1 Dynamics a branch of mechanics that deals with the “motion” of bodies under the action of...

1

Dynamics

a branch of mechanics that deals with the “motion” of bodies under the action of forces.

Two parts:

1. Kinematics

Study of motion without reference to the forces that cause the motion

2. Kinetics

Study of motion under the action of forces on bodies for resulting motions.

2

Kinematics Kinetics

study of motion under the action of forces on bodies about the bodies’ motions.

study of objects’ motion without reference to the forces that cause the motion

How does wAB related with another w ＣＤ (kinematics)

If you want aCD = 36.5 rad/s2, how much torques do you apply to AB

(kinetics)

kinematics relation is necessary to solve kinetics problem

3

Kinematics KineticsP

arti

cles

Rig

id B

odie

s

Before Midterm

After Midterm

4

s

Kinematics of particles How to describe the position?

Frame: Ref-Point + Coordinate

O

A

Path of the particle

'Ar

Ar

= “position vector” (start from some convenient point: reference point)

r

A'

from A to A' takes t secondr

Displacement ( in t ):

Distant traveled:

measured along the path, scalar

Coordinate

(x,y) coord

r

q

(r,q) coord

Reference Point

r

: ( ) at any tme Ause r t t

(x,y) coord n-t coord

relative frame

s

rv

t

sv

t

r-q coord

You are going to learn:

time-related info: velocity , acceleration

How to describe the motion?

: Frameuse

Motion: time-related position

50lim

t

s dsv s

t dt

“instantaneous” velocity :

The particle is at point P at time t and point P at time t+t with a moving distance of s.

2/2 Rectilinear Motion (1D-motion)

“average” velocity:

Particle moving on straight line path.

“displacement, velocity, acceleration” can be considered as scalar quantity.

Positive v is defined in the same direction as positive s.

i.e.) positive v implies that s is increasing, and negative v implies that s is deceasing.

By defining the axis according to the moving direction

reference point

reference axis

P: t P’: t+Dt

D s+s

average

sv

t

6

( similarly as )Rectilinear Motion

dva v

dt

2

2

d sa s

dt

or

vdv ads sds sdsor

Positive a is defined in the same direction as positive v (or s).

Ex. positive a implies that the particle is speeding up (accelerating)

negative a implies that the particle is slowing down (decelerating).

dsv

dt

• Eliminating dt, we have

“instantaneous” acceleration

ds

v sdt

( )

dsdva

3 Equations, but only 2 are independent

8

Formula Interpretations of a,v,s,t

(1) Constant acceleration (a= constant). Find v(t), s(t)

dva

dt

dsv

dt

vdv ads

0 0

v t

v

dv a dt 0v v at

0 0

v t

v

v dv a ds 0

2 202 ( )v v a s s

0 0

v s

v s

v dt ds 20 0

1

2s s v t at

oa atime : 0

displacement :

velocity: o

o

s

v

time :

displacement :

velocity:

t

s

v

where is the function of a a

or ( ) a t or ( ) a v or ( ) a s

or ( , ) or ( , , ) a v s a v s t

Ans

asvv 220

2

hgvv

6

122

02

581.96

12222

v

smv /51.4dv

adt

ds

vdt

vdv ads

2

9.81

6 6 (constant)

g ma

s

2 m/sov

?v 5 m

11

1ts2

s1 =3

1 2

3 0.39 2.61 m

h s s

20 0

1( ) ( )

2s s v t a t

213 1 2 9.81 t 1 0.782t s

Time required for 3-m falling.

21

2s gt

2 0.782 0.5 0.282 st

Thus, the 2nd ball has time to fall:

2

2

19.81 0.282 0.39 m

2s

Therefore, the 2nd ball travels:

a=g (const)1 2

1 (sec)

2t t

2t

1st

2nd

12

Formula Interpretations of a,v,s,t

(2) Acceleration given as a function of time,

a = f(t) . Find v(t), s(t)dv

adt

dsv

dt vdv ads

0 0

( ) v t

v

dv f t dt 0

0

( )t

v v f t dt

ds vdts(t)

v(t)

Or by solving the differential equation:

2

2( ) ( )

d sa v s f t

dt s(t) v(t)

23 2a t t

0 3 ( 0)v t

13

(3) Acceleration given as a function of velocity,

a = f(v)

00

( )

t v

v

dvdt

f v

0 0

( )

s v

s v

vds dv

f v s(v)

t(v)dv

adt

vdv ads

dsv

dt

0 0

v s

v s

v dt ds

v(t)

s(t)

Find s(v), t(v)

Formula Interpretations of a,v,s,t24 0.03a v

0 3 ( 0)v t

14

Formula Interpretations of a,v,s,t(4) Acceleration as function of displacement:

a = f (s) Find v(s), t(s)

v(s)

t(s) s(t)

v(t)

dsv

dt

vdv ads

,dv

adt

a(t)

Find s(t), v(t), a(t)

( ) o o

v s

v s

vdv f s ds 1

( )

o o

t s

t s

dt dsv s

4 0.1 lna s

0 3 ( 0)v t

15

Graphical Interpretations of a, v, s, t• The familiar slopes and areas

vdt ds

2 2

1 1

2 1

v s

v s

vdt ds s s

Area under v-t curve from t1-t2

= s(t2) − s(t1)

Area under a-t curve from t1-t2

= v(t2) − v(t1)

adt dv2 2

1 1

2 1

t v

t v

adt dv v v s-t curve

v-t curve

a-t curve

16

2

1

2 22 1

2 2

v

v

v vvdv ads vdv

2

1

t

t

ads dv

a vds

Graphical Interpretations of a, v, s, t• Less familiar interpretations

v-s curve

a-s curve

Find a !

q

q

tana v

Find v !

