Dynamics (Meriam and Kraige, Ed. ,2013) Chapter 1. Introduction

Transcription

Dynamics
th
(Meriam and Kraige, 7 Ed. ,2013)
Chapter 1. Introduction
Engineering Mechanics
Statics:
Statics
Dynamics
Strength of Materials
Vibration
 (F
+ Fr ) = 0 ,distribution of reaction force Fr from the applied
force Fa
Dynamics:
a
 (F
a
+ Fr ) = mx , x(t)= f(F(t)) displacement as a function of
time and applied force
Strength of Materials: δ = f(P) deflection and applied force on deformable
bodies
Vibration: x(t) = f(F(t)) on particles and rigid bodies
1
Newtonian Dynamics
‧Kinematics: the relation among
dx(t )
d 2 x(t )
x(t ), x (t ), and 
x(t ), x 
and 
x
dt
dt 2
without reference to applied force
‧Kinetics: the relation between

x(t ) and F (t )
Terms to Know
‧Reference frame: Coordinate system
‧Inertial System: Newton’s 2nd Law of motion
‧Particle and Rigid body
‧Scalar and Vector
2
Chap. 2 Kinematics of Particles
Rectangular Coordinates r ( x, y, z )
3
Cylindrical Coordinates
r (r , , z )
Spherical Coordinates
r ( R,  ,  )
Displacement, Velocity, and Acceleration

s
2
s

1
v
v

x
x
4
ds 
dv 
1
2
1
2
dx 
t

2
t
1

t
t

t
t
vdt
adt
1
2
1
2
 
xdt

t
t
2
1
t
 )dt
(  2 xdt
t
1
Velocity and Acceleration Vector
dr
r 
v
dt

r
5
dr
a
dt
Rectangular Coordinates
r
r  xi  yj  zk
x
x 
 x 
 


r  xi  yj  zk r   y  r   y  

r

y
 
 
 

 z 
 z 
 
r  
xi  
yj  
zk
z 
6
Centrifugal and Tangential Acceleration
dr
r 
 v  v et  vet
dt
Time derivative of a vector
dr

r
dt
 vet  ve t
v
 vet  v en

 : radius of curvature
7
Time Derivative of the Unit Vectors
in Polar (Cylindrical) Coordinates (2D)
de r
 e
d
de
 e r
d
de r d  de r 
e r 

  e
dt
dt d
de d de
e  

 e r
dt
dt d
8
Cylindrical Coordinates (3D)
r  re r  ze z
r  re r  re r  ze z  ze z
 rer  re  ze z

r  
re r  re r  re  re  re   
ze z
 (
r  r 2 )e r  ( r  2r)e  
ze z
9
r  Re r
r  R e  Re
r
r
ω  e z  e
   sin  e r  cos  e   e
  sin e r  e   cos  e
e r  ω  e r
 o  e   cos  e
 r  R e r  R cos  e  Re
10
Velocity and acceleration in
r  v  v e  v e  v e
v  R
R

R



R
v  R cos 
v  R



r aa e a e a e
  R  R cos 
a R
R

R
2


2
2

R
cos  d
  sin 
a 
 R    2 R
R dt
1 d
a 
R    R sin  cos 

R dt
2

2

11
2
Chap.3 Kinetics of Particles
3.1
3.2
3.3
3.4
Force, Mass, and Acceleration
Work and Energy
Impulse and Momentum
Impact and Orbital Mechanics
3.1 Force, Mass, and Acceleration
‧Newton’s Second Law Equation of Motion F  mr
‧Inertial System: A coordinate system where F  mr
‧Free-body diagram
12
Engineering Mechanics Dynamics -- IAA
Sample 3/1
A 75-kg man stands on a spring
scale in an elevator. The tension
T in the hoisting cable is 8300 N.
Find the reading R of the scale in
newtons and the velocity υ of the
elevator after 3 seconds. The total
mass of the elevator, man, and
scale is 750kg.
 Fy  may 
T  m1g  m1 
y
8300  7360  750ay
ay  1.257 m s
2
 Fy  may 
R  m2 g  m2 
y
R  736  75(1.257)
R  830 N
  a dt 
 

3
  0   1.257 dt
0
  3.77 m s
13
Sample 3/3
The 250-lb concrete block A is released from rest in the position shown and
pulls the 400-lb log up the 30° ramp. If the coefficient of kinetic friction
between the log and the ramp is 0.5, determine the velocity of the block as it
hits the ground at B.
y1  0
 N  m1g cos  m1 
 N  2T  m g sin  m 

1
1 x1
, 4 equation for 4 unknowns

y2
m2 g  T  m2 
2 x1  y2  constant
14
y2
y1
x1
x2
B
30 30
A  2B
a
y
x
The steel ball is suspended from the
accelerating frame by the two cords A
and B. Determine the acceleration of
the frame which will cause the tension in
A to be twice that in B
mg
 Fx  max
2B sin30  B sin30  mx
 Fy  0
2B cos30  B cos30  mg  my  0
xa
Eliminate B and get 
15
g
3 3
problem 03/19
The 10-kg sphere is suspended from the 15-kg frame sliding
down the 20° incline. If the coefficient of kinetic friction
between the frame and incline is 0.15, compute each tension
TA
of wires A and B
25(9.81) N
y
45
45
TB
y
20
N
x
0.15 N
Sphere alone：
 Fy  0
x
20 10(9.81) N
(TA  TB )cos45 10(9.81)cos20  0
Frame and sphere as an unit：
 Fy  N  mg cos  my  0
TA  TB  130.4 N
 Fx  max
N  25(9.81)cos20  0
N  230 N
 Fx  mg cos   N  mx
(TB  TA )sin 45  9.81sin 20  10(1.973)
TB  TA  19.56 N
25(9.81)sin20  0.15(230)  25a
Solution：
TA  75.0 N, TB  55.4 N
a  1.973 m/s2
16
The system is released from rest with
the cable taut. Neglect the small
mass and friction of the pulley and
calculate the acceleration of each
body and the cable tension T upon
release if (a) μs = 0.25, μk = 0.2 and (b)
μs = 0.15, μk = 0.1
T
B
Check for motion. Assume static equilibrium.
From B：T  196.2 N
Mass A：
 Fx  0
m2 g
y
x
T
196.2  F  (60)(9.81)sin30  0
F  9.81 N
A
F
N
m1g
x
Fmax  s N
 (0.25)(60)(9.81)cos30  127.4 N (a)
No motion for (a)： a  0, T  196.2 N
17
The system is released from rest with
the cable taut. Neglect the small
mass and friction of the pulley and
calculate the acceleration of each
body and the cable tension T upon
release if (a) μs = 0.25, μk = 0.2 and (b)
μs = 0.15, μk = 0.1
T
B
Fmax  (0.15)(60)(9.81)cos30  76.5 N
m2 g
y
motion for (b)
x
T
A
F
N
x
A： Fx  max
T  (60)(9.81)sin30  (0.1)(60)(9.81)cos30  60a
B：  Fy  max  (20)(9.81)  T  20a
2
Solution：a  0.589 m/s , T  208 N
m1g
18
Non-constant Acceleration
The chain is released from rest with
the length b of overhanging links just
sufficient to initiate motion. The
coefficients of static and kinetic fiction
between the links and the horizontal
surface have essentially the same
value μ. Determine the velocity υ of
the chain when the last link leaves the
edge. Neglect any friction at the
corner.
 ( L  b) g
T0
F
N   ( L  b) g
 gb
Let  = mass / length ：
F   N   g  ( L  b)
F  0
T0   g  ( L  b)  0  T0   gb
L
Solve to obtain： b 
1 
19
 g ( L  x)
T
 g ( L  x)
 g ( L  x)
 gx
 F  ma
T   g  ( L  x)   ( L  x)a
 gx  T   xa
Eliminate T to obtain：
g
a  
x  [ x(1   )   L]
L
 d  
xdx
g
[ x (1   )   L ]dx
0
b L
1 2 g x2
  [ (1   )   Lx]bL
2
L 2
Substitute b and simplify：


