Summary: Three Coordinates (Tool)
Velocity
Reference
Frame
Acceleration
vy
vn
r
(n,t) coord
velocity meter
q
(r,q)
coord
vx  x
at
ar
an
vx
x
y
r
Observer
Path
ay
aq
O
(x,y)
coord
r
Reference
Frame
vr
vq
x
Observer’s
measuring
tool
vt
Path
Observer
y
ax
vy  y
ax  x
ay  y
vn  0
vt  v
v2
at  v
vr  r
vq  rq
an 

ar  r  rq 2
aq  2rq 1 rq
Choice of Coordinates
Velocity
Reference
Frame
Acceleration
vy
vn
r
(n,t) coord
velocity meter
q
(r,q)
coord
vx  x
at
ar
an
vx
x
y
r
Observer
Path
ay
aq
O
(x,y)
coord
r
Reference
Frame
vr
vq
x
Observer’s
measuring
tool
vt
Path
Observer
y
ax
vy  y
ax  x
ay  y
vn  0
vt  v
v2
at  v
vr  r
vq  rq
an 

ar  r  rq 2
aq  2rq 2 rq
Translating
Observer
“Translating-only Frame”
will be studied today
No!
Observer’s
Measuring tool
(x,y)
coord
Path
Observer B
(moving)
(n,t) coord
velocity meter
r
q
(r,q)
coord
Rotating
Two observers (moving and
not moving) see the particle
moving the same way?
Observer O
(non-moving)
Which observer sees
the “true” velocity?
A
both! It’s matter of viewpoint.
This particle
path, depends
on specific
observer’s
viewpoint
“relative” “absolute”
Two observers (rotating and non
rotating) see the particle moving
the same way?
Point: if O
understand B’s
motion, he can
describe the velocity
which B sees.
No!
Observer
(non-rotating)
“Rotating axis”
will be studied later.
“translating”
“rotating” 4
2/8 Relative Motion (Translating axises)
 Sometimes it is convenient to describe motions of a particle “relative” to
a moving “reference frame” (reference observer B)
 If motions of the reference axis is known, then “absolute motion” of the
particle can also be found.
 A = a particle to be studied
Reference frame O
Reference frame B
A

rA

rB

rA / B
B
O
frame work O is considered
as fixed (non-moving)
 B = a “(moving) observer”
 Motions of A measured by the observer
at B is called the “relative motions of A
with respect to B”
 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”;
5
i.e. non-moving.
Relative position
Ĵ
If the observer at B use the x-y **
coordinate system to describe the
position vector of A we have
ĵ
Y
y
A

rA

rA / B

rB
O
x
iˆ

rA / B  xiˆ  yˆj
B
X
where
Here we will consider only the case
where the x-y axis is not rotating
(translate only)
Iˆ

rA / B = position vector of A relative to B (or with respect to B),
iˆ and ĵ are the unit vectors along x and y axes
(x, y) is the coordinate of A measured in x-y frame
** other coordinates systems can be used; e.g. n-t.
6
Relative Motion (Translating Only)
 x-y frame is not rotating
(translate only)
ĵ
y
A
Y

rA
rA / B

rB
x
O
B
X
iˆ  0
ˆj  0
iˆ
rA  rB  rA / B
0
rA  rB  ( xiˆ  yjˆ)  ( xiˆ  yjˆ)
vA/ B
xiˆ  yjˆ
Note: Any 3 coords
can be applied to
Both 2 frames.
Direction of frame’s
unit vectors do not
change

 
a A  aB  a A / B
Notation using when
B is a translating frame.

 
v A  vB  v A / B
rA  rB  xiˆ  yjˆ  ( xiˆ  yjˆ)70
aA/ B
Path
Understanding the equation
Translation-only Frame!
Observer B
A
O & B has a “relative” translation-only motion

 
v A  vB  v A / B
This particle
path, depends
on specific
observer’s
viewpoint
Observer O
reference
reference
framework O
vA/O
frame work B
vB / O
A

rA

rB
O

rA / B
B
Observer O
Observer O
Observer B
(translation-only
Relative velocity with O)
This is an equation of adding vectors8
of different viewpoint (world) !!!
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 ?
v A B  v A  vB
Solution
vB  800
vA B 
v A  1200
q
vA  1200 ˆj
y
vA B
800  1200
tan q 
2
800
1200
vB  800 iˆ

v A B  800 î  1200 ĵ
x
9
2
Translational-only relative velocity
vA 
18 ˆ
i  5iˆ m / s
3.6
a A  3iˆ m / s 2
vA B
aA B
v A B  v A  vB
a A B  a A  aB
q  2  3 
1


rad/s
60 10
q 0
You can find v and a of B
10
v2
an  rq 
r
at  rq  0
2
v
vA  5iˆ m / s
q

