fluid

StaticEquilibrium body

DynamicsAccelerated motion particle/body

Kinematics(Geometric aspect of motion)

Kinetics(Analysis of force causing the motion)

INTRODUCTION• Mechanics – the state of rest of motion of

bodies subjected to the action of forces• Static – equilibrium of a body that is either at

rest or moves with constant velocity• Dynamics – deals with accelerated motion of

a body1) Kinematics – treats with geometric aspects of the motion2) Kinetics – analysis of the forces causing the motion

Chapter 2 : Kinematics Of

Particle in Plane Motion

Chapter Objectives

• To introduce the concepts of position, displacement,velocity, and acceleration.

• To study particle motion along a straight line and represent this motion graphically.

• To investigate particle motion along a curved path using different coordinate systems.

• To present an analysis of dependent motion of two particles.

• To examine the principles of relative motion of two particles using translating axes.

Chapter Outline• Introduction

• Rectilinear Kinematics: Continuous Motion

• Rectilinear Kinematics: Erratic Motion

• Curvilinear Motion: Rectangular Components

• Motion of a Projectile

• Curvilinear Motion: Normal and Tangential Components

•Curvilinear Motion: Cylindrical Components

•Absolute Dependent Motion Analysis of Two Particles

•Relative Motion Analysis of Two Particles Using Translating Axes

KINEMATICS OF PARTICLESKINEMATICS OF PARTICLES

Kinematics of particles

Rectilinear motion Rectilinear motion

Curvilinear motionCurvilinear motion

x-y coord.x-y coord. n-t coord.n-t coord. r-θ coord.r-θ coord.

Dependent motion

Dependent motion

Relative motionRelative motion

Projectile MotionProjectile Motion

Rectilinear Kinematics: Continuous Motion

• Rectilinear Kinematics – specifying at any instant, the particle’s position, velocity, and acceleration

• Position1) Single coordinate axis, s2) Origin, O3) Position vector r – specific location of particle P at any instant

4) Algebraic Scalar s in metres

Note : - Magnitude of s = Dist from O to P- The sense (arrowhead dir of r) is defined

by algebraic sign on s=> +ve = right of origin, -ve = left of origin


• Displacement – change in its position, vector quantity



• If particle moves from P to P’=>

is +ve if particle’s position is right of its initial positionis -ve if particle’s position is left of its initial position

sss −′=∆

rrr −′=∆

s∆

s∆


• VelocityAverage velocity,

Instantaneous velocity is defined as

trvavg ∆

∆=

( )trvtins ∆∆=→∆

/lim0

dtdrvins =⇒


Representing as an algebraic scalar,

Velocity is +ve = particle moving to the rightVelocity is –ve = Particle moving to the leftMagnitude of velocity is the speed (m/s)

insv

dtdsv = ⎟

⎠⎞

⎜⎝⎛→

+

Average speed is defined as total distance traveled by a particle, sT, divided by the elapsed time .

The particle travels alongthe path of length sT in time

=>

t∆( )

tsv T

avgsp ∆=

t∆( )

tavg ∆sv

tsv T

avgsp

∆−=

∆=



• Acceleration – velocity of particle is known at points P and P’ during time interval ∆t, average acceleration is

• ∆v represents difference in the velocity during the time interval ∆t, ie

tvaavg ∆

∆=

vvv −=∆ '

tvaavg ∆

∆=


Instantaneous acceleration at time t is found by taking smaller and smaller values of ∆t and corresponding smaller and smaller values of ∆v, ( )tva

t∆∆=

→∆/lim

0

⎟⎠⎞

⎜⎝⎛→

+

⎟⎠⎞

⎜⎝⎛→

+

2

2

dtsda

dtdva

=

=


• Particle is slowing down, its speed is decreasing => decelerating => will be negative.

• Consequently, a will also be negative, therefore it will act to the left, in the opposite sense to v

• If velocity is constant, acceleration is zero

vvv −=∆ '


• Velocity as a Function of TimeIntegrate ac = dv/dt, assuming that initially v = v0 when t = 0.

