Dynamics (Meriam and Kraige, Ed. ,2013) Chapter 1. Introduction Engineering Mechanics Statics Dynamics Strength of Materials Vibration Statics: ,distribution of reaction force from the applied force Dynamics: , 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 a r (F +F )=mx a r (F +F )=0 r F a F th 7
126
Embed
Dynamics - WordPress.com · Time derivative of a vector. Time Derivative of the Unit Vectors in Polar (Cylindrical) Coordinates ... Engineering Mechanics Dynamics --IAA ...
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Dynamics (Meriam and Kraige, Ed. ,2013)
Chapter 1. Introduction
Engineering Mechanics
Statics Dynamics Strength of Materials Vibration
Statics: ,distribution of reaction force from the appliedforce
Dynamics: , 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 deformablebodies
Vibration: x(t) = f(F(t)) on particles and rigid bodies1
a r(F +F )=mx
a r(F +F )=0 rFaF
th7
Newtonian Dynamics‧Kinematics: the relation among
‧Kinetics: the relation between ( ) and ( )x t F t
2
2
( ) ( )( ), ( ), and ( ), and dx t d x tx t x t x t x xdt dt
without reference to applied force
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
Cylindrical Coordinates
Spherical Coordinates
( , , )x y zr
( , , )r zr
( , , )R r
3
Displacement, Velocity, and Acceleration
2 2
1 1
s t
s tds vdt
2 2
1 1
v
v
t
tdv adt
2 2 2 2
1 1 1 1
( )x t t t
x t t tdx xdt xdt dt
4
Velocity and Acceleration Vector
ddt
rr v d
dt
rr a
5
Rectangular Coordinates
x y zx y zx y z
r i j kr i j kr i j k
r
x x xy y yz z z
r r r
6
t td vdt
rr v v e e
: radius of curvature
t t
t n
ddtv v
vv v
rr
e e
e e
Centrifugal and Tangential Acceleration
7
Time derivative of a vector
Time Derivative of the Unit Vectors in Polar (Cylindrical) Coordinates (2D)
r
r
r rr
r
dddd
d d ddt dt dd dddt dt d
e e
e e
e ee e
e ee e
8
r z
r r z z
r r z
r zr r z z
r r r r r z
r e er e e e e
r e e e e e e
Cylindrical Coordinates (3D)
9
r zr r z e e e
2( ) ( 2 )r zr r r r z e e e
Spherical Coordinates
sin cos
sin cos
cos
cos
r
r r
z
r
r
r r
r
RR R
o
R R R
r er e e
ω e e
e e e
e e e
e ω e
e e
r e e e
10
2 2 2
2
2 2
cos
coscos 2 sin
1 sin cos
R R
R
R R
R
v v v
v Rv Rv R
a a a
a R R Rda R R
R dtda R R
R dt
r v e e e
r a e e e
Velocity and acceleration in Spherical Coordinates
11
Engineering Mechanics Dynamics -- IAA12
Chap.3 Kinetics of Particles
3.1 Force, Mass, and Acceleration3.2 Work and Energy3.3 Impulse and Momentum3.4 Impact and Orbital Mechanics
3.1 Force, Mass, and Acceleration
‧Newton’s Second Law Equation of Motion‧Inertial System: A coordinate system where ‧Free-body diagram
mF rmF r
Engineering Mechanics Dynamics -- IAA13
1 1
2
2 2
3
0
8300 7360 750
1.257 m s
736 75(1.257) 830 N
0 1.257
y y
y
y
y y
F ma
T m g m ya
a
F ma
R m g m yRR
a dt
dt
3.77 m s
Sample 3/1A 75-kg man stands on a springscale in an elevator. The tensionT in the hoisting cable is 8300 N.Find the reading R of the scale innewtons and the velocity υ of theelevator after 3 seconds. The totalmass of the elevator, man, andscale is 750kg.
Engineering Mechanics Dynamics -- IAA14
Sample 3/3The 250-lb concrete block A is released from rest in the position shown andpulls the 400-lb log up the 30° ramp. If the coefficient of kinetic frictionbetween the log and the ramp is 0.5, determine the velocity of the block as ithits the ground at B.
