ch12

Chapter 12 : Kinematics Of A Particle

©2007 Pearson Education South Asia Pte Ltd

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

Chapter Outline


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


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

t

rvavg

trvt

ins

/lim0

dt

drvins



Representing as an algebraic scalar,

Velocity is +ve = particle moving to the right

Velocity is –ve = Particle moving to the left

Magnitude of velocity is the speed (m/s)

insv

dt

dsv



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

The particle travels along

the path of length sT in time

=>

t

t

sv T

avgsp

t

t

sv

t

sv

avg

Tavgsp



• Acceleration – velocity of particle is known at points P and P’ during time interval Δt, average acceleration is

• Δv represents difference in the velocity during the time interval Δt, ie

t

vaavg

vvv '

t

vaavg



Instantaneous acceleration at time t is found by taking smaller and smaller values of Δt and corresponding smaller and smaller values of Δv,

tvat

/lim0

2

2

dt

sda

dt

dva



• 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 Time

Integrate 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 Time

Integrate 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 Position

Integrate v dv = ac ds, assuming that initially v = v0 at s = s0

020

2 2

00

ssavv

dsavdv

c

s

s c

v

v




PROCEDURE FOR ANALYSIS

1) 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 kinematic equations

• 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.



The car moves in a straight line such that for a short time its velocity is defined by v = (0.9t2 + 0.6t) m/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.

EXAMPLE 12.1


Solution:

Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right.

Position. Since v = f(t), the car’s position can be determined from v = ds/dt, since this equation relates v, s and t. Noting that s = 0 when t = 0, we have

ttdt

dsv 6.09.0 2

EXAMPLE 12.1


23

0

23

0

0

2

0

3.03.0

3.03.0

6.09.0

tts

tts

dtttds

ts

ts

When t = 3s,

s = 10.8m

EXAMPLE 12.1


Acceleration. Knowing v = f(t), the acceleration is determined from a = dv/dt, since this equation relates a, v and t.

6.08.1

6.09.0 2

t

ttdt

d

dt

dva

When t = 3s,

a = 6m/s2

EXAMPLE 12.1


A small projectile is forced downward into a

fluid medium with an initial velocity of 60m/s.

Due to the resistance of the fluid the

projectile experiences a deceleration equal to a =

(-0.4v3)m/s2, where v is in m/s2.

Determine the projectile’s

velocity and position 4s

after it is fired.

EXAMPLE 12.2


Solution:

Coordinate System. Since the motion is

downward, the position coordinate is downwards

positive, with the origin located at O.

Velocity.Here a = f(v), velocity is a function of

time using a = dv/dt, since this equation relates v,

a and t.

34.0 vdt

dva

EXAMPLE 12.2


smtv

tv

dtv

dtv

dv

tv

tv

sm

/8.060

1

60

11

8.0

1

1

2

1

4.0

14.0

2/1

2

22

0602

0/60 3

When t = 4s,

v = 0.559 m/s

EXAMPLE 12.2


Position. Since v = f(t), the projectile’s position can be determined from v = ds/dt, since this equation relates v, s and t. Noting that s = 0 when t = 0, we have

2/1

2 8.060

1

t

dt

dsv

t

ts

ts

dttds

0

2/1

2

0

2/1

20

8.060

1

8.0

2

8.060

1

EXAMPLE 12.2


When t = 4s,

s = 4.43m

mts

60

18.0

60

1

4.0

12/1

2

EXAMPLE 12.2


A rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails. Determine max height sB reached by the rocket and its speed just before it hits the ground.

EXAMPLE 12.3


Solution:Coordinate System. Origin O for the position coordinate at ground level with positive upward.Maximum Height. Rocket traveling upward, vA = +75m/s when t = 0. s = sB when vB = 0 at max ht. For entire motion, acceleration aC = -9.81m/s2 (negative since it act opposite sense to positive velocity or positive displacement)

EXAMPLE 12.3


)(222ABCAB ssavv

sB = 327 m

Velocity.

smsmv

ssavv

C

BCCBC

/1/80/1.80

)(22

22

The negative root was chosen since the rocket is moving downward

EXAMPLE 12.3


A metallic particle travels downward through a fluid that extends from plate A and plate B under the influence of magnetic field. If particle is released from rest at midpoint C, s = 100 mm, and acceleration, a = (4s) m/s2, where s in meters, determine velocity when it reaches plate B and time need to travel from C to B

