ENGR 3423 Dynamics - Arkansas State Universitymyweb.astate.edu/sharan/Dynamics/Lecture Notes/4 Chapt...

ENGR 3423 Dynamics ENGR 3423 Dynamics sharansharan 11

ENGR 3423 DynamicsENGR 3423 Dynamics

Dr. Shivan Haran, Assistant ProfessorDr. Shivan Haran, Assistant Professor

Office: LSW Room 244Office: LSW Room 244Phone: (870) 972 3413Phone: (870) 972 3413

Email: Email: [email protected]@astate.edu

Office HoursOffice Hours: M, T, W, : M, T, W, ThTh, F , F 9:30 AM 9:30 AM 11:00 AM11:00 AM

Text:Text: Engineering Mechanics Engineering Mechanics DynamicsDynamicsby R. C. by R. C. HibbelerHibbeler, 10th Ed., Prentice Hall, 2004, 10th Ed., Prentice Hall, 2004

OR LaterOR Later


General General PrinciplesPrinciples


MechanicsMechanics::

Branch of physical sciences Branch of physical sciences concerned with the state of rest or concerned with the state of rest or

motion of bodies subjected to motion of bodies subjected to forcesforces


Rigid Body MechanicsRigid Body Mechanics

StaticsStatics Bodies at restBodies at restDynamicsDynamics Accelerated motion Accelerated motion

of bodies of bodies


Statics:Statics: the study the study of bodies in of bodies in equilibrium equilibrium

Dynamics:Dynamics:1. Kinematics 1. Kinematics concerned concerned

with the geometric aspects of with the geometric aspects of motionmotion

2. Kinetics 2. Kinetics -- concerned with concerned with the forces causing the motion the forces causing the motion

1) Particles, Rigid Bodies 2) Deformable Bodie1) Particles, Rigid Bodies 2) Deformable Bodiess

Mechanics:Mechanics: the study of how bodies react to forces acting on themthe study of how bodies react to forces acting on them


Continuum Mechanics:Continuum Mechanics:

-- ElasticityElasticity

-- Fluid MechanicsFluid Mechanics






Gives rise to Gives rise to Partial Partial Differential EquationsDifferential Equations

Gives rise to Gives rise to Ordinary Ordinary Differential EquationsDifferential Equations


APPLICATIONS OF DYNAMICS: ParticlesAPPLICATIONS OF DYNAMICS: Particles

The motion of large objects, such as The motion of large objects, such as rockets, airplanes, or cars, can often be rockets, airplanes, or cars, can often be analyzed as if they were particles.analyzed as if they were particles.

Why?Why?

Typical question:Typical question:Can we predict the altitude, Can we predict the altitude, velocity and acceleration of velocity and acceleration of this rocket as a function of this rocket as a function of time?time?


APPLICATIONSAPPLICATIONS (continued)(continued)

A train travels along a straight length of track.A train travels along a straight length of track.

Can we treat the train as a particle?Can we treat the train as a particle?

Typical Question:Typical Question:If the train accelerates at a constant rate, how can we If the train accelerates at a constant rate, how can we determine its position and velocity at some instant?determine its position and velocity at some instant?



What about a baseball? What about a baseball? What is the What is the correctcorrect model: particle, rigid body or model: particle, rigid body or deformable body?deformable body?

Depends:Depends: A particle model will describe the approximate A particle model will describe the approximate displacement when thrown or hit.displacement when thrown or hit. A rigid body model will describe the effect of spin.A rigid body model will describe the effect of spin. A deformable body model will describe the transfer A deformable body model will describe the transfer of energy when hit by a bat.of energy when hit by a bat.


Goals of CourseGoals of Course

Develop modeling skills for realistic Develop modeling skills for realistic problemsproblems

Apply a logical, mathematical Apply a logical, mathematical framework to realistic problemsframework to realistic problems

Understand the role of force in motion Understand the role of force in motion


Basic Quantities Basic Quantities -- UnitsUnits

LengthLengthmetermeterfootfoot

TimeTimesecondsecond

MassMasskilogram kilogram slugslug

ForceForceNewton Newton poundpound


LengthLength

Needed to locate the position of a Needed to locate the position of a point in space and describe the size point in space and describe the size of a physical systemof a physical system


TimeTime

Conceived as a succession of eventsConceived as a succession of events Concepts of STATICS are time Concepts of STATICS are time independentindependent


MassMass

A property of matter by which we can A property of matter by which we can compare the action of one body to anothercompare the action of one body to another This property manifests itself as a This property manifests itself as a gravitational attraction between two bodies gravitational attraction between two bodies and provide a qualitative measure of the and provide a qualitative measure of the resistance of matter to a change in velocityresistance of matter to a change in velocity


IdealizationsIdealizations

Particle Particle -- an object having mass but the an object having mass but the size is neglected size is neglected

Rigid BodyRigid Body -- a combination of a large a combination of a large number of particles which remain in a number of particles which remain in a fixed position relative to each other, both fixed position relative to each other, both before and after the application of a forcebefore and after the application of a force


ForceForce

Generally considered as a push or a pull Generally considered as a push or a pull exerted by one body on anotherexerted by one body on another Interaction occurs when there is direct Interaction occurs when there is direct contact between the bodiescontact between the bodies Gravitational, electrical and magnetic Gravitational, electrical and magnetic forces do not require direct contactforces do not require direct contact Force is characterized by magnitude, Force is characterized by magnitude, direction and point of applicationdirection and point of application


NewtonNewtons Laws of Motions Laws of Motion

First LawFirst LawA particle originally at rest, or moving in a straight line withA particle originally at rest, or moving in a straight line with constant constant velocity, will remain in this state provided the particle is velocity, will remain in this state provided the particle is notnotsubjected to unbalanced forces.subjected to unbalanced forces.


Second LawSecond LawA particle acted upon by an unbalanced force A particle acted upon by an unbalanced force FF experiences an acceleration experiences an acceleration aa that has the same direction as the force and a magnitude that isthat has the same direction as the force and a magnitude that is directly directly proportional to the force. If proportional to the force. If FF is applied to a particle of mass is applied to a particle of mass mm then: then: F = F = mmaa

OR another version:OR another version:

The rate of change of momentum of a body is directly proportionaThe rate of change of momentum of a body is directly proportional to l to the force and acts in the direction of the force.the force and acts in the direction of the force.




Third LawThird LawThe mutual forces of action and reaction between two particles aThe mutual forces of action and reaction between two particles are equal, re equal, opposite and collinear.opposite and collinear.


NewtonNewtons Laws of Gravitational Attractions Laws of Gravitational Attraction

221

rmmGF =

Where:Where:FF = force of gravitation= force of gravitationGG = universal constant of gravitation= universal constant of gravitationmm11, m, m22= mass of the two particles= mass of the two particlesrr = distance between the two particles= distance between the two particles


WeightWeight

1 22

22

22

m mW Gr

mmW Gr

mW mGr

=

=

=

mm = mass of object= mass of objectmm22 = mass of earth= mass of earthrr = distance from center of earth = distance from center of earth

to particleto particle


UnitsUnits

1.1. Basic quantities (force, mass, length, time) are Basic quantities (force, mass, length, time) are related by Newtonrelated by Newtons second law.s second law.

2.2. Units used to measure quantities are not all Units used to measure quantities are not all independent.independent.

3.3. Three of four units, called Three of four units, called basebase units, are arbitrarily units, are arbitrarily defined and the fourth is derived.defined and the fourth is derived.