1 1 1
1 1 1
2 2 2
1 2
cos 02 sin
, 4 equation for 4 unknowns
2 constant
N m g m yN T m g m xm g T m yx y
x1
y1y2
x2
Engineering Mechanics Dynamics -- IAA15
2 sin30 sin300
2 cos30 cos30 0
x x
y
F ma
B B mxF
B B mg my
Eliminate B and get3 3gx a
30302A BB
mg
a
x
y
The steel ball is suspended from theaccelerating frame by the two cords Aand B. Determine the acceleration ofthe frame which will cause the tension inA to be twice that in B
Engineering Mechanics Dynamics -- IAA16
cos 0
25(9.81)cos20 0230 N
yF N mg my
NN
2
cos
25(9.81)sin20 0.15(230) 25
1.973 m/s
xF mg N mx
a
a
Frame and sphere as an unit:
Sphere alone:0
( )cos45 10(9.81)cos20 0130.4 N
y
A B
A B
F
T TT T
( )sin45 9.81sin20 10(1.973)19.56 N
x x
B A
B A
F ma
T TT T
75.0 N, 55.4 NA BT T Solution:
45AT
10(9.81) N
x
45BT
20
y
x
y20
25(9.81) N
0.15 NN
problem 03/19The 10-kg sphere is suspended from the 15-kg frame slidingdown the 20° incline. If the coefficient of kinetic frictionbetween the frame and incline is 0.15, compute each tensionof wires A and B
Engineering Mechanics Dynamics -- IAA17
Check for motion. Assume static equilibrium.
From B: 196.2 NT Mass A:
max
0
196.2 (60)(9.81)sin30 09.81N
(0.25)(60)(9.81)cos30 127.4 N ( )
x
s
F
FFF N
a
xy
1m gN
T
AF
2m g
B
T
x
The system is released from rest withthe cable taut. Neglect the smallmass and friction of the pulley andcalculate the acceleration of eachbody and the cable tension T uponrelease if (a) μs = 0.25, μk = 0.2 and (b)μs = 0.15, μk = 0.1
No motion for (a): 0, 196.2 Na T
Engineering Mechanics Dynamics -- IAA18
20.589 m/s , 208 Na T Solution:
max (0.15)(60)(9.81)cos30 76.5 NF
motion for (b)
(60)(9.81)sin30 (0.1)(60)(9.81)cos30 60T a
A:
B: (20)(9.81) 20y xF ma T a
xy
1m gN
T
AF
2m g
B
T
xx xF ma
The system is released from rest withthe cable taut. Neglect the smallmass and friction of the pulley andcalculate the acceleration of eachbody and the cable tension T uponrelease if (a) μs = 0.25, μk = 0.2 and (b)μs = 0.15, μk = 0.1
Engineering Mechanics Dynamics -- IAA19
0 0
( )0
( ) 0
F N g L bFT g L b T gb
Let = mass / length :
Solve to obtain:1Lb
The chain is released from rest withthe length b of overhanging links justsufficient to initiate motion. Thecoefficients of static and kinetic fictionbetween the links and the horizontalsurface have essentially the samevalue μ. Determine the velocity υ ofthe chain when the last link leaves theedge. Neglect any friction at thecorner.