20
 d  
L
gL
1 
Another approach:
 xg    L  x  g    Lx
xg   g  L  x   Lx
g
x 1      L 

L
g 1   

x
x  g
L

x
t
2
2
x e ,  a 0 a 
at
x  c1e  c2e
 at
g 1   
L
L

1 
21
Curvilinear Motion in Polar
Coordinates
F  mr
 
 Fr 
r  r 2 
 F   m     
 
 r  2r 
r  re r  ze z
r  re r  re r  ze z  ze z
 re r  re  ze z

r  
re r  re r  re  re  re  
ze z
 (
r  r 2 )e r  (r  2r)e  
ze z
22
Sample 3/10
Tube A rotates about the vertical O-axis
with a constant angular rate    and
contains a small cylindrical plug B if mass
m whose radial position is controlled by
the cord wound around the drum of
radius b. Determine the tension T in the
cord and the horizontal force Fθ exerted
by tube on the plug if the constant angular
rate of rotation of the drum is ω0 first in
the direction for case (a) and second in
the direction for case (b). Neglect friction.
23
 Fr  mar 
 T  m(r  r2 )
 F  ma 
F  m(r  2r)
case (a) T  mr 2
F  2mb0
case (b) T  mr 2
F  2mb0
If the 2-kg block passes over the top B of the circular
portion of the path with a speed of 3.5 m/s, calculate the
magnitude NB of the normal force exerted by the path on
the block. Determine the maximum speed υ which the
block can have at A without losing contact with the path.
2
 Fr  m(r  r )  m r
(3.5)2
 2(9.81)  N B  2
2.4
N B  9.41 N
2
r
Loss of contact at A： N A  0
mg
mg
FB
tθ
NB
nr
FA
2
 Fr  m r
θt
NA  0
30 rn
24
2

 mg cos30  m
2.4
  4.52 m/s
problem 03/66
The small sphere of mass m is suspended initially at rest by the two
wires. If one wire is suddenly cut, determine the ratio k of the
tension in the remaining wire immediately after the other wire is cut
to the initial equilibrium tension.
30
30
T1
Equilibrium：
 F  0  T1  mg
T1
mg
30
Motion：
 Fn  man  0  T2  mg sin 30  0
T2 mg sin 30
k 
 0.5
T1
mg
T2

mg 30
w try using x-y coordinates
25
n
t
A small bead of mass m is carried by a circular hoop of radius r which
rotates about a fixed vertical axis. Show how one might determine
the angular speed ω of the hoop by observing the angle θ which
locates the bead. Neglect friction in your analysis.
 Fy  0
N cos  mg  0

y

n

2

F

m
(
r

r

)
r
 N sin  m(r sin ) 2
r
N
N  mg / cos
(
mg
)sin  mr sin 2
cos

mg
g
r cos
g
cos


1
Note that
2
r
g
2 
is a restriction.
r
26
problem 03/76
Determine the speed υ at which the race car will have no reliance
on friction to the banked track. In addition, determine the minimum
and maximum speeds, using the coefficient of static friction μs =
0.9.
For no slipping tendency, set F to zero on

F

0
N
cos30
 mg  0
 y
r
2
2

 Fr  m r N sin 30  m 1200
Solve： N  1.155 mg ,   149.4 ft/sec
min  0 as
 max  tan 1  s  tan 1 (0.9)  42.0  30
y
For max , set F  Fmax   s N
mg


F

0
N
cos30

mg


N
sin
30
0
 y
s
F
30
n
N
2
max
2


 Fr  m r   s N cos30  N sin 30  m r
with  s  0.9 N  2.40 mg
max  345 ft/sec
27
A small vehicle enters the top A of the circular path with a horizontal
velocity υ0 and gathers speed as it moves down the path. Determine an
expression for the angle β which locates the point where the vehicle
leaves the path and becomes a projectile. Evaluate your expression for
υ0 = 0. Neglect friction.
 F  ma ,
mg sin   ma , a  g sin 


 d  a ds,   d  0 g sin  ( Rd )
0
 2  02  2 gR (1  cos )
mg
0

N
R
t

n
2
 Fr  mar ,  mg cos  N  m R
02
N  mg cos   m  2mg (1  cos  )
R
02
 mg (3cos   2 
)
gR
θ
02
02
1 2
When N  0, so 3cos   2 
  cos ( 
)
gR
3 3gR
2
For 0  0,   cos 1 ( )  48.2
3
28
3.2 Work and Energy
Work and Kinetic Energy
U   F T dr
  m
r T dr
  mr T dr
1
 mr T r
2
29
Work and Potential Energy
2
2
U12   F  dr   (mgj)  (dxi  dyj)
1
1
y2
  mg  dy  mg ( y2  y1 )
y1
Gme m
e r  dre r
2
1
1
r
r2 dr
 Gme m  2
r1 r
1 1
 Gme m(  )
r2 r1
2
U12   F  dr  
 mgR 2 (
30
2
1 1
 )
r2 r1
Sample 3/15
A satellite of mass m is put into an elliptical orbit around the earth. At point A, its
distance from the earth is h1 = 500 km and it has a velocity υ1 = 30000 km/h.
Determine the velocity υ2 of the satellite as it reaches point B, a distance h2 = 1200
km from the earth.
 1 1
U1-2  mgR2   
 r2 r1 
1 2 1
1 1
2 1
m1  mgR     m22
2
2
 r2 r1  2
 1 1
2    2gR   
 r2 r1 
2
2
1
2
2
103
103 
 30 000 
3 2
2  

  2(9.81) (6371)(10)  

6371

1200
6371

500
 3.6 


2
 69.44(106 ) 10.72(106 )  58.73(106 ) (m/s)2
2  7663 m/s
31
Potential Energy
V   F T dr
GMm
dr
2
r
mgR 2
   2 dr
r
 
V   F T dr
mgR 2

r
  mgdy
r2
r1
 mgh
32
Conservative Force
Principle of Work and Energy
T  W  V  0
V  W  T
‧Kinetic energy
dU FT dr
dr
‧Power P=