1.1. Modern version of metric systemModern version of metric system2.2. Base units are length, time and mass, meter (m), Base units are length, time and mass, meter (m),

second (s), and kilogram (kg)second (s), and kilogram (kg)3.3. Acceleration of gravity:Acceleration of gravity:

SI UnitsSI Units

2mg 9.81s

=


4.4. Force is Force is derived quantityderived quantity measured in unit called measured in unit called a a newtonnewton

SI UnitsSI Units

2smkg1N1 =


U.S. Customary Units (fps)U.S. Customary Units (fps)

1.1. Base units are length, time and force.Base units are length, time and force.2.2. feet (ft), second (s), and pound (lb)feet (ft), second (s), and pound (lb)3.3. Acceleration of gravity: Acceleration of gravity:

2ftg 32.2

s=


U.S. Customary Units (fps)U.S. Customary Units (fps)

4.4. Mass is Mass is derived quantityderived quantity measured in a unit called measured in a unit called a a slug:slug:

ftslb1slug1

2=


Systems of UnitsSystems of Units

pound(lb)

slug(lb s2/ft

second(s)

foot(ft)

US Customary

newton(N)

kilogram(kg)

second(s)

meter(m)

SI

ForceMassTimeLengthName


Converting between different systems of unitsConverting between different systems of units

Useful Conversion factors:Useful Conversion factors:1 inch1 inch = 2.54 cm= 2.54 cm1 m 1 m = 3.28 ft= 3.28 ft1 mile1 mile = 5280 ft = 5280 ft 1 mile 1 mile = 1.61 km= 1.61 km

Example: convert miles per hour to meters per second:Example: convert miles per hour to meters per second:

mi mi 5280 ft 1 m 1 hr1 1hr hr 1 mi 3.28 ft 3600 s

= m0.447s

=


Force : 1 lb 4.4482NMass : 1 slug 14.5938 kgLength : 1 ft 0.3048 m

===

Unit ConversionsUnit ConversionsUnit Conversions


Prefixes for SI unitsPrefixes for SI units

Exponential form Prefix SI symbol

Multiple 1,000,000,000 109 giga G 1,000,000 106 mega M 1,000 103 kilo k Submultiple 0.001 10-3 milli m 0.000001 10-6 micro 0.00000001 10-9 nano n


Dimensional HomogeneityDimensional Homogeneity

Each of the terms in an equation must be expressed in the same Each of the terms in an equation must be expressed in the same units units

s = v t + 1/2 a ts = v t + 1/2 a t22

ss is position in metersis position in meters vv is velocity in m/sis velocity in m/s aa is acceleration in m/sis acceleration in m/s22

tt is time in secondsis time in seconds m = m/s m = m/s * * s + m/ss + m/s22 * * ss2 2 = m= m


CalculationsCalculations

When performing calculations When performing calculations retain a greater number of digits retain a greater number of digits than the problem data.than the problem data.Engineers usually round off Engineers usually round off final answerfinal answer to three significant to three significant figures. Intermediate figures. Intermediate calculations are usually done to calculations are usually done to four significant figures.four significant figures.Answer can never have more Answer can never have more significant figures than given significant figures than given data!data!


Convert 2 km/h to m/s. How many ft/s is this?SOLUTION:Since 1 km = 1000 m and 1 h = 3600 s, the conversion factors are arranged so that a cancellation of units can be applied.

sm556.05555.0

sm

36002000

hkm2

s3600h1

kmm1000

hkm2h

km2

===

=

Recall that 1 ft = 0.3038 mm m 1 ft ft ft0.556 0.556 1.824 1.82s s 0.3048 m s s

= = =


Procedure for AnalysisProcedure for Analysis

1.1. Read the problem carefully and correlate Read the problem carefully and correlate the actual physical situation with the theory the actual physical situation with the theory studied.studied.

2.2. Draw necessary diagrams and tables.Draw necessary diagrams and tables.3.3. Apply relevant principles, generally in Apply relevant principles, generally in

mathematical form.mathematical form.


Procedure for AnalysisProcedure for Analysis

4.4. Solve the equations algebraically (without Solve the equations algebraically (without numbers) as far as possible, then obtain a numbers) as far as possible, then obtain a numerical answer.numerical answer.

5.5. Be sure to use a consistent set of units.Be sure to use a consistent set of units.6.6. Report the answer with no more significant Report the answer with no more significant

figures than the accuracy of the given data.figures than the accuracy of the given data.7.7. Decide if answer seems reasonable.Decide if answer seems reasonable.8.8. Think about what the problem taught you!Think about what the problem taught you!


Important PointsImportant Points

1.1. Statics is the study of bodies at rest or Statics is the study of bodies at rest or moving with constant velocitymoving with constant velocity

2.2. A particle has mass but a size that can be A particle has mass but a size that can be neglectedneglected

3.3. A rigid body does not deform under loadA rigid body does not deform under load4.4. Concentrated forces are assumed to act at Concentrated forces are assumed to act at

a point on a bodya point on a body



1.1. NewtonNewtons three laws of motion must be s three laws of motion must be memorized!memorized!

2.2. Mass is a property of matter that does not Mass is a property of matter that does not change from one location to anotherchange from one location to another

3.3. Weight is the gravitational attraction of the Weight is the gravitational attraction of the earth on a body or quantity of mass. Its earth on a body or quantity of mass. Its magnitude depends on the location of the magnitude depends on the location of the massmass



1.1. Mass, length and time are the base quantitiesMass, length and time are the base quantities2.2. In the SI system the unit of force is the In the SI system the unit of force is the

Newton. It is a derived quantity.Newton. It is a derived quantity.3.3. In the SI system prefixes are used to denote In the SI system prefixes are used to denote

large or small numerical quantities of a unit large or small numerical quantities of a unit 4.4. Perform numerical calculation to several Perform numerical calculation to several

significant figures and report answers to three significant figures and report answers to three significant figuressignificant figures

5.5. Be sure that all equations are dimensionally Be sure that all equations are dimensionally homogeneoushomogeneous


INTRODUCTION & RECTILINEAR KINEMATICS: INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION (Sections 12.1 CONTINUOUS MOTION (Sections 12.1 -- 12.2)12.2)

ObjectivesObjectives::Introduce the kinematic quantities Introduce the kinematic quantities (position, displacement, velocity, and acceleration) (position, displacement, velocity, and acceleration) of a particle traveling along a of a particle traveling along a straightstraight pathpath..


DynamicsDynamics


DynamicsDynamics


Displacement: Change in Displacement: Change in position (position (s) on the paths) on the path

s = ss = s -- s s


Motion in 1 dimensionMotion in 1 dimension(1(1--D Kinematics)D Kinematics)

In 1In 1--D, we usually write position as D, we usually write position as s(ts(t))..

Since itSince its in 1s in 1--D, all we need to indicate D, all we need to indicate directiondirection is is + + or or ..

DisplacementDisplacement in a time in a time t = tt = t22 -- tt11 isis

ss = s(t= s(t22) ) -- s(ts(t11) = x) = x22 -- xx11

t

xsome particlesome particles trajectorys trajectory

in 1in 1--DD

t1 t2

t

ss1

s2


Displacement vs. DistanceDisplacement vs. Distance

DisplacementDisplacement is the net change in position of the particle (can is the net change in position of the particle (can be positive or negative).be positive or negative).DistanceDistance is how far the particle traveled (can only be positive).is how far the particle traveled (can only be positive).Example:Example:

I travel from the I travel from the enggengg. building to home (a distance . building to home (a distance LL) and ) and come half way back.come half way back.

ss1 1 = = initial positioninitial position ((enggengg. building); s. building); s22 = = final position final position

distancedistance = 1.5 = 1.5 LL

displacementdisplacement = = ss22 -- ss11= 0.5 = 0.5 LL

my trajectorymy trajectoryin 1in 1--DD

tt

xx

tt11 tt22

tt

LLss11ss22

homehome

EnggEngg..

1/2 way1/2 way


vdsdt

st

st

= =

lim &

0


VelocityVelocity

2 1

2 1

( ) ( ) =

avgx t x t x

t tv

t

Velocity Velocity vv is the is the rate of change of positionrate of change of positionAverageAverage velocity velocity vvavgavg in the time in the time t = tt = t22 -- tt11 is is

VVavgavg == slope of line connecting slope of line connecting xx11 and and ss22..

tt

xxtrajectorytrajectory

tt11 tt22

tt

xxxx11

xx22


Instantaneous VelocityInstantaneous Velocity

ConsiderConsider the the limit limit

( ) ( )2 10

( )lim( )t

x t x t dx tt d

vt

t

= =

Instantaneous velocityInstantaneous velocity v v is definedis defined as:as:1 2 0 i.e., t t t

soso v(tv(t22)) == slope of line tangent to path at slope of line tangent to path at tt22..

trajectorytrajectory

tt

xx

tt11 tt22

xx11xx22

0 t

0 x


Velocity vs. SpeedVelocity vs. Speed

Instantaneous Instantaneous velocityvelocity is is how fasthow fast a particle moves at a given a particle moves at a given instant instant ANDAND the direction in which it is moving at that instant the direction in which it is moving at that instant (can be positive or negative).(can be positive or negative).InstantaneousInstantaneous speedspeed is is how fasthow fast a particle moves at a given a particle moves at a given instant, i.e., the magnitude of instant, i.e., the magnitude of velocityvelocity (can only be positive).(can only be positive).Example (my trip from Example (my trip from enggengg. to home and 1/2 way back):. to home and 1/2 way back):

I travel to and from home at a constant I travel to and from home at a constant speedspeed, , ss. . Let Let LL = 1000 m and = 1000 m and tt = 200s (for this L). Thus, = 200s (for this L). Thus, ss = 5 m/s.= 5 m/s.

slope = 5 m/s

t

x

t1 t2

t

Lx1x2

home

Engg.