∫∫ =t

c

v

vdtadv

00

⎟⎠⎞

⎜⎝⎛→

+

tavv c+= 0

Constant Acceleration


• Position as a Function of TimeIntegrate v = ds/dt = v0 + act, assuming that initially s = s0 when t = 0

( )

200

0 0

21

0

tatvss

dttavds

c

tc

s

s

++=

+= ∫∫⎟⎠⎞

⎜⎝⎛→

+



• Velocity as a Function of PositionIntegrate v dv = ac ds, assuming that initially v = v0 at s = s0

( )020

2 200

ssavv

dsavdv

c

s

s c

v

v

−+=

= ∫∫⎟⎠⎞

⎜⎝⎛→

+



PROCEDURE FOR ANALYSIS1) Coordinate System• Establish a position coordinate s along the

path and specify its fixed origin and positive direction.

• The particle’s position, velocity, and acceleration, can be represented as s, v and a respectively and their sense is then determined from their algebraic signs.

• The positive sense for each scalar can be indicated by an arrow shown alongside each kinematics eqn as it is applied


2) Kinematic Equation• If a relationship is known between any two of the four

variables a, v, s and t, then a third variable can be obtained by using one of the three the kinematicequations

• When integration is performed, it is important that position and velocity be known at a given instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits of integration if a definite integral is used

• Remember that the three kinematics equations can only be applied to situation where the acceleration of the particle is constant.

Curvilinear motion

General Curvilinear Motion

Curvilinear motion occurs when the particle moves along a curved path

Position. The position of the particle, measured from a fixed point O, is designated by the position vector r = r(t).


Displacement. Suppose during a small time interval ∆t the particle moves a distance ∆s along the curve to a new position P`, defined by r` = r + ∆r. The displacement ∆r represents the change in the particle’s position.


Velocity. During the time ∆t, the average velocityof the particle is defined as

trvavg ∆

∆=

The instantaneous velocity is determined from this equation by letting ∆t 0, and consequently the direction of ∆r approaches the tangent to the curve at point P. Hence,

dtdrvins =


• Direction of vins is tangent to the curve

• Magnitude of vins is the speed, which may be obtained by noting the magnitude of the displacement ∆r is the length of the straight line segment from P to P`.

dtdsv =


Acceleration. If the particle has a velocity v at time t and a velocity v` = v + ∆v at time t` = t + ∆t. The average acceleration during the time interval∆t is

tvaavg ∆

∆=

2

2

dtrd

dtdva ==


a acts tangent to the hodograph, therefore it is not tangent to the path

Curvilinear Motion: Rectangular Components

Position. Position vector is defined by

r = xi + yj + zk

The magnitude of r is always positive and defined as

222 zyxr ++=

The direction of r is specified by the components of the unit vector ur = r/r


Velocity.

zvyvxv

kvjvivdtdrv

zyx

zyx

&&&

rrr

===

++==

where

The velocity has a magnitude defined as the positive value of

222zyx vvvv ++=

and a direction that is specified by the components of the unit vector uv=v/v and is always tangent to the path.


Acceleration.

zva

yvaxva

kajaiadtdva

zz

yy

xx

zyx

&&&

&&&

&&&

rrr

==

====

++==

where

The acceleration has a magnitude defined as the positive value of

222zyx aaaa ++=


• The acceleration has a direction specified by the components of the unit vector ua = a/a.

• Since a represents the time rate of change in velocity, a will not be tangent to the path.


PROCEDURE FOR ANALYSISCoordinate System

• A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y and zcomponents.


Kinematic Quantities

• Since the rectilinear motion occurs along each coordinate axis, the motion of each component is found using v = ds/dt and a = dv/dt, or a ds = v ds

• Once the x, y, z components of v and a have been determined. The magnitudes of these vectors are found from the Pythagorean theorem and their directions from the components of their unit vectors.