 FT
 FT r
dt
dt
dt



i  j k
x y
z
F  V

33
A satellite is put into an elliptical orbit around the earth and has a
velocity υP at the perigee position P. Determine the expression for
the velocity υA at the apogee position A. The radii to A and P are,
respectively, rA and rP. Note that the total energy remains constant.
Constant total energy is E  TA  VA  Tp  V p
Thus
1 2 mgR 2 1 2 mgR 2
m A 
 m p 
2
rA
2
rp
 A2   p2  2 gR 2 (
1 1
 )
rp rA
 A   p2  2 gR 2 (
34
1 1
 )
rp rA
The chain starts from rest with a sufficient number of links hanging
over the edge to barely initiate motion in overcoming friction
between the remainder of the chain and the horizontal supporting
surface. Determine the velocity υ of the chain as the last link leaves
the edge. The coefficient of kinetic friction is μk. Neglect and friction
at the edge.
 = mass per unit length
k L
b


gb



g
(
L

b
)
For equil. at start
k
1 
k
U  T  Vg
(L  b)2
U    dF  x     k  gxdx    k  g
0
2
1
Lb
 T   L 2
 V g    g ( L  b )(
)
2
2
(L  b)2
1
L2  b 2
2
  L   g
Thus   k  g
L b
Lb
dF
dx
2
x
b
b
Lb
Lb 2
2
Lb
2
2
2
b
)( L  b   k [ L  b ]) Now substitute
L
k
k b
k L
2
)( L [1 
]  k [L 
])
So   g (1 
1  k
1  k
1  k
 2  g (1 

gL
1  k
35
 
gL
1  k
3.3 Impulse and Momentum
Linear momentum
G  m r
F  m
r
d
( m r )
dt
d
 G
dt

G

Impulse
r2



F
dt

m
r
dt

md
r

m
r
|
r1  G



Conservation of Linear Momentum
 Fdt  G
if F  0  G  0
36
Sample 3/19
The horizontal velocities of the ball just before and after impact are separately υ1 =
50 ft/sec and υ2 = 70 ft/sec. If the 4-oz ball is in contact with the racket for 0.02 sec,
determine the magnitude of the average force R exerted by the racket on the ball
and the angle β made by R with the horizontal
m( )  t2 F dt  m( ) 
x 2
 x 1 t1  x

4 /16
4 /16

(50)  Rx (0.02) 
(70cos15 )
32.2
32.2
m( )  t2 F dt  m( ) 
y 1
y 2
t1  y


t2
t
4 /16
4 /16
(0)  Ry (0.02) 
(70sin15 )
32.2
32.2
Rx  45.7 lb, Ry  7.03 lb
y
mgdt
1
mv1
+
t2
t
=
Rx dt
15 x
R  Rx 2  Ry 2  45.72  7.032  46.2 lb
1
t2
t
  tan
Ry dt
1
37
1
Ry
Rx
 tan1
7.03
 8.75
45.7
problem 03/207
The 1.62-oz golf ball is struck by the five-iron and acquires the velocity shown in a
time period of 0.001 sec. Determine the magnitude R of the average force exerted
by the club on the ball. What acceleration magnitude ɑ does this force cause, and
what is the distance d over which the launch velocity is achieved, assuming
constant acceleration?
mg  0
150 ft / sec
25


RT  m : R(0.001)=
1.62/16
(150) , R =472 lb
32.2
1.62/16
a , a  150,000 ft / sec2 (4660 g )
32.2
 2  02  2ad : 1502  02  2(150,000)d , d  0.075 ft or 0.900in
R  ma : 472=
38
Angular Impulse and Momentum
H O  r  mv
HO  r  mv
 m(vz y  vy z)i  m(vx z  vz x) j  m(vy x  vx y)k
i
HO  x
j
y
vx vy
39
k  H x  m(vz y  vy z) 

  
z   H y   m(vx z  vz x) 
vz  H z  m(vy x  vx y)
Time Derivative of Angular Momemtum
HO  r  mv
0

HO  r  mv  r  mv  v  mv  r  mv  r F  MO

M  H
O
O
Conservation of Angular Momentum

 MO  H
O

t2
t1
 M O dt  ( H O ) 2  ( H O )1  H O
The total angular impulse on a particle of mass m
about a fixed point O equals the corresponding
change in angular momentum about that point.
t2
(H O )1    M O dt  ( H O ) 2
t1
Principle of Conservation of Angular Momentum
if  M O  0 , then H O  0 or (H O )1 =(H O ) 2
40
Sample 3/25 Molynia Orbit
A comet is in the highly eccentric orbit shown in the figure. Its speed ate the most
distant point A, which is at the outer edge of the solar system, is υA = 740 m/s .
Determine its speed at the point B of closest approach to the sun.
(HO ) A  (HO )B
mrA A  mrB B
rA A 6(109 )740
B 

rB
75(106 )
 B  59 200 m/s
41
The central attractive force F on an earth
satellite can have no moment about the center
O of the earth. For the particular elliptical orbit
with major and minor axes as shown, a satellite
will have a velocity of 33880 km/h at the perigee
altitude of 390 km. Determine the velocity of the
satellite at point B and at apogee A. The radius
of the earth is 6371 km.
B
11720 km / h
M
B
r

O
 H 0  0 so H O  constant
rmin  6371  390  6761 km
33880 km / h
rmax  2(13520)  6761  20279 km
o
For H O constant
A
6371(33880)  11720 B  20279 A
 A  19540km / h
 B  11300km / h
A
rmax
2(13520) km
rmin
42
problem 03/228
The two spheres of equal mass m are able to slide along the horizontal rotating
rod. If they are initially latched in position a distance r from the rotating axis with
the assembly rotating freely with an angular velocity ω0 , determine the new
angular velocity ω after the spheres are released and finally assume positions
at the ends of the rod at a radial distance of 2r. Also find the fraction n of the
initial kinetic energy of the system which is lost. Neglect the small mass of the
rod and shaft.
H  0 ; 2mr0 (r )  2m(2r ) (2r )  0   0 / 4

1
1
T  2( m[r0 ]2 )  2( m[2r 0 ]2 )  mr 202 (3 / 4)
2
2
4
3
n  T / T  mr 202 / mr 202  3 / 4
4
43
 M 0  H 0
mgl cos  
The simple pendulum of mass m and length l
is released from rest at θ = 0. Using only the
principle of angular impulse and momentum,
determine the expression for  in terms of θ
and the velocity υ of the pendulum at θ = 90°.
Compare this approach with a solution by the
work-energy principle.
O

T
d
(ml 2)  ml 2
dt
g
  cos 
l
From  d   d
2   g
|  cos d
2 0 0 l
g
 2  sin 
l
2g
 90 
l
so at   90
  l  2 gl
l
By work-energy
V  T
1
mgl  m 2
2
  2 gl
mg
44
Direct Central Impact
45
Coefficient of Restitution

e


e

t
t0
t0
0
t
t0
t0
0
Fr dt
m1[1'  (0 )] 0' 1


Fd dt m2[0  (1 )] 1 0
Fr dt
m2 (2' 0 ) 2' 0


Fd dt m2 (0 2 ) 0 2
2' 1' relative velocity of separation
e

1 2 relative velocity of approach
46
47
Oblique Central Impact
m1 (1 )n  m2 (2 )n  m1 (1' )n  m2 (2' )n
m1 (1 )t  m1 (1' )t
m2 (2 )t  m2 (2' )t
(2' )n  (1' )n
Vn
e