( )L b g
( )N L b g F
gb
0T
Non-constant Acceleration
Engineering Mechanics Dynamics -- IAA20
1gL
Eliminate T to obtain:
0
22
[ (1 ) ]
1 [ (1 ) ]2 2
L
b
Lb
gd x L dxL
g x LxL
[ (1 ) ]ga x x LL
d xdx
( ) ( )F maT g L x L x agx T xa
Substitute b and simplify:
( )g L x
( )g L x gx
T( )g L x
Non-constant Acceleration
Engineering Mechanics Dynamics -- IAA21
2 2
1 2
1
1
1, 0
1
t
at at
xg L x g Lx
xg g L x Lxgx x LLg
x x gL
gx e a a
LLx c e c e
Non-constant AccelerationAnother approach:
Engineering Mechanics Dynamics -- IAA22
Curvilinear Motion in Polar Coordinates
2
2r
mF r r
mF r r
F r
r z
r r z z
r r z
r zr r z z
r r r r r z
r e er e e e e
r e e e e e e
r zr r z e e e
2( ) ( 2 )r zr r r r z e e e
Engineering Mechanics Dynamics -- IAA23
2 ( )
( 2 )
r rF ma T m r r
F ma F m r r
case (a)case (b)
20
20
2
2
T mr F mb
T mr F mb
Sample 3/10Tube A rotates about the vertical O-axiswith a constant angular rate andcontains a small cylindrical plug B if massm whose radial position is controlled bythe cord wound around the drum ofradius b. Determine the tension T in thecord and the horizontal force Fθ exertedby tube on the plug if the constant angularrate of rotation of the drum is ω0 first inthe direction for case (a) and second inthe direction for case (b). Neglect friction.
Engineering Mechanics Dynamics -- IAA24
22
2
( )
(3.5)2(9.81) 22.4
9.41 N
r
B
B
F m r r mr
N
N
Loss of contact at A:
2
2cos30
2.44.52 m/s
rF mr
mg m
0AN
BN
t
n
mg
BF
0AN
t
n
mg
AF
30
If the 2-kg block passes over the top B of the circularportion of the path with a speed of 3.5 m/s, calculate themagnitude NB of the normal force exerted by the path onthe block. Determine the maximum speed υ which theblock can have at A without losing contact with the path.
r
rr
θ
θ
Engineering Mechanics Dynamics -- IAA25
Equilibrium:
10F T mg
20 sin30 0n nF ma T mg Motion:
2
1
sin30 0.5T mgkT mg
3030
1T
mg
1T
30
mg
2T
n
30 t
problem 03/66
The small sphere of mass m is suspended initially at rest by the twowires. If one wire is suddenly cut, determine the ratio k of thetension in the remaining wire immediately after the other wire is cutto the initial equilibrium tension.
w try using x-y coordinates
Engineering Mechanics Dynamics -- IAA26
0
cos 0/cos
yF
N mgN mg
2
2
2
( )
sin ( sin )
( )sin sincos
cos
rF m r r
N m rmg mr
gr
Note that2 g
r
2cos 1gr
is a restriction.
N
y
n
mg
r
A small bead of mass m is carried by a circular hoop of radius r whichrotates about a fixed vertical axis. Show how one might determinethe angular speed ω of the hoop by observing the angle θ whichlocates the bead. Neglect friction in your analysis.
Engineering Mechanics Dynamics -- IAA27
2 2
0 cos30 0
sin301200
y
r
F N mg
F m N mr
Solve:
For no slipping tendency, set F to zero on
0.9s with
1.155 , 149.4 ft/secN mg
min 0 as1 1
max tan tan (0.9) 42.0 30s
For max , set max sF F N
22max
0 cos30 sin30 0
cos30 sin30
y s
r s
F N mg N
F m N N mr r
max
2.40345 ft/sec
N mg
N
F
mg
n
y
30
problem 03/76Determine the speed υ at which the race car will have no relianceon friction to the banked track. In addition, determine the minimumand maximum speeds, using the coefficient of static friction μs =0.9.
r
Engineering Mechanics Dynamics -- IAA28
0 0
2 20
, sin , sin
, sin ( )
2 (1 cos )
F ma mg ma a g
d a ds d g Rd
gR
When
2
20
20
, cos
cos 2 (1 cos )
(3cos 2 )
r rF ma mg N mR
N mg m mgR
mggR
0,N so2 2
10 023cos 2 cos ( )3 3gR gR
For 10
20, cos ( ) 48.23
R N
n
t
0 mg
A small vehicle enters the top A of the circular path with a horizontalvelocity υ0 and gathers speed as it moves down the path. Determine anexpression for the angle β which locates the point where the vehicleleaves the path and becomes a projectile. Evaluate your expression forυ0 = 0. Neglect friction.
Sample 3/15A satellite of mass m is put into an elliptical orbit around the earth. At point A, itsdistance 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 = 1200km from the earth.
