12.1 Introduction
Mechanics
of rigid
body body
mechanics
of rigid
Galilei
Statics
Statics
Dynamics
Dynamics
Newton
Euler
Lagrange
Equilibrium
ΣF=0
Kinematics
v=ds/dt
a=dv/dt
Kinetics
ΣF=ma
12.2 Rectilinear Kinematics: Continuous motion
1.
Kinematics
Analysis of the geometric aspects of motion.
2.
Particle
A particle has a mass but negligible size and shape.
3.
Rectilinear Kinematics
Kinematics of objects moving along straight path and
characterized by objects position, velocity and
acceleration.
4.
Position
(1) position vector r
A vector used to specify the location of particle P at any
instant from origin O.
(2) position coordinate , S
An algebraic scalar used to represent the position coordinate of particle P
from O to P.
S  r
scalar
vector
r’
o
r
r
s
P s
P’
s’
5. Displacement
Change in position of a particle , vector
(1) Displacement
r  r 'r
or
S  S 'S
(2) Distance
Total length of path traversed by the particle.
A positive scalar.
6. Velocity
(1) Average velocity
V avg 
r
t
(2) Instantaneous Velocity


r dr
V  lim


t
dt
t 0
S ds

t 0 t
dt
v  lim

(

)
(3) Speed
speed = magnitude of velocity = | v |
ST
Average speed = Total distance/elapsed time = t
7. Acceleration
(1) Average acceleration
a avg 
V
t
(2) (Instantaneous) acceleration



v dv
a  lim

t 0 t
t
dV d ds d 2 S
a
 ( ) 2
dt dt dt
dt

(

)
8. Relation involving a , s and v
v=ds/dt, dt=ds/v
a=dv/dt, dt=dv/a
so, ds/v=dv/a
vdv=ads
9. Constants acceleration a = ac
ac 
dv
dt
c
v  v o  at
v v  act 0
v
t
0
ds
v
dt
S
 dv   a dt
dv  ac dt
s
t
 ds   (v
 act )dt
S  s0  v0 t 
1
2
2 ac t
s0
S 
1
2 t
  v0 t  a c t 
So 
2
0
0
v
c
v0
v
 a cs s
s
v0
s
 vdv   a ds
vdv  a c ds
1 2
v
2
0
s0
v2  v0  2a c (s  s0 )
2
0
10. Analysis Procedure
(1)Coordinate System
A. Establish a position coordinate s along the path.
B. Specify the fixed origin and positive direction of the coordinate.
(2) Kinematic Equations
A. Know the relationship between any two of the
four variables a, v, a and t.
B. Use the kinematic equations to determine the
unknown varaibles
12.3 Rectilinear Kinematics : Erratic Motion
S
V
t
a
t
t
Given
method
Kinematics egn
Find
S-t graph
Measure slope
V=ds/dt
V-t graph
V-t graph
Measure slope
A=dv/dt
a-t graph
A-t graph
Area integration
t
v  v  v0   adt
v-t graph
0
v-t graph
Area integration
s  s  s0   vdt
s-t graph
1
2 2
v  (2 ads  v0 )
v-s graph
A=v(dv/ds)
a-s graph
t
0
A-s graph
Area integration
s
s0
v-s graph
Measure slope
12-4 General Curvilinear Motion
1. Curvilinear motion
The particle moves along a curved path.
p
Vector analysis will be used to formulate the
particle’s position, velocity and acceleration.
2. Position
 
r  r (t)
p
s
r (t)
s
o

3. Displacement  r
  
 r  r ' r
= change in position of particle form p to p’
p
r
o
s
r
r’
p’
4. Velocity
(1) average velocity 平均


r
Vavg  
t
(2) Instantaneous velocity 瞬時



r dr
V  lim
 
dt
t 0 t
= “tangent” to the curve at Pt .p
= “tangent” to the path of motion
p v
(3) Speed

v v
 lim
t 0
r

r
t
 lim
t 0
s ds

t dt
o
r
r’
p’
5. Acceleration
(1) Average acceleration

a avg




v v
v


t
t
= time rate of change of velocity vectors

v

v'

v
Hodograph
Hodogragh is a curve of the locus of
points for the arrowhead of velocity vector.
(2) Instantaneous acceleration



2
v dv d  d r  d r

a  lim

  
dt dt  dt  dt
t 0 t
which is not tangent to the curve of motion, but
tangent to the hodograph.
12-5 Curvilinear Motion : Rectangular components
s
z
path
r
p

 

r  xi  yj  zk
y
θ
x
xyz : fixed rectangular coordinate system
1. Position vector




r (t )  x(t )i  y (t ) j  z (t )k



x
y
z

 x y z
i
j
2
2
2
2
2
2
 x2  y2  z 2
x

y

z
x

y

z


 ru r
2
2
2

k


Here

2
2
2
r r  x y z

r

ur 
r
= magnitude of
= unit vector = direction of

r

r
2. Velocity


dr (t)
v( t ) 
dt
d   

x i  y j  zk
dt

0
0
0
dx 
d i dy 
d j dz 
dk

i x 
jy  kz
dt
dt dt
dt dt
dt
dx  dy  dz 

i
j k
dt
dt
dt



 vx i  vy j  vzk

 vu v


v  vx2  v y2  vz2

v

uv 
v
tangent to the path
3. Acceleration

 dV
a
dt



d

Vx i  Vv j  Vz k
dt
dVx  dVy  dVz 

i
j
k
dt
dt
dt
d 2 x  d 2 y  d 2z 
 2 i  2 j 2 k
dt
dt
dt



 a x i  a y j  a zk

 au a

a  ax2  a y2  az2

 a
ua 
a

12.6 Motion of a projectile
a= - g j
(v0)y v0
y
(v0)x
v0
v
x

V 0 : initial velocity

a : Constant downward acceleration

V : velocity at any instant
Position Vector
(x,y components)



r =x i +y j
initial position



=
x
o
+
y
o
r0
i
j
Velocity Vector




 dv
=
=
x
+
y
=
Vx
+
Vy
i
i
j
j
V
dt
Acceleration Vector

dv
a=
dt

dVx  dVy 
i +
j = -g j
=
dt
dt


V0 = (Vx)o i + (Vy)o j (known)