1/2 way

slope = 5 m/s


Velocity vs. Speed (Velocity vs. Speed (concontt))

My trip from My trip from enggengg. to home:. to home:instantaneous instantaneous speedspeed = 5 m/s= 5 m/sinstantaneous instantaneous velocityvelocity = 5 m/s= 5 m/s

My trip back:My trip back:instantaneous instantaneous speedspeed = 5 m/s= 5 m/sinstantaneous instantaneous velocityvelocity = = --5 m/s5 m/s

slope = 5 m/s

t

x

t1 t2

t

Lx1x2

home

Engg.

1/2 way

slope = 5 m/s


Velocity vs. Speed (Velocity vs. Speed (concontt))

Total trip:average speed = 5 m/saverage velocity = 1.7 m/s

.2 12 1

500 m 1 7 m/s300 savg

x x xt t t

v = = =

Note: in general, average speed is NOT the magnitude of average velocity

slope = 5 m/s

t

x

t1 t2

t

Lx1x2

home

Engg.

1/2 way

slope = 5 m/sxx


Velocity vs. SpeedVelocity vs. SpeedDetailsDetails

= =dxs v vdt

= t

avgt

v vdtT

2

1

1

= = = t t

avgt t

s sdt v dt T t tT T

2 2

1 1

2 11 1

avg avgs v in general

t t

t t

v dt vdt2 2

1 1

since in general

InstantaneousInstantaneous speedspeedand velocity:and velocity:

AverageAverage speedspeedand velocity:and velocity:


advdt

vt

v st

= = =

lim & &&

0


AccelerationAcceleration

( ) ( )2 12 1

avgv t v t va

t t t

=

Acceleration, Acceleration, aa, is the , is the rate of change of velocityrate of change of velocityAverageAverage acceleration, acceleration, aaavgavg, , in the time in the time t = tt = t22 tt11 is is

AndAnd instantaneousinstantaneous acceleration, acceleration, aa, is defined, is defined as:as:

2

20

( ) ( )( ) limt

v dv t d x ta tt dt dt

= = =

dt)t(dx)t(v =usingusing


AccelerationAcceleration

advdt

vt

v st

= = =

lim & &&

0

vdsdt

st

st

= =

lim &

0

Eliminating Eliminating dtdt from the two equations above, from the two equations above, we getwe get

a a dsds = v = v dvdv

This equation is useful when we donThis equation is useful when we dont have t have tt


If the position If the position ss is known as a function of time, is known as a function of time, then we can find both velocity then we can find both velocity vv and acceleration and acceleration aa as a function of time!as a function of time!

a dvdt

d sdt

= =22

v dsdt=

s s t= ( )

v

t

s

t

a

t

Recap Recap


ExampleExampleConsider a particle whose position Consider a particle whose position xx is given as a function is given as a function of time as shown below. Motion starts at point of time as shown below. Motion starts at point AAIn terms of time In terms of time tt answer the following:answer the following:

When is its When is its positionposition negative?negative?

When is its velocityWhen is its velocitynegative?negative?zero?zero?positive?positive?

Starting at Starting at tt = 0, when is its = 0, when is its displacementdisplacementnegative?negative?

Starting at Starting at tt = 0, when is its = 0, when is its average average velocityvelocity zero?zero?

2 4t< t = 3 st = 3 s

1) 1) Integrate accelerationIntegrate acceleration to determine the velocity.to determine the velocity.

a = a = dvdv / / dtdt => => dvdv = a = a dtdt

=> v => v vvoo = = --3t3t22 => v = => v = --3t3t22 + + vvoo

==tt

oo

vv

vv

dtdtttdvdvoo

))66((=>=>


EXAMPLEEXAMPLE (continued)(continued)

Solution:Solution:

3)3) Now Now calculate the distancecalculate the distance traveled in traveled in 3s3s by integrating the by integrating the velocity using velocity using ssoo = 0= 0::

v = v = dsds / / dtdt => => dsds = v = v dtdt

=> =>

=> s => s ssoo = = --tt33 + + vvoott

=> s => s 0 = 0 = -- (3)(3)33 + (27)(3) => + (27)(3) => s = 54 ms = 54 m

++==tt

oooo

ss

ss

dtdtvvttdsdsoo

))33(( 22


GravityGravityGravity (near the earthGravity (near the earths surface) is a nice example of s surface) is a nice example of constant acceleration.constant acceleration.In this case, acceleration is caused by the In this case, acceleration is caused by the forceforce of gravity.of gravity.

Usually pick Usually pick yy--axis axis upwardupwardAcceleration due to gravity is Acceleration due to gravity is downdown::

vt

y

t

at

g

ya g=

210 0 2yy y v t gt= + 0y yv v gt=

y

ya g=


Gravity factsGravity facts

gg does not depend on the nature (e.g., mass) of the does not depend on the nature (e.g., mass) of the material!material! Galileo (1564Galileo (1564--1642) figured this out without fancy 1642) figured this out without fancy

clocks & rulers!clocks & rulers!

Nominally, Nominally, gg = 9.81 m/s= 9.81 m/s22

At the equatorAt the equator gg = 9.78 m/s= 9.78 m/s22

At the North poleAt the North pole gg = 9.83 m/s= 9.83 m/s22

Hence difference in Hence difference in weightsweights !!


ExampleExample1D free fall1D free fall

Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of heightHH.. Both throw a ball with initial speedBoth throw a ball with initial speed vv00, Alice , Alice straightstraight downdown and Bill straightand Bill straight upup. The speed of the . The speed of the balls when they hit the ground areballs when they hit the ground are vvAA andand vvBBrespectively.respectively. Which of the following is true:Which of the following is true:

(a)(a) vvAA < < vvBB (b) (b) vvAA = = vvBB (c) (c) vvAA > > vvBB

vv00vv00

BillBillAliceAlice

HHvvAA vvBB


For Bill, the ballFor Bill, the balls motion up and back down to him is symmetric. s motion up and back down to him is symmetric. Thus, your intuition tells you that Thus, your intuition tells you that v = v = vv00

We can prove that your intuition is correct:We can prove that your intuition is correct:

This looks just like Bill threw the This looks just like Bill threw the ball down with ball down with speed speed vv00, so, sothe speed at the bottom shouldthe speed at the bottom shouldbe the same as Alicebe the same as Alices ball.s ball.

vv00BillBill

HH

vv == vv00

y = 0y = 0

( )( )2 2 20 02v v g H H v= + =Equation:Equation: ( )2 20 02v v a y y= +

0v v=

ExampleExample1D free fall1D free fall


We can also just use the equation directly:We can also just use the equation directly:

same equations !!same equations !!

Alice:Alice: ( ) ( )( )22 0 2 0v v g H= + Bill:Bill: ( ) ( )( )22 0 2 0v v g H= + +

vv00BillBill

HH

vv == vv00

y = 0y = 0

ExampleExample

1D free fall1D free fall


ExampleExample

The pilot of a hovering The pilot of a hovering helicopter drops a lead brick helicopter drops a lead brick from a height of from a height of 1000 m1000 m..

How long does it take to How long does it take to reach the ground and reach the ground and How fast is it moving when How fast is it moving when it gets there? (neglect air it gets there? (neglect air resistance)resistance)

1000 m


First choose coordinate system.First choose coordinate system. Origin and Origin and ++yy directiondirection

20y0 gt2

1tvyy +=

20 gt2

1yy =

1000 m

y = 0y = 0

yy

Next write down position equation:Next write down position equation:

Realize that Realize that vv0y0y = 0= 0 ::

ExampleExample


yy00 ==

y

y = 0

1000 m

Solve for time Solve for time tt (= (= ttff) when ) when y = 0y = 0given that given that yy00 = 1000 m.= 1000 m.