18

Given the acceleration-displacement plot as shown.

Determine the velocity when x = 1.4 m, assuming that the velocity is 0.8 m/s at x = 0

ads vdv

o

v

v

v dv a dx Area under curve = 0.16 + 0.12 + 0.08 + 0 = 0.36

36.02

vv

v

2

o

72.0vv 2o

2

36.164.072.0v2 s/m17.1v ANS

a-s curve

+ or -?

: is always + in that range

:o

f

v a

v

1.17 m/sv

1.4

21

v0 v

s

a=a(v)

Fig1

dsavdv 0

savv 020

2

2

1

221 1000

250 0 22 3600

s

1206 ms

Case (a)

Case (b) dskvavdv 20

0

2

20 0

( )

2

v s

v

d vds

a kv

vdv ads

sv

v

dsavdv0

0

0

1268 ms

2502

0

1ln( )

2 os a kvk

22

2a k s

Find v(t), s(t) of the mass if s start at zero and

0( 0, 0, 0)t s v v

dva

dt

dsv

dt vdv ads

2( )vdv ads k s ds

2 2 2

12 2

v k sC

2 2 20v v k s

20

1 2

vC

2 2 20v v k s

2 2 20

dsv k s

dt 2 2 2

0

1ds dt

v k s

12

0

1sin

kst C

k v

0 sin( )v

s ktk

0 cos( )ds

v v ktdt

0v v

+ or - ?max (when )

2ov

s tk k

22

2( )

dsa k s

dt

22

20

dsk s

dt

Altenative Solution: Differential equation

112 2

sindx x

Caa x

s

0

initial condition

( 0, )t v v

initial condition

( 0, 0)t s

max (when 0, ,...)ov v tk

23

0.006k

9.81g

2( )vdv ads g kv ds

dva

dt

dsv

dt

vdv ads

2( )

vdv s h

g kv

2

02 00

1 1ln( ) ln( )

2 2ov

g kvh g kv

k k g

Upward:

downward:

2( )vdv ads g kv ds

0

20

( )fv

h

vdv ds h

kv g

2 22

0

1 1 1ln( ) ln( ) ln(1 )

2 2 2fv f fkv g kv

kv gk k g k g

2(1 )khf

gv e

k = 24.1 m/s

= 36.5 m

Find h and v when theball hits the ground.

ov

24

SP2/1 Given:

3( ) 2 24 6x t t t

2( ) 6 24xv t x t

( ) 12xa t v t

1) t when v=+72

72 4vt 72 4vt

32 3vt 3 36ta

4 1t tx x 38 ( 16) 54 m

net displacement

!= total distance traveled

1: 16t s 4 : 38t x 18v 102v

At the turn 0v

2 1 4 2| - | | - |t t t tx x x x

| 26 ( 16) |

+ | (38) ( 26) |

| 10 | + | 64 | 74 m

total distance traveledCorrect?

x

2 to make =0

t=-2 is not possible,

*only* 1 U-turn

t v

2) a when v=+32

3) total distance traveled from t=1 to t=4

Find

How many turns?2 : 26

0

t x

v

25

2/44 The electronic trottle control of a model train is programmed so that the train speed varies with position as shown. Determine the time t required for the train to complete one lap.

Rectilinear motion?

Rectilinear equations can be used for curved motion if s, v, a are measured along the curve(more on this soon)

v, a

s

v, a

s

dva

dt

dsv

dt

vdv ads

a along the path

not total a

26

2/44 The electronic trottle control of a model train is programmed so that the train speed varies with position as shown. Determine the time t required for the train to complete one lap.

Area = vds Slope = dv/ds

dv a

ds v = Slope = Constant. =C

dv

dt

1 1( )dt dv

C v

0.125

0.25

1 1 0.125ln( ) ln

0.25t v

C C

0( )V V const1 0.125

ln ln 20.25 0.25

0.125 0.25

2

t

2 42( ln 2 ln 2) (1 ln 2)

0.25 0.25 0.25 0.25t

0 2

2

0.25T

a Cv

2 (2 )2

ln 2

0.25T

,dv

adt

dsv

dt ,vdv ads

= 50.8 s

0.125 0.25( 0)

2

C

27

Recommended Problem

2/29 2/36 2/46 2/58 a g cy

2a kv

2 30 1 2 3v c c t c t c t

slope =0 at both end

29

s

Kinematics of particles How to describe this particle’s motion?

Framework for describing the motion

O

A


'Ar

Ar


r

A'

from A to A' takes t secondr

displacement ( in t ):

Distant traveled:

measured along the path, scalar

Reference Frame

(x,y) coord

r

q

(r,q) coord

Reference Point

r

Recording ( )Ar t

(x,y) coord n-t coord

"time"-related info

relative frames

rv

t

sv

t

r-q coord

You are going to learn:

30

s

2/3 Plane Curvilinear Motion

Motion of a particle along a curved path in a single plane (2D curve)

O

A


rr

r


r

t

rvavg

.

“Average Velocity” “(Instantaneous) Velocity”

t

rv

t

0lim r

dt

rdv

Basic Concept: “time derivative of a vector”

A'

from A to A' takes t second

r

r

displacement of the particle

during time t :

Distant traveled = s, measured along the path, scalar

Reference Frame

(x,y) coord

r

q

(r,q) coord

v

rdt

rdv

Speed and Velocity“(Instantaneous) Velocity”

“(Instantaneous) Speed”

dsv s

dt

0 :

| |

t

s r

Velocity vector is tangent to the curve path

0

1lim | |

tr

t

0lim

t

r

t

0lim

t

r

t

r

O

A

0lim

t

s

t

drv

dt


sr

r

O

A

A'


rr

32

Instantaneous Speed

drr

dt

dr ds

dt dt

| |dr d r

dt dt

r

O

A

A'


rr

r

rdt

rdv

s

dsv s

dt

speed

Time rate of change of length of the position vector r

r

q

(r,q) coord| |

dsv

dt

33

Acceleration

1v

A

A'

Path of the particle2v

v

t

va

t

0lim

t

vaavg

.

