Chapter 12 Kinematics of a Particle

CHAPTER-12

KINEMATICS OF A PARTICLE

INTRODUCTIONMECHANICS

(concerned with the state of rest or motion of bodies subjected to the

action of forces )

STATICS(concerned with equilibrium of a body that is either at rest or moves with constant velocity)

DYNAMICS(Deals with the accelerated

motion of a body)

KINEMATICS(deals only with the geometric aspects of motion without regard to the forces which

produce that motion)

KINETICS(analysis of forces causing the

motion)

In this chapter we will study the geometric aspects of the motion of a particle

RECTILINEAR KINEMATICS: CONTINUOUS MOTION

• A particle has a mass but negligible size and shape• Thus our applications will be limited to those objects that have

dimensions that are of no consequence in the analysis of motion• In most problems, one is interested in bodies of finite size, such as

rockets, projectiles or vehicles. • Such objects may be considered as particles provided motion of the

body is characterized by motion of its mass center and any rotation of the body is neglected.

• A particle can move along either a straight or a curved path.• Rectilinear motion is the motion of a body along a straight line• The kinematics of this motion is characterized by specifying, at any

given instant, the particle’s position, velocity and acceleration.


POSITION:

• The straight line path of particle will be defined using a single coordinate axis, s

• Origin ‘O’ on the path is a fixed point, and from this fixed point the position vector r is used to specify the location of the particle P at any given instant.

• Since r is always along the s-axis and so its direction never changes, only its magnitude changes, so for analytical work it is often convenient to represent r by an algebraic scalar s, representing the position coordinate of the particle

• The magnitude of s (and r) is the distance from O to P, and the sense (or arrowhead direction of r) is defined by the algebraic sign of s.


DISPLACEMENT:

• The displacement of the particle is defined as the change in its position. e.g. if the particle moves from P to P’ the displacement is ∆r = r’ – r. Using algebraic scalars to represent ∆r, we have : ∆s = s’ – s

• Here ∆s is positive since the particle’s final position is to the right of its initial position i.e. s’>s, likewise, if the final position is to the left of its initial position, ∆s is negative

• Displacement of a particle must be distinguished from the distance traveled by the particle, which is always a positive scalar quantity that represents the total length of path over which a particle travels.


VELOCITY:• If the particle moves through a displacement ∆r from P to P’ during time interval ∆t, the average velocity of the particle is:

vavg = ∆r/∆t

• If we take smaller and smaller values of ∆t, the magnitude of ∆r becomes smaller and smaller. Consequently the instantaneous velocity is defined by :

v = lim∆t-0 (∆r/∆t) or v = dr/dt

• Representing v as an algebraic scalar, we can also write:

v = ds/dt ------------------------------(1)

• Since ∆t or dt is always positive, the sign used to define the sense of velocity is the same as that of ∆s or ds. For example, if the particle is moving to the right, the velocity is positive, whereas if it is moving to the left, the velocity is negative.


Difference in Average Speed & Average Velocity:

• The magnitude of velocity is known as speed.• The Average Speed is always a positive scalar and is defined as the total

distance traveled by a particle, sT, divided by the elapsed time Δt, i.e.(vsp)avg = sT / Δt

• Whereas the average velocity for the above mentioned particle as it has traveled from P to P’ will be:

vavg = - Δs / Δt

Os

sT

ΔsPP’


ACCELERATION:• The average acceleration of a particle during the time interval ∆t is defined by:

aavg = ∆v/∆t

• Here ∆v represents the difference in the velocity during the time interval ∆t i.e.∆v = v’ – v

• If we take smaller and smaller values of ∆t, the magnitude of ∆v becomes smaller and smaller. Consequently the instantaneous acceleration is defined by :

a = lim∆t-0 (∆v/∆t) or a = dv/dt or a = d2s / dt2 ------------------(2)

• Both the average and instantaneous acceleration can be either positive or negative.


VELOCITY: v = ds/dt ------------------------------(1)ACCELERATION: a = dv/dt ------------------------------(2)

• A differential relation involving the displacement, velocity and acceleration along the path may be obtained by eliminating the time differential dt between equation (1) and (2).

a ds = v dv ------------------------------(3)


CONSTANT ACCELERATION: When the acceleration is constant, each of the three kinematic equations v = ds/dt, a = dv/dt, a ds = v dv, may be integrated to obtain formulas that

relate ac, v, s and t.