(1 )n  (2 )n
Vn
48
Sample 3/29
A ball is projected onto the heavy plate with a velocity of 50 ft/sec at the
30° angular shown. If the effective coefficient of restitution is 0.5, compute
the rebound velocity υ′ and its angle θ′.
(2' )n  (1' )n
e
(1 )n  (2 )n
0  (1' )n
0.5 
50sin30  0
(1' )n  12.5 ft/sec
m(1 )t  m(1' )t
(1' )t  (1 )t  50cos30  43.3 ft/sec
 '  (1' )n2  (1' )t 2  12.52  43.32
 45.1 ft/sec
 (1' )n 
 12.5 

  tan  '   tan1 

16.10

 43.3 
 (1 )t 
'
49
1
Sample 3/30
Spherical particle 1 has a velocity υ1 = 6 m/s in the
direction shown and collides with spherical particle
2 of equal mass and diameter and initially at rest. If
the coefficient of restitution for these conditions
is e  0.6 , determine the resulting motion of each
particle following impact.
Also calculate the
percentage loss of energy due to the impact.
m1 (1 )n  m2 (2 )n  m1 (1' )n  m2 (2' )n
5.20  0  (1' )n  (2' )n
(2' )n  (1' )n
e
(1 )n  (2 )n
(2' )n  (1' )n
0.6 
5.20  0
(1' )n  1.039 m/s
(2' )n  4.16 m/s
m1 (1 )t  m1 (1' )t
(1' )t  (1 )t  3 m/s
m2 (2 )t  m2 (2' )t
(2' )t  (2 )t  0
50
As a check of the basketball before the start of a game, the referee
releases the ball from the overhead position shown, and the ball
rebounds to about waist level. Determine the coefficient of
restitution and the percentage n of the original energy lost during
the impact.
  2gh ,  '  2gh'
'
h'
1100
e 

 0.724

h
2100
mgh  mgh'
n
mgh
2100  1100

2100
 47.6%
51
Central-Force Motion
F  mr
 Gmm0 
2 



r

r

 r2  

 r  2r
 0 
 mr 2  h
52
Orbital Mechanics
Gm
1
 C cos  2 0
r
h
53
D’Alembert’s Principle and Inertia Force
F  ma  0
F  mr  0 mr inertia force
D’Alembert’s Principle using inertia force
to treat dynamics by statics
54
Chap. 4 Kinetics of
Systems of Particles
Mass center
mrc   mi ri
Equation of motion
 F   f   mi ri
  F  mrc
Principle of motion of the mass
center the resultant of the
external forces on any system of
masses equal the total mass
times
the
mass
center
acceleration.
fig_04_001
55
Linear Momentum
G   miri
  mi (rc  ρ i )
d
  mirc  ( miρi )
dt
 mrc
56
Kinetic Energy
ri  rc  ρi
ri  rc  ρ i
1
T   mi  riT ri
2
1
 (mircT rc  2mircT ρ i  miρ Tiρ i )
2
1 T
1


 mrc rc   miρ iT ρ i
2
2
fig_04_003
57
Angular Momentum about a Fixed
Point
  (r  m r )
H
O
i
i i
  (r  m r  r  m r )
H
O
i i
i
i i
 O   ri  Fi
 MO
Angular Momentum about c.g.
   ρ  m (r  ρ )   ρ  m r
H
G
i
i C
i
i
i i
HG   ρi  miri
  ρi  mi (rc  ρ i )
  ρi  Fi
 (ρi  mirc  ρi  mi ρ i )
 (ρi  miρ i )
58
  MG
problem 04/22
The man of mass m1 and the woman of mass m2 are
standing on opposite ends of the platform of mass m0 which
moves with negligible friction and is initially at rest with s = 0.
The man and woman begin to approach each other. Derive
an expression for the displacement s of the platform when
the two meet in terms if the displacement x1 of the relative to
s
the platform.
l
m1 x
1
x2 m2
m0
B
A
C
p_04_020
With respect to C,mi xi  constant
l
l
m1l  m2 (0)  m0  m1(s  l  x1)  m2 (s  x2 )  m0 (s  )
2
2
m x  m2 x2
Simplify and get s  1 1
m0  m1  m2
(m  m2 ) x1  m2l
But they meet whenx2  x1  l so s  1
m0  m1  m2
59
Sample 4/4
sp_04_03_01
A shell with a mass of 20 kg is fired from
point O, with a velocity u = 300 m/s in the
vertical x-z plane at the inclination shown.
When it reaches the top of its trajectory at P,
it explodes into three fragments A, B, and C.
Immediately after the explosion, fragment A
is observed to rise vertically a distance of 500
m above P, and fragment B is seen to have a
horizontal velocity vB and eventually lands at
point Q. When recovered, the masses of the
fragments A, B, and C are found to be 5, 9,
and 6 kg, respectively. Calculate the velocity
which fragment C has immediately after the
explosion. Neglect atmosphere resistance.
60
Sample 4/4
t  uz / g  300(4/5)/9.81  24.5s
uz2 [(300)(4/5)]2
h

 2940m
2g
2(9.81)
A  2ghA  2(9.81)(500)  99.0m/s
B  s / t  4000/ 24.5  163.5m/s
[G1  G2 ]
sp_04_03_01
mv  mAv A  mB vB  mC vC
3
20(300)( )i  5(99.0k)  9(163.5)(i cos45  jsin 45 )  6vC
5
6vC  2560i 1040j  495k
vC  427i 173.4j  82.5k m/s
C  (427)2  (173.4)2  (82.5)2  468m/s
61
Steady Mass Flow
1 A11  2 A22  m
G  (m)v2  (m)v1  m(v2  v1 )
F  mv
62
problem 04/35
The jet aircraft has a mass of 4.6 Mg and a drag (air
resistance) of 32 kN at a speed of 1000 km/h at a particular
altitude. The aircraft consumes air at the rate of 106 kg/s
through its intake scoop and uses fuel at the rate of 4 kg/s.
If the exhaust has a rearward velocity of 680 m/s relative to
the exhaust nozzle, determine the maximum angle of
elevation at which the jet can fly with a constant speed of
1000 km/h at the particular altitude in question.
T  ma (u )  mf u
 106(680  1000/3.6)  4(680)
mg  4.6(9.81) kN
T
 45400 N  45.4 kN
R  32 kN
x
 Fx  max  0
45.4  32  4.6(9.81)sin  0