EXAMPLE 12.4


Solution:Coordinate System. It is shown that s is taken positive downward, measured from plate AVelocity. Since a = f(s), velocity as a function of position can be obtained by using v dv = a ds. Realising v = 0 at s = 100mm = 0.1m

EXAMPLE 12.4


21

2

1.0

2

0

2

1.00

01.02

2

4

2

1

4

sv

sv

dssdvv

dsadvv

Sv

sv

At s = 200mm = 0.2m,

smmsmvB /346/346.0

EXAMPLE 12.4


tss

tss

dts

ds

dts

dtvds

ts

ts

233.201.0ln

201.0ln

201.0

01.02

2

01.0

2

01.0 5.02

5.02

At s = 200mm = 0.2m, t = 0.658s

EXAMPLE 12.4


A particle moves along a horizontal path with a velocity of v = (3t2 – 6t) m/s. if it is initially located at the origin O, determine the distance traveled in 3.5s and the particle’s average velocity and speed during the time interval.

EXAMPLE 12.5


Solution:Coordinate System. Assuming positive motion to the right, measured from the origin, ODistance traveled. Since v = f(t), the position as a function of time may be found integrating v = ds/dt with t = 0, s = 0.

EXAMPLE 12.5

View Free Body Diagram


mtts

tdtdttds

dttt

vdtds

s tt

23

0 00

2

2

3

63

63

EXAMPLE 12.5


mtts

tdtdttds

dttt

vdtds

s tt

23

0 00

2

2

3

63

63

0 ≤ t < 2 s -> -ve velocity -> the particle is moving to the left, t > 2a -> +ve velocity -> the particle is moving to the right

msst

125.65.3

ms

st0.4

2

0

0

ts

EXAMPLE 12.5


The distance traveled in 3.5s is

sT = 4.0 + 4.0 + 6.125 = 14.125m

Velocity. The displacement from t = 0 to t = 3.5s is Δs = 6.125 – 0 = 6.125m

And so the average velocity is

smt

saavg /75.1

05.3

125.6

Average speed, smt

sv T

avgsp /04.405.3

125.14

EXAMPLE 12.5


Rectilinear Kinematics: Erratic Motion

• When particle’s motion is erratic, it is best described graphically using a series of curves that can be generated experimentally from computer output.

• a graph can be established describing the relationship with any two of the variables, a, v, s, t

• using the kinematics equations a = dv/dt, v = ds/dt, a ds = v dv


Given the s-t Graph, construct the v-t Graph

•The s-t graph can be plotted if the position of the particle can be determined experimentally during a period of time t.

•To determine the particle’s velocity as a function of time, the v-t Graph, use v = ds/dt

•Velocity as any instant is determined by measuring the slope of the s-t graph



vdt

ds

Slope of s-t graph = velocity



Given the v-t Graph, construct the a-t Graph

•When the particle’s v-t graph is known, the acceleration as a function of time, the a-t graph can be determined using a = dv/dt

•Acceleration as any instant is determined by measuring the slope of the v-t graph



adt

dv

Slope of v-t graph = acceleration



• Since differentiation reduces a polynomial of degree n to that of degree n-1, then if the s-t graph is parabolic (2nd degree curve), the v-t graph will be sloping line (1st degree curve), and the a-t graph will be a constant or horizontal line (zero degree curve)



EXAMPLE 12.6

A bicycle moves along a straight road such that it position is described by the graph as shown. Construct the v-t and a-t graphs for 0 ≤ t ≤ 30s.


Solution:v-t Graph. The v-t graph can be determined by differentiating the eqns defining the s-t graph

6306;3010

6.03.0;100 2

dt

dsvtssts

tdt

dsvtsst

The results are plotted.

EXAMPLE 12.6


We obtain specify values of v by measuring the slope of the s-t graph at a given time instant.

smt

sv /6

1030

30150

a-t Graph. The a-t graph can be determined by differentiating the eqns defining the lines of the v-t graph.

EXAMPLE 12.6


06;3010

6.06.0;100

dt

dvavst

dt

dvatvst

The results are plotted.