“Average acceleration” “(Instantaneous) Acceleration”

vdt

vda

C

1v2v

2a

*** The velocity is always tangent to the path of the particle (frequently used in problems) while the acceleration is tangent to the hodograph (not very important) ***

O1r

2r

Hodograph

C

1v2v

1aHodograph

34

Vector Equation and reference frame

- Rectangular: x-y

- Normal-Tangent: n-t

- Polar: r-

A


rr

r

A'

r

s

Reference Frame

OVector equation is in general form, not depending on used coordinate.

rdt

rdv

vdt

vda

ReferenceFrame(coordinate)

Usage will depend on selection.More than one can be used At the same time.

ˆ ˆr xi yj

tv ve

rectangular n-t

ˆrr re

r-q

35

Derivatives of Vectors

Derivatives of Vectors: Obey the same rules as they do for scalars

( )d uPuP uP

dt

( )d P QP Q P Q

dt

( )d P QP Q P Q

dt

Basic Agreement:

Direction ofreference axis {x,y} do not change on time variation.

2/4 Rectangular Coordinates

r

O x

y

v

j

i

Path

ˆ ˆ r x i y j

ˆ ˆ v r x i y j

ˆ ˆ a v x i y j

ˆ ˆ ( )xi yj

Correct?

00

xv yv

xa ya

Reference Frame

O

rdt

rdv

vdt

vda

( , )A x y

Rectilinear Motion in 2 perpendicular & independent axes.

Both “divided” particles, are moving in rectilinear motion

x

y

xx v xx a

yy v

yy a

O

x

x

x

v

a

r

O x

y

xr x

yr y

v

j

i

xv x

yv y

xa x

Path

a

rdt

rdv

vdt

vda

( , )A x y

r

O x

y

j

i

ya y

y

y

y

v

a

t

dsv

dtdv

adt

vdv ads

rectilinear in x-axis

rectilinear in y-axis

rectilinear in 2 dimension, related which other via time.

If given ( , , )x xa f x v t

( , , )y ya f y v t

can you find

( ) ( )

( ) ( )

x x y y

x x y y

v v t v v t

s s t s s t

Common Cases

Rectangular coordinates are usually good for problems where x and y variables can be calculated independently!

jyixr ˆˆ

jyixva ˆˆ

ˆˆ jyixrv

Ex1) Given ax = f1(t) and ay = f2(t)

1( )xs f t 1( )xv f t 1 ( )xa f t

2( )ys f t 2( )yv f t 2 ( )ya f t1( )a f t

2( )a f tEx2) Given x = f1(t) and y = f2(t)

1( )xa f t 1( )t

x

o

v f t dt

2( )ya f t2( )

t

y

o

v f t dt

( )t

x x

o

s v t dt

( )t

y y

o

s v t dt

From this, you can find “path” of particles

40

Projectile motion

• The most common case is when ax = 0 and ay = -g (approximation)

• x and y direction can be calculated independently

(vo)x = v0 cos

vx vx

x

y

g

vy

vy

vo

v

v

(vo)y =

v0 sin

( )y o yv v gt

2 2( ) 2 ( )y o y ov v g y y

21( )

2o o yy y v t gt

x-axis

vx = (vo)x

x = xo + (vo)xt

y-axis

0

22 2

(tan )( ) ( )2 coso o o

gy y x x x x

v

0

22 2

(tan )2 cos

gy x x

v

a=0

( )eliminate , tan

( )o yo

o x o x

vx - xt t =

(v ) v

Note: a = const

ov v at

2 2 2 ( )o ov v a s s

21

2o os s v t at

ds dv

v a ads vdvdt dt

( in case of )0o ox y

Determine the minimum horizontal velocity u that a boy can throw a rock at A to just pass B.

2gt2

1y 2t)81.9(

2

110

.sec43.1t

it can be applied in both x and y direction

2yyo ta

2

1tvyy

If we use eq. (3) in the y-direction :2xxo ta

2

1tvxx

x u t

40 (1.43)u

28 m/sec.u

g

Note: rectilinear (a=const)

atvv o

)ss(a2vv o2o

2

2oo at

2

1tvss

(1)

(2)

(3)

x

y

0

0x

x o

a

v u x

42

Find x-,y-component of velocity and displacement as function of time , if the drag on the projectile results in an acceleration term as specified. Include the gravitational acceleration.

Da kv

k : const

ˆDa a gj

0

1x

x

v

xxv

dv ktv

0 0 coskt ktx xv v e v e

00

0

coscos (1 )

tkt ktv

x v e dt ek

xx x

dva kv

dt y

y y

dva kv g

dt

0

1y

y

v

y

vy

dv ktg

vk

0 0( ) ( sin )kt kt

y y

g g g gv v e v e

k k k k

0

0

0

( sin )

1 ( sin )(1 )

tkt

kt

g gy v e dt

k k

g gv e t

k k k

ˆ ˆ( )x yk v i v j ˆ ˆ ( )x ykv i kv g j

:t 0 x y

gv v

k

0

max

cos

vx y

k

ˆ ˆx ya i a j

43

2/95 Find q which maximizes R (in term of v-zero and a)

0( cos )x v t

20

1( sin )

2y v t gt

0cos ( cos ) BR v t

0

cos

cosB

Rt

v

20

cos 1 cossin ( sin )( ) ( )

cos 2 coso o

R RR v g

v v

2 202 cos

(tan tan )cos

vR

g

2

2 202( 2cos sin )(tan tan ) cos (sec ) 0

cos

v

g

(sin 2 )(tan tan ) 1 0 2(sin 2 )(tan ) 2sin 1 0

sin 2 (tan ) (1 cos2 ) 1 0 1tan 2

tan

1 11 12 tan ( ) tan ( ) ( )

tan tan 2 2

1( )

2 2

R=R(q)

0dR

d

Find Rmax

amax 00 ( : fixed)

2v

x

y Find R=R(q)

44

H12-96 A boy throws 2 balls into the air with a speed v0 at the different angles {1, 2} (1 2). If he want the two ball collide in the mid air, what is the time delay between the 1st throw and 2nd throw.