Velocity as a function of Time:v = vo + act --------------------------------(4)

Position as a function of Time:s = so + vot + (1/2)act2 ---------------------(5)

Velocity as a function of Position:v2 = vo

2 + 2ac(s – so) -----------------------(6)

• It is important to remember that equations (4), (5) and (6) are useful only when the acceleration is constant and when t=0, s=so, v = vo


Examples:

12.1, 12.2, 12.3, 12.4 and 12.5

Fundamental Problems:

F12-1, F12-6

Practice Problems:

12.7, 12.17, 12.21, 12.29, 12.31

EXAMPLE 12-1

During a test, the car shown below moves in a straight line such that for a short time its velocity is defined by :

v = (3t2+2t) ft/s, where t is in seconds. Determine its position and acceleration when t = 3s. When t=0, s=0.

EXAMPLE 12-3

During a test, a rocket is traveling upward at 75m/s, and when it is 40m from the ground its engine fails. Determine the maximum height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81m/s2 due to gravity. Neglect the effect of air resistance.

FUNDAMENTAL PROBLEM F12-6

PROBLEM 12-7

PROBLEM 12-29

RECTILINEAR KINEMATICS: ERRATIC MOTION

• When a particle’s motion during a time period is erratic (inconsistent, irregular) it may be difficult to obtain a continuous mathematical function to describe its position, velocity or acceleration.

• Instead, the motion may best be described graphically using a series of curves that can be generated experimentally from computer output.

• If the resulting graph describes the relationship between any two of the variables, a, v, s, t, a graph describing the relationship between the other variables can be established by using the kinematic equations 1, 2 and 3.

• The concept is explained with the help of the following examples:

EXAMPLE 12-6(Given the s-t graph, construct the v-t and a-t graphs)

A bicycle moves along a straight road such that its position is described by the graph shown in figure. Construct the v-t and a-t graphs for

st 300

EXAMPLE 12-6 contd. (Given the s-t graph, construct the v-t and a-t graphs)

EXAMPLE 12-7(Given the a-t graph, construct the v-t and s-t graphs)

The rocket sled shown in figure starts from rest and travels along a straight line such that it accelerates at a constant rate for 10s and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’ needed to stop the sled. How far has the sled traveled?

EXAMPLE 12-7 contd.(Given the a-t graph, construct the v-t and s-t graphs)

EXAMPLE 12-8(Given the a-s graph, construct the v-s graph)or (Given the v-s graph, construct the a-s graph)

The v-s graph describing the motion of a motorcycle is shown. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 400 ft.

EXAMPLE 12-8 contd.(Given the a-s graph, construct the v-s graph)or (Given the v-s graph, construct the a-s graph)

RECTILINEAR KINEMATICS: ERRATIC MOTION

Examples:

12.6, 12.7, 12.8


F12-9, F12-14

Practice Problems:

12.47, 12.53, 12.57, 12.65,

12.67

PROBLEM 12-57

GENERAL CURVILINEAR MOTION• Curvilinear motion occurs when the particle moves along a curved path• Since this path is often described in three dimensions vector analysis

will be used to formulate the particle’s position, velocity and acceleration

POSITION:• Consider a particle located at point P on a

space curve defined by the path function s

• The position of the particle, measured form a fixed point O, will be designated by the position vector r = r(t)

• This vector is a function of time since, in general, both its magnitude and direction changes as the particle moves along the curve.

GENERAL CURVILINEAR MOTION

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 and is determined by vector subtraction:

∆r = r’ – r


VELOCITY:• During the time interval ∆t, the average velocity of the particle is defined

by:

vavg = ∆r / ∆t

• 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,

v = lim∆t-0 (∆r/∆t) or v = dr/dt

• Since dr will be tangent to the curve at P, the direction of v is also tangent to the curve


ACCELERATION:• If the particle has a velocity v at time t and a velocity v’ = v + ∆v at t + ∆t, then

the average acceleration of the particle during the time interval ∆t is

aavg = ∆v / ∆t where ∆v = v’ - v

• The two velocity vectors v’ and v are plotted in figure such that their tails are located at the fixed point O’ and their arrowheads touch points on the curve. This curve is called a Hodograph, and when constructed, it describes the locus of points for the arrowhead of the velocity vector in the same manner as the path s describes the locus of points for the arrowhead of the position vector.

• To obtain the instantaneous acceleration, let ∆t – 0, so

a = lim∆t-0 (∆v/∆t) or a = dv/dt or a = d2r / dt2

• By definition of the derivative, a acts tangent to the hodograph, and therefore, in general, a is not tangent to the path of motion.

