ch11 cinematica de particulas

7/27/2019 Ch11 Cinematica de Particulas

1/34

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2007 The McGraw-Hill Companies, Inc. All rights reserved.

11Kinematics of Particles


2/34


Vector Mechanics for Engineers: DynamicsEighth

Edition

11 - 2

Contents: KINEMATICS OF PARTICLES

Introduction

Rectilinear Motion: Position,Velocity & Acceleration

Determination of the Motion of a

Particle

Sample Problem 11.2

Sample Problem 11.3

Uniform Rectilinear-Motion

Uniformly Accelerated Rectilinear-

Motion

Motion of Several Particles: RelativeMotion

Sample Problem 11.4

Motion of Several Particles:

Dependent Motion

Sample Problem 11.5

Graphical Solution of Rectilinear-MotionProblems

Other Graphical Methods

Curvilinear Motion: Position, Velocity &

Acceleration

Derivatives of Vector Functions

Rectangular Components of Velocity and

Acceleration

Motion Relative to a Frame in Translation

Tangential and Normal Components

Radial and Transverse Components

Sample Problem 11.10

Sample Problem 11.12


3/34



Edition

11 - 3

Introduction

Dynamics includes:

- Kinematics: study of the geometry of motion. Kinematics is used to

relate displacement, velocity, acceleration, and time without reference to

the cause of motion.

- Kinetics: study of the relations existing between the forces acting on a

body, the mass of the body, and the motion of the body. Kinetics is usedto predict the motion caused by given forces or to determine the forces

required to produce a given motion.

Rectilinearmotion: position, velocity, and acceleration of a particle as it

moves along a straight line.

Curvilinearmotion: position, velocity, and acceleration of a particle as it

moves along a curved line in two or three dimensions.


4/34



Edition

11 - 4

Rectilinear Motion: Position, Velocity & Acceleration

Particle moving along a straight line is said

to be in rectilinear motion.

Position coordinate of a particle is defined by

positive or negative distance of particle from

a fixed origin on the line.

The motion of a particle is known if theposition coordinate for particle is known for

every value of time t. Motion of the particle

may be expressed in the form of a function,

e.g.,32

6 ttx or in the form of a graphx vs. t.

EE


5/34



Edition

11 - 5


Instantaneous velocity may be positive or

negative. Magnitude of velocity is referred

to asparticle speed.

Consider particle which occupies positionP

at time tand Pat t+Dt,

t

xv

t

x

t D

D

D

D

D 0lim

Average velocity

Instantaneous velocity

From the definition of a derivative,

dtdx

txv

t

DD

D 0lim

e.g.,

2

32

312

6

tt

dt

dxv

ttx

EE


6/34



Edition

11 - 6

Rectilinear Motion: Position, Velocity & Acceleration Consider particle with velocity v at time tand

vat t+Dt,

Instantaneous accelerationt

va

t D

D

D 0lim

t

dt

dva

ttv

dt

xddtdv

tva

t

612

312e.g.

lim

2

2

2

0

DD

D

From the definition of a derivative,

Instantaneous acceleration may be:

- positive: increasing positive velocity

or decreasing negative velocity

- negative: decreasing positive velocity

or increasing negative velocity.

EE


7/34 2007 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

11 - 7


Consider particle with motion given by

326 ttx

2312 ttdt

dxv

tdtxd

dtdva 612

2

2

at t= 0, x = 0, v = 0, a = 12 m/s2

at t= 2 s, x = 16 m, v = vmax = 12 m/s, a = 0

at t= 4 s, x =xmax = 32 m, v = 0, a = -12 m/s2

at t= 6 s, x = 0, v = -36 m/s, a = 24 m/s2

EE




Edition

11 - 8

Determination of the Motion of a Particle

Recall, motion of a particle is known if position is known for all time t.

Typically, conditions of motion are specified by the type of acceleration

experienced by the particle. Determination of velocity and position requires

two successive integrations.

Three classes of motion may be defined for:

- acceleration given as a function oftime, a =f(t)- acceleration given as a function ofposition, a = f(x)

- acceleration given as a function ofvelocity, a = f(v)

V M h i f E i D iEE




Edition

11 - 9


Acceleration given as a function oftime, a =f(t):

tttx

x

tttv

v

dttvxtxdttvdxdttvdxtvdt

dx

dttfvtvdttfdvdttfdvtfadt

dv

00

0

00

0

0

0

Acceleration given as a function ofposition, a =f(x):

x

x

x

x

xv

v

dxxfvxvdxxfdvvdxxfdvv

xfdx

dvva

dt

dva

v

dxdt

dt

dxv

000

202

12

21

oror

V t M h i f E i D iEE




Edition

11 - 10


Acceleration given as a function of velocity, a =f(v):

tv

v

tv

v

tx

x

tv

v

ttv

v

vf

dvvxtx

vf

dvvdx

vf

dvvdxvfa

dx

dvv

t

vf

dv

dtvf

dvdt

vf

dvvfa

dt

dv

0

00

0

0

0

0





Edition

11 - 11

Sample Problem 11.2

Determine:

velocity and elevation above ground attime t,

highest elevation reached by ball and

corresponding time, and

time when ball will hit the ground and

corresponding velocity.