= ax i + ay j
1. Horizontal motion, ax=0
dv x
0
dt
Vx = (Vx)0 + axt = (Vx)0
dx
vx 
 (v x )
dt
X = X0 + (Vx)0t
vx  (vx )0  2a x (x  x 0 )  (vx )0
2
2
One independent eqn
X = X0 + (Vx)0t
2
Same as 1st Eq.
2. Vertical motion, ay=-g constant
v y  (v y )0  a y t  (v y )0  gt
1
2
y  y0  ( v y ) 0 t  a y t
2
1 2
y  y 0  ( v y ) 0 t  gt
2
2
2
v y  ( v y ) 0  2g( y  y 0 )
two independent eqns
Can be derived from
above two Eqs.
12-7 Curvilinear Motion:Normal and Tangential components.
Path of motion of a particle is known.
1. Planar motion
ρ
o
s
n
o’
un
p u
t
path
t
Here:
t (tangent axis ): axis tangent to the curve at P and positive in
the direction of increasing S; ut: unit vector
n (normal axis ): axis perpendicular to t axis and directed from P
toward to the center of curvature o’; un: unit vector
o’ = center of curvature
r = radius of curvature
p = origin of coordinate system tn
(1) Path Function
st   s (known)
(2) Velocity
ds
v
dt
ρ



 v  vu t  st u t
un dθ
p
ds
ut
un
(3) Acceleration

du t
dv d 

a
 su t   su t  s
dt dt
dt
 

ut '  ut  dut


dut  ut d  d

 

 dut  dut ut  dut
 
dut // un
o’
ut’
ut’
p
dθ
ut
dut

dut d 


u n  u n
dt
dt
rd  ds
  
s
r

du t s 

 un
dt r
  s 2 


 a  su t  u n  a t u t  a n u n
r
at: Change in magnitude of velocity
an: Change in direction of velocity
at ds  vdv
 v2 
dv  v 2 

 a  v u t  u n  v u t  u n
r
ds
r
  dy  2 
1   dx  

r 
2
d y
dx 2
3/ 2
If the path in y = f ( x )
12-8 Curvilinear Motion：Cylindrical Components
1. Polar coordinates
(1) coordinates (r,)

r：radial coordinate ,ur

：transverse coordinate ,u (c.c.w)


u
r
o

p

ur
r
Reference line
(2) Position


r  rur
(3) Velocity

  dr d 



v r 
 (rur )  rur  rur
dt dt



ur '  u  ur




ur  ur u  u

u r  

u
t
t

u

u r
 

u
lim
lim

t

t
t 0
t 0

ur ' u
r
ur


u 1
d  1



u r  u 




 v  rur  ru


 vr ur  v u
v  (r ) 2  (r ) 2
r 
 
rate of change of the length of the radial coordinate.
angular velocity (rad/s)
(4) Acceleration

 dv 


a
 v  (vr u r  v u  )
dt



u  '  u   u 


u   u    



u   (u r )  u r


u  ur





a  rur  ru  (r)u  rur


 (r  r 2 )ur  (r  2r)u


 ar ur  a u

a  a  (a 2 )  (a  ) 2
  angular acceleration




 v r u r  v r u r  v  u   v  u 

u 
 

(

)u r
lim
lim
t
t 0 t
t 0
2. Cylindrical coordinates

uz  3D
z
u

rp
v
y
x


ur
r
Position vector



rp  rur  zu z
Velocity


  




v p  rp  (ru r )  zu z  ru r  ru   z u z
Acceleration



  


a p  v p  (rur  ru )  zu z
 (r  r 2 )ur  (r  2r)u  zu z
12.9 Absolute Dependent Motion Analysis of
Two Particles
1. Absolute Dependent Motion
The motion of one particle depends on the corresponding motion of
another particle when they are interconnected by inextensible cords
which are wrapped around pulleys.
A
B
A
B
2.Analysis procedure
(1)position-coordinate equation
A. Specify the location of particles using position
coordinates having their origin located at a fixed point
or datum line.
B. Relate coordinates to the total length of card lT
(2)Time Derivatives
Take time derivatives of the position-coordinate
equation to yield the required velocity and
acceleration equations.
3. Example
Datum
Datum
SB
SA
B
A
(1)position-coordinate equation

lT  S A  CD  S B
(2) Time Derivatives

l  S  CD
 S
T
A
B
0  S A  SB  VA  VB
VA  VB
aA  aB
12.10 Relative-Motion Analysis of Two Particles
1. Translating frames of reference
A frame of reference whose axes do not rotate and are only permitted to
translate relative to the fixed frame.
z’
A
z
rA
o
y
y’
rB/A
x’
rB
B
x
xyz:fixed frame
x’y’z’:translating frame moving with particle A
rA、rB : absolute positions of particle A & B
rB/A : relative position of B with respect to A
2. position vector
rA
rB
A
rA
rB/A
3. velocity Vector

drB d  
 (rA  rB / A )
dt dt
  
VB  VA  VB / A
O
rB/A
rB
B
VB/A : relative velocity observed from the translating frame.
4. acceleration vector
d 
d  
(VB )  (VA  VB / A )
dt
dt
  
a B  a A  a B/ A
aB/A:acceleration of B as seen by an observer located at A and
translating with x’y’z’ frame.