CURVILINEAR MOTION: RECTANGULAR COMPONENTS

222 zyxr

POSITION:• If at a given instant, the particle P is at a point (x,y,z) on the curved path

s, its location is then defined by the position vector

r = xi +yj + zk

• The magnitude of r is always positive and is defined as:

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

VELOCITY:• The first time derivative of r yields the velocity v of the particle,

Hence


)()()( zkdt

dyj

dt

dxi

dt

d

dt

drv

000

,, zvyvxv zyx

kvjvivdt

drv zyx

• The velocity has a magnitude defined as the positive value of:

v = √(vx2+vy

2+vz2)

• The direction of v is specified by the components of the unit vector,

uv = v/v

• This direction is always tangent to the path.

ACCELERATION:• The first time derivative of v yields the acceleration a of the particle,

Hence


000000000

,, zvayvaxva zzyyxx

kajaiadt

dva zyx

• The acceleration has a magnitude defined as the positive value of:

a = √(ax2+ay

2+az2)

• The direction of a is specified by the components of the unit vector,

ua = a/a

EXAMPLE 12-9

At any instant the horizontal position of the weather balloon is defined by x=(8t) ft where t is given in seconds. If the equation of the path is y=x2/10, determine:

(a) the distance of the balloon from the station at A when t=2s, (b) the magnitude and direction of the velocity when t=2s, (c) the magnitude and direction of the acceleration when t=2s

MOTION OF A PROJECTILE

• The free flight motion of a projectile is often studied in terms of its rectangular components, since projectile’s acceleration always acts in the vertical direction

• Consider a projectile launched at point (x0,y0)

• The path is defined in the x-y plane such that the initial velocity is v0, having components (vx)0 and (vy)0

• When air resistance is neglected, the only force acting on the projectile is its weight, which causes the projectile to have a constant downward acceleration of 9.8m/s2 (32.3ft/s2)


Horizontal Motion:Since ax=0, application of the constant acceleration equation yields: v = v0 + act; vx = (vx)0

x = x0 + v0t +(1/2)act2; x = x0 + (vx)0 t v2 = v0

2 + 2ac(s-s0); vx = (vx)0

Vertical Motion:Since positive y-axis is directed upward, then ax= -g, application of the constant

acceleration equation yields:v = v0 + act; vy = (vy)0 - gt

y = y0 + v0t +(1/2)act2; y = y0 + (vy)0 t – (1/2)gt2

v2 = v02 + 2ac(y-y0); vy

2 = (vy)02 – 2g(y-y0)


Examples:

12.9, 12.10, 12.11, 12.12, 12.13


F12-17, F12-22

Practice Problems:

12.97, 12.98, 12.105, 12.107,

12.109

EXAMPLE 12-11

A sack slides off the ramp, as shown in figure, with a horizontal velocity of 12m/s. If the height of the ramp is 6m from the floor, determine the time needed for the sack to strike the floor and the range R where sacks begin to pile up.

EXAMPLE 12-13

The track for the racing event is designed so that the rider jumps off the slope at 30º, from a height of 1m. During a race it was observed that the rider remained in mid air for 1.5 s. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground, and the maximum height he attains. Neglect the size of the bike and the rider.

PROBLEM 12-105

The boy at A attempts to throw a ball over the roof of a barn with an initial speed of vA = 15 m/s. Determine the angle ӨA at which the ball must be thrown so that it reaches its maximum height at C. Also, find the distance d where he should stand to make the throw.

PROBLEM 12-109

Determine the horizontal velocity vA of a tennis ball at A so that it just clears the net at B. Also, find the distance s where the ball strikes the ground.

CURVILINEAR MOTION: Normal & Tangential Coordinates

When the path along which a particle is moving is known, it is often convenient to describe the motion using n and t coordinates which act normal and tangent to the path.

Planar Motion:• Consider the particle P shown, which is moving in a plane along a fixed curve,

such that at a given instant it is at the position s, measured from point O.

• The t axis is tangent to the curve at P and is positive in the direction of increasing s. This positive direction is designated by the unit vector ut.

• The normal axis n is perpendicular to the t-axis and is directed from P towards the center of curvature O’. The positive direction, which is always on the concave side of the curve, will be designated by the unit vector un.

• The plane which contains the n and t axis is referred to a s the osculating plane.


Velocity:• Since the particle is moving, s is a function of time.