Ball tossed with 10 m/s vertical velocity

from window 20 m above ground.

SOLUTION:

Integrate twice to find v(t) andy(t).

Solve fortat which velocity equals

zero (time for maximum elevation)

and evaluate corresponding altitude.

Solve fortat which altitude equals

zero (time for ground impact) and

evaluate corresponding velocity.





Edition

11 - 12

Sample Problem 11.2

tvtvdtdv

adt

dv

ttv

v

81.981.9

sm81.9

0

0

2

0

ttv

2s

m81.9

s

m10

2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdt

dy

tty

y

22

s

m905.4

s

m10m20 ttty

SOLUTION:

Integrate twice to find v(t) andy(t).





Edition

11 - 13

Sample Problem 11.2

Solve fortat which velocity equals zero and evaluate

corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

Solve fortat which altitude equals zero and evaluate

corresponding velocity.

22

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y





Edition

11 - 14

Sample Problem 11.2

Solve fortat which altitude equals zero and

evaluate corresponding velocity.

0s

m905.4

s

m10m20

2

2

ttty

s28.3

smeaningless243.1

t

t

s28.3s

m

81.9s

m

10s28.3

s

m81.9

s

m10

2

2

v

ttv

s

m2.22v

V t M h i f E i D iE

E




Edition

11 - 15

Sample Problem 11.3

Brake mechanism used to reduce gunrecoil consists of piston attached to barrel

moving in fixed cylinder filled with oil.

As barrel recoils with initial velocity v0,

piston moves and oil is forced through

orifices in piston, causing piston andcylinder to decelerate at rate proportional

to their velocity.

Determine v(t),x(t), and v(x).

kva

SOLUTION:

Integrate a = dv/dt= -kv to find v(t).

Integrate v(t) = dx/dt to findx(t).

Integrate a = v dv/dx = -kv to find

v(x).

V t M h i f E i D iEi

Ed




Edition

11 - 16

Sample Problem 11.3

SOLUTION:

Integrate a = dv/dt= -kv to find v(t).

ktv

tvdtk

v

dvkv

dt

dva

ttv

v

00

ln

0

ktevtv 0

Integrate v(t) = dx/dt to findx(t).

tkt

tkt

tx

kt

ek

vtxdtevdx

evdt

dxtv

00

00

0

0

1

ktek

vtx

10

V t M h i f E i D iEi

Ed




Edition

11 - 17

Sample Problem 11.3

Integrate a = v dv/dx = -kv to find v(x).

kxvv

dxkdvdxkdvkvdx

dvvaxv

v

0

00

kxvv 0

Alternatively,

0

0 1v

tv

k

vtx

kxvv 0

0

0 orvtveevtv ktkt

ktek

vtx

10with

and

then

Vector Mechanics for Engineers: DynamicsEi

Ed



Vector Mechanics for Engineers: Dynamicsighth

dition

11 - 18

Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero and

the velocity is constant.

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant


Ed



Vector Mechanics for Engineers: Dynamicsghth

dition

11 - 19

Uniformly Accelerated Rectilinear Motion

For particle in uniformly accelerated rectilinear motion, the acceleration of

the particle is constant.

atvv

atvvdtadvadt

dv tv

v

0

000

constant

221

00

221

000

00

0

attvxx

attvxxdtatvdxatvdt

dx tx

x

020

2

020

2

21

2

constant

00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v


Ed




dition

11 - 20

Motion of Several Particles: Relative Motion

For particles moving along the same line, time

should be recorded from the same startinginstant and displacements should be measured

from the same origin in the same direction.

ABAB xxx relative position ofBwith respect toA

ABAB xxx

ABAB vvv relative velocity ofBwith respect toA

ABAB vvv

ABAB aaa relative acceleration ofBwith respect toA

ABAB

aaa

Vector Mechanics for Engineers: DynamicsEig

Ed




dition

11 - 21

Sample Problem 11.4

Ball thrown vertically from 12 m level

in elevator shaft with initial velocity of

18 m/s. At same instant, open-platform

elevator passes 5 m level moving

upward at 2 m/s.

Determine (a) when and where ball hits

elevator and (b) relative velocity of ball

and elevator at contact.

SOLUTION:

Substitute initial position and velocityand constant acceleration of ball into

general equations for uniformly

accelerated rectilinear motion.

Substitute initial position and constant

velocity of elevator into equation for

uniform rectilinear motion.

Write equation for relative position of

ball with respect to elevator and solve

for zero relative position, i.e., impact.

Substitute impact time into equation

for position of elevator and relative

velocity of ball with respect to

elevator.


Ed




dition

11 - 22

Sample Problem 11.4SOLUTION:

Substitute initial position and velocity and constantacceleration of ball into general equations for

uniformly accelerated rectilinear motion.