  1000 km/h
N
63
sin  0.296    17.22
problem 04/38
The jet water ski has reached its maximum velocity of 70 km/h
when operating in salt water. The water intake is in the
horizontal tunnel in the bottom of the hull, so the water enters
the intake at the velocity of 70 km/h relative to the ski. The
motorized pump discharge water from the horizontal exhaust
nozzle of 50-mm diameter at the rate of 0.082 m3/s. Calculate
the resistance R of the water to the hull at the operating speed.
Resistance R equals net trust T
where T  m(u  )
Density of salt water,
Nozzle velocity u  Q / A 
  1030 kg/m3
m  Q  1030(0.082)  84.5 kg/s,   70
0.082
 41.8 m/s
2
 (0.050)
4
1000
 19.44 m/s
3600
R  T  84.5(41.8 19.44)  1885 N
64
problem 04/45
The jet-engine thrust reverser to reduce an aircraft speed of 200
km/h after landing employs folding vanes which deflect the
exhaust gases in the direction indicated. If the engine is
consuming 59 kg of air and 0.65 kg of fuel per second, calculate
the braking thrust as a fraction n of the engine thrust without the
deflector vanes. The exhaust gases have a velocity of 650 m/s
relative to the nozzle.
 F  mu
With reversers in place
TR  mg u sin30  ma
 (50  0.65)(650)sin30  50(55.6  0)
 16460  2780  19240 N
30
Without reversers
650 m/s
T  mg u  ma
TR
 (50  0.65)650  50(55.6)
x
 32900  2780  30100 N


30
650 m/s
so n 
  200/3.6  55.6 m/s
65
19240
 0.638
30100
Variable Mass system

R  m( 0 )  mu
F  R  m

  F  m  mu
66
Rocket Propulsion
mu  pA  mg  R  m
67
Sample 4/12
A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuel
is exhausted. The relative nozzle velocity of the exhaust gas has a constant value u at
atmospheric pressure throughout the flight. If the residual mass of the rocket structure
and machinery is mb when burnout occurs, determine the expression for the maximum
velocity reached by the rocket. Neglect atmospheric resistance and the variation of
gravity with altitude.
Solution I (F=ma solution)
   mu  mg  m
T  mg  m, T  mu  mu

m dm
t
dm
 gdt   d  u 
 g  dt
0
m0 m
0
m
m
  u ln 0  gt
m
let mb：mass of rocket when burnout occurs
d  u
tb  (mb  m0 )/ m
max  u ln
m0 g
 (m0  mb )

mb m
68
Sample 4/12
A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuel
is exhausted. The relative nozzle velocity of the exhaust gas has a constant value u at
atmospheric pressure throughout the flight. If the residual mass of the rocket structure
and machinery is mb when burnout occurs, determine the expression for the maximum
velocity reached by the rocket. Neglect atmospheric resistance and the variation of
gravity with altitude.
Solution II (Variable-Mass solution)
   mg  m  mu

 F  mg,  F  m  mu
  mu  T  T  mg  m
mu
69
same as Solution I
Chap.5 Planar Kinematics
fig_05_001
70
Z
Fixed Axis Rotation
Y
X
displacement r
velocity r  ω  r
fig_05_00
4
acceleration r  ω  (ω  r)  ω  r
71
Sample 5/3
The right-angle bar rotates
clockwise with an angular
acceleration
  4k rad/s2
Write the vector expressions for the
velocity and acceleration of point A when
  2k rad/s
sp_05_03_01
[v    r ]
[an    (  r )]
[at    r ]
[a  an  at ]
v  2k  (0.4i  0.3j)  0.6i  0.8j m / s
an  2k  (0.6i  0.8j)  1.6i  1.2j m / s2
at  4k  (0.4i  0.3j)  1.2i  1.6j m / s
a  2.8i  0.4j m / s2
2
  0.62  0.82  1m/s
a  2.82  0.42  2.83 m/s
72
Determine the velocity and acceleration
of (a) point A and (b) point B with
  6 rad/s,   4rad/s2
(a) A    r A  (6k  45 j)
 270i mm/s
aA    r A   2 r A
 4k  45 j  62 (45 j)
 180i 1620 j mm/s2
(b) B    r B  6k  (30i  45 j)
 270i 180 j mm/s
aB    r B   2 r B
 4k  (30i  45 j)  62 (30i  45 j)
 900i 1740 j mm/s2
p_05_002
73
Sample 5/4
A wheel of radius r rolls on a flat surface without
slipping. Determine the angular motion of the
wheel in terms of the linear motion of its center
O.
s  r
O  r, aO  r
where O  s, aO  O  
s,
  , and     
sp_05_04_01
Determine the acceleration of a point on the rim of the wheel
as the point comes into contact with the surface on which the
wheel rolls.
y  r  r cos  r (1  cos )
x  s  r sin  r (  sin )
x  r(1  cos )  O (1  cos ) y  r sin  O sin


y   sin    cos
x   (1  cos )   sin
O
O
 aO (1  cos )  r 2 sin
O
O
 aO sin  r 2 cos
74

x  0 and 
y  r2
Sample 5/7
Calculate the velocity of point A on the
wheel without slipping for the instant
represented.
v A  vO  v A/O  vO  ω r0
ω = -10k rad/ s
r0  0.2(i cos30  jsin30 )  0.1732i  0.1j m
vO  3i m/ s
v A  3i 
i
j
0
0
0.1732 0.1
sp_05_07_0
1
k
10  3i  1.732j  i
0
 4i  1.732j m/s
A  42  (1.732)2  19  4.36 m/s
75
Sample 5/11
Locate the instantaneous center of
zero velocity and use it to find the
velocity of point A for the position
indicated.
[   / r]
  O / OC  3 / 0.300  10 rad / s
AC  (0.300)2  (0.200)2  2(0.300)(0.200)cos120
 0.436 m
[  r]
 A  AC  0.436(10)  4.36 m/s
76
Sample 5/8
r2
r3
r1
r  r1  r2  r3
r  ω1  r1  ω2  r2  ω3  r3
r
0  ω1k 100j  ω2k  (175i  50j)  2k  75i
100ω1  50ω2 



175
ω

150

2

sp_05_08_0
1
3
6
ω1   , ω2  
7
7
77
Sample 5/14
r2
r  r1  r2  r3
r  ω1  r1  ω2  r2  ω3  r3
r3
r1
r  ω 1  r1  ω1  (ω1  r1 )
ω 2  r2  ω2  (ω2  r2 )
r
ω 3  r3  ω3  (ω3  r3 )
sp_05_08_0
1
3
3
r  1k 100j  ( k )  ( k 100j)   2k  (175i  50j)
7
7
6
6
( )k  [( k)  (175i  50j)]  0  2k  (2k [75i])
7
7
1  0.1050 rad/s2
2  4.34 rad/s2
78
Sample 5/9
sp_05_09_01
v A  vB  v A/ B
[  r]
5 1500(2 )
65.4 ft / sec
12 60
5
14