Motion of a Projectile• Free-flight motion studied in terms of rectangular components since projectile’s acceleration always act vertically• Consider projectile launched at (x0, y0)• Path defined in the x-y plane• Air resistance neglected• Only force acting on the projectile is its weight, resulting in constant downwards acceleration• ac = g = 9.81 m/s2

Motion of a Projectile


Horizontal Motion Since ax = 0,

);(2

;21

;

020

2

200

0

ssavv

tatvxx

tavv

c

c

c

−+=

++=

+=⎟⎠⎞

⎜⎝⎛→

+

xx

x

xx

vvtvxx

vv

)()(

)(

0

00

0

=+=

=⎟⎠⎞

⎜⎝⎛→

+

⎟⎠⎞

⎜⎝⎛→

+

Horizontal component of velocity remain constant during the motion


Vertical. Positive y axis is directed upward, then ay = - g

)(2)(21)(

)(

02

0

200

0

yygvv

gttvyy

gtvv

yx

y

yy

−−=

−+=

−=( )↑+

);(2

;21

;

020

2

200

0

yyavv

tatvyy

tavv

c

c

c

−+=

++=

+=

( )↑+

( )↑+


• Problems involving the motion of a projectile have at most three unknowns since only three independent equations can be written:- one in the horizontal direction

- two in the vertical direction• Velocity in the horizontal and vertical direction are used to obtain the resultant velocity• Resultant velocity is always tangent to the path


PROCEDURE FOR ANALYSISCoordinate System• Establish the fixed x, y, z axes and sketch the trajectory of the particle• Specify the three unknowns and data between any two points on the path• Acceleration of gravity always acts downwards• Express the particle initial and final velocities in the x, y components


• Positive and negative position, velocity and acceleration components always act in accordance with their associated coordinate directionsKinematics Equations• Decide on the equations to be applied between the two points on the path for the most direct solution


Horizontal Motion• Velocity in the horizontal or x directions is constant (vx) = (vo)x

x = xo + (vo)x t

Vertical Motion• Only two of the following three equations should be used


)(2)(21)(

)(

02

0

200

0

yygvv

gttvyy

gtvv

yx

y

yy

−−=

−+=

−=

• Eg: if final velocity is not needed, first and third of the equations would not be needed

Curvilinear Motion: Normal and Tangential Components

• When the path of motion of a particle is known, describe the path using n and t coordinates which act normal and tangent to the path• Consider origin located at the particle

Planar Motion• Consider particle P which is moving in a plane along a fixed curve, such that at a given instant it is at position s, measured from point O


• Consider a coordinate system that has origin at a fixed point on the curve on the curve, and at the instant, considered this origin happen to coincide with the location of the particle

• t axis is tangent to the curve at P and is positive in the direction of increasing s


• Designate this positive position direction with unit vector ut• For normal axis, note that geometrically, the curve is constructed from series differential arc segments• Each segment ds is formed from the arc of an associated circle having a radius of curvature ρ(rho) and center of curvature O’


• Normal axis n is perpendicular to the t axis and is directed from P towards the center of curvature O’• Positive direction is always on the concave side of the curve, designed by un• Plane containing both the n and t axes is known as the oscillating plane and is fixed on the plane of motion


Velocity.• Since the particle is moving, s is a function of time• Particle’s velocity v has direction that is always tangent to the path and a magnitude that is determined by taking the time derivative of the path function s = s(t)

svuvv t

&

rr

==

where


Acceleration• Acceleration of the particle is the time rate of change of velocity

tt uvuvvar&

r&

r&

r +==


• As the particle moves along the arc ds in time dt, ut preserves its magnitude of unity• When particle changes direction, it becomes ut’

ut’ = ut + dut• dut stretches between the arrowhead of ut and ut’, which lie on an infinitesimal arc of radius ut = 1

nnnt uvusuu rr&r&r&

ρρθ ===


where

ρ

2va

vdvdsavauauaa

n

tt

nntt

=

==+=

&

rrr

and

• Magnitude of acceleration is the positive value of

22nt aaa +=


Consider two special cases of motion• If the particle moves along a straight line, then ρ → ∞ and an = 0. Thus , we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of velocity.