EXAMPLE 12.6



Given the a-t Graph, construct the v-t Graph

• When the a-t graph is known, the v-t graph may be constructed using a = dv/dt

dtav

Change in velocity

Area under a-t graph

=


• Knowing particle’s initial velocity v0, and add to this small increments of area (Δv)

• Successive points v1 = v0 + Δv, for the v-t graph

• Each eqn for each segment of the a-t graph may be integrated to yield eqns for corresponding segments of the v-t graph



Given the v-t Graph, construct the s-t Graph

• When the v-t graph is known, the s-t graph may be constructed using v = ds/dt

dtvs

Displacement Area under v-t graph

=



• knowing the initial position s0, and add to this area increments Δs determined from v-t graph.

• describe each of there segments of the v-t graph by a series of eqns, each of these eqns may be integrated to yield eqns that describe corresponding segments of the s-t graph



EXAMPLE 12.7

A test car starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 s and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’ needed to stop the car. How far has the car traveled?


Solution:v-t Graph. The v-t graph can be determined by integrating the straight-line segments of the a-t graph. Using initial condition v = 0 when t = 0,

tvdtdvasttv

10,10;1010000

EXAMPLE 12.7


When t = t’ we require v = 0. This yield t’ = 60 s

s-t Graph. Integrating the eqns of the v-t graph yields the corresponding eqns of the s-t graph. Using the initial conditions s = 0 when t = 0,

2

005,10;10;100 tsdttdstvst

ts

When t = 10s, v = 100m/s, using this as initial condition for the next time period, we have

1202,2;2;1010100

tvdtdvattstv

EXAMPLE 12.7


When t’ = 60s, the position is s = 3000m

When t = 10s, s = 500m. Using this initial condition,

600120

1202;1202;6010

2

10500

tts

dttdstvststs

EXAMPLE 12.7



Given the a-s Graph, construct the v-s Graph

• v-s graph can be determined by using v dv = a ds, integrating this eqn between the limit v = v0 at s = s0 and v = v1 at s = s1

1

0

20

212

1 s

sdsavv

Area under a-s graph


• determine the eqns which define the segments of the a-s graph

• corresponding eqns defining the segments of the v-s graph can be obtained from integration, using vdv = a ds



Given the v-s Graph, construct the a-s Graph

• v-s graph is known, the acceleration a at any position s can be determined using a ds = v dv

ds

dvva

Acceleration = velocity times slope of v-s graph



• At any point (s,v), the slope dv/ds of the v-s graph is measured

• Since v and dv/ds are known, the value of a can be calculated



EXAMPLE 12.8

The v-s graph describing the motion of a motorcycle is shown in Fig 12-15a. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 120 m.


Solution:a-s Graph. Since the eqns for the segments of the v-s graph are given, a-s graph can be determined using a ds = v dv.

0

;15;12060

6.004.0

32.0;600

ds

dvva

vmsm

sds

dvva

svms

EXAMPLE 12.8



Time. The time can be obtained using v-s graph and v = ds/dt. For the first segment of motion, s = 0 at t = 0,

3ln5)32.0ln(532.0

32.0;32.0;600

0

sts

dsdt

ds

v

dsdtsvms

st

o

At s = 60 m, t = 8.05 s

EXAMPLE 12.8


For second segment of motion,

05.415

15

15;15;12060

6005.8

s

t

dsdt

ds

v

dsdtvms

st

At s = 120 m, t = 12.0 s

EXAMPLE 12.8


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 Δt the particle moves a distance Δs along the curve to a new position P`, defined by r` = r + Δr. The displacement Δr represents the change in the particle’s position.



Velocity. During the time Δt, the average velocity of the particle is defined as

t

rvavg

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

dt

drvins



• 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 Δr is the length of the straight line segment from P to P`.

dt

dsv



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

t

vaavg

2

2

dt

rd

dt

dva



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

kvjvivdt

drv

zyx

zyx

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

yva

xva

kajaiadt

dva

zz

yy

xx

zyx

The acceleration has a magnitude defined as the positive value of

222zyx aaaa

where



• 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 ANALYSIS

Coordinate 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 z components.



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.



EXAMPLE 12.9

At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in second. If the equation of the path is y = x2/30, determine the distance of the balloon from the station at A, the magnitude and direction of the both the velocity and acceleration when t = 2 s.


