Vector Mechanics for Engineers: Dynamics · 2014-06-03 · Vector Mechanics for Engineers: Dynamics...

Vector Mechanics for Engineers: Dynamics

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Introduction to the course

• Scope

• Key concepts

Dynamics

Particles

Rigid body

Deal

with Motions under

applied force

Kinematics

Kinetics

Particle A moving body, on which two arbitrary points keeps fixed relative

position. (A geometric point with mass. The position of an arbitrary

point represents the whole body’s position)

Rigid body A moving body, on which the relative position of two arbitrary points

changes.

Kinematics Geometry of motion (The relation between displacement, velocity,

acceleration). Don’t refer to the cause of motion.

Kinetics (1) Relation between motion and applied forces.

(2) To predict the motion by given forces or to determine the forces to

produce a given motion

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

Dr. Gangyi Zhou

UC Irvine

Jul 3rd, 2006

CHAPTER

11 Kinematics of Particles

Part 1: Rectilinear motion


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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 (Displacement) of a particle is

known if the position 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., 326 ttx

or in the form of a graph x vs. t.


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• Instantaneous velocity may be positive or

negative. Magnitude of velocity is referred

to as particle speed.

• Consider particle which occupies position P

at time t and P’ at t+Dt,

t

xv

t

x

t D

D

D

D

D 0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dt

dx

t

xv

t

D

D

D 0lim

e.g.,

2

32

312

06

ttdt

dxv

ttx


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Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with velocity v at time t and

v’ at t+Dt,

Instantaneous acceleration t

va

t D

D

D 0lim

tdt

dva

ttv

dt

xd

dt

dv

t

va

t

612

312e.g.

lim

2

2

2

0

D

D

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.


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• Consider particle with motion given by

326 ttx

2312 ttdt

dxv

tdt

xd

dt

dva 612

2

2

Dis

pla

cem

ent

Vel

oci

ty

Acc

eler

atio

n

Po

siti

on

0 7 7 -40

20

0 40

20

40

-30


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Concept comparison

Displacement

Velocity

Acceleration

Average

velocity

Average

acceleration

Instant Course

X

X

X

X

X


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Determination of the Motion of a Particle

Motion of

a particle

Acceleration

experienced

by the particle

a function

of time: a = f(t)

a function

of displacement:

a = f(x)

a function

of velocity:

a = f(v)

Displacement

function

Expressed by


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• Acceleration given as a function of time, a = f(t):

tfadt

dvBy the definition of acceleration: dtdt

yields: dttfdv

Integrating both sides yields:

ttv

v

dfdv00

0vtv

So, velocity can be expressed by:

t

dfvtv0

0

Because of , by the same procedure, we can get: dt

dxv

t

dvxtx0

0


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• Acceleration given as a function of position, a = f(x):

By the definition of acceleration: xfadt

dv

dt

dx

dx

dv

=v

yields: xfdx

dvv

Multiplying dx to both sides, yields: dxxfvdv

Integrating both sides, we get:

x

x

xv

xv

xdxfvdv

00

~~ 2

0

2

2

1

2

1xvxv

So, velocity can be expressed by:

x

xdxfxvxv0

2

0

2 ~~2

Not uniquely determined


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

By the definition of acceleration: vfadt

dv

vfdt

1

dt

vf

1

Integrating both sides, yields

yields:

dtvf

dv

tv

v

t

dvf

dv

0 0

t

For the displacement, because of , we have : vfadx

dvv

dx

vf

vdv

Integrating both sides, yields:

000

xtxdxvf

vdvtx

x

tv

v


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t

dfvtv0

0

Determination of the Motion of a Particle: summary

a = f(t):

a = f(x):

a = f(v)

Velocity Displacement

NA

t

dvxtx0

0

tv

vvf

vdvxtx

0

0

x

xdxfxvxv0

2

0

2 ~~2

Not uniquely determined

tv

v

t

dvf

dv

0 0


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Sample Problem 11.2

Determine:

• velocity and elevation above ground at

time 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: (a problem of a=f(t))

• Integrate twice to find v(t) and y(t).

• Solve for t at which velocity equals

zero (time for maximum elevation)

and evaluate corresponding altitude.

• Solve for t at which altitude equals

zero (time for ground impact) and

evaluate corresponding velocity.


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Sample Problem 11.2

tvtvdtdv

adt

dv

ttv

v

81.981.9

sm81.9

00

2

0

ttv

2s

m81.9

s

m10

2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdt

dy

tty

y

2

2s

m905.4

s

m10m20 ttty

SOLUTION:

• Integrate twice to find v(t) and y(t).


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Sample Problem 11.2

• Solve for t at which velocity equals zero and evaluate

corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

• Solve for t at 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


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Sample Problem 11.2

• Solve for t at which altitude equals zero and

evaluate corresponding velocity.