The first throw should be 1q or 2q ?

0

22 2

(tan )2 cos

gy x x

v

0

21 2 2

1

(tan )2 cos

gy x x

v

0

22 2 2

2

(tan )2 cos

gy x x

v

1q

21 2

2 21 2

0,

2 (tan tan )1 1

( )cos cos

oc

x

vx

g

10 1cos

cxt

v 2

0 2coscx

tv

1 20 1 2

1 1( )cos cos

cxt t t

v

2 1

0 1 2

cos cos( )

cos coscx

v

1 2

1 2

2 sin( )=

(cos cos )ov

g

21 2 1 2

c 2 22 1

2 sin( )cos cosx =

(cos cos )ov

g

intersected point

46

0( cos )x v t

20

1( sin )

2y v t gt

0

22 2

(tan )2 cos

gy x x

v

h

R = R(h)

2/84 Determine the maximum horizontal range R of the projectile and the corresponding launch angle q.

How do you throw the ball? q=45 ?

4515.928 moh

2200 sin 2( )v g h 2 2sin

2ov

hg

this way?

Ceil’s height (5 m) is our limitation!

20

20

2(sin )(cos )

sin 2

vR

g

v

g

Yes! but why?

h(q)

R(q)

: 04

Range

h R

max maxR h

47

0( cos )x v t

20

1( sin )

2y v t gt

0

22 2

(tan )2 cos

gy x x

v

2/84 Determine the maximum horizontal range R of the projectile and the corresponding launch angle q.

this way?

2 2sin

2ov

hg

20

20

2(sin )(cos )

sin 2

vR

g

v

g

23.3

to make h=5

o

23.346.3oR m

1 2sin

o

gh

v

48

H12/92 The man throws a ball with a speed v=15 m/s. Determine the angle q at which he should released the ball so that it strikes the wall at the highest point possible. The room has a ceiling height of 6m.

50

and positive n toward the center of the curvature of the path

2/5 Normal and Tangential Coordinates (n-t)

Path: known

t

O

n

t

nO

n

t

Os

Curves can be considered as many tangential circular arcs takes positive t in the direction of

increasing s

the origin and the axes move (and rotate) along with the path of particle

Forward velocity and forward acceleration make more sense to the driver

The driver is only aware of forward direction (t) and lateral direction (n).

Brake and acceleration force are often more convenient to describe relative to the car (t-direction).

Turning (side) force also easier to describe relative to the car (n-direction)

Fixed point on curve

51

Normal and Tangential Coordinates (n-t)

v,a

ss

dva

dt

dsv

dt

vdv ads

ˆt t tv v v e v

ˆ ?t ta a e

ˆ ˆt t n nv v e v e tevv ˆ

t ta v s tv s

s measured along the path

Consider: scalar variables (s) along the path (t direction)

similar to rectilinear motion

ˆ ˆ( )t na ve e

v0 why?

The reason why we define this coordinate

tv

generally, not total aRectilinear Similarity

, taPath: known

t

O

n

t

nO

n

t

Os ˆne

te

52

= (d)tv ve

)( v

** The velocity is always tangent to the path **

C

Velocity

tv v

v

Path

dsA

Ate

ˆne

Speed:

Velocity ( ):v

d

dt

d

Small curves can be considered as circular arcs

Fixed point on curve

32 2

2

2

1dfdx

d fdx

x

y path

ˆ( ) te

dss

dt

( )y f x

53

ˆ ˆt ne e ˆne

Acceleration

dva

dt

teˆ

d

tde

2

ˆ ˆt n

vve e

C

v

v

d

Path

n

t

ds = (d)A

A

ˆ

ˆNeed: tt

dee

dt

ˆ ˆ ˆ(| | )t t nde e d e

ta v s 2

2n

va v

2 2t na a a

teˆ

ˆne

tt eevv ˆˆ )( v

te

t

n

te

ˆ( )ˆ ˆt

t t

d veve ve

dt

ˆ ˆ( )t na ve e

0

ˆ ˆ t na ve v e

db

ˆˆn

t

dee

dt (similarly)