Solution:Position. When t = 2 s, x = 9(2) m = 18 m and

y = (18)2/30 = 10.8 mThe straight-line distance from A to B is

218.1018 22 r m

Velocity.

smxdt

dyv

smtdt

dxv

y

x

/8.1030/

/99

2

EXAMPLE 12.9


When t = 2 s, the magnitude of velocity is

smv /1.148.109 22

The direction is tangent to the path, where

2.50tan 1

x

yv v

v

Acceleration.

2/4.5

0

smva

va

yy

xx

EXAMPLE 12.9


222 /4.54.50 sma

The direction of a is 90

0

4.5tan 1

a

EXAMPLE 12.9


The motion of box B is defined by the position vector r = {0.5sin(2t)i + 0.5cos(2t)j – 0.2tk} m, where t is in seconds and the arguments for sine and cosine are in radians (π rad = 180°). Determine the location of box when t = 0.75 s and the magnitude of its velocity and acceleration at his instant.

EXAMPLE 12.10


Solution:Position. Evaluating r when t = 0.75 s yields

mkjradiradr st })75.0(2.0)5.1cos(5.0)5.1sin(5.0{75.0

mkji }150.00354.0499.0{

The distance of the box from the origin is

mr 522.0)150.0()0354.0()499.0( 222

EXAMPLE 12.10



The direction of r is obtained from the components of the unit vector,

107

1.86

2.17)955.0(cos

287.00678.0955.0

522.0

150.0

522.0

0352.0

522.0

499.0

1

kji

kjir

rur

EXAMPLE 12.10


Velocity.

smkjtitdt

rdv /}2.0)2sin(1)2cos(1{

Hence at t = 0.75 s, the magnitude of velocity, is

smvvvv zyx /02.1222

Acceleration. The acceleration is not tangent to the path. 2/})2cos(2)2sin(2{ smjtit

dt

vda

At t = 0.75 s, a = 2 m/s2

EXAMPLE 12.10


• 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

;2

1

;

020

2

200

0

ssavv

tatvxx

tavv

c

c

c

xx

x

xx

vv

tvxx

vv

)(

)(

)(

0

00

0

Horizontal component of velocity remain constant during the motion



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

);(2

;2

1

;

020

2

200

0

yyavv

tatvyy

tavv

c

c

c

)(2)(

2

1)(

)(

02

0

200

0

yygvv

gttvyy

gtvv

yx

y

yy



• 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 directions

Kinematics 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)(

2

1)(

)(

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



EXAMPLE 12.11

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


Coordinate System. Origin of the coordinates is established at the beginning of the path, point A.Initial velocity of a sack has components (vA)x = 12 m/s and (vA)y = 0Acceleration between point A and B ay = -9.81 m/s2Since (vB)x = (vA)x = 12 m/s, the three unknown are (vB)y, R and the time of flight tAB

EXAMPLE 12.11


Vertical Motion. Vertical distance from A to B is known

st

tatvyy

AB

ABcABy

11.1

;2

1)( 2

00

The above calculations also indicate that if a sack is released from rest at A, it would take the same amount of time to strike the floor at C

EXAMPLE 12.11


Horizontal Motion.

mR

tvxx ABx

3.13

)( 00

EXAMPLE 12.11


The chipping machine is designed to eject wood at chips vO = 7.5 m/s. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.

EXAMPLE 12.12


Coordinate System. Three unknown h, time of flight, tOA and the vertical component of velocity (vB)y. Taking origin at O, for initial velocity of a chip,

(vA)x = (vO)x = 6.5 m/s and ay = -9.81 m/s2

smv

smv

yO

xO

/75.3)30sin5.7()(

/5.6)30cos5.7()(

EXAMPLE 12.12


Horizontal Motion.

st

tvxx

OA

OAxA

9231.0

)( 00

Vertical Motion.Relating tOA to initial and final elevation of the chips,

mh

tatvyhy OAcOAyOA

38.12

1)(1.2 2

0

EXAMPLE 12.12


The track for this racing event was designed so that the riders jump off the slope at 30°, from a height of 1m. During the 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 rider.

EXAMPLE 12.13


Coordinate System. Origin is established at point A. Three unknown are initial speed vA, range R and the vertical component of velocity vB.

