KINEMATICS OF A PARTICLE

1 Introduction:

In this chapter, we study the kinematics of a particle - recall that a particle has a mass but negligiblesize and shape. Therefore, we limit the discussion to those objects that have dimensions that are of noconsequence in the analysis of the motion. Such objects may be considered as particles, provided motionof the body is characterized by motion of its center of mass and any rotation of the body is neglected.

By Curvilinear motion we refer to the motion of an object along a curve. The first step in describing themotion of a particle is to identify its position at any time t. This can be done by specifying its rectangularcoordinates (x, y, z), or in the case of plane motion (x, y) coordinates, its cylindrical coordinates (r, θ, z),or in the case of plane motion polar coordinates (r, θ). The motion of the particle can also be described bymeasurement along the tangent t and normal n to the curve as shown in the figure below. The directionst and n lie in the local plane of the curve. In the next few paragraphs we will discuss the kinematics of aparticle motion in all these different frames.

2 Rectilinear Kinematics: Continuous Motion:

• Rectilinear kinematics refers to straight line motion. The kinematics of a particle is characterized byspecifying at any given time t, the particle’s position, velocity, and acceleration.

– Position. The position of the particle is represented by a position vector ~r starting from theorigin O of the axis of the motion. If s is the algebraic scalar that represents the positioncoordinate of the particle, then

~r = s u

where u is a unit vector pointing toward the positive direction of the axis of the motion. Themagnitude of ~r is (r =| s |). The direction of ~r is the same as u if s > 0 and opposite to u ifs < 0. For analytical work it is therefore convenient to represent ~r by the algebraic scalar s.

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– Displacement. The displacement of a particle is a vector ∆~r defined as the change in heparticle’s position vector ~r. Algebraically the displacement of a particle from P of coordinate sto point P ′ of coordinate s′ is represented by

∆s = s′ − s

– Velocity. The velocity of a particle is a vector

∗ The average velocity is the displacement divided by time(~Vav = ∆~r/∆t

)or algebraically

as

Vav =∆s

∆t

∗ The instantaneous velocity (or velocity) is

(~V = lim

∆t→0

~Vav = lim∆t→0

∆~r

∆t=d~r

dt

)or algebraically

as

V =ds

dt= s

∗ Instantaneous speed (or speed) is a positive scalar and refers to the magnitude of theinstantaneous velocity

v =| ~V |=∣∣∣∣dsdt∣∣∣∣

∗ The average speed is a scalar and refers to the total distance divided by the total time

– Acceleration. The acceleration of a particle is a vector

(~a = lim

∆t→0

∆~V

∆t=d~V

dt

)or algebraically

as

a =dV

dt= V =

d2s

dt2= s

∗ In rectilinear kinematics, the acceleration has the same direction as the velocity when theparticle is speeding and opposite when the particle is slowing down.

∗ A particle can have an acceleration and yet have zero velocity

∗ One additional thing is that

a =dV

dtor dt =

dV

a

V =ds

dtor dt =

ds

Vds

V=dV

aor a ds = V dV

∗ The last equality can be integrated to solve for the speed as an example if one can expressa as a function of s.

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2.1 Constant acceleration

• Let a = ac = constant. Assume that V = Vo and s = so at time t = 0. Then

– Velocity as a function of time∫ V

Vo

dV =

∫ t

o

ac dt or V = Vo + ac t

– Position as a function of time∫ s

so

ds =

∫ t

o

(Vo + ac t) dt or s = so + Vo t+1

2ac t

2

– Speed as a function of position∫ V

Vo

V dV =

∫ s

so

(ac) ds or v2 = v2

o + 2 ac (s− so)

2.2 Procedure for Solving Problems

The equations of rectilinear kinematics should be applied as follow

• Coordinate System

– Establish a position coordinate s along the linear path and specify its fixed origin and positivedirection

– Since the motion os along a straight line, the particle’s position, velocity, and acceleration can berepresented as algebraic scalars. For analytical work, the sense of s, V , and a is then determinedfrom their algebraic signs

• Kinematic Equations

– If a relationship is known between any two or the four variables a, V , s, and t, then a third

variable can be obtained by using one of the kinematic equations

(a =

dV

dt

),

(V =

ds

dt

), or

(a ds = V dV ), which relates all the variables.