ˆne

ˆneˆnde

d

54

Alternative Proof of ˆ ˆ t ne e

C

v

te

Path

n

t

A

O

y

x

jidt

ed t ˆ)cos(ˆ)sin(ˆ

ˆ ˆ{( sin ) (cos ) }i j

nt e

dt

edˆ

ˆ

jietˆ)(sinˆ)(cosˆ

jienˆ)(cosˆ)(sinˆ

Using x-y coordinate

ˆne

ˆˆn

t

dee

dt

55

tvd )(nvd )(

Understanding the equation ˆ ˆt na ve v e

vd

tn

v

vd

vdvd n |)(|

| ( ) |tdv dv

na

vdtvda nn /|)(|

an comes from changes in the direction of v

| ( ) |tt

dv dva v s

dt dt

ta

at comes from changes in the magnitude of v

ˆne

C

v

v

d

Path

n

t

ds = (d)A

A

teˆ

ˆne te

( ) ( )t ndv dv dv

ˆ ˆ| ( ) | | ( ) |t t n ndv e dv e

What to remember

ˆ ˆt t n nv v e v e

v

dt

vda

nt ev

eva ˆ ˆ 2

by definition of n-t axis ˆ t tv v e

ˆ tv e

nntt eae aa ˆ ˆ

dt

edvev t

t

ˆ ˆ

dt

evd t )ˆ(

ˆ ˆ nt evev

nt e

dt

edˆ

ˆ

nt ev

ev ˆ ˆ2

v,a

ss

dva

dt

dsv

dt

vdv ads

tv

generally, not total a

Rectilinear Similarity

, ta

32 2

2

2

1dfdx

d fdx

x

y path

( )y f x

tt

dva

dt

t

dsv

dt

t t tv dv a ds

57

32 2

2

2

1dydx

d ydx

x

y

tv ve

)( v

ntnt ev

eveveva ˆˆˆˆ2

bdy

+b db

ds

d

( )d ds

tandy

dx 2(sec ) ( )

dyd d

dx

2

2

d ydx

dx

22

2(cos )

d yd dx

dx

22

2

d y dxdx

dx ds

22 2 ( 1 )

dyds dx dy dx

dx

32

2

1( )ds

d y dxdx

2

32

2

1( 1 )

dyd y dxdx

x

y path

dx

ds

dbr

Proof

59

Inflection pointInflection point

Direction of vt an

2

ˆ ˆt n

va ve e

a

v

B

A

Speed

Increasing

an is always plus. Its direction is toward the center of curvature.

a

v

B

A

Speed Decreasing

60

n-t coordinates are usually good for problems where

“ Curvature path is known ”

s

, t tv a

tevv ˆ ta v s

tv s v

v s 2

ˆ ˆt n

va ve e

22

n

va v

1) distance along the curvature path (s) is concerned

2) curvature radius (r) is concerned.

62

A rocket is traveling above the atmosphere such that g = 8.43 m/sec2. However because of thrust, the rocket has an additional acceleration component a1 = 8.80 m/sec2 and the velocity v = 8333.33 m/sec. Compute the radius of curvature and the rate of change of the speed

Here : 215.4)5.0()43.8(60cosga n

2

n

va

233.8333215.4

.km470,16m537,475,16

30cosgaav 1t

30cos)43.8(80.8

2sec/m5.1v ANS

g = 8.43 m/sec2

30°

60°an

=>

a1

Find r Find na v ta

t

n

63

: constta

2 21( ) 2.41

2t C Aa v vs

t tvdv a ds a ds

Condition at A:2 2 2 1.785n ta a a

2 2(27.8)432

1.785n

vm

a

Condition at B: 0na 2.41ta

Condition at C: 2.41ta 2 2(13.89)1.286

150n

va

The driver applies her brakes to produce a uniform deaccceleration. Her speed is 27.8 m/s at A and 13.89 m/s at C. She experience a acceleration of 3 m/s^2 at A. Calculate 1) the radius of curvature at A 2) the acceleration at the inflection point B 3) the total acceleration at C

, , ?t na a a

, , ?t na a a

3a

64

Circular Motion (special case)

n

t

r

v

at

an

r

22v

r vr

r

Particle is moving clockwise

direction n? t?

tevv ˆ

ta v s tv s v

v s 2

ˆ ˆt n

va ve e

22

n

va v

(const.)r

with speed increasing.v

ta

na

65

2/123 Determine the velocity and the acceleration of guide C for a given value of angle q if

t

( , )y y

(I) 0

v r

n

C and P shares the vertical velocity, acceleration.

( )ta v r 2

2( )n

va r

r

q

(II) 0

v r r sin siny v r

0ta 2na r

q q

2

cos sin

cos

n ty a a

r

0v r sin 0y v

ta r0na cos sin

sinn ty a a

r

q q

66

The motorcycle starts from A with speed 1 m/s, and increased its speed along the curve at

20.1 m/s (const.)v

Determine its velocity and acceleration at the instant t = 5s .

( )t

dv

dt

1 (0.1)tv t

22 2( ) ( ) 1

dyds dx dy dx

dx 5| 3.184tx

21s x dx

... ... C Cs x

5| 1.5 m/stv velocity is the vector (magnitude + direction)!

t

ds

dt

20.1( )

2s t t 5| 6.25 mts

6.25

0

5| ....tx

15| tan (3.184)

72.564

t

o

dyx

dx

2 21

1( 1 ln( 1 ))

2x x x x C

s=0, x=0

ta tan

dy

dx

v 1

5tan ( | )tx

0

1

t

v

5

...

t

v

0s

b

what is x , when t = 5 ?

6.25

NumericalMethod

3.184

67

The motorcycle starts from A with speed 1 m/s, and increased its speed along the curve at

Determine its velocity and acceleration at the instant t = 5s .

1 (0.1)tv t

21.50.061

37.171na

ˆ ˆt t n na a e a e 5 5| | 1.5 m/st t tv v

t

32 2(1 )

1

x

32 2| (1 3.184 )

37.171cx x

6.25

3.184cx

b

n

2

n

va

32 2

2

2

1dydx

d ydx

dyx

dx

5| 72.564ot

20.1 m/s (const.)v ta

0.1

b

69

ˆ ˆ r r

drv r e r e

dt

2/6 Polar Coordinates ( r - )

ˆ rr re

Ar

reference axis

r

Path

dva

dt

re

direction of = direction of positive r ˆre

e direction of = direction of positive e

reference point

, , , r r Detect

ˆ ˆ ˆ re e e

ˆ ˆ ˆ(?) (?) r re e e

Find: , v a

t r q0.0 25.1 32.00.1 26.2 35.00.2 26.1 39.00.3 24.8 40.00.4 23.2 37.00.5 25.2 35.0

Radar Coordinate

r r

What is the velocity and acceleration of the plane?