• If the particle moves along the curve with a constant speed, then and

vaa t &==

0== vat & ρ/2vaa n ==


• Normal component of acceleration represents the time rate of change in the direction of the velocity.Since an always acts towards the center of curvature, this component is sometimes referred to as the centripetal acceleration• As a result, a particle moving along the curved path will have accelerations directed as shown


Three Dimensional Motion• If the particle is moving along a space curve, at a given instant, t axis is completely unique• An infinite number of straight lines can be constructed normal to tangent axis at P


• For planar motion, - choose positive n axis directed from P towards path’s center of curvature O’- The above axis also referred as principle normal to curve at P-ut and un are always perpendicular to one another and lies in the osculating plane


• For spatial motion, a third unit vector ub, defines a binormal axis bwhich is perpendicular to ut and un

• Three unit vectors are related by vector cross product

ub = ut X un• un is always on the concave side


PROCEDURE FOR ANALYSISCoordinate System• When path of the particle is known, establish a set of n and t coordinates having a fixed originwhich is coincident with the particle at the instant• Positive tangent axis acts in the direction of the motion and the positive normal axis id directed toward the path’s center of curvature• n and t axes are advantageous for studying the velocity and acceleration of the particle


Velocity• Particle’s velocity is always tangent to the path• Magnitude of the velocity is found from the derivative of the path function

Tangential Acceleration• Tangential component of acceleration is the result of the time rate of change in the magnitude of velocity

sv &=


• Tangential component acts in the positive s direction if the particle’s speed is increasing and in the opposite direction if the seed is decreasing• For rectilinear motion,

vdvdsava tt == &

• If at is constant,

)()(2

)(

)(21

020

20

200

ssavv

tavv

tatvss

cc

cc

cc

−+=

+=

++=


Normal Acceleration• Normal component of acceleration is the result of the time rate of change in the direction of the particle’s velocity• Normal component is always directed towards the center of curvature of the path along the positive n axis• For magnitude of the normal component,

ρ

2van =


• If the path is expressed as y = f(x), the radius of the curvature ρ at any point on the path is determined from

22

2/32

/])/(1[

dxyddxdy+

=ρ

Curvilinear Motion: Cylindrical Components

• If motion is restricted to the plane, polar coordinates r and θ are used

Polar Coordinates• Specify the location of P using both the radial coordinate r, which extends outward from the fixed origin O to the particle and a transverse coordinateθ, which is the counterclockwise angle between a fixed reference line and the r axis


• Angle usually measured in degrees or radians, where 1 rad= 180°• Positive directions of the r and θ coordinates are defined by the unit vectors ur and uθ• ur extends from P along increasing r, when θ is held fixed


• uθ extends from P in the direction that occurs when r is held fixed and θ is increased• Note these directions are perpendicular to each other

Position• At any instant, position of the particle defined by the position vector

rurr rr =


Velocity• Instantaneous velocity v is obtained by the time derivative of r

rr ururrvr&

r&

r&

r +==

• To evaluate , note that ur changes only its direction with respect to time since magnitude of this vector = 1• During time ∆t, a change ∆r will not cause a change in the direction of u

rur&


• However, a change ∆θ will cause ur to become ur’where

ur’ = ur + ∆ur• Time change is ∆ur• For small angles ∆θ, vector has a magnitude of 1 and acts in the uθ direction

θ

θ

θ

θ

uu

utt

uu

r

tr

tr

r&r&

rrr

&

=

⎟⎠⎞

⎜⎝⎛

∆∆

=∆∆

=→∆→∆ 00

limlim

θθ

θθ

&

&

rrr

rvrv

uvuvv

r

rr

=

=+=

where


Velocity

• Radical component vr is a measure of the rate of increase or decrease in the length of the radial coordinate