Vertical Motion. Since time of flight and the vertical distance between the ends of the paths are known,

smv

tatvss

A

ABCAByAyAyB

/4.132

1)()()( 2

EXAMPLE 12.13



Horizontal Motion

•For maximum height h, we consider path AC•Three unknown are time of flight, tAC, horizontal distance from A to C and the height h•At maximum height (vC)y = 0

m

R

tvss ABAxAxB

4.17

)5.1(30cos38.130

)()()(

EXAMPLE 12.13


•Since vA known, determine h using the following equations

•Show that the bike will strike the ground at B with velocity having components of

smvsmv

mh

h

ssavv

yBxB

yAyCcAc yy

/02.8)(,/6.11)(

28.3

]0)1)[(81.9(2)30sin38.13()0(

])()[(2)()(

22

22

EXAMPLE 12.13


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)

sv

uvv t

where



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

tt uvuvva



• 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 uv

us

uu



2va

vdvdsava

uauaa

n

tt

nntt

• Magnitude of acceleration is the positive value of

22nt aaa

where

and



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

0vat /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 b which 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 origin which 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

2

0

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,

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[

dxyd

dxdy



EXAMPLE 12.14

When the skier reaches the point A along the parabolic path, he has a speed of 6m/s which is increasing at 2m/s2. Determine the direction of his velocity and the direction and magnitude of this acceleration at this instant. Neglect the size of the skier in the calculation.


Coordinate System. Establish the origin of the n, t axes at the fixed point A on the path and determine the components of v and a along these axes.

Velocity. The velocity is directed tangent to the path.

1,201

10

2 xdx

dyxy

v make an angle of θ = tan-1 = 45° with the x axis

smvA /6

EXAMPLE 12.14


Acceleration. Determined from nt uvuva)/( 2

mdxyd

dxdy28.28

/

])/(1[22

2/32

The acceleration becomes

2

2

/}273.12{ smuu

uv

uva

nt

ntA

EXAMPLE 12.14


5.57

327.12

tan

/37.2237.12

1

222

sma

Thus, 57.5° – 45 ° = 12.5 °

a = 2.37 m/s2

EXAMPLE 12.14


Race car C travels round the horizontal circular track that has a radius of 90 m. If the car increases its speed at a constant rate of 2.1 m/s2, starting from rest, determine the time needed for it to reach an acceleration of 2.4 m/s2. What is its speed at this instant?

EXAMPLE 12.15


Coordinate System. The origin of the n and t axes is coincident with the car at the instant. The t axis is in the direction of the motion, and the positive n axis is directed toward the center of the circle.

Acceleration. The magnitude of acceleration can be related to its components using 22

nt aaa

t

tavv ct

1.2

)(0

EXAMPLE 12.15


222

/049.0 smtv

an

Thus,

The time needed for the acceleration to reach 2.4m/s2 is 22

nt aaa

Solving for t = 4.87 s

Velocity. The speed at time t = 4.87 s is

smtv /2.101.2

EXAMPLE 12.15


The boxes travels alone the industrial conveyor. If a box starts from rest at A and increases its speed such that at = (0.2t) m/s2, determine the magnitude of its acceleration when it arrives at point B.

EXAMPLE 12.16


Coordinate System. The position of the box at any instant is defined by s, from the fixed point A. The acceleration is to be determined at B, so the origin of the n, t axes is at this point.

Acceleration. Since vA when t = 0

2

00

1.0

2.0

2.0

tv

dttdv

tvatv

t

(1)

(2)

EXAMPLE 12.16



The time needed for the box to reach point B can be determined by realizing that the position of B is sB = 3 + 2π(2)/4 = 6.142 m, since sA = 0 when t = 0

st

dttds

tdtds

v

B

tB

690.5

1.0

1.0

02142.6

0

2

EXAMPLE 12.16


Substituting into eqn (1) and (2),

smv

smva

B

BtB

/238.3)69.5(1.0

/138.1)690.5(2.0)(2

2

At B, ρB = 2 m, 2

2/242.5)( sm

va

B

BnB

2

22

/36.5

)242.5()138.1(

sm

aB

EXAMPLE 12.16


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 traverse coordinate θ, 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



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

rr ururrv

• 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

ru



• 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

uu

utt

uu

r

t

r

tr

00limlim



rv

rv

uvuvv

r

rr

where

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 /

22 rrv



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

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



rra

rra

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



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

222 2 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

)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

,,,rr



The amusement park consists of a chair that is rotating in a horizontal circular path of radius r such that the arm OB has an angular velocity and angular acceleration. Determine the radial and transverse components of velocity and acceleration of the passenger.