– Whenever integration is performed, it is important that the position and velocity be known ata given instant in order to evaluate either the constant of integration if an indefinite integral isused, or the limit of integration if a definite integral is used.

– The equation developed in section 2-1 apply only when the acceleration is constant

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3 Rectilinear Kinematics: Erratic Motion

When a particle’s motion during a time period is erratic or there is a discontinuity in the motion, it maybe difficult to obtain a continuous function to describe its position, velocity, or acceleration. Instead, themotion may best be described graphically using a series of curves that can be generated experimentally bycomputer. There are several frequently occurring situations:

• Given s-t graph, construct V-t graph. The position of a particle can be plotted over time (s-tgraph), the particle’s velocity as a function of time (V-t graph)can be obtained by measuring theslope of the s-t graph.

ds

dt= V

slope of s-t graph = velocity

• Given v-t graph, construct a-t graph. When the particle’s V-t graph is known, the particle’sacceleration as a function of time (a-t graph)can be obtained by measuring the slope of the V-t graph.

dV

dt= a

slope of V-t graph = acceleration

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• Given a-t graph, construct V-t graph. When the particle’s a-t graph is given, the particle’svelocity as a function of time (V-t graph)may be constructed by:

∆V =

∫a dt

change in velocity = area under a-t graph

• Given V-t graph, construct s-t graph. When the particle’s V-t graph is given, the particle’sposition as a function of time (s-t graph)may be constructed by:

∆s =

∫V dt

displacement = area under V-t graph

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• Given a-s graph, construct V-s graph. When the particle’s a-s graph is given, the particle’svelocity as a function of position (V-s graph)may be constructed by:

∫ V2

V1

V dV =1

2(V2

2 − V12) =

∫ s1

so

a ds

1

2(V2

2 − V12) = area under a-s graph

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• Given V-s graph, construct a-s graph. When the particle’s V-s graph is given, the particle’sacceleration as a function of position (a-s graph)may be constructed by:

a = VdV

ds

acceleration=Velocity × slope of V-s graph

4 General Curvilinear Motion

Curvilinear motion occurs when the particle moves along a curved path. Since this path is often described inthree dimensions, vector analysis will be used to formulate the particle’s position, velocity, and acceleration.In this section the general aspects of curvilinear motion are discussed, and in the subsequent sections wewill consider three types of coordinate systems often used to analyze this motion: rectangular, Normal andTangential, and Cylindrial.

• Position. The position of the particle is measured from a fixed point and described by the positionvector ~r(t). This vector is a function of time since both its magnitude and direction change as theparticle moves along its path (described by the path function s(t)).

• displacement. The displacement ∆~r = ~r′ − ~r represents the change in the particle’s position.

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• Velocity.

– The average velocity of the particle is defined as ~Vav =∆~r

∆t

– The velocity of the particle is described by the vector ~V (t) = lim∆t→0

∆~r

∆t=d~r

dt= ~r. The velocity

is tangent to the curve.

– As ∆t ≈ 0, | d~r |≈| ds | and the speed v =

∣∣∣∣d~rdt∣∣∣∣ =

ds

dt= s: the speed can be obtained by

differentiating the path function s(t) with respect to time.

• Acceleration.

– The Average acceleration ~aav is defined as ~aav =∆~V

∆t

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– The acceleration of the particle is described by the vector ~a = lim∆t→0

∆~V

∆t=d~V

dt=d2~r

dt2

– ∆~V and consequently ~a must account for changes made in both the magnitude and directionof the velocity ~V . as the particle moves from P1 to P2. Just a magnitude change increases (or

decreases) the ”length” of ~V , and this in itself would allow ~a to remain tangent to the path.However, in order for the particle to follow the path, the directional change always ”swings” thevelocity vector toward the ”inside” or ”concave side” of the path, and therefore ~a cannot remaintangent to the path. In fact ~a points toward the concave side of the path.

4.1 Curvilinear Motion: Rectangular Components

When the position of a particle P is defined at any instant by its rectangular coordinates x, y, z, it isconvenient to resolve the velocity ~V and acceleration ~a of the particle into rectangular components

• Position.

~r(t) = x(t)i+ y(t)j + z(t)k

Magnitude = r =√x2 + y2 + z2

Direction given by =~r

r

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• Velocity.