For final velocity:For final velocity:

Solve for Solve for vvyy (= (= vvfyfy) when ) when yy = 0:= 0:

..

02

2 2 10009

14 3 s81f

y mtg m s

= = =

02 140 m/sfyv gy= =

( )2 20 02y yv v g y y=

20 gt2

1yy =

0y yv v gt= 140 m/sfy fv gt = =

OROR

ExampleExample


ENGR 3423 Dynamics ENGR 3423 Dynamics sharansharan 8181 ENGR 3423 Dynamics ENGR 3423 Dynamics sharansharan 8282

Acceleration is given, hence:Acceleration is given, hence:

a = a = dv/dtdv/dt

(1)(1)

(2)(2) Motion stops when velocity is zero, Motion stops when velocity is zero, hence:hence:


Velocity is known, hence:Velocity is known, hence:(3)(3)

v = v = ds/dtds/dt (4)(4) Now, determine the positions of Now, determine the positions of the particlesthe particles


(5)(5) Now, determine the positions of Now, determine the positions of the particle Bthe particle B

(6)(6) Determine the Distances traveledDetermine the Distances traveled


TipsTipsRead the problem!Read the problem!

Before you start work on a problem, read the problem Before you start work on a problem, read the problem statement thoroughly. Make sure you understand what statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning information is given, what is asked for, and the meaning of all the terms used in stating the problem.of all the terms used in stating the problem.

Watch your units !Watch your units !Always check the units of your answer and carry the Always check the units of your answer and carry the units along with your numbers during the calculation.units along with your numbers during the calculation.

Understand the limits !Understand the limits !Many equations we use are special cases of more Many equations we use are special cases of more general laws. Understanding how they are derived will general laws. Understanding how they are derived will help you recognize their limitations (for example, help you recognize their limitations (for example, constant acceleration).constant acceleration).


-- Useful for analyzing motion that changes Useful for analyzing motion that changes erraticallyerratically

-- Done using slopes and areas under curves of Done using slopes and areas under curves of motion variables (motion variables (s; v; a; ts; v; a; t))

-- There are two types of curvesThere are two types of curves

-- time dependent curvestime dependent curves

-- position dependent curvesposition dependent curves

12.312.3 Graphical analysisGraphical analysis


12.3 pg. 1712.3 pg. 17

Rectilinear KinematicsRectilinear Kinematics Erratic MotionErratic Motion

When motion is erratic (not a When motion is erratic (not a straight line essentially) straight line essentially) graphical methods are more graphical methods are more convenientconvenient

ss--t and vt and v--t graphs t graphs

-- slopeslope, , ds/dtds/dt = = velocity, velocity, vv


vv--t and at and a--t graphs t graphs

Time dependent curvesTime dependent curves

-- slopeslope, , dv/dtdv/dt = = acceleration, acceleration, aa


Example 12Example 12--6 pg. 206 pg. 20

Q:Q:

-- Draw Draw vv--tt and and aa--tt graphsgraphs


vv--tt graphgraph

aa--tt graphgraph


aa--t to vt to v--t graphs t graphs


area under the curve = velocity area under the curve = velocity


vv--t to st to s--t graphs t graphs

area under the curve = area under the curve = displacementdisplacement



Q:Q:

-- Draw Draw vv--tt and and ss--tt graphsgraphs

-- Determine time, tDetermine time, t, needed to stop the car, needed to stop the car

-- How far has the car traveledHow far has the car traveled

Example 12Example 12--7 pg. 227 pg. 22


Example 12Example 12--7 pg. 22 contd.7 pg. 22 contd.


aa--s to vs to v--s graphs s graphs

Position dependent curvesPosition dependent curves

area under the curve = area under the curve =


vv--s to as to a--s graphs s graphs

Slope: Slope:

Position dependent curvesPosition dependent curves


(1)(1) (2)(2)


(1)(1)

(2)(2)


Vectors (review)Vectors (review)

ScalarsScalars (magnitude only):(magnitude only):time, speed, mass, energy, charge, temperature, ...time, speed, mass, energy, charge, temperature, ...

Vector Vector (magnitude and direction):(magnitude and direction):position, displacement, velocity, acceleration, position, displacement, velocity, acceleration, momentum, force, ...momentum, force, ...


VectorsVectors

In In 11 dimension, we can specify direction with a dimension, we can specify direction with a ++ or or -- signsignIn In 22 or or 33 dimensions, we need more than a sign to specify the dimensions, we need more than a sign to specify the direction of somethingdirection of somethingTo illustrate this, consider the To illustrate this, consider the position vectorposition vector,, rr,, in in 22 dimensions.dimensions.

Example: Example: Where is Duluth, Ga.?Where is Duluth, Ga.? N

S

EWChoose coordinates of distance Choose coordinates of distance

(miles)(miles), and direction , and direction (N,S,E,W)(N,S,E,W)

In this case In this case rr is a vector that is a vector that points points 20 miles20 miles northnorth..

Atlanta

Duluth

r r Choose origin at Choose origin at AtlantaAtlanta


VectorsVectors

An arrow is commonly used to represent a vector An arrow is commonly used to represent a vector quantity; moreover, there are two common ways quantity; moreover, there are two common ways to symbolize a vector quantity:to symbolize a vector quantity:

Boldface notation: Boldface notation: AA

rAArrowArrow notation:notation:

AA A=r


Vectors and their componentsVectors and their components

The The componentscomponents of of are its are its (x,y,z)(x,y,z) coordinates:coordinates:rr

Consider this in 2Consider this in 2--D (since itD (since its easier to draw):s easier to draw):

( ) ( ), , , ,x y zr r r r x y z= =r

yy

xx

(x,y)(x,y)

rr

( )arctan y x =

( ) ( ), ,x yr r r x y= =r

cossin

x

y

r x rr y r

= == =

where r r=r

2 2r r x y= = +r

The The magnitudemagnitude (length) of (length) of is found using the Pythagorean theorem:is found using the Pythagorean theorem:rr

The length of a vector clearly The length of a vector clearly does does notnot depend on its depend on its direction.direction.


Unit VectorsUnit VectorsA A Unit Vector Unit Vector is a vector having unit length and is a vector having unit length and no unitsno unitsIt is used to specify a It is used to specify a directiondirection

Unit vector Unit vector points in the direction of points in the direction of u ur

In Cartesian coordinates, unit vectors In Cartesian coordinates, unit vectors in +in +xx, +, +yy, & +, & +zz directions are often directions are often given the symbols respectively given the symbols respectively , ,i j k

x

y

z

i j

k

u

ur

u uuu u

= =r r

r

( ) , ,x y z x y zA A A A A i A j A k= = + +r

Ar

xA

yA

zA

A vector in terms of unit vectors isA vector in terms of unit vectors isAr


Vector additionVector addition

We can arrange the vectors as we want, as long as we maintain their length and direction!

Consider the vectors and . Find Consider the vectors and . Find Ar

Br

A B+r r

Ar

Br

x

y

z

i j

k

Ar

xA

yA

zA yA j

xA i

zA k

( ) x y zA i A j A k= + +

zA A k= +r

A=r

Ar

Ar

Br

C A B= +r r r

Ar

Br

head to tailmethod

Ar

Br

Ar

Br

Ar

Br

C A B= +r r r

parallelogram method

x y zA i A j A k+ +ConsiderConsider

x y zA A i A j A k= + +

r


Vector addition using componentsVector addition using components

Bx

Ax

By

Ay

( ) ( ) x y x yC A i A j B i B j= + + +r

( ) ( ) (a ) x x y yA B i A B j= + + + (b) x yi jC C= +

(in 2D)C A B= +r r r

Consider

Comparing components in (a) and (b):

xx x

y yy

CC

A BA B

= += +

Cr

Br

Ar yC

xC


ExampleExample

(a)(a) (3,5,(3,5,--1)1) (b)(b) (4,(4,--2,5)2,5) (c)(c) (5,(5,--2,4)2,4)

Vector Vector = (0,2,1)= (0,2,1)Vector = (3,0,2)Vector = (3,0,2)Vector Vector = (1,= (1,--4,2)4,2)C

rBrAr

D A B C= + +rr r r

What is the resultant vector ?