• The particle’s velocity v has a 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), i.e. v = ds/dt. Hence:

v = v ut where v = 0

s


Acceleration:• The acceleration of the particle is the time rate of change of velocity. Thus

------------------(1)

• In order to determine the time derivative of unit vector ut note that as the particle moves along the arc ds in time dt, ut preserves its magnitude of unity, however, its direction changes and becomes ut’


• As shown in the figure, ut’ = ut + dut

• dut has a magnitude of dut = (1) dӨ, and its direction is defined by un

• So dut = dӨ un

• Therefore the time derivative becomes


SPECIAL CASES:

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

• If the particle moves along the curve with a constant speed, then at= 0 and a = an = v2/ρ. Therefore, the normal component of acceleration represents the time rate of change in the direction of the velocity. Since an always acts towards the center of the curvature, this component is sometimes referred to as the centripetal acceleration.

Three Dimensional Motion

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


22

2/32

/

])/(1[

dxyd

dxdy


Examples:

12.14, 12.15, 12.16


F12-28, F12-31

Practice Problems:

12.116, 12.118, 12.138,

12.140, 12.141

EXAMPLE 12-15

A race car C travels around the horizontal circular track that has a radius of 300ft. If the car increases its speed at a constant rate of 7ft/s, starting from rest, determine the time needed for it to reach an acceleration of 8 ft/s2. What is its speed at this instant?

EXAMPLE 12-16 The box travels along the industrial conveyor as shown. If it starts

from rest at A and increases its speed such that at = (0.2t) m/s2, where t is in seconds, determine the magnitude of its acceleration when it arrives at point B.

Problem 12-138

Problem 12-141 The truck travels along a circular road that has a radius of 50m at a

speed of 4m/s. For a short distance when t=0, its speed is then increased by v0=(0.4t) m/s2, where t is in seconds. Determine the speed and the magnitude of the truck’s acceleration when t=4s.

CURVILINEAR MOTION: Cylindrical Components

Path of motion of a particle can sometimes be expressed in terms of cylindrical components (r, Ө, z). If the motion is restricted to the plane, the polar coordinates r and Ө are used.

Polar Coordinates:• The location of the particle P can be specified by using both the radial

coordinate r, which extends outward from the fixed origin O to the particle, and a transverse coordinate θ, which is the counterclockwise angle between a fixed reference line and the r axis.

• The angle is generally measured in degrees or radians

• The positive directions of r and θ coordinates are defined by the unit vectors ur and uθ respectively.


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

r = r ur

Velocity:

• The instantaneous velocity v is obtained by taking the time derivative of position vector r. Thus we have


• To evaluate the time derivative of ur, it must be noticed that ur changes only its direction with respect to time, since by definition the magnitude of this vector is always one unit.

• During the time Δt, a change Δθ will cause ur to become ur’ where ur’ = ur + Δur

• The time change in ur is then Δur

• For small angles Δθ this vector has a magnitude Δur ≈ 1(Δθ) and acts in the uθ direction.

• Hence Δur = Δθ uθ • Thus,

• Therefore the velocity vector v can be written as:

where:

• These components are shown graphically in the above figure• Since vr and vθ are mutually perpendicular, the magnitude of velocity is defined

by:

• And the direction of v is tangent to the path at P.


Acceleration:

• Taking the time derivative of velocity vector, the particle’s instantaneous acceleration is obtained, which is:

• To evaluate the time derivative of uθ, it must be noticed that uθ changes only its direction with respect to time, since by definition the magnitude of this vector is always one unit.

• During the time Δt, a change Δθ will cause uθ to become uθ’ where uθ’ = uθ + Δ uθ. The time change in uθ is then Δ uθ

• For small angles Δθ this vector has a magnitude Δuθ ≈ 1(Δθ) and acts in the -ur direction. Hence Δuθ = -Δθ ur

• Thus,


• Therefore the acceleration vector a can be written as:

where:

• These components are shown graphically in the above figure• Since ar and aθ are mutually perpendicular, the magnitude of acceleration is

defined by:



Cylindrical Coordinates:• If the particle P moves along a space curve as shown, then its location may be

specified by the three cylindrical coordinates, r, θ and z.• The z coordinate is identical to that used for rectangular coordinates.


Examples:

12.17, 12.18, 12.19, 12.20


F12-34, F12-36

Practice Problems:

12.165, 12.167, 12.169, 12-186

12.187

Example 12-18 The rod OA is rotating in the horizontal plane such that θ=(t3) rad.