sin  sin60
A
65.4

sin78.0 sin72.0
A/ B
65.4

sin30 sin72.0
[   / r ]
AB 
A  67.3 ft /sec
A/ B  34.4 ft /sec
A/ B 34.4

 29.5 rad /sec
AB 14/12
vG  v B  vG / B
B 
  sin1 0.309  18.02
G / B  GBAB 
GB
4
 A/ B  (34.4)  9.83 ft / sec
14
AB
G  64.1ft /sec
79
y
Sample 5/9
r1
r2
r
x
r  r1  r2
r  ω1  r1  ω2  r2
sp_05_09_01
r  ω 1  r1  ω1  (ω1  r1 )
ω 2  r2  ω2  (ω2  r2 )
80
Sample 5/15
aA  10280cos60 1015cos18.02  ( A/ B )t sin18.02
0  10280sin60 1015cos18.02  ( A/ B )t cos18.02
sp_05_15_01
aA  aB  (aA/ B )n  (aA/ B )t
aB 
5 1500[2 ] 2
(
)
12
60
2
 10280 ft / sec
 A  3310 ft /sec2
[an  r2 ]
[  at / r]
[an  r2 ]
14
(29.5)2
12
 1015 ft / sec2
(aA/ B )n 
( A/ B )t  9030 ft /sec2
 A/ B  9030/(14/12) ft /sec2  7740 rad/sec2
 A  3310 ft / sec2
81
( A/ B )t  9030 ft /sec2
General Motion: Rotation + Translation
fig_05_005
fig_05_006
82
Instantaneous Center
fig_05_007
83
Body-Fixed Coordinates in Rotation
Coriolis Acceleration
rA
fig_05_011
rA  rB  ρ
rA  rB  ω  ρ+ ρ
rA  rB  ω   ω  ρ  ω  ρ  ρ+
 2ω ρ
rA  rB  ω   ω  ρ  ω  ρ  ω ρ+
 ω ρ+
 ρ


 rB  ω   ω  ρ  ω  ρ+ 2ω ρ  ρ

Coriolis acceleration
84
Sample 5/16
The motion of slider A is separately controlled, and at this
2
instant, r = 6 in., r =5 in./sec, and r =81 in./sec . Determine the
absolute velocity and acceleration of A for this position.
y
v A    ρ  ρ
v A  4k  6i  5i  24j  5i in./sec
A  (24)2  (5)2  24.5 in./sec
a A    (  ρ)    ρ  2  ρ  
ρ
x
sp_05_16_01
  (  ρ)  4k  (4k  6i)  4k  24j  961 in./sec2
  ρ  10k  6i  60j in./sec2
2  ρ  2(4k)  5i  40j in./sec2

ρ  81i in./sec2
a A  (81  96)i  (40  60) j  15i  20j in./sec2
aA  (15)2  (20)2  25 in./sec2
85
With and without Body-Fixed Coordinate
Let r  5i,   2k,   3k
r    r  10j

r      r     r
= -20i + 30j
displacement r
velocity r  ω  r
acceleration r  ω  (ω  r)  ω  r
86
problem 05/163
A vehicle A travels with constant speed v along a
north-south track. Determine the Coriolis
acceleration aCor as a function of the latitude θ at (a)
the equator and (b) the north pole.
z
B

   ( sin j  cos k)
acor  2 
A


 2k ( sin j  cos k)
y
 2 sin i (west)
For   500 km/h
(a) Equator,   0  acor  0
(b) North pole,
500
 0.0203 m/s2
3.6
The track provides the necessary westward
acceleration so that the velocity vector is properly
rotated and reduced in magnitude.
  90  acor  2(7.292 105 )
p_05_161
87
Chap. 6 Dynamics of Planar Rigid Body
Equation of Motion
‧The resultant of the external forces equals to the inertia of the mass center
‧The resultant moment about C.G. of the external forces equals to the rate change
of the angular moment about C.G.
F  mrG

M  H
G
88
G
Equation of Motion in 2D
Angular momentum
HG   ρi  miρ i
  ρi   miω  ρi 

  ω dm
2
 Iω
F  mrc

MG  Iω
89
Moment about a Fixed Point
  ρ  mr
MP  H
G
C
  ρ  mrC
 I Gω
  ρ  mrP
 I Pω
if rP  0, P is fixed
  I Oω

 I Pω
90
Determine the value of the force P which
would cause the cabinet to begin to tip.
What coefficient μs of static friction is
necessary to ensure that tipping occurs
without slipping?
0.8m
50(9.81)N
1.2m G 
FA
NA
when tipping impends N A  mg

MG  mg(0.4)  mg(0.6)  I  0

2
3
 F  mg  mx
2

x  g  g
3
As a whole :
P  (m  m1)
x  (50  10)
x  60
x  40g
91
Center of Percussion
F  mrC

MG  IGω
or
MO  IO
ko
ko
Radius of gyration about point O
The resultant force at the
center of percussion is zero.
92
Center of Percussion
-F  RX  mx
mg  RY  my
F  h  I 
F
O
y  0
1
x  
2
2
h 
3
h
mg
RX
RY
93
problem 06/34
The 20-kg uniform steel plate is freely hinged about the z-axis as
shown. Calculate the force supported by each of the bearings at A
and B an instant after the plate is released from rest in the
horizontal y-z plane.
MO  IO
1
 20(9.81)(0.2)  20(0.4)2
3
  36.8rad/s2

x  r
2F
0.2m
G

x  0.2  36.8  7.36m/s2
 Fx  mx
0.2m
O
20(9.81)  2F  20(7.36)
20(9.81) N
1
2F  49.0  mg
4
FA  FB  F  24.5N
t
94
problem 06/38
Determine the angular acceleration and the force on the bearing
at O for (a) the narrow ring of mass m and (b) the flat circular
disk of mass m immediately after each is released from rest in
the vertical plane with OC horizontal.
(a)
  mgr  2mr 2
M

I

 O O
  g / 2r
 Fy  my  mg  O  mr(
g
)
2r
O  mg / 2
O
O
O
r
G
O
r
1 2

M

I


mgr

(
mr  mr 2 )
(b)  O O
2
  2g / 3r
G
 Fy  my  mg  O  mr(
y
mg
mg
2g
)
3r
O  mg / 3
95
Sample 6/5
A metal hoop with a radius r = 6 in. is
released from rest on the 20° incline. If the
coefficients of static and kinetic friction are
μs = 0.15 and μk = 0.12, determine the
angular acceleration α of the hoop and the
time for the hoop to move a distance of 10 ft
down the incline.
96
Sample 6/5
[ Fx  max ]
mg sin20  F  ma
[ Fy  may  0] N  mg cos20  0
Fr  mr 2
[ MG  I  ]
Assume pure rolling a  r 4 equations for 4 unknowns
F  0.1710mg
and
N  mg cos20  0.940mg
Check if the assumption valid. The friction force be bounded by  N
[ Fmax  s N ]
Fmax  0.15(0.940mg )  0.1410mg
[ Fmax  k N ]
F  0.12(0.940mg )  0.1128mg
So it is slipping f  k N.
[ Fx  max ]
[ MG  I  ]
1
[ x  at 2 ]
2
Solve again the 4 unknowns
mg sin20  0.1128mg  ma a  0.229(32.2)  7.38 ft/sec2
0.1128(32.2)
0.1128mg (r )  mr 2  
 7.26 rad/sec2
6 /12
2x
2(10)
t