~V (t) =d~r

dt=dx

dti+

dy

dtj +

dz

dtk = Vxi+ Vy j + Vz k

Magnitude = v =√Vx

2 + Vy2 + Vz

2

Direction: =~V

valways tangent to path

• Acceleration.

~a(t) =d~V

dt=dVxdt

i+dVydt

j +dVzdt

k = axi+ ay j + az k = xi+ yj + zk

Magnitude = a =√ax2 + ay2 + az2

Direction: =~a

atoward the concave part of the path

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4.1.1 Application: Motion of a Projectile

The free-flight motion of a projectile is often studied in terms of its rectangular components, since theprojectile’s acceleration always acts in the vertical direction. Free flight projectile motion can be solvedusing the following procedure:

• Coordinate system

– Establish the fixed x, y coordinate axes and sketch the trajectory of the particle. Betweenany two points on the path specify the problem data and the three unknowns. In all casesthe acceleration of gravity acts downward. The particle’s initial and final velocities should berepresented in terms of their x and y components

– Remember that positive and negative position, velocity and acceleration components always actin accordance with their associated coordinate directions.

• Kinematic Equations

– Depending upon the known data and what is to be determined, a choice should be made as towhich three of the following five equations should be applied between the two points on the pathto obtain the most direct solution to the problem.

∗ Horizontal Motion

Vx = (Vo)x

x = xo + (Vo)x t

∗ Vertical Motion

Vy = (Vo)y − g t

y = yo + (Vo)y t−1

2g t2

V 2y = (Vo)

2y − 2g (y − yo)

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4.2 Curvilinear Motion: Normal and Tangential Components

When the path along which the particle is moving is known, then it is often convenient to describe themotion using n and t coordinates which act normal and tangent to the path, respectively, and at the instantconsidered have their origin located at the particle.

• Coordinate system

– The positive tangent axis always acts in the direction of motion and the positive normal axis isdirected toward the path’s center of curvature. Geometrically the curve is constructed from aseries of differential arc segments ds. Each segment ds is formed from the arc of an associatedcircle having a radius of curvature ρ and center of curvature O′.

• Velocity and acceleration

– The n and t axes are particulary advantageous for studying the velocity and acceleration of theparticle, because the velocity ~V and the acceleration ~a are expressed by the equations:

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~V = v u = s u

~a = at ut + an un

Consider a small displacement of the particle in a time dt. Assume ds to be small so that theradius of curvature ρ stays the same.

We have the following relations

~a =d~V

dt=d(vu)

dt=dv

dtu+ v

du

dt

dut =| ut | dθ un or ˙ut = θ un

replacing ˙ut with its present expression in the acceleration found earlier.

~a = vu+ v θ un

on the other hand we have

ds = ρ dθ or s = ρ θ and θ =s

ρ

so

~a = vu+s2

ρun

and

at = v = s

an =s2

ρ

∗ The velocity is always tangent to the path, its magnitude (speed) is the time derivative ofthe path function v = s

∗ The tangential component of the acceleration is he result of the time rate of change of thespeed. This component acts in the positive s-direction if the particle’s speed is increasingor in the opposite direction if the speed is decreasing.

∗ The relation between at, v and s are the same as for rectilinear motion

at = v =dv

dt=dv

ds

ds

dt

at ds = v dv

· If at is constant at = (at)c, the above equations, when integrated yield

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s = so + vo t+1

2(at)c t

2

V = Vo + (at)c t

v2 = v2o + 2 (at)c (s− so)

∗ The normal component of the acceleration is the result of change in the direction of theparticle velocity. This component is always directed toward the center of curvature of thepath

· The magnitude of this component is determined from an =v2

ρ

• As a result of these interpretations, a particle moving along a curved path will have accelerationsdirected as shown in the figure below

• A special case of the plane curvilinear motion is the circular motion where the radius of curvaturebecomes constant and equal to the radius of the circle ρ = r

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

ρ =

[1 + (dy/dx)2

]3/2∣∣∣∣d2y

dx2

∣∣∣∣

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4.3 Curvilinear Motion: Cylindrical Components

Some engineering problems involve angular position and a radial distance. for these cases it is oftenconvenient to express the path of motion in terms of cylindrical coordinates. r, θ, z. If the motion isrestricted to a plane, The polar coordinates r, θ are used.