At the same time, the collar B is sliding outward along OA so that r=(100t2) mm. If in both cases t is in seconds, determine the velocity and acceleration of the collar when t=1s.

Example 12-18 contd

Example 12-20 Due to the rotation of the forked rod, the cylindrical peg A travels

around the slotted path, a portion of which is in the shape of a cardioid, r=0.5(1-Cosθ) ft, where θ is in radians. If the peg’s velocity is v=4 ft/s and its acceleration is a=30 ft/s2 at the instant θ=180º, determine the angular velocity and angular acceleration of the fork.

Problem 12-169

Problem 12-187

Absolute Dependent Motion Analysis of Two Particles

• In some types of 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 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.

• We can show this mathematically by first specifying the location of the blocks using position coordinates sA and sB

• Each of the coordinate axes is:i) referenced from a fixed point (O) or fixed datum lineii) measured along each incline plane in the direction of motion of

block A and block Biii) has a positive sense from C to A and D to B.

Absolute Dependent Motion Analysis of Two Particles

• If the total cord length is lT, the position coordinates are related by the equation:

sA + lCD + sB = lT

• Taking the time derivative of this expression and noting that lCD and lT reamin constant, we have:

(d sA/dt) + (d sB/dt) = 0 or vB = -vA

• The negative sign indicates that when block A has a velocity downward, i.e. in the direction of positive sA, it causes a corresponding upward velocity of block B; i.e. B moves in negative sB direction.

• Similarly, time differentiation of the velocities yields the relation between the accelerations, i.e.

aB = -aA

Absolute Dependent Motion Analysis of Two Particles• Consider a more complicated example

involving dependent motion of two blocks.

• In this case the position of A is specified by sA, and the position of the end of the cord from which block B is suspended is defined by sB

• Coordinate axes are chosen such that they are:i) referenced from fixed points or datumsii) measured in the direction of motion of each block, andiii) positive to the right (sA) and positive downward (sB)

• If l represents the total length of the cord, then: 2sB+h+sA = l

2vB = -vA

2aB = -aA

Absolute Dependent Motion Analysis of Two Particles• Same example can also be worked out

by defining the position of block B from the center of the bottom pulley.

• In this case:

2(h-sB) + h + sA = l

2vB = vA

2aB = aA

Example 12.21

Determine the speed of block A, if block B has an upward speed of 6ft/s.

Example 12.23

Determine the speed with which block B rises if the end of the cord at A is pulled down with a speed of 2 m/s.

Relative Motion Analysis of Two Particles Using Translating Axes

• Uptil now, absolute motion of a particle has been determined using a single fixed reference frame for measurement

• However, when the path of motion for a particle is complicated, sometimes it is feasible to analyze the motion in parts by using two or more frames of reference.

• At this instant, only translating frames of reference will be considered for analysis.

Relative Motion Analysis of Two Particles Using Translating AxesPosition:• Consider particles A and B which move 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, z reference frame.

• The origin of a second frame of reference x’, y’, z’is attached to and moves with particle A.

• The relative position of “B with respect to A” is designated by a relative position vector rB/A, such that:

rB = rA + rB/A

Velocity:• An equation that relates the velocities of the

particles can be determined by taking the time derivative of above equation:

vB = vA + vB/A

Acceleration:• The time derivative of velocity equation yields a

similar vector relationship between the absolute and relative accelerations of particles.

aB = aA + aB/A

Examples: 12.21, 12.22, 12.23, 12.24,

12.25, 12.26, 12.27Fundamental Problems:

F12-39, F12-42Practice Problems:

12.202, 12.207, 12.217, 12.223, 12.225

Relative Motion Analysis of Two Particles Using Translating Axes

Example 12-25 A train traveling at a constant speed of 60 mi/h, crosses over a road

as shown. If the automobile A is traveling at 45 mi/h along the road, determine the magnitude and direction of the relative velocity of the train with respect to the automobile.

Example 12-26 Plane A is flying along a straight line path, whereas plane B is flying

along a circular path having a radius of curvature of ρB=400 km. Determine the velocity and acceleration of B as measured by the pilot of A.

Example 12-27 At the instant shown cars A and B are traveling at speeds of 18 m/s

and 12 m/s respectively. Also at this instant, A has a decrease in speed of 2 m/s2 and B has an increase in speed of 3 m/s2. Determine the velocity and acceleration of B with respect to A.

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Chapter 12 Kinematics of a Particle