10
a A  3iˆ m / s 2
q 0
rad/s
vA
2
B
v
9
a

B
vB  rq  
R
10
9
vB  (  )  cos 45o i  sin 45o j   2iˆ  2 ˆj
10
vA / B  v A  vB  3iˆ  2 ˆj m / s
2
B


v
aB 
 cos 45o iˆ  sin 45o ˆj  0.628iˆ  0.628 ˆj
R
aA / B  aA  aB  3.628iˆ  0.628 ˆj m / s
y
vB
vA/B
x
Velocity Diagram
y
aB
aA
aA/B
x
Acceleration
Diagram 11
Is observer B a translating-only observer
B
relative with O
v A  vB  v A / B
?
vB  ? v A  vB / A
Yes
Yes
O
v A  vB  v A / B
?
vB  v A  vB / A
?
Yes
No
vB  vA  vrel:B / A   ? r
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
q = 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 q
v  rq  10 q
Relative Motion:
AB
(Cicular Motion)
20
v A B : atobserber
Consider
point A and B,
B
as r-q coordinate system
  50
v A  vB  v A B
M
?
?
Point: Most 2 unknowns can
be solved with
1 vector (2D) equation.
A
q  30
16.67
sin 40
60
O.K.

10 mtranslating?
vB / A : obserber A,
B
20
60
30
 30
D
translating?
vA B
40
60
vA
vA
vA
sin 120

 22.5 m s
v A B  16.67
120 
20
vB 
60
 16.67 m s
3.6
sin 20
 10 q
sin 40
q  0.88713rad s
2/206 A skydriver B has reached a terminal
vB  50 m / s
speed vB  50 m/s . The airplane has the constant speed
v A  50 m/s and is just beginning to follow the circular path v  50 ˆj
B
shown of curvature radius = 2000 m
aB  0
Determine
(a) the vel. and acc. of the airplane relative to skydriver.
(b) the time rate of change of the speed vr of the
v A  50iˆ
airplane and the radius of curvature  r of its path, both
aA  0
observed by the nonrotating skydriver.
 a A  x  0  ( a A )t
 a A  y  (a A )n 
v A2
A
a A  (a y ) ˆj  1.250 ˆj m / s 2
v A / B = v A - vB , a A / B  a A - a B
rB / A ,q B / A
v A / B  50iˆ  50 ˆj
a A / B  1.250 ˆj
14
(b) the time rate of change of the speed vr of the
airplane and the radius of curvature  r of its path,
both observed by the nonrotating skydriver.
vB  50 ˆj
aB  0
v A  50iˆ
a A  1.250 ˆj m / s 2
vA/ B , aA/ B
t
aA/ B
n
vr  r
n  t coord
45o
vA/ B
45o
v A / B  50iˆ  50 ˆj
a A / B  1.250 ˆj
vr  (a A/ B )t  a A/ B sin 45o
vA2 / B
r
 (a A/ B )n  a A/ B cos 45o
15
vA 
1000 ˆ
i m/s
3.6
a A  1.2iˆ m / s 2
1500 ˆ
vB 
i m/s
3.6
aB  0 m / s 2
r ,q : relative world
rB / A ,q B / A
r q
coord
vB / A , a B / A
16
vA 
1000 ˆ
i m/s
3.6
a A  1.2iˆ m / s 2
 30o
vB 
q
r
v
a
vB / A
500 ˆ

i
3.6
q
r  q coord
r
( vB / A ) r  r
 v cosq
(vB / A )q  rq
 v sin q
(aB / A )r  r  rq 2
(aB / A )q  rq  2rq
aB / A  1.2iˆ
 a cosq
 a sin q
1500 ˆ
i m/s
3.6
aB  0 m / s 2
1800  1200
 1200
sin 30o
r  v cosq  120.3
q  0.00579
r  0.637
q  0.166 103
17