 1.646 sec
7.38

97
Sample 6/7
The slender bar AB weighs 60 lb and moves in the vertical plane, with its ends
constrained to follow the smooth horizontal and vertical guides. If the 30-lb force is
applied at A with the bar initially at rest in the position for which θ = 30°, calculate
the resulting angular acceleration of the bar and the forces in the small end rollers
at A and B.
98
Sample 6/7
ax  a cos30  2 cos30  1.732 ft/sec2
ay  a sin30  2 sin30  1.0 ft/sec2
1 60 2
[ MG  I  ]
30(2cos30 )  A(2sin30 )  B(2cos30 ) 
(4 )
12 32.2
60
30  B 
(1.732 )
[ Fx  max ]
32.2
60
A  60 
(1.0 )
[ Fy  may ]
32.2
2
A  68.2 lb B  15.74 lb   4.42 rad/sec
1 60 2
4.39  9.94
[ MC  I   m d ] 30(4cos30 )  60(2sin30 ) 
(4 )
12 32.2
2


4.42
rad/sec
60
60

(1.732 )(2cos30 ) 
(1.0 )(2sin30 )
32.2
32.2
[ Fy  may ]
[ Fx  max ]
60
A 60 
(1.0)(4.42)
32.2
60
30  B 
(1.732)(4.42)
32.2
99
A  68.2 lb
B  15.74 lb
problem 06/82
Determine the angular acceleration of each of the two wheels
as they roll without slopping down the inclines. For wheel A
investigate the case where the mass of the rim and spokes is
negligible and the mass of the bar is concentrated along its
centerline. For wheel B assume that the thickness of the rim is
negligible compared with its radius so that all of the mass is
concentrated in the rim. Also specify the minimum coefficient of
static friction μs required to prevent each wheel from slopping.
MO  I   I  0 M  0
A
mg
G 
r
x
Hence no friction force and s  0
g
F

ma

mg
sin


mr

x x
A   A  sin
r
 N
100
problem 06/82
Determine the angular acceleration of each of the two wheels
as they roll without slopping down the inclines. For wheel A
investigate the case where the mass of the rim and spokes is
negligible and the mass of the bar is concentrated along its
centerline. For wheel B assume that the thickness of the rim is
negligible compared with its radius so that all of the mass is
concentrated in the rim. Also specify the minimum coefficient of
static friction μs required to prevent each wheel from slopping.
MC  IC  mgr sin  2mr2B
B 

M  I   Fr  mr 2
B
mg
G 
r
x
C
F
g
sin
2r
N
g
sin
2r
1
 F  mg sin
2
F 1
s   mg sin / mg cos
N 2
1
s  tan
2
101
problem 06/84
The uniform 12-kg square panel is suspended from point C
by the two wires at A and B. If the wire at B suddenly
breaks, calculate the tension T in the wire at A an instant
after the break occurs.
M A  I   mad
C

T
A
mgb 1 2
b
b
 mb   m 
2
6
2
2
3g

4b
 MG  I 
45
I
r
G
m(aG / A )t
 r
T
b
b 1 2 3g
 mb ( )
4b
2 6
T
mg maA
2
2
mg 
(12)(9.81)  20.8 N
8
8
b
102
Kinetic Energy
1
T  m 2
2
1
T  IO2
2
1
1
T  m 2  IC 2
2
2
103
problem 06/116
A
mg
x

B
mg
x
U  T
U  mgx sin
1
1
T  m 2  I 2
2
2
1
case A： T  m 2  0
2
1
1

case B： T  m 2  mr 2 ( )2  m 2
2
2
r
1
case A： mgx sin  m 2   A  2gx sin
2
case B： mgx sin  m 2  B  gx sin

104
For the pivoted slender rod of length l, determine the distance x
for which the angular velocity will be a maximum as the bar
passes the vertical position after being released in the
horizontal position shown. State the corresponding angular
velocity.
T1  V12  T2
l
g(  x)
2  2 2
l lx x2
 
6 2 2
l
 l  1 1

mg   x    ml 2  m(  x)2 2
2
 2  2 12

d2
set
 0 obtain
dx
l
g(  0.211l )
g
2
max  x0.211  2

1.861
l
l 0.211l 2 (0.211l)2


6
2
2
105
x  0.789l
or
x  0.211l
The solution x  0.789l
would yield the same max
only then the motion is CCW.
Linear Momentum
Angular Momentum
HO  IO
MO  H O
t2
(HO )1   MOdt  (HO )2

F  G
t1
t2
G1   Fdt  G2
t1
106
Sample 6/16
The uniform rectangular block of dimensions shown is sliding to the left
on the horizontal surface with a velocity v1 when it strikes the small step
at O. Assume negligible rebound at the step and compute the minimum
value of v1 which will permit the block to pivot freely about O and just
reach the standing position A with no velocity. Compute the percentage
energy loss n for b = c.
107
Sample 6/16
[HO  IO]
[(HO )1  (HO )2 ]
1
c
b
m
(HO )2  { m(b2  c2 )  m[( )2  ( )2 ]}2  (b2  c2 )2
12
2
2
3
b m
31b
m1  (b2  c2 )2 2 
2 3
2(b2  c2 )
2
2
1
b c b
[T2  V2  T3  V3 ] IO22  0  0  mg[       ]
2
 2  2 2
31b 2 mg
1m 2 2
(b  c )[ 2 2 ] 
( b2  c2  b)
23
2
2(b  c )
g
c2
1  2( (1  2 )( b2  c2  b)
3
b
1 2 1
2
2
m


I

2 2
2
2 
1
O
2


E 2
k 
b c
3b
3
2
n

 1  O 2 2  b2  c2 

1

 2 2 
1 2
E
3
 c2 

1

  2(b  c ) 
m1
41  2 
2
 b 
n  62.5% b  c
108
problem 06/188
The homogeneous sphere of mass m and radius r is projected along the
incline of angle θ with an initial speed v0 and no angular velocity (ω0 =
0). If the coefficient of kinetic friction is μk, determine the time duration t
of the period of slopping. In addition, state the velocity v of the mass
center G and the angular velocity ω at the end of the period of slipping.
109
problem 06/188
t
0  Fydt  m(y  y )  0  N  mg cos
t
0  Fxdt  m(x x )
y
0
x
0
(k mg cos  mg sin )t  m( 0 )
(1)
t
mg
0 MGdt  I (  0 )
G
2
(k mg cos r )t  mr 2
(2)
5
We desire the time t when   r (3)
r
N
Solution of Eqs. (1)-(3)：
20
t
g (7k cos  2sin )
k N
For slipping to cease,
50 k