ExampleExample(formal solution)(formal solution)

( ) ( ) ( ) x y x z xz yy zB B CA AD i j k i j k CA B Ci j k= + + + + + + + +r

( ) ( ) ( ) x yx yx y zzzCA A Ai j CB B kCB= + + + + + + + +( ) ( ) ( ) 0 3 1 2 0 4 1 2 2i j k= + + + + + + +

4 2 5i j k= +

( ), ,4 2 5=


ExampleExample

A golfer takes three putts to get the ball into the A golfer takes three putts to get the ball into the hole. The first putt displaces the ball 3.0m hole. The first putt displaces the ball 3.0m North, the second 2.0m Southeast, and the third North, the second 2.0m Southeast, and the third 1.0m Southwest.1.0m Southwest.

What displacement would be needed to get the What displacement would be needed to get the ball into the hole in the first putt ?ball into the hole in the first putt ?


ExampleExample


ExampleExample


ExampleExample


ExampleExample


ExampleExample


ExampleExample


Magnitude:Magnitude: ( ) ( ). . .2 20 7 0 9 1 1 mA B C+ + = + =rr r

Direction:Direction: .tan.

1 0 9 52 degrees north of east0 7

=

ExampleExample


Curvilinear Motion occurs when a particle moves along Curvilinear Motion occurs when a particle moves along a curved patha curved path

Path usually described in 3Path usually described in 3--D: D: KinematicKinematic quantities quantities change by both change by both directiondirection and and magnitudemagnitude

THREE types of coordinate systems will be introduced THREE types of coordinate systems will be introduced to describe the motion:to describe the motion:

Rectangular; NormalRectangular; Normal--Tangential; CylindricalTangential; Cylindrical

We use vector analyses to formulate the particleWe use vector analyses to formulate the particles s positionposition, , velocityvelocity and and accelerationacceleration

Motion along a curved path defined by a path Motion along a curved path defined by a path function function ss

12.4 General Curvilinear Particle Kinematics12.4 General Curvilinear Particle Kinematics


Position >>Position >>

Velocity >>

Average VelocityAverage Velocity is given byis given by

= = rr / / tt

Instantaneous VelocityInstantaneous Velocity (or just velocity) (or just velocity) is obtained by is obtained by


Velocity >>Velocity >>



Acceleration >>Acceleration >>


vv is not tangent to the path, is not tangent to the path, therefore therefore aa is also is also not tangent to the not tangent to the pathpath

aa is tangent to the is tangent to the hodographhodograph..

aa = = ddvv//dtdt

vv = = ddrr//dtdt

aa = d= d22rr//dtdt22

&&



Frames of reference for vector analysis Frames of reference for vector analysis

Rectangular Rectangular cartesiancartesian ((x x -- y y -- zz) frame) frame

Normal and tangential (Normal and tangential (n n -- t t -- zz) frame) frame

Cylindrical polar (Cylindrical polar (r r -- -- zz) frame) frame

Acceleration, Velocity & Acceleration, Velocity & Displacement Displacement


12.5 Curvilinear Motion 12.5 Curvilinear Motion -- Rectangular CoordinatesRectangular Coordinates

-- Motion is described using space variables of Motion is described using space variables of fixed directions (fixed directions (xx--yy--zz) with unit vectors ) with unit vectors ii, , jj and and kk

-- Origin is fixedOrigin is fixed

-- DDirections of axes are fixedirections of axes are fixed


Cartesian CoordinatesCartesian Coordinates

Position >> Position >>



12.6 Projectile Motion12.6 Projectile Motion Horizontal motion with influence of vertical Horizontal motion with influence of vertical

gravitygravity

Is the most direct application of the Is the most direct application of the cartersiancartersiancoordinates in analysis of curvilinear coordinates in analysis of curvilinear motionmotion


Projectile MotionProjectile Motion

g = 9.81 m/sg = 9.81 m/s22 or 32.2 ft/sor 32.2 ft/s22



(Ref. (Ref. EqnsEqns. 12. 12--4 to 124 to 12--6 on page 8)6 on page 8)

Derived from the Derived from the two two eqnseqns. above. above

Horizontal component of Velocity is always constantHorizontal component of Velocity is always constant



Solving problems Solving problems

problems have, at the most, three problems have, at the most, three unknowns unknowns since onlysince onlythree independent equationsthree independent equations

one one eqneqn. in horizontal direction. in horizontal direction two two eqnseqns. in vertical direction. in vertical direction resultant velocity is always tangenresultant velocity is always tangential totial to the paththe path

is the vector sum of is the vector sum of vvxx and and vvyy see Fig. 12see Fig. 12--2020


ExampleExampleProjectile Motion: Basketball Free ThrowsProjectile Motion: Basketball Free Throws

If the basketball is thrown at 55 degrees If the basketball is thrown at 55 degrees above the horizontal, what must its initial above the horizontal, what must its initial speedspeed be for the foul shot to go in?be for the foul shot to go in?


210 2 = = +ox xx x x v t a t

( ) 0= =ox xv t a21

0 2 = = +oy yy y y v t a t

Constant Acceleration ProblemConstant Acceleration Problem

Choose Choose yy axis upaxis upChoose Choose xx axis parallel to the floor in the direction of the basketaxis parallel to the floor in the direction of the basketChoose the origin Choose the origin (0,0)(0,0) to be at the ballto be at the ball

Example Example -- 11

y

x(0,0)

( )212 = = oy yv t gt a g


( )0cos5514 ft ov t=

00 0 sin 55=yv v055

0rv

x

y

= oxx v t

212 = oyy v t gt

00 0 cos55=xv v

Find the x and y Find the x and y components of components of initial velocity:initial velocity:

y

x(0,0)

( ) ( )20 2ft12 ssin 554 ft 32o t tv=



014 cos55= ov t0 24 sin 55 16= ov t t

2 equations, 2 unknowns ( ),ov t

( )014 / cos55 24.4 /= =o ot v v

0 24.4 ft/s=v

( ) ( )20 24.4 /4 24.4s /in 55 16= oo ovv v( )7.44 or 16.5 mphms




Barry Bonds clobbers a fastball toward centerBarry Bonds clobbers a fastball toward center--field. The ball is field. The ball is hit hit 1m (1m (d d )) above the plate, and its initial velocity is above the plate, and its initial velocity is 36.5m/s36.5m/s ((vv00 ))at an angle of at an angle of 3030oo (( )) above horizontal. The centerabove horizontal. The center--field wall is field wall is 113m113m ((DD)) from the plate and is from the plate and is 3m3m ((hh)) high.high.

How much time does it take for the ball reach the fence?How much time does it take for the ball reach the fence?Does Barry get a home run?Does Barry get a home run?

vv00h

D

d


y oyv v gt= = + 210 2yy d v t gt

x oxv v=

oxx v t=

Kinematics equations are:

vv00h

D

d

y

x(0,0)

Choose Choose yy axis vertically upaxis vertically upChoose Choose xx axis along the ground in the direction of the hitaxis along the ground in the direction of the hitChoose the origin Choose the origin (0,0)(0,0) to be at the home plateto be at the home plateThe ball is hit at The ball is hit at t = 0t = 0, , x = xx = x00 = 0= 0 and and y = yy = y00 = d= d..



Use geometry to figure out Use geometry to figure out vv0x0x and and vv0y0y ::

sin0yv v =r

cos0xv v =r

y

x

g

v0x

v0yy0

vr



The time to reach the wall is: The time to reach the wall is: t = D / vt = D / v0x0x (easy!)(easy!)We have an equation that tells us We have an equation that tells us y(t) = d + vy(t) = d + v0y 0y t t -- g tg t22/ 2/ 2So, weSo, were done....now we just plug in the numbers:re done....now we just plug in the numbers:

Find:Find:vv0x0x = 36.5 cos(30= 36.5 cos(30oo) m/s = 31.6 m/s) m/s = 31.6 m/svv0y0y = 36.5 sin(30= 36.5 sin(30oo) m/s = 18.25 ) m/s = 18.25 m/sm/s

tt = (113 m) / (31.6 m/s) = 3.58 s= (113 m) / (31.6 m/s) = 3.58 s

y(t)y(t) = (1.0 m) + (18.25 m/s)(3.58 s) = (1.0 m) + (18.25 m/s)(3.58 s) -- (0.5)(9.8 m/s(0.5)(9.8 m/s22)(3.58 s))(3.58 s)22

= (1.0 + 65.3 = (1.0 + 65.3 -- 62.8) m = 62.8) m = 3.5 m3.5 m

Since the wall is Since the wall is 3.0 m3.0 m high, Barry gets the homer !!high, Barry gets the homer !!



