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
10
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
20
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
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
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
Path of the particle
rr
r
= “position vector” (start from some convenient point: reference point)
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
Path of the particle
sr
r
O
A
A'
Path of the particle
rr
32
Instantaneous Speed
drr
dt
dr ds
dt dt
| |dr d r
dt dt
r
O
A
A'
Path of the particle
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
Path of the particle
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)
errerra r ˆ)2(ˆ)( 2
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
errerra r ˆ)2(ˆ)( 2
Find: , , , r r
Detect , v a
75
Circular Motion (Special Case)
ererv r ˆˆ
errerra r ˆ)2(ˆ)( 2
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
errerra r ˆ)2(ˆ)( 2
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
errerra r ˆ)2(ˆ)( 2
( ), ( )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
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°
86
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
(n,t) coordvelocity meter
Choice of CoordinatesVelocity Acceleration
Observer
Observer’smeasuringtool
Observer
95
Path
(x,y) coord
r
q
(r,q) coord
(n,t) coordvelocity meter
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