7k  2tan
7k cos  2sin
2
or k  tan
7
50 k

7k r  2r tan
110
Chap.7 3D Kinematics and Kinetics
Translation
Rotation
v   r
a    r    (  r)
111
Sample 7/1
The 0.8-m arm OA for a remote-control mechanism is
pivoted about the horizontal x-axis of the clevis, and
the entire assembly rotates about the z-axis with a
constant speed N = 60 rev/min. Simultaneously, the
arm is being raised at the constant rate  4 rad/s/ for

the position where   30 , determine (a) the angular
velocity of OA, (b) the angular acceleration of OA, (c)
the velocity of point A, and (d) the acceleration of
point A. If, in addition to the motion described, the
vertical shaft and point O had a linear motion, say, in
the z-direction, would that motion change the angular
velocity or angular acceleration of OA?
112
Sample 7/1
(a)   x  z  4i  6.28k rad/s
(b)
     x  z  x   z
z  x  6.28k  4i  25.1j rad/s2
  25.1j  0  25.1j rad/s2
i
j
k
(c) v A    r  4
0
6.28  4.35i  1.60j  2.77k m/s
0 0.693 0.4
(d) a A    r    (  r)    r    v
i
0
j
25.1
k
i
j
k
0 
4
0
6.28
0 0.693 0.4
4.35 1.60 2.77
 (10.05i)  (10.05i  38.4j  6.40k )
 20.1i  38.4j  6.40k m/s2
113
Body-Fixed Coordinate Translation
114
Rotation
Inertial coordinates X-Y-Z
Body fixed coordinates x-y-z
A: point of interest
B: origin of body-fixed coordinates,
often
the mass center A
Ω: angular velocity of the rigid
angular
velocity of x-y-z about X-Y-Z
i    i j    j k    k
v A  vB    rA / B  vrel
  r    (  r
a a 
A
B
A/ B
A/ B
)  2  vrel  arel
or rA  rB      
            2    

rA  
rB  
115
Sample 7/3,7/4
Crank CB rotates about the horizontal axis
with an angular velocity ω1 = 6 rad/s which is
constant for a short interval of motion which
includes the position shown. The link AB has
a ball-and-socket fitting on each end and
connects crank DA with CB. For the instant
shown, determine the angular velocity ω2 of
crank DA and the angular velocity ωn of link
AB.
116
Sample 7/3,7/4
v A  v B  n  rA/ B
[  r] v A  502 j, vB  100(6)i  600i mm/s
i
j
k
502 j  600i  nx ny
nz
6 
50 100 100
 ny  nz
2  2nx
 nz
2  6 rad/s
0  2nx  ny
[n rA / B  0] 50nx 100ny 100nz  0
4
8
10
nx   rad/s ny   rad/s nz  rad/s
3
3
3
2
2
n  (2i  4j  5k) rad/s
n  22  42  52  2 5 rad/s
3
3
117
Angular velocity
r3
r = r1  r2  r3
r
r2
r = 1  r1  2  r2  3  r3  0
 T
2 r2  0
4 equations for 4 unknows; 2 and 3
r1
if r1, r2 , and r3 are general vectors,
r1  0

then 2Tr2  0
 T
3 r3  0
118
Angular Acceleration
r3
r
r2
r = r1  r2  r3
r = 1  r1  2  r2  3  r3  0
r1

r = 1  1  1  r1 

   2  2  2  r2 

  3  3  3  r3   0
 T
2 r2  0
119
Angular Momentum
x-y-z Body-fixed coordinates at C.G.
ω: angular velocity of the rigid body
HG   [  (   )]dm
HO   [r  (  r)]dm
HG  IG 
(a)
33
31
 I xx I xy I xz 


 I yx I yy I yz   
 I zx I zy I zz 


 I xxx  I xyy  I xzz 


 I yxx  I yyy  I yzz 
I   I   I  
 zx x zy y zz z 
(b)
120
Inertia Matrix
 I xx I xy I xz 



I
I

I
yy
yz 
 yx
 I zx I zy I zz 


I xx   ( y 2  z 2 )dm
I xy   xydm
I yy   ( z 2  x2 )dm
I xz   xzdm
I zz   ( x2  y2 )dm
I yz   yzdm
Principal Axes
 I xx 0
0 I
yy

 0 0
0
0

I zz 
H  I xxx i  I yyy j  I zzzk
121
problem 07/64
Introduce axes
The rectangular plate, with a




0,


,


mass of 3 kg and a uniform small
x
y
z
2
2
thickness, is welded at the 45° angle to
1
1 2
2
I xy  0, I yy  m(2a)  ma
the vertical shaft, which rotates with the
12
3
angular velocity of 20π rad/s. Determine
I yz  0, I xz  0
the angular momentum H of the plate
1
1
about O and find the kinetic energy of the
I zz  m[(2a)2  (2b)2 ]  m(a2  b2 )
12
3
plate.
Eq. 7/11 applied tox  y  z gives
H  jI yyy  kI zzz
z
1

1

 j( ma2 )
 k[ m(a2  b2 )]
3
3
2
2
z
1


y
But j  jcos45  k sin45 
( j  k)
a
2
a
1
k  jsin 45  k cos45 
(j  k)
y
2
1
3
b
H  m[b2 j  (2a2  b2 )k]  20 (0.04j  0.06k)
6
6

x
b
  (0.4j  0.6k) Nms
1
1
T  T H  (20 k) (0.4j  0.6k )  6.0 2  59.2 J
2
2
122
problem 07/64
HG  IG
about x-y-z
HG  IG about x′ -y′ -z′
z
z
y
a
a
 I xx 0
I'G   0 I yy

y
0
 0
b
x
b
2
1

m
2
a
0
0


12

0 

1
2

0 
0
m 2b
0
 

12
I zz  
2
1
2
2

0
0
m 4a  4b  
12


0
0  0  0 
1
'  Α  0 cos sin    0    sin  

  

0  sin  cos     cos 
z

45
Mx
HG  AHG  AIGA  IG
y

Note that IG  A IG A
1
2
1
2
Kinetic energy T  TIG   TIG
z  20 rad/s
123
Euler’s Equation
F  G
M
M
M
M
M
M
M  H
dH
)xyz    H
dt
 ( H x i  H y j  H z k)    H
M  (
M  (H x  H yz  Hzy )i
x
 H x  H yz  H zy
y
 H y  H zx  H xz
z
 H z  H xy  H yx
x
 I xx x  (I yy  I zz )yz
y
 I yy y  (I zz  I xx )zx
z
 I zz z  (I xx  I yy )xy
 (H y  H zx  H xz ) j
 (H z  H xy  H yx )k
124
problem 07/82
The plate has a mass of 3 kg and is
welded to the fixed vertical shaft, which
rotates at the constant speed of 20π
rad/s. Compute the moment M applied
to the shaft by the plate due to dynamics
imbalance.
Eq. 7/23
2
M

I

 x yz z
2
I yz   yzdm   y dl  
b/ 2
b / 2
z
y 2  2dy
 2 b3
b3
mb2 3(0.2)2

(

)

 0.02 kgm2
3 2 2 2 2
6
6
45
y
Mx
z  20 rad/s
M x  0.02(20 )2  79.0 Nm
on plate, Mx  79.0i Nm
but acting on shaft, M  79.0i Nm
125
problem 07/82
d
d
H

 G   HG 
dt
dt
 IG     IG   
MG 
 0     IG   
 0 z x  


  z 0 x  

y z 0  


 0  0 
  0 0 

 
 0 0 0 
z
45
y
Mx
z  20 rad/s
126

Dynamics (Meriam and Kraige, Ed. ,2013) Chapter 1. Introduction

Transcription

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