12.7 Curvilinear Kinematics 12.7 Curvilinear Kinematics Normal & Tangential CoordinatesNormal & Tangential Coordinates Motion is described using path variablesMotion is described using path variables

Origin is fixed to and moves with the particleOrigin is fixed to and moves with the particle

The The first axisfirst axis (tangential axis) points (tangential axis) points tangentialtangential to the path in the to the path in the direction of motion. Unit vector in this direction is denoted bydirection of motion. Unit vector in this direction is denoted by uutt The The second axissecond axis (normal axis) points (normal axis) points normalnormal to the path towards to the path towards the center of curvature the center of curvature OO of the path. Unit vector in this direction of the path. Unit vector in this direction is denoted is denoted uun n The third axis is perpendicular to the plane of the path and is known as the binomal axis with unit vector uubb


Each segment is Each segment is dsds formed formed from the arc of an associated from the arc of an associated circle, having a radius of circle, having a radius of curvature curvature and center of and center of curvature Ocurvature O

Positive direction is always on Positive direction is always on the concave side of the curvethe concave side of the curve

The plane containing the nThe plane containing the n--t t axes is called the axes is called the osculatingosculatingplane plane




Remember, If Remember, If uutt is constant (not changing), then is constant (not changing), then dduutt/dt/dt = 0 = 0 in the expression abovein the expression above


Page 49 Page 49 HibbelerHibbeler ::

From geometry, arc length From geometry, arc length dsds = = dd

Now, Now,

Magnitude: Magnitude:

dudutt = 1.d= 1.d (radius = unity)(radius = unity)

Direction is: Direction is: uunn

Which leads to:Which leads to:


We now need to simplify We now need to simplify dduutt/dt/dt


Acceleration >>Acceleration >>Which leads to:Which leads to:


To determine the angular velocity, recall that at every segment To determine the angular velocity, recall that at every segment of a curved path is part a circle.of a curved path is part a circle.

Any point along the path can be fitted with a circle of radius Any point along the path can be fitted with a circle of radius and center and center OO::


If the particle moves along a straight line, then radius If the particle moves along a straight line, then radius infinity, from infinity, from EqEq. 12. 12--20 20 aann = 0. Thus a = a= 0. Thus a = att = = dv/dtdv/dt

Thus, Tangential component of acceleration represents Thus, Tangential component of acceleration represents time rate of change of the velocitytime rate of change of the velocity

If it moves with a constant speed, then If it moves with a constant speed, then dv/dtdv/dt = 0 and = 0 and a = aa = ann = v= v2

2//

Thus, Normal component acts towards the center Thus, Normal component acts towards the center called called Centripetal accelerationCentripetal acceleration


ThreeThree--Dimensional MotionDimensional Motion

uubb = = uutt x x uunn

Use this relation to establish the Use this relation to establish the direction of the axisdirection of the axis

uunn is always pointed inwards or is always pointed inwards or (concave side of the motion)(concave side of the motion)

uutt & & uunn are always perpendicular are always perpendicular to each other and lie in the to each other and lie in the osculating plane of motionosculating plane of motion


aatt

aann


aatt

aann

Magnitude of accelerationMagnitude of acceleration


aann

aatt


In In tt = 2s := 2s :


12-125 Continued.


12.8 Curvilinear Kinematics 12.8 Curvilinear Kinematics Cylindrical (Polar) CoordinatesCylindrical (Polar) Coordinates

Motion represented in terms of Motion represented in terms of rr, , and and zz cylcyl. co. co--odsods..


dd

dd

dduurr = = dduu


dd

dd

dduu = = -- dduurr


Position and Velocity >>Position and Velocity >>

((EqnsEqns. 12. 12--24/25)24/25)


Position and Velocity >>Position and Velocity >>

The term The term dd/dt/dt is sometimes is sometimes called the called the angular velocityangular velocity as it as it indicates the time rate of change of indicates the time rate of change of the angle the angle . Units . Units radrad/sec./sec.



Time derivative of the Velocity Time derivative of the Velocity vector:vector:



We can write it in this form:We can write it in this form:


Cylindrical Coordinates:Cylindrical Coordinates: If motion is as shown, then the zIf motion is as shown, then the z--axis must also be axis must also be included, it is specified by the included, it is specified by the cylindrical coordinatescylindrical coordinates

((EqnsEqns. 12. 12--31/32)31/32)


APPLICATIONSAPPLICATIONS CCylindrical coordinatesylindrical coordinates are are commonly used in cases commonly used in cases where the particle moves where the particle moves along a helical curve.along a helical curve.

z z const: const: Polar coordinatesPolar coordinates

Typical Problem:Typical Problem:The boy slides down the The boy slides down the slide at a constant speed of 2 slide at a constant speed of 2 m/s. How fast is his m/s. How fast is his elevation from the ground elevation from the ground changing (i.e., what is changing (i.e., what is z z )?)?

..

CURVILINEAR MOTION: CYLINDRICAL/POLAR CURVILINEAR MOTION: CYLINDRICAL/POLAR COMPONENTS (Section 12.8)COMPONENTS (Section 12.8)


POSITION (POLAR COORDINATES)POSITION (POLAR COORDINATES)

The The locationlocation of P in polar coordinates is of P in polar coordinates is rr = = rruurr..The radial direction, The radial direction, rr, extends outward from the fixed origin, , extends outward from the fixed origin, OO, , and the transverse coordinate, and the transverse coordinate, ,, is measured counteris measured counter--clockwise clockwise from the horizontal.from the horizontal.


VELOCITY (POLAR COORDINATES)VELOCITY (POLAR COORDINATES)

The instantaneous velocity is defined as:The instantaneous velocity is defined as:vv = d= drr/dt = /dt = d(rd(ruurr)/dt)/dtvv = = rruurr + r+ r

dduurrdtdt

..

Using the chain rule:Using the chain rule:dduurr/dt/dt = (= (dduurr/d/d)(d)(d/dt/dt))

dduurr/d/d = = uu so so dduurr/dt/dt = (= (dd//dtdt)) uuTherefore: Therefore: vv = = rruurr + + rruu

.. ..

....

.. ..

Thus, the velocity vector has two components: r, Thus, the velocity vector has two components: r, called the called the radial componentradial component, and , and rr,, called the called the transverse componenttransverse component. The speed of the particle at . The speed of the particle at any given instant is the square root of the sum of any given instant is the square root of the sum of the squares of both components orthe squares of both components or

vv == ((r r ) )22 + ( + ( r )r )22


ACCELERATION (POLAR COORDINATES)ACCELERATION (POLAR COORDINATES)

But: But: dduu/dt/dt= = -- dd/dt/dt uurr..So the acceleration can be expressed asSo the acceleration can be expressed as

aa = (r = (r rr22))uurr + (+ (rr + 2r+ 2r))uu.... .. .... .. ..

..

The magnitude of acceleration is a = (r The magnitude of acceleration is a = (r rr22))22 + (+ (rr + 2r+ 2r))22

The term (r The term (r rr22)) is the is the radial accelerationradial accelerationor or aarr..

The term (The term (rr + 2r+ 2r) is the ) is the transverse transverse acceleration or acceleration or aa

....

.... .. ..

.. .. ...... ....

The instantaneous acceleration is defined as:The instantaneous acceleration is defined as:

aa = = ddvv/dt/dt = (= (d/dt)(d/dt)(rruurr + + rruu)).. ..


CYLINDRICAL COORDINATESCYLINDRICAL COORDINATES

If the particle P moves along a space If the particle P moves along a space curve, its position can be written as curve, its position can be written as

rrPP = = rruurr + + zzuuzz

Taking time derivatives and using Taking time derivatives and using the chain rule (the chain rule (uuzz constant):constant):

Velocity:Velocity: vvPP = = rruurr + + rruu + z + z uuzz

Acceleration:Acceleration: aaPP = (r = (r rr22))uurr + (+ (rr + 2r+ 2r))uu + z + z uuzz.... .. .... .. ..