70d

Polar Coordinates ( r - )

eˆ reˆred ˆ

ed ˆ

d

-r

ˆ ˆ(1 )rde d e ˆ ˆ(1 )( )rde d e

ˆ re e ˆ re e re

e

ˆ ˆ r r

drv r e r e

dt

ˆ rr re

Ar

reference axis

r

Path

dva

dt

re

direction of = direction of positive r ˆre

e direction of = direction of positive e

reference point

ˆ ˆ ˆ re e e

ˆ ˆ ˆ(?) (?) r re e e

71

Velocity and Acceleration

ˆ ˆ ,re e ˆ ˆ re e

rr ererdt

rdv

ˆˆ dv

adt

22vvv r

rvr ……….. the change of the length of the vector r

rv ………. the rotation of

the vector r

ˆ ˆ rv r e r e

errerra r ˆ)2(ˆ)( 2

2 rrar

rra 2

22aaa r

Physical meaningwill be discussednext page

rq

ˆ ˆ r rr e r e

ˆ ˆ

ˆ

r e r e

r e

21 ( )

dr

r dt

72

Understanding the acceleration equation

dd

rvv

Magnitude change of rv

r (in r direction)

Direction change of

rv

r (in direction)

Magnitude change of

v

rr (in direction)

Direction change of

v

2r (in -r direction)


Let’s look at how the velocities change

ererv r ˆˆ

rd

dr

drrd)( rd

( )d r dr

r d

rvr

rv

r

, , , r r Detect

Find: , v a

t r q0.0 25.1 32.00.1 26.2 35.00.2 26.1 39.00.3 24.8 40.00.4 23.2 37.00.5 25.2 35.0

Radar Coordinate

r r

What is the velocity and acceleration of the plane?

ˆ ˆ rv r e r e


Find: , , , r r

Detect , v a

75

Circular Motion (Special Case)

ererv r ˆˆ


r

a

ar

vr

v

r

0rvr

rv

2 2ra r r r

ra

r = const

n

t

ˆ ˆn re e ˆ ˆte e

76

• direction depends on its curvature path.

• r- coord depend on { ref point, ref axis }

r- coordn-t coord

t ta v s tv s

tevv ˆ

2

ˆ ˆt n

va ve e

rerr ˆ

ererv r ˆˆ


Usually, Path need to be known

no r

dva

dt

rr

dva

dts

tv , ta

// // path (tangent line)t ta v v

77

Problem typesrerr ˆ

ererv r ˆˆ


( ), ( )r r t t , ,r v a

oinstant t : , , , , , r r r o, , (at t=t )r v a

o, , (at t=t )r v a

oinstant t : , , , , , r r r

ˆre

2/145 The angular position of the arm is given by the shown function, where is in radians and t is in seconds.The slider is at r = 1.6 m (t = 0) and is drawn inward at the constant rate of 0.2 m/s. Determine the magnitude and direction (expressed by the angle relative to the x-axis) of the velocity and acceleration of the slider when t = 4. 2

0.820

tt

1.6 0.2r t 2.0r 0r

0.810

t 1

10

At t = 4s: 8.0r 2.0r 0.0r

4.2 4.0 1.0

ˆ ˆ( ) ( )rv r e r e

2 ˆ ˆ2ra r r e r r e

q

v

a

x

y1 0.32 180

tan (2.4) 2600.2

ov

180242 (2.4) 310.75o

1 00.24

tan 2420.12

( ), ( )r r t t , ,r v aˆ ˆ ˆ(cos ) (sin )

ˆ ˆ ˆ( sin ) (cos )

r x y

x y

e e e

e e e

Anse 1 0.32tan 302

0.2o

b

ˆ ˆ0.2 0.32re e

ˆ ˆ0.128 0.24re e q

Ans

81

At the bottom of loop, airplane P has a horizontal velocity of 600 km/h and no horizontal acceleration. The radius of curvature of loop is 1200 m. determine the record value of for this instant.

oinstant t : , , , , , r r r o, , (at t=t )r v a

, r

te

ˆne

v

a 2 /v

v

a

2 ˆ ˆ ( ) ( 2 ) ra r r e r r e

ˆ ˆ rv r e r e

2 2400 1000r

1 400tan 21.8

1000o

600166.7 m/s

3.6v

2

23.1 m/sv

a

cos 154.7 m/srr v v

sin0.575 rad/s

v v

r r

2sin 12.15 m/sr a r

2cos 20.0365 rad/s

a r

r

Use n-t coord to find v,a

ˆree

2sin ( )ra a r r

cos ( ) 2a a r r

The piston of the hydraulic cylinder gives pin A a constant velocity v = 1.5 m/s in the direction shown for an interval of its motion. For the instant when = 60°, determine OArwhereand,,r,r

150 mm

O

A

o60=q

150 mm

O

A

o60=q

r v

rv

= 60°

From viewpoint of From viewpoint of

v

ˆ1.5

0

v i

a

x-y coord

r- q coord

rv r

rv r

2ra r r

2a r r

x-y coordr- q coord

A Av v

150 mm

O

A

o60=q

150

A

mm

Oo

60=q

v

r v

rv

2D vector equation

q

-1.5cos60 - 0.75 /or m s

sinv r v

cosrv r v

1.5 sin 60 1.299 m/s

The piston of the hydraulic cylinder gives pin A a constant velocity v = 1.5 m/s in the direction shown for an interval of its motion. For the instant when = 60°, determine OArwhereand,,r,r