2

2

221

00

20

s

m905.4s

m18m12

s

m81.9

s

m18

ttattvyy

tatvv

B

B

Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear motion.

ttvyy

v

EE

E

s

m2m5

s

m2

0


Ed




dition

11 - 23

Sample Problem 11.4

Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

025905.41812 2 ttty EB

s65.3

smeaningless39.0

t

t

Substitute impact time into equations for position of elevator

and relative velocity of ball with respect to elevator.

65.325Eym3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv


Ed




dition

11 - 24

Motion of Several Particles: Dependent Motion

Position of a particle may dependon position of one

or more other particles.

Position of blockB depends on position of blockA.

Since rope is of constant length, it follows that sum of

lengths of segments must be constant.

BA

xx 2constant (one degree of freedom)

Positions of three blocks are dependent.

CBA xxx 22 constant (two degrees of freedom)

For linearly related positions, similar relations hold

between velocities and accelerations.

022or022

022or022

CBACBA

CBACBA

aaa

dt

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx


Ed


25/34



ition

11 - 25

Sample Problem 11.5

PulleyD is attached to a collar which

is pulled down at 3 cm/s. At t= 0,

collarA starts moving down fromK

with constant acceleration and zero

initial velocity. Knowing that velocity

of collarA is 12 cm/s as it passesL,

determine the change in elevation,

velocity, and acceleration of blockB

when blockA is atL.

SOLUTION:

Define origin at upper horizontal surfacewith positive displacement downward.

CollarA has uniformly accelerated

rectilinear motion. Solve for acceleration

and time tto reachL.

PulleyD has uniform rectilinear motion.

Calculate change of position at time t.

BlockB motion is dependent on motions

of collarA and pulleyD. Write motion

relationship and solve for change of blockB position at time t.

Differentiate motion relation twice to

develop equations for velocity and

acceleration of blockB.


Ed


26/34



ition

11 - 26

Sample Problem 11.5SOLUTION:

Define origin at upper horizontal surface with

positive displacement downward.

CollarA has uniformly accelerated rectilinear

motion. Solve for acceleration and time tto reachL.

2

2

0

2

0

2

s

cm9cm82

s

cm12

2

AA

AAAAA

aa

xxavv

s333.1

s

cm9

s

cm12

2

0

tt

tavv AAA


Edi


27/34



ition

11 - 27

Sample Problem 11.5 PulleyD has uniform rectilinear motion. Calculate

change of position at time t.

cm4s333.1s

cm30

0

DD

DDD

xx

tvxx

BlockB motion is dependent on motions of collarA and pulleyD. Write motion relationship and

solve for change of blockB position at time t.

Total length of cable remains constant,

0cm42cm8

02

22

0

000

000

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

cm160 BB xx


Edi


28/34



ition

11 - 28

Sample Problem 11.5

Differentiate motion relation twice to develop

equations for velocity and acceleration of blockB.

0

s

cm32

s

cm12

02

constant2

B

BDA

BDA

v

vvv

xxx

s

cm

18Bv

0

s

cm9

02

2

B

BDA

v

aaa

2s

cm9Ba


Edi


29/34



tion

11 - 29

Graphical Solution of Rectilinear-Motion Problems

Given thex-tcurve, the v-tcurve is equal to

thex-tcurve slope.

Given the v-tcurve, the a-tcurve is equal to

the v-tcurve slope.


Edi


30/34


Vector Mechanics for Engineers: Dynamicshthtion

11 - 30

Graphical Solution of Rectilinear-Motion Problems

Given the a-tcurve, the change in velocity between t1 and t2 is

equal to the area under the a-tcurve between t1 and t2.

Given the v-tcurve, the change in position between t1 and t2 is

equal to the area under the v-tcurve between t1 and t2.


Edit


31/34



11 - 31


Moment-area methodto determine particle position at

time tdirectly from the a-tcurve:

1

0

110

01 curveunderarea

v

v

dvtttv

tvxx

using dv = a dt ,

1

0

11001

v

v

dtatttvxx

1

01

v

v dtatt first moment of area undera-tcurvewith respect to t= t1 line.

Ct

tta-ttvxx

centroidofabscissa

curveunderarea 11001

Vector Mechanics for Engineers: DynamicsEigh

Edit


32/34



11 - 32


Method to determine particle acceleration

from v-x curve:

BC

AB

dx

dvva

tan

subnormalto v-x curve


Edit


33/34



11 - 33

Curvilinear Motion: Position, Velocity & Acceleration

Particle moving along a curve other than a straight line

is in curvilinear motion.

Position vectorof a particle at time tis defined by a

vector between origin O of a fixed reference frame and

the position occupied by particle.

Consider particle which occupies position Pdefinedby at time tand Pdefined by at t + Dt,r

r

D

D

D

D

D

D

dt

ds

t

sv

dt

rd

t

rv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)


Edit


34/34

Vector Mechanics for Engineers: DynamicshthtionCurvilinear Motion: Position, Velocity & Acceleration

D

D

D dt

vd

t

va

t

0lim

instantaneous acceleration (vector)

Consider velocity of particle at time tand velocity

at t +D

t,

v

v

In general, acceleration vector is not tangent to

particle path and velocity vector.

ch11 cinematica de particulas

Documents