0s

m905.4

s

m10m20 2

2

ttty

s28.3

smeaningles s243.1

t

t

s28.3s

m81.9

s

m10s28.3

s

m81.9

s

m10

2

2

v

ttv

s

m2.22v


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Sample Problem 11.3

Brake mechanism used to reduce gun

recoil 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 and

cylinder to decelerate at rate proportional

to their velocity.

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

kva

SOLUTION: (a problem of a=f(v))

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

• Integrate v(t) = dx/dt to find x(t).

• Integrate a = v dv/dx = -kv to find

v(x).


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Sample Problem 11.3

SOLUTION:

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

kt

v

tvdtk

v

dvkv

dt

dva

ttv

v

00

ln

0

ktevtv 0

• Integrate v(t) = dx/dt to find x(t).

tkt

tkt

tx

kt

ek

vtxdtevdx

evdt

dxtv

00

00

0

0

1

ktek

vtx 10


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Sample Problem 11.3

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

kxvv

dxkdvdxkdvkvdx

dvva

xv

v

0

00

kxvv 0

• Alternatively,

0

0 1v

tv

k

vtx

kxvv 0

00 or

v

tveevtv ktkt

ktek

vtx 10with

and

then


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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


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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

221

2

constant

00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v


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Motion of Several Particles: Relative Motion

• For particles moving along the same line, time

should be recorded from the same starting

instant and displacements should be measured

from the same origin in the same direction.

ABAB xxx relative position of B

with respect to A ABAB xxx

ABAB vvv relative velocity of B

with respect to A ABAB vvv

ABAB aaa relative acceleration of B

with respect to A ABAB aaa


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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 velocity

and 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.


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Sample Problem 11.4 SOLUTION:

• Substitute initial position and velocity and constant

acceleration of ball into general equations for

uniformly accelerated rectilinear motion.

2

2

221

00

20

s

m905.4

s

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


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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

smeaningles s39.0

t

t

• Substitute impact time into equations for position of elevator

and relative velocity of ball with respect to elevator.

65.325Ey

m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv


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Motion of Several Particles: Dependent Motion

• Position of a particle may depend on position of one

or more other particles.

• Position of block B depends on position of block A.

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

lengths of segments must be constant.

BA xx 2 constant (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

aaadt

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx


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Sample Problem 11.5

Pulley D is attached to a collar which

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

collar A starts moving down from K

with constant acceleration and zero

initial velocity. Knowing that velocity

of collar A is 12 in./s as it passes L,

determine the change in elevation,

velocity, and acceleration of block B

when block A is at L.

SOLUTION:

• Define origin at upper horizontal surface

with positive displacement downward.

• Collar A has uniformly accelerated

rectilinear motion. Solve for acceleration

and time t to reach L.

• Pulley D has uniform rectilinear motion.

Calculate change of position at time t.

• Block B motion is dependent on motions

of collar A and pulley D. Write motion

relationship and solve for change of block

B position at time t.

• Differentiate motion relation twice to

develop equations for velocity and

acceleration of block B.


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Sample Problem 11.5 SOLUTION:

• Define origin at upper horizontal surface with

positive displacement downward.

• Collar A has uniformly accelerated rectilinear motion.

Solve for acceleration and time t to reach L.

2

2

020

2

s

in.9in.82

s

in.12

2

AA

AAAAA

aa

xxavv

s 333.1s

in.9

s

in.12

2

0

tt

tavv AAA


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Sample Problem 11.5 • Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

in. 4s333.1s

in.30

0

DD

DDD

xx

tvxx

• Block B motion is dependent on motions of collar

A and pulley D. Write motion relationship and

solve for change of block B position at time t.

Total length of cable remains constant,

0in.42in.8

02

22

0

000

000

BB

BBDDAA

BDABDA

xx

xxxxxx

xxxxxx

in.160 BB xx


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Sample Problem 11.5

• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

0s

in.32

s

in.12

02

constant2

B

BDA

BDA

v

vvv

xxx

s

in.18Bv

0s

in.9

02

2

B

BDA

v

aaa

2s

in.9Ba


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Motion

determination

Uniform

rectilinear

motion and

Uniform

accelerated

motion

Particle and

rigid body

Relative

motion and

Dependent

motions

Course summary

Position

(displacement)

, velocity, and

acceleration


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Homework

• 11.5(7th) or 11.3(8th);

• 11.20(7th) or 11.21(8th);

• 11.21(7th) or 11.25(8th);

• 11.39(7th) or 11.43(8th),

• 11.45(7th) or 11.47(8th),

• 11.48(7th) or 11.50(8th),

Due, next Wednesday, 07/12/2006!

Vector Mechanics for Engineers: Dynamics · 2014-06-03 · Vector Mechanics for Engineers: Dynamics...

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Transcript of Vector Mechanics for Engineers: Dynamics · 2014-06-03 · Vector Mechanics for Engineers: Dynamics...