• Transverse component vθ is the rate of motion along the circumference of a circle having a radius r


• Angular velocity indicates the rate of change of the angle θ• Since vr and vθ are mutually perpendicular, the magnitude of the velocity or speed is simply the positive value of

• Direction of v is tangent to the path at P

dtd /θθ =&

( ) ( )22 θ&& rrv +=


Acceleration• Taking the time derivatives, for the instant acceleration,

θθθ θθθ urururururva rr &&&&&&&&&&& ++++==

• During the time ∆t, a change ∆r will not change the direction uθ although a change in ∆θ will cause uθ to become uθ’• For small angles, this vector has a magnitude = 1 and acts in the –ur direction

∆uθ= - ∆θur

θθθ

θ

θθ

&&&&

&&&

rrarra

uauaa

r

rr

2

2

+=

−=

+=

where

• The term is called the angular acceleration since it measures the change made in the angular velocity during an instant of time

• Use unit rad/s2

22 / dtd θθ =&&


Acceleration


• Since ar and aθ are always perpendicular, the magnitude of the acceleration is simply the positive value of

( ) ( )222 2 θθθ &&&&&&& rrrra ++−=

• Direction is determined from the vector addition of its components • Acceleration is not tangent to the path


Cylindrical Coordinates• If the particle P moves along a space, then its location may be specified by the three cylindrical coordinates r, θ, z• z coordinate is similar to that used for rectangular coordinates


• Since the unit vector defining its direction, uz, is constant, the time derivatives of this vector are zero• Position, velocity, acceleration of the particle can be written in cylindrical coordinates as shown

zr

zr

zrp

uzurrurra

uzururv

uzurr

r&&

r&&&&r&&&r

r&

r&r&

r

rrr

+++−=

++=

+=

θ

θ

θθθ

θ

)2()( 2


Time DerivativesTwo types of problems usually occur1) If the coordinates are specified as time

parametric equations, r = r(t) and θ = θ(t), then the time derivative can be formed directly.

2) If the time parametric equations are not given, it is necessary to specify the path r = f(θ) and find the relationship between the time derivatives using the chain rule of calculus.


PROCEDURE FOR ANALYSISCoordinate System• Polar coordinate are used to solve problem involving angular motion of the radial coordinate r, used to describe the particle’s motion• To use polar coordinates, the origin is established at a fixed point and the radial line r is directed to the particle• The transverse coordinate θ is measured from a fixed reference line to radial line


Velocity and Acceleration• Once r and the four time derivatives have been evaluated at the instant considered, their values can be used to obtain the radial and transverse components of v and a• Use chain rule of calculus to find the time derivatives of r = f(θ)• Motion in 3D requires a simple extension of the above procedure to find

θθ &&&&&& ,,,rr

Absolute Dependent Motion Analysis of Two Particles

• Motion of one particle will depend on the corresponding motion of another particle• Dependency occur when particles are interconnected by the inextensible cords which are wrapped around pulleys• For example, the movement of block A downward along the inclined plane will cause a corresponding movement of block B up the other incline• Specify the locations of the blocks using position coordinate sA and sB


• Note each of the coordinate axes is (1) referenced from a fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B and (3) has a positive sense from C to A and D to B• If total cord length is lT, the position coordinate are elated by the equation

TBCDA lsls =++


• Here lCD is the length passing over arc CD• Taking time derivative of this expression, realizing that lCD and lT remain constant, while sA and sBmeasure the lengths of the changing segments of the cord

ABBA vv

dtds

dtds

−==+ 0 or

• The negative sign indicates that when block A has a velocity downward in the direction of position sA, it causes a corresponding upward velocity of block B; B moving in the negative sB direction


• Time differentiation of the velocities yields the relation between accelerations

aB = - aA

• For example involving dependent motion of two blocks• Position of block A is specified by sA, and the position of the end of the cord which block B is suspended is defined by sB