.. .. ..

....



.

..

.

..

11 22

33

44


1122

33

44


EXAMPLE (12EXAMPLE (12--164)164)

Given:Given: r = 5 cos(2r = 5 cos(2) [m]) [m] = 3t= 3t22 [[rad/srad/s]]oo = 0= 0

Find:Find: Velocity and acceleration at Velocity and acceleration at = 30= 30..Plan:Plan: Apply chain rule to determine r and r Apply chain rule to determine r and r

and evaluate at and evaluate at = 30= 30..

......

..

00

tttt

00oo ==

Solution:Solution: = = dtdt = 3t= 3t22 dtdt = t= t33

At At = 30= 30, , = = t= = t33.. Therefore: t = 0.806 s.Therefore: t = 0.806 s.

= 3t= 3t22 = 3(0.806)= 3(0.806)22 = 1.95 = 1.95 rad/srad/s

tt

66

..

..


EXAMPLE (continued)EXAMPLE (continued)

Substitute in the equation for velocitySubstitute in the equation for velocityvv = = rruurr + + rruuv v = = --16.8816.88uurr + 2.5(1.95)+ 2.5(1.95)uu

vv = (16.88)= (16.88)22 + (4.87)+ (4.87)22 = 17.57 m/s= 17.57 m/s

.. ..

= 6t = 6(0.806) = 4.836 rad/s= 6t = 6(0.806) = 4.836 rad/s22

rr = 5 cos(2= 5 cos(2) = 5 cos(60) = 2.5m) = 5 cos(60) = 2.5m

rr = = --10 sin(210 sin(2)) = = --10 sin(60)(1.95) = 10 sin(60)(1.95) = --16.88 m/s16.88 m/s

rr = = --20 cos(220 cos(2))22 10 sin(210 sin(2))

= = --20 cos(60)(1.95)20 cos(60)(1.95)22 10 sin(60)(4.836) = 10 sin(60)(4.836) = --80 m/s80 m/s22

....

..

.... ....

..

..



Substitute in the equation for acceleration:Substitute in the equation for acceleration:

aa = (r = (r rr22))uurr + (+ (rr + 2r+ 2r))uu

aa = [= [--80 80 2.5(1.95)2.5(1.95)22]]uurr + [2.5(4.836) + 2(+ [2.5(4.836) + 2(--16.88)(1.95)]16.88)(1.95)]uu

aa = = --89.589.5uurr 53.753.7uu m/sm/s22

a = (89.5)a = (89.5)22 + (53.7)+ (53.7)22 = 104.4 m/s= 104.4 m/s22

.... .. .... .. ..


12.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF 12.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLESTWO PARTICLES

APPLICATIONSAPPLICATIONSThe cable and pulley system shown The cable and pulley system shown here can be used to modify the speed here can be used to modify the speed of block B relative to the speed of the of block B relative to the speed of the motor. It is important to relate the motor. It is important to relate the various motions in order to determine various motions in order to determine the power requirements for the motor the power requirements for the motor and the tension in the cable.and the tension in the cable.

Typical Problem:Typical Problem:If the speed of the cable coming onto If the speed of the cable coming onto the motor pulley is known, how can the motor pulley is known, how can we determine the speed of block B?we determine the speed of block B?



Rope and pulley arrangements Rope and pulley arrangements are often used to assist in lifting are often used to assist in lifting heavy objects. The total lifting heavy objects. The total lifting force required from the truck force required from the truck depends on the acceleration of depends on the acceleration of the cabinet.the cabinet.

Typical Problem:Typical Problem:How can we determine the How can we determine the acceleration and velocity of acceleration and velocity of the cabinet if the acceleration the cabinet if the acceleration of the truck is known?of the truck is known?


DEPENDENT MOTIONDEPENDENT MOTION

In many kinematics problems, the motion of one object will In many kinematics problems, the motion of one object will dependdepend on the motion of another object.on the motion of another object.

The blocks in this figure are The blocks in this figure are connected by an connected by an inextensible cordinextensible cordwrapped around a pulley. If block wrapped around a pulley. If block A moves downward along the A moves downward along the inclined plane, block B will move inclined plane, block B will move up the other incline.up the other incline.

The motion of each block can be related mathematically by The motion of each block can be related mathematically by defining defining position coordinatesposition coordinates, , ssAA and and ssBB. Each coordinate axis is . Each coordinate axis is defined from a defined from a fixed point or datum linefixed point or datum line, measured , measured positive positive along along each plane in the each plane in the direction of motiondirection of motion of each block.of each block.


DEPENDENT MOTIONDEPENDENT MOTION (continued)(continued)

Position coordinates Position coordinates ssAA and and ssBBcan be defined from fixed can be defined from fixed datum lines extending from the datum lines extending from the center of the pulley along each center of the pulley along each incline to blocks A and B.incline to blocks A and B.

If the If the cord has a fixed lengthcord has a fixed length, the position coordinates , the position coordinates ssAA and and ssBB are are relatedrelated by the equationby the equation

ssAA + + llCDCD + + ssBB = = llTT

Here Here llTT is the is the total cord lengthtotal cord length and and llCDCD is the is the length of cordlength of cordpassing over arc CD on the pulleypassing over arc CD on the pulley..


DEPENDENT MOTIONDEPENDENT MOTION (continued)(continued)

The negative sign indicates that as A The negative sign indicates that as A moves downmoves down the incline the incline (positive (positive ssAA direction), B direction), B moves upmoves up the incline (negative the incline (negative ssBBdirection).direction).AccelerationsAccelerations can be found by can be found by differentiating differentiating the velocity the velocity expression. expression. IsIs aaBB = = --aaAA ??

The The velocitiesvelocities of blocks A and B of blocks A and B can be related by can be related by differentiatingdifferentiatingthe position equation. Note that the position equation. Note that llCDCD and and llTT remain constantremain constant, so , so dldlCDCD/dt/dt = = dldlTT/dt/dt = 0= 0

dsdsAA/dt/dt + + dsdsBB/dt/dt = 0 => = 0 => vvBB = = --vvAA


DEPENDENT MOTION DEPENDENT MOTION AN EXAMPLEAN EXAMPLE

Here, position coordinates (Here, position coordinates (ssAA and and ssBB) are defined from fixed datum ) are defined from fixed datum lines measured along the direction lines measured along the direction of motion of each block.of motion of each block.

Note that Note that ssBB is only defined to the is only defined to the center of the pulley above block center of the pulley above block B, since this block moves with the B, since this block moves with the pulley. Also, pulley. Also, hh is a constantis a constant..

The The red coloredred colored segments of the cord segments of the cord remain constantremain constant in length in length during motion of the blocks.during motion of the blocks.


DEPENDENT MOTION EXAMPLEDEPENDENT MOTION EXAMPLE (continued)(continued)

The position coordinates are related The position coordinates are related by the equationby the equation

2s2sBB + h + + h + ssAA = l= lWhere l is the Where l is the total cord length minus total cord length minus the lengths of the red segmentsthe lengths of the red segments..

Since l and h remain constant during Since l and h remain constant during the motion, the velocities and the motion, the velocities and accelerations can be related by two accelerations can be related by two successive time derivatives:successive time derivatives:

2v2vBB = = --vvAA and and 2a2aBB = = --aaAAWhen block B moves downward (+When block B moves downward (+ssBB), block A moves to the ), block A moves to the left (left (--ssAA). Remember to be ). Remember to be consistent with the sign conventionconsistent with the sign convention!!


DEPENDENT MOTION EXAMPLEDEPENDENT MOTION EXAMPLE (continued)(continued)

This example can also be worked This example can also be worked by defining the position coordinate by defining the position coordinate for B (for B (ssBB) from the bottom pulley ) from the bottom pulley instead of the top pulley. instead of the top pulley.

The position, velocity, and The position, velocity, and acceleration relations then becomeacceleration relations then become

2(h 2(h ssBB) + h + ) + h + ssAA = l= l

and 2vand 2vBB = = vvAA 2a2aBB = = aaAA

Prove to yourself that the Prove to yourself that the results are the sameresults are the same, even if the sign , even if the sign conventions are different than the previous formulation.conventions are different than the previous formulation.