EXAMPLE 12.17


Coordinate System. Since the angular motion of the arm is reported, polar coordinates are chosen for the solution. θ is not related to r, since radius is constant for all θ.

Velocity and Acceleration. Since r is constant,

00 rrrr

EXAMPLE 12.17


rrra

rrra

rv

rv

r

r

2

0

22

This special case of circular motion happen to be collinear with r and θ axes

EXAMPLE 12.17


The rob OA is rotating in the horizontal plane such that θ = (t3) rad. At the same time, the collar B is sliding outwards 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


Coordinate System. Since time-parametric equations of the particle is given, it is not necessary to relate r to θ.

Velocity and Acceleration.

22

2

31

2

/66/200200

/23/200200

73.51100100

11

11

1

sradtsmmr

sradtsmmtr

radtmmtr

stst

stst

stst

EXAMPLE 12.18


smmuu

ururv

r

r

/}300200{

The magnitude of v is

1143.57

3.56200300

tan

/361300200

1

22

smmv

EXAMPLE 12.18


2

2

/}1800700{

)2()(

smmuu

urrurra

r

r

The magnitude of a is

1693.57)180(

7.687001800

tan

/19301800700

1

222

smma

EXAMPLE 12.18


The searchlight casts a spot of light along the face of a wall that is located 100m from the searchlight. Determine the magnitudes of the velocity and acceleration at which the spot appears to travel across the wall at the instant θ = 45°. The searchlight is rotating at a constant rate of 4 rad/s

EXAMPLE 12.19


Coordinate System. Polar coordinates will be used since the angular rate of the searchlight is given. To find the time derivatives, it is necessary to relate r to θ.

r = 100/cos θ = 100sec θ


)tan(sec100sec100tansec100

)tan(sec1002322

r

r

EXAMPLE 12.19


Since = 4 rad/s = constant, = 0, when = 45°,

2.6788

7.565

4.141

r

r

r

smv

smuu

ururv

r

r

/800

/}7.5657.565{

EXAMPLE 12.19


2

2

2

/6400

/}5.45255.4525{

)2()(

smma

smmuu

urrurra

r

r

EXAMPLE 12.19


Due to the rotation of the forked rod, ball A travels across the slotted path, a portion of which is in the shape of a cardioids, r = 0.15(1 – cos θ)m where θ is in radians. If the ball’s velocity is v = 1.2m/s and its acceleration is 9m/s2 at instant θ = 180°, determine the angular velocity and angular acceleration of the fork.

EXAMPLE 12.20


Coordinate System. For this unusual path, use polar coordinates.


)(sin15.0)()(cos15.0

)(sin15.0

)cos1(15.0

r

r

r

Evaluating these results at θ = 180°215.003.0 rrmr

EXAMPLE 12.20



Since v = 1.2 m/s

srad

rrv

/4

22

2

222

/18

)2()(

srad

rrrra

EXAMPLE 12.20


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 sB measure the lengths of the changing segments of the cord

ABBA vvdtds

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



EXAMPLE 12.21

Determine the speed of block A if block B has an upward speed of 2 m/s.


Position Coordinate System. There is one cord in this system having segments which are changing length. Position coordinates sA and sB will be used since each is measured from a fixed point (C or D) and extends along each block’s path of motion. In particular, sB is directed to point E since motion of B and E is the same. The red colored segments of the cords remain at a constant length and do not have to be considered as the block move.

EXAMPLE 12.21


The remaining length of the cord, l, is also considered and is related to the changing position coordinates sA and sB by the equation

lss BA 3

Time Derivative. Taking the time derivative yields

03 BA vv

so that when vB = -2m/s (upward)vA = 6m/s ↓

EXAMPLE 12.21


Determine the speed of block A if block B has an upward speed of 2m/s.

EXAMPLE 12.22


Position Coordinate Equation. Positions of A and B are defined using coordinates sA and sB.Since the system has two cords which change length, it is necessary to use a third coordinate sC in order to relate sA to sB. Length of the cords can be expressed in terms of sA and sC, and the length of the other cord can be expressed in terms of sB and sC. The red colored segments are not considered in this analysis.