• Chose coordinate axes which are (1) referenced from fixed points and datums, (2) measured in the direction of motion of each block, (3) positive to the right (sA) and positive downward (sB)• During the motion, the red colored segments of the cord remain constant• If l represents the total length of the cord minus these segments, then the position coordinates can be related by

lshs AB =++ 22


Since l and h are constant during the motion, the two time derivatives yields

ABAB aavv −=−= 22

• When B moves downward (+sB), A moves to left (-sA) with two times the motion• This example can also be worked by defining the position of block B from the center of the bottom pulley ( a fixed point)


ABAB

AB

aavv

lshsh

==

=++−

22

)(2• Time differentiation yields


PROCEDURE FOR ANALYSISPosition-Coordinate Equation• Establish position coordinates which have their origin located at a fixed point or datum• The coordinates are directed along the path of motion and extend to a point having the same motion as each of the particles• It is not necessary that the origin be the same for each of the coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particle


• Using geometry or trigonometry, relate the coordinates to the total length of the cod, lT, or to that portion of cord, l, which excludes the segments that do not change length as the particle move –such as arc segments wrapped over pulleys• For problem involving a system of two or more cords wrapped over pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure• Separate equations must be written for a fixed length of each cord of the system.


Time Derivatives• Two successive time derivatives of the position-coordinates equations yield the required velocity and acceleration equations which relate motions of the particles• The signs of the terms in these equations will be consistent with those that specify the positive and negative sense of the position coordinates

Relative Motion Analysis of Two Particles Using Translating Axes

• There are many cases where the path of the motion for a particle is complicated, so that it may be feasible to analyze the motions in parts by using two or more frames of reference• For example, motion of an particle located at the tip of an airplane propeller while the plane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particle measured from a reference attached to the airplane


Position.• Consider particle A and B, which moves along the arbitrary paths aa and bb, respectively• The absolute position of each particle rA and rB, is measured from the common origin O of the fixed x, y, zreference frame


• Origin of the second frame of reference x’, y’ and z’ is attached to and moves with particle A• Axes of this frame only permitted to translate relative to fixed frame• Relative position of “B with respect to A” is designated by a relative-position vector rB/A• Using vector addition

ABAB rrr /rrr +=


Velocity.• By time derivatives,

• Here refer to absolute velocities, since they are observed from the fixed frame• Relative velocity is observed from the translating frame

ABAB vvv /rrr +=

dtrdvanddtrdv AABB // rrrr ==

dtrdv ABAB ///rr =


• Since the x’, y’ and z’ axes translate, the components of rB/A will not change direction and therefore time derivative o this vector components will only have to account for the change in the vector magnitude• Velocity of B is equal to the velocity of A plus (vectorially) the relative velocity of “B relative to A” as measured by the translating observerfixed in the x’, y’ and z’ reference


Acceleration.• The time derivative yields a similar relationship between the absolute and relative accelerations of the particles A and B

• Here aB/A is the acceleration of B as seen by the observer located at A and translating with the x’, y’and z’ reference frame

ABAB aaa /rrr +=


PROCEDURE FOR ANALYSIS

• When applying the relative position equations, rB= rA + rB/A, it is necessary to specify the location of the fixed x, y , z and translating x’, y’ and z’• Usually, the origin A of the translating axes is located at a point having a known position rA• A graphical representation of the vector addition can be shown, and both the known and unknown quantities labeled on this sketch


• Since vector addition forms a triangle, there can be at most two unknowns, represented by the magnitudes and/or directions of the vector quantities• These unknown can be solved for either graphically, using trigonometry, or resolving each of the three vectors rA, rBand rB/A into rectangular or Cartesian components, thereby generating a set of scalar equations• The relative motion equations vB = vA + vB/A and aB = aA + aB/A are applied in the same manner as explained above, except in this case, origin O of the fixed axes x, y, z axes does not have to be specified

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Kinematics of Particle