• Coordinate system

– Polar coordinates are particularly suitable for solving problems for which data regarding theangular motion of the radial coordinate r is given to describe the path’s motion.

– To use coordinates, the origin is established at a fixed point, and the radial line r is directed tothe particle

– The transverse coordinate θ is measured counterclockwise from a fixed reference line to theradial line

• Position, Velocity and Acceleration

– The position of the particle is defined by the position vector ~r = r ur

– The velocity of the particle is given by

~V = ~r = ˙r ur = r ur + r ˙ur = r ur + rdθ

dtuθ

˙ur = lim∆t→0

∆ur∆t

= lim∆t→0

∆θ

∆tuθ

– The acceleration of the particle is given by

~a = ~V =d

dt

(r ur + r θuθ

)= r ur + r ˙ur + rθ uθ + rθ uθ + rθ ˙uθ

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˙uθ = lim∆t→0

∆uθ∆t

= − lim∆t→0

∆θ

∆tur

~a = (r − rθ2) ur + (rθ + 2rθ) uθ

• If the particle travels in a circular path, s = constant so that r = r = 0 the previous equationssimplify considerably

• If the particle moves along a space curve as shown in teh figure below, then its location may bespecified by the three cylindrical coordinates r, θ, z. The z coordinate is identical to that usedfor rectangular coordinates. Since the unit vector defining its direction, ur is constant, the timederivatives of this vector are zero and

~rp = r ur + z uz

~V = r ur + rθ uθ + z uz

~a = (r − rθ2) ur + (rθ + 2rθ) uθ + z uz

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5 Absolute dependent motion. Analysis of two particles

In some problems, the motion of one particle will depend on the corresponding motion of another particle.This dependency commonly occurs if the particles are interconnected by inextensible cords which arewrapped around pulleys. When each particle moves along a rectilinear path, the magnitude of the velocityand acceleration of the particles change, not their line of direction.

The following procedure can be used:

• Position-Coordinate equations

– Establish position coordinates which have their origin located at fixed or datums

– The coordinates are directed along the path of motion and extend to a point having the samemotion as each of the particles

– it is not necessary that the origin be the same for each of the coordinates; however, it is importantthat 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 cord, lT orto that portion of the cord, l, which excludes the segments that do not change length as theparticles move, such as arc segments wrapped over pulleys.

– If a problem involves a system of two or more cords wrapped around pulleys, then the positionof a point one one cord must be related to the position of a point on another cord. Separateequations are written for a fixed length of each cord of the system and teh position of the twoparticles are then related by these equations

Example:

In the figure above the position coordinate equations are:

sA + lCD + sB = lT

• Time derivatives

– Two successive time derivatives of the position-coordinate equations yield the required velocity,and acceleration equations which relate the motions of the particles

– The signs of the terms in these equations will be consistent with those that specify the positiveand negative sense of the position coordinates

Example:

In the figure above the position coordinate equations are:

dsAdt

+dsBdt

= 0 or VA = −VB

d2sAdt2

+d2sBdt2

= 0 or aA = −aB

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6 Relative motion. Analysis of two particles using translatingaxes

Throughout this chapter the absolute motion of a particle has been determined using a single fixed ref-erence frame for measurement. There are many cases, however, where the path of motion for a particleis complicated, so that it may be feasible to analyze the motion in parts by using two or more frames ofreference. In this section translating frames of reference will be considered for the analysis.

• When applying the relative-position equation ~rB = ~rA + ~rB/A, it is first necessary to specify thelocation of the fixed frame x, y, z and the translating frame x′, y′, z′

• Usually the origin A of the translating axes is located at a point having known position, ~rA.

• Since vector addition form a triangle, there can be at most two unknowns.

• these unknowns can be solved for either graphically, using trigonometry, or by resolving each of thethree vectors ~rA, ~rB and ~rb/A into rectangular or cartesian components, thereby generating a set ofscalar equations.

• The relative-motion equations ~VB = ~VA + ~VB/A and ~aB = ~aA +~aB/A are applied in teh same manneras mentioned above except that in this case teh origin of the fixed x, y, z axes does not have to bespecified.

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