EXAMPLE 12.22


For the remaining cord length,

21 )(2 lssslss CBBCA

Eliminating sC yields,

1224 llss BA

Time Derivative. The time derivative gives04 BA vv

so that vB = -2m/s (upward)

smsmvB /8/8

EXAMPLE 12.22


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

EXAMPLE 12.23


Position-Coordinate Equation. The position of A is defined by sA, and the position of block B is specified by sB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates sA and sB cannot be related directly. By establishing a third position coordinate, sC, and the length of the other cord in terms of sA, sB and sC.

EXAMPLE 12.23


Excluding the red colored segments of the cords, the remaining constant cord lengths l1 and l2 (along the hook and link dimensions) can be expressed as

12 24 llss BC

2

1

lsssss

lss

BCBCA

BC

Eliminating sC yields

EXAMPLE 12.23


Time Derivative. The time derivative give

smsmv

vv

B

BA

/5.0/5.0

04

when vA = 2m/s (downward)

EXAMPLE 12.23


A man at A s hoisting a safe S by walking to the right with a constant velocity vA = 0.5m/s. Determine the velocity and acceleration of the safe when it reaches the elevation at E. The rope is 30m long and passes over a small pulley at D.

EXAMPLE 12.24


Position Coordinate System. Rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the position of S and A, are specified by means of the x and y coordinates measured from a fixed point and directed along the paths of motion of the ends of the rope. The x and y coordinates may be related since the rope has a fixed length l = 30m, which at all times is equal to the length of the segment DA plus CD.

EXAMPLE 12.24



Using Pythagorean Theorem,

yIxI CDDA 1515 22

15225

151530

2

22

xy

yx

lll CDDA

(1)

EXAMPLE 12.24


Time Derivative. Taking time derivative, using the chain rule where, vS = dy/dt and vA = dx/dt

A

S

vx

x

dtdx

x

xdtdy

v

2

2

225

225

221

At y = 10 m, x = 20 m, vA = 0.5 m/s, vS = 400mm/s ↑

(2)

EXAMPLE 12.24


The acceleration is determined by taking the time derivative of eqn (2),

2/32

2

2

22/322

2

225

225

225

1

225

1

)225(

)/(

x

vdtdvx

x

vdtdx

xxv

x

dtdxx

dt

yda

AA

AAS

At x = 20 m, with vA = 0.5 m/s,

2/6.3 smmaS

EXAMPLE 12.24


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, z reference 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 /



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 /

dtrdvanddtrdv AABB //

dtrdv ABAB ///



• 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 observer fixed 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 /



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, rB and 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



A train, traveling at a constant speed of 90km/h, crosses over a road. If automobile A is traveling t 67.5km/h along the road, determine the magnitude and direction of relative velocity of the train with respect to the automobile

EXAMPLE 12.25


Vector Analysis. The relative velocity is measured from the translating x’, y’ axes attached to the automobile. Since vT and vA are known in both magnitude and direction, the unknowns become the x and y components of vT/A. Using the x, y axes and a Cartesian vector analysis,

hkmjiv

vjii

vvv

AT

AT

ATAT

/)~7.47~3.42{

)~45sin5.67~45cos5.67(~90

/

/

/

EXAMPLE 12.25


The magnitude of vT/A is

hkmv AT /8.63)7.473.42( 222/

The direction of vT/A defined from the x axis is

40.48

3.427.47

tan/

/

xAT

yAT

v

v

EXAMPLE 12.25


Plane A is flying along a straight-line path, while 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 12.26


Velocity. The x, y axes are located at an arbitrary fixed point. Since the motion relative to plane A is to be determined, the translating frame of reference x’. y’ is attached to it. Applying the relative-velocity equation in scalar form since the velocity vectors of both plane are parallel at the instant shown,

hkmhkmv

v

vvv

AB

AB

ABAB

/100/100

700600

/

/

/)(

EXAMPLE 12.26


Acceleration. Plane B has both tangential and normal components of acceleration, since it is flying along a curved path. Magnitude of normal acceleration,

22

/900 hkmv

a BnB

Applying the relative-acceleration equation,

2/

/

/

/~150~900

~50~100~900

hkmjia

ajji

aaa

AB

AB

ABAB

EXAMPLE 12.26


From the figure shown, the magnitude and direction of ABa /

46.9900150

tan/912 12/ hkma AB

EXAMPLE 12.26


At the instant, car A and B are traveling with the speed 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.