2 2 20.15(7.5) 9.74 /

sin 60r r m s

22 2( 0.75) (7.5)65.0 /

0.15 sin 60

rrad s

r

150 mm

O

A

o60=q

150 mm

O

A

o60=q

r v

rv

= 60°

From viewpoint of From viewpoint of

v

ˆ1.5

0

v i

a

x-y coord

r- q coord

rv r

rv r

2ra r r

2a r r

x-y coordr- q coord

A Aa a

150 mm

O

A

o60=q

150 mm

O

A

o60=q

a=0

r a

ra

2D vector equation

q2 0ra r r

2 0a r r

The piston of the hydraulic cylinder has a constant velocity v = 1.5 m/s

For the instant when = 60°, determine OArwhereand,,r,r

ˆ1.5 0A Av i a

ˆPoint A : 1.5 0 ?A Av i a

, ,ˆˆ 1.5 0 A x A xv i a

ˆ ˆˆ ˆ 1.5 rre r e i j

Only max 2 unknown to be solved

r 0(also, 0)r

mag?mag?

mag?

= 60°

x-y coordr-q coord

= 60°

mag=0

x-y coordr- q coord

A Av v

85

ˆ ˆ ˆ ˆ1.5 = ri j re r e

mag? mag?mag?

x-y coord r-q coord

ˆ ˆˆ (cos ) (sin )re i j

ˆ ˆˆ ( sin ) (cos )e i j

1.510

sinr

cos 1.732r

mag?mag?

mag?

x-y coord r-q coord

??

q

qq

1.5

1.5

sinv r

2ˆ ˆ ˆ = ( ) ( 2 )r rj r r e r r e

: 2 unknowns

Alternate Solution

velocity

Acceleration

= 60°

87

22 2x yr s s

1 2tan y

x

s

s

cos sinr x yr v v v

sin cosx yr v v v 0( cos30 )o

xs v t

20

1( sin 30 )

2o

ys v t gt

0 cos30oxv v

0 sin 30oyv v gt

0xa

gay

2 cos sinr x yr r a a a

2/35.3 smr

20.717 rad/s

2

sy

sx

vx

vy

x

y

rq

= 12.990 m

= 6.274 m

At t = 0.5 s

= 25.981 m/s

= 10.095 m/s

= 15.401 m

= 32.495 deg

= 27.337 m/s

2 cos sinx yr r a a a

= -0.353 rad/s

x-y coord

89

The slotted link is pinned at O, and as a result of the constant angular velocity 3 rad/s, it drives the peg P for a short distance along the spiral guide r = 0.4 q m, where q is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e. when r = 0.5 m

Use r-q where reference-origin is at O, and reference axis is horizontal line.

ˆ ˆ( ) ( )rv r e r e

2 ˆ ˆ2ra r r e r r e

Constrained motion

0.4r

0.4r

0.4r

(for all time)0.4(0.3) 0.12r

(for all time)0.4(0) 0r

at r=0.5 ˆ ˆ(0.12) (0.5 0.3)rv e e

2 ˆ ˆ(0 0.5 0.3 ) (2 0.12 0.3 0.5 0)ra e e

92

O

PathReference Frame

(x,y) coord

r

q

(r,q) coord

xv

yv

x

y

xv xyv y

nv

0nv tv v

r

rvv

v r rv r

PathReference Frame

xa

ya

x

y

ta

na

r

ra

a

tv

xa x ya y2

n

va

ta v

2a r r 2ra r r

(n,t) coordvelocity meter

Summary: Three Coordinates (Tool)

Velocity Acceleration

Observer

Observer’smeasuringtool

Observer

93

O

PathReference Frame

(x,y) coord

r

q

(r,q) coord

xv

yv

x

y

xv xyv y

nv

0nv tv v

r

rvv

v r rv r

PathReference Frame

xa

ya

x

y

ta

na

r

ra

a

tv

xa x ya y2

n

va

ta v

2a r r 2ra r r


Choice of CoordinatesVelocity Acceleration

Observer

Observer’smeasuringtool

Observer

95

Path

(x,y) coord

r

q

(r,q) coord


Translating Observer

Two observers (moving and not moving) see the particle moving the same way?

Observer O(non-moving)

Observer’sMeasuring tool

Observer (non-rotating)

Two observers (rotating and non rotating) see the particle moving the same way?

Observer B (moving)

Rotating

No!

No!

“Translating-only Frame” will be studied today

Which observer sees the “true” velocity?

both! It’s matter of viewpoint.

“Rotating axis” will be studied later.

Point: if O understand B’s motion, he can describe the velocity which B sees.

This particle path, depends on specific observer’s viewpoint

“relative” “absolute”

A

“translating” “rotating”

96

2/8 Relative Motion (Translating axises)

A = a particle to be studied

Ar

A

Reference frame O

frame work O is considered as fixed (non-moving)

Br

If motions of the reference axis is known, then “absolute motion” of the particle can also be found.

O

Motions of A measured using framework O is called the “absolute motions”

For most engineering problems, O attached to the earth surface may be assumed “fixed”; i.e. non-moving.

Sometimes it is convenient to describe motions of a particle “relative” to a moving “reference frame” (reference observer B)

B

Reference frame B B = a “(moving) observer”

BAr /

Motions of A measured by the observer at B is called the “relative motions of A with respect to B”

97

Relative position

If the observer at B use the x-y ** coordinate system to describe the position vector of A we have

jyixr BAˆˆ

/

where

= position vector of A relative to B (or with respect to B),

and are the unit vectors along x and y axes

(x, y) is the coordinate of A measured in x-y frame

i jBAr /

** other coordinates systems can be used; e.g. n-t.

Br

B

Ar

BAr /

A

X

Y

x

y

O

j

i

Here we will consider only the case where the x-y axis is not rotating (translate only)

J

I

98

ˆ ˆ ˆ ˆ( ) ( )A Br r xi yj xi yj

ˆ ˆ ˆ ˆ( )A Br r xi yj xi yj

Br

B

Ar

/A Br

A

X

Y

x

y

O

j

i

x-y frame is not rotating (translate only)

Relative Motion (Translating Only)

Direction of frame’s unit vectors do not change

ˆ 0i

ˆ 0j

0

/A Bv

/A Ba

0

/A B A Br r r

ˆ ˆxi yjNotation using when

B is a translating frame.