EXAMPLE PROBLEMEXAMPLE PROBLEM

Given:Given: In the figure on the In the figure on the left, the cord at A is left, the cord at A is pulled down with a pulled down with a speed of 8 ft/s.speed of 8 ft/s.

Find:Find: The speed of block B.The speed of block B.

Plan:Plan: There are There are two cordstwo cords involved in the motion in this involved in the motion in this example. The position of a point on one cord must be example. The position of a point on one cord must be related to the position of a point on the other cord. There related to the position of a point on the other cord. There will be will be two position equationstwo position equations (one for each cord).(one for each cord).



1)1) Define the position coordinatesDefine the position coordinates from a fixed datum line. Three from a fixed datum line. Three coordinates must be defined: one for point A (coordinates must be defined: one for point A (ssAA), one for block B ), one for block B ((ssBB), and one relating positions on the two cords. Note that pulle), and one relating positions on the two cords. Note that pulley y C relates the motion of the two cords.C relates the motion of the two cords.

Solution:Solution:

Define the datum line through the top Define the datum line through the top pulley (which has a fixed position).pulley (which has a fixed position).

ssAA can be defined to the center of the can be defined to the center of the pulley above point A.pulley above point A.

ssBB can be defined to the center of the can be defined to the center of the pulley above B.pulley above B.

ssCC is defined to the center of pulley C.is defined to the center of pulley C. All coordinates are defined as All coordinates are defined as

positive down and along the direction positive down and along the direction of motion of each point/object.of motion of each point/object.

ssAA ssCC ssBB

DATUMDATUM



3)3) Eliminating Eliminating ssCC between the two between the two equations, we getequations, we get

2s2sAA + 4s+ 4sBB = l= l11 + 2l+ 2l22

2)2) Write position/length equations Write position/length equations for each cordfor each cord.. Define lDefine l11 as the as the length of the first cord, minus length of the first cord, minus any segments of constant length. any segments of constant length. Define lDefine l22 in a similar manner for in a similar manner for the second cord:the second cord:

4)4) Relate velocities by Relate velocities by differentiatingdifferentiating this expression. Note that lthis expression. Note that l11 and land l22are constant lengths.are constant lengths.

2v2vAA + 4v+ 4vBB = 0 => = 0 => vvBB = = -- 0.5v0.5vAA = = -- 0.5(8) = 0.5(8) = -- 4 ft/s4 ft/sThe velocity of block B is 4 ft/s up (negative The velocity of block B is 4 ft/s up (negative ssBB direction).direction).

ssAAssCC ssBB

DATUMDATUM

Cord 1: 2sCord 1: 2sAA + 2s+ 2sCC = l= l11Cord 2: Cord 2: ssBB + (+ (ssBB ssCC) = l) = l22


APPLICATIONSAPPLICATIONS

Typical Problems:Typical Problems:If the aircraft carrier travels at a forward velocity of 50 km/hIf the aircraft carrier travels at a forward velocity of 50 km/hr and r and plane A takes off at a horizontal air speed of 200 km/hr (measurplane A takes off at a horizontal air speed of 200 km/hr (measured ed by someone on the water), how do we find the velocity of the plaby someone on the water), how do we find the velocity of the plane ne relative to the carrier?relative to the carrier?

How would you find the same thing for airplane B?How would you find the same thing for airplane B?

How does the wind impact this sort of situation?How does the wind impact this sort of situation?

When fighter jets take off When fighter jets take off or land on an aircraft or land on an aircraft carrier, the velocity of the carrier, the velocity of the carrier becomes an issue.carrier becomes an issue.

12.10 RELATIVE MOTION ANALYSIS12.10 RELATIVE MOTION ANALYSIS


RELATIVE POSITIONRELATIVE POSITION

The The absolute positionabsolute position of two of two particles A and B with respect to particles A and B with respect to the fixed x, y, z reference frame the fixed x, y, z reference frame are given by are given by rrAA and and rrBB.. The The position of B relative to Aposition of B relative to A is is represented by represented by

rrB/AB/A = = rrBB rrAA

Therefore, if Therefore, if rrBB = (10 = (10 ii + 2 + 2 jj ) m) m

andand rrAA = (4 = (4 ii + 5 + 5 jj ) m,) m,

thenthen rrB/AB/A = (6 = (6 ii 3 3 jj ) m.) m.


RELATIVE VELOCITYRELATIVE VELOCITY

To determine the To determine the relative velocityrelative velocity of B of B with respect to A, the time derivative with respect to A, the time derivative of the relative position equation is of the relative position equation is taken. taken.

vvBB/A/A == vvBB vvAAoror

vvBB == vvAA ++ vvBB/A/AIn these equations,In these equations, vvBB and and vvAA are called are called absolute velocitiesabsolute velocitiesand and vvBB/A/A is the is the relative velocityrelative velocity of B with respect to A. of B with respect to A.

Note that Note that vvBB/A/A = = -- vvAA/B/B ..


RELATIVE ACCELERATIONRELATIVE ACCELERATION

The time derivative of the relative The time derivative of the relative velocity equation yields a similar velocity equation yields a similar vector relationship between the vector relationship between the absoluteabsolute and and relative accelerationsrelative accelerations of of particles A and B.particles A and B.

aaB/AB/A = = aaBB aaAAoror

aaBB = = aaAA + + aaB/AB/A


EXAMPLEEXAMPLE

Given:Given: vvAA = 600 km/hr= 600 km/hrvvBB = 700 km/hr= 700 km/hr

Find:Find: vvBB/A/A

Plan:Plan: Write vectors Write vectors vvAA and and vvBB in Cartesian form, then in Cartesian form, then determine determine vvBB vvAA



vvBB/A/A = = vvBB vvAA = (= (-- 11911191..5 5 ii + 344+ 344..1 1 j j )) km/hrkm/hr

hrhrkmkmvv AABB 22..12401240))11..344344(())55..11911191((

2222// ==++==

wherewhere ==== 11..1616))55..1191119111..344344((tantan 11

Solution:Solution:

vvAA = 600 = 600 coscos 35 35 ii 600 sin 35 600 sin 35 jj= (491= (491..5 5 ii 344344..11 jj ) ) km/hrkm/hr

vvBB = = --700 700 i i km/hrkm/hr


EXAMPLEEXAMPLE

Given:Given: vvAA = 10 m/s= 10 m/svvBB = 18.5 m/s= 18.5 m/s

((aat t ))AA = 5 m/s= 5 m/s22aaBB = 2 m/s= 2 m/s22

Find:Find: vvAA/B/BaaAA/B/B

Plan:Plan: Write the velocity and acceleration vectors for A and B Write the velocity and acceleration vectors for A and B and determine and determine vvAA/B/B and and aaAA/B/B by using vector equations.by using vector equations.

Solution:Solution:

The velocity of A is:The velocity of A is:

vvAA = 10 cos(45)= 10 cos(45)ii 10 sin(45)10 sin(45)jj = (7.07= (7.07ii 7.077.07jj) m/s) m/s

yy

xx


The velocity of B is:The velocity of B is:

vvBB = 18.5= 18.5ii (m/s)(m/s)

The relative velocity of A with respect to B is (The relative velocity of A with respect to B is (vvAA/B/B):):

vvAA/B/B = = vvAA vvBB = (7.07= (7.07ii 7.077.07jj) ) (18.5(18.5ii) = ) = --11.4311.43ii 7.077.07jj

or or vvBB/A/A = (11.43)= (11.43)22 + (7.07)+ (7.07)22 = 13.4 m/s= 13.4 m/s

= tan= tan--11(( ) = 31.73) = 31.73 7.077.0711.4311.43

yy

xx

EXAMPLE contd..EXAMPLE contd..


The acceleration of B is:aB = 2i (m/s2)

The acceleration of A is:aA = (at)A + (an)A = [5 cos(45)i 5 sin(45)j]

+ [-( ) sin(45)i ( ) cos(45)j]

aA = 2.83i 4.24j (m/s2)

102100

102100

The relative acceleration of A with respect to B is:aA/B = aA aB = (2.83i 4.24j) (2i) = 0.83i 4.24j

aA/B = (0.83)2 + (4.24)2 = 4.32 m/s2

= tan-1( ) = 78.9

4.240.83

EXAMPLE contd..



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