EXAMPLE 12.27


Velocity. The fixed x, y axes are established at a point on the ground and the translating x’, y’ axes are attached to car A. Using Cartesian vector analysis,

smv

smjiv

vjij

vvv

AB

AB

AB

ABAB

/69.9588.39

/~588.3~9

~60sin18~60cos18~12

22/

/

/

/

Thus,

EXAMPLE 12.27



Its direction is

7.21

9588.3

tan/

/

xAB

yAB

v

v

Acceleration. The magnitude of the normal component is

22

/440.1 smv

a BnB

EXAMPLE 12.27


Applying the equation for relative acceleration yields

2

/

/

/

/~732.4~440.2

~60sin2~60cos2~3~440.1

smjia

ajiji

aaa

AB

AB

ABAB

Magnitude and direction is

7.62

/32.5 2/

sma AB

EXAMPLE 12.27


Chapter Review

• Rectilinear Kinematics. Rectilinear kinematics refers to motion along a straight line. A position coordinate s specifies the location of the particle on the line and the displacement ∆s is the change in this position.

The average velocity is a vector quantity, defined as the displacement divided by the time interval.

tr

vavg


This is the different than the average speed, which is a scalar and is the total distance traveled divided by the time of travel.

ts

v Tavgsp

The time, position, instantaneous velocity and instantaneous acceleration are related by the differential equations

dvvdsadtdvadtdsv //

Chapter Review


If the acceleration is known to be constant, then the integration of these equations yields

tavv c 02

00 21tatvss c

020

2 2 ssavv c

Chapter Review


• Graphical Solutions. If the motion is erratic, then it can be described by a graph. If one of these graphs is given, then the others can be established using the differential relations, v = ds/dt, a = dv/dt, or a ds = v dv.

• Curvilinear Motion, x, y, z. For this case, motion along the path is resolved into rectilinear motion along the x, y, z axes. The equation of the path is used to relate the motion along each axis.

Chapter Review


• Projectile Motion. Free flight motion of a projectile follows a parabolic path. It has a constant velocity in the horizontal direction and constant acceleration of g = 9.81 m/s2 in the vertical direction. Any two of the three equations for constant acceleration apply in the vertical direction, and in the horizontal direction only

tvxx x)( 00

Chapter Review


• Curvilinear Motion n, t. If normal and tangential axes are used for the analysis, then v is always in the positive t direction. The acceleration has two components. The tangential components, at, accounts for the change in the magnitude of the velocity; a slowing down is in the negative t direction, and a speeding up is in the positive t direction. The normal component, an accounts for the change in the direction of velocity. The component is always in the positive n direction.

Chapter Review


• Curvilinear Motion r, θ, z. If the path of motion is expressed in polar coordinates, then the velocity and acceleration components can be written as

rrarv

rrarv rr

2

2

To apply these equations, it is necessary to determine at the instant considered. If the path r

= f(θ) is given, then the chain rule of calculus must be used to obtain the time derivatives.

,,,, rrr

Chapter Review


Once the data is substituted into the equations, then the algebraic sign of the results will indicate the direction of components of v or a along each axis.

• Absolute Dependent Motion of Two Particles. The dependent motion of blocks that are suspended from pulleys and cables can be related by the geometry of the system.

Chapter Review


This is done by first establishing position coordinates, measured from a fixed origin to each block so that they are directed along the line of motion of the blocks. Using geometry and/or trigonometry, the coordinates are then related to the cable length in order to formulate a position coordinate equation. The first time derivative of this equation gives a relationship between the velocities of the blocks, and a second time derivative gives the relationship between their accelerations.

Chapter Review


• Relative Motion Analysis Using Translating Axes. If two particles A and B undergo independent motions, then these motions can be related to their relative motion. Using a translating set axes attached to one of the particles (A), the velocity and acceleration equations become

ABAB

ABAB

aaa

vvv

/

/

Chapter Review


For planar motion, each of these equations produces two scalar equations, one in the x, and the other in the y direction. For solution, the vectors can be expressed in Cartesian form or the x and y scalar components can be written directly.

Chapter Review

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