BABA vvv /

BABA aaa /

Note: Any 3 coords can be applied toBoth 2 frames.

99

Understanding the equation

BABA vvv /

Translation-only Frame!Path

Observer O

Observer B

This particle path, depends on specific observer’s viewpoint

Ar

A

reference

framework O

Br

O

B

reference

frame work B

BAr /

A

/A Ov

/B Ov

Observer O Observer O

Observer B (translation-onlyRelative velocity with O)

This is an equation of adding vectors of different viewpoint (world) !!!

O & B has a “relative” translation-only motion

100

The passenger aircraft B is flying with a linear motion to theeast with velocity vB = 800 km/h. A jet is traveling south with velocity vA = 1200 km/h. What velocity does A appear to a passenger in B ?

A B A Bv v v Solution

800Bv

1200Av

x

y

A Bv

j1200i800v BA

2 2800 1200A Bv

1200

800tan

ˆ1200Av j ˆ800Bv i

101

A B A Bv a

A B A Bv v v

A B A Ba a a

18 ˆ ˆ5 /3.6Av i i m s

2ˆ3 /Aa i m s

12 3 rad/s

60 10

0

Translational-only relative velocity

You can find v and a of B

102

vA

vB vA/B

Velocity Diagramx

y

aAaB

aA/B

Acceleration Diagramx

y

9 ˆ ˆ( ) cos45 sin 45 2 210

o oBv i j i j

/ˆ ˆ3 2 /A B A Bv v v i j m s

2

ˆ ˆ ˆ ˆcos45 sin 45 0.628 0.628o oBB

va i j i j

R

/ˆ ˆ3.628 0.628 /A B A Ba a a i j m s

v rad/s

10

9

10Bv r

0

2B

B

va

R

ˆ5 /Av i m s 2ˆ3 /Aa i m s

0ta r 2

2n

va r

r

: /B A rel B Av v v r

?

?/A B A Bv v v

?/B A B Av v v

B

?/A B A Bv v v

?/B A B Av v v

Yes

Yes

Yes

No

O

Is observer B a translating-only observer

relative with O

104

50 : obserber B, translating?A Bv

/ : obserber A, translating?B Av

BAv

To increase his speed, the water skier A cuts across the wake of the tow boat B, which has velocity of 60 km/h. At the instant when = 30°, the actual path of the skier makes an angle = 50° with the tow rope. For this position determine the velocity vA of the skier and the value of

Relative Motion:(Cicular Motion)

m10A

B

A

B

10A Bv r

sm67.166.3

60vB

Av

60120 20

40

120sin

v

40sin

67.16 A

sm5.22vA sin 20

16.67 10sin 40A Bv

0.887 rad s

o30

D

M ? ?O.K.

Point: Most 2 unknowns canbe solved with 1 vector (2D) equation.

A B A Bv v v

20

2060

60

30

30

Consider at point A and B as r- coordinate system

105

2/206 A skydriver B has reached a terminal speed . The airplane has the constant speed and is just beginning to follow the circular path shown of curvature radius = 2000 mDetermine (a) the vel. and acc. of the airplane relative to skydriver. (b) the time rate of change of the speed of the airplane and the radius of curvature of its path, both observed by the nonrotating skydriver.

0Ba

ˆ50Av i

0Aa

/ /, B A B Ar

50 /Bv m s

ˆ50Bv j

0 ( )A A txa a

2

( ) AA A ny

A

va a

2ˆ ˆ( ) 1.250 /A ya a j j m s

/ /= - , - A B A B A B A Bv v v a a a

50 m/sAv

rrv

50 m/sBv

/ˆ ˆ50 50A Bv i j

/

ˆ1.250A Ba j

106

/ A Bv/ A Ba

(b) the time rate of change of the speed of the airplane and the radius of curvature of its path, both observed by the nonrotating skydriver.

rrv

n

t

/ /( ) sin 45or A B t A Bv a a

2/

/ /( ) cos 45oA BA B n A B

r

va a

/ / , A B A Bv a

/ˆ ˆ50 50A Bv i j

/ˆ1.250A Ba j

0Ba

ˆ50Av i

ˆ50Bv j

2ˆ1.250 /Aa j m s

coordn t

rv r45o

45o

107

1000 ˆ m /3.6Av i s

2ˆ1.2 /Aa i m s

1500 ˆ /3.6Bv i m s

20 /Ba m s

, : relative worldr

/ /, B A B Ar

coord r

/ /, B A B Av a

108

/

500 ˆ3.6B Av i

/

ˆ1.2B Aa i

va

/( )B A rv r cos 120.3r v

/( )B Av r 0.00579

2/( )B A ra r r

/( ) 2B Aa r r

0.637r

30.166 10

r

1000 ˆ m /3.6Av i s

2ˆ1.2 /Aa i m s

1500 ˆ /3.6Bv i m s

20 /Ba m s

cosv

sinv

cosa

sina

30o

coord r

1800 12001200

sin 30or

1 Dynamics a branch of mechanics that deals with the “motion” of bodies under the action of...

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Transcript of 1 Dynamics a branch of mechanics that deals with the “motion” of bodies under the action of...