Engineering Mechanics Dynamics -- IAA32
Potential Energy
TV d
mgdy
mgh
F r 2
1
2
2
2
2
T
r
r
V d
GMm drr
mgR drr
mgRr
F r
Engineering Mechanics Dynamics -- IAA33
Principle of Work and Energy
V W T
TT TdU d dP
dt dt dt
F r rF F r=
‧Kinetic energy
‧Power
0T W V
Conservative Force
x y zV
i j k
F
Engineering Mechanics Dynamics -- IAA34
Constant total energy is A A p pE T V T V
Thus2 2
2 21 12 2A p
A p
mgR mgRm mr r
2 2 2 1 12 ( )A pp A
gRr r
2 2 1 12 ( )A pp A
gRr r
A satellite is put into an elliptical orbit around the earth and has avelocity υP at the perigee position P. Determine the expression forthe 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.
Engineering Mechanics Dynamics -- IAA35
= mass per unit lengthFor equil. at start ( )kgb g L b
1k
k
Lb
gU T V 2
0
( )2
L b
k kL bU dF x gxdx g
212
T L ( )( )2g
L bV g L b
L b
bdFdx x b
2L b 2
L b
2L b
The chain starts from rest with a sufficient number of links hangingover the edge to barely initiate motion in overcoming frictionbetween the remainder of the chain and the horizontal supportingsurface. Determine the velocity υ of the chain as the last link leavesthe edge. The coefficient of kinetic friction is μk. Neglect and frictionat the edge.
Thus2 2 2
2( ) 12 2 2
k
L b L bg L g
2 (1 )( [ ])kbg L b L bL
Now substitute b
So 2 (1 )( [1 ] [ ])1 1 1
1
k k kk
k k k
k
Lg L L
gL
1 k
gL
Engineering Mechanics Dynamics -- IAA36
3.3 Impulse and Momentum
( )
mmd mdtddt
G rF r
r
G
G
Linear momentum
Impulse
2
1|dt m dt md m rrF r r r G
Conservation of Linear Momentumdt F G if 0 0 F G
Engineering Mechanics Dynamics -- IAA37
1mv
2
1
t
tmgdt
2
1
txtR dt
2
1
tytR dt
15
y
x=+
Sample 3/19
2
1
2
1
1 2
1 2
2 2 2 2
1
( ) ( )
4 /16 4 /16(50) (0.02) (70cos15 )32.2 32.2
( ) ( )
4 /16 4 /16(0) (0.02) (70sin15 )32.2 32.2
45.7 lb, 7.03 lb
45.7 7.03 46.2 lb
tan
t
x x xt
x
t
y y yt
y
x y
x y
y
m F dt m
R
m F dt m
R
R R
R R R
R 1 7.03tan 8.7545.7
xR
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 balland the angle β made by R with the horizontal
Engineering Mechanics Dynamics -- IAA38
R T m
R ma
2 20 2ad
:
:
:
1.62/16(0.001)= (150)32.2
R =472 lbR
1.62/16472=32.2
a
,
2150,000 / sec (4660 )a ft g,
2 2150 0 2(150,000)d , 0.075d ft or 0.900in
25
0mg 150 / secft
problem 03/207
The 1.62-oz golf ball is struck by the five-iron and acquires the velocity shown in atime period of 0.001 sec. Determine the magnitude R of the average force exertedby the club on the ball. What acceleration magnitude ɑ does this force cause, andwhat is the distance d over which the launch velocity is achieved, assumingconstant acceleration?
Engineering Mechanics Dynamics -- IAA39
Angular Impulse and Momentum
O m H r v
( ) ( ) ( )
( )
( )( )
O
z y x z y x
z yx
O y x z
x y z y xz
mm v y v z m v z v x m v x v y
m v y v zHx y z H m v z v xv v v m v x v yH
H r vi j k
i j kH
Engineering Mechanics Dynamics -- IAA40
O
O O
O O
mm m m m
H r vH r v r v v v r v r F M
M H
Time Derivative of Angular Momemtum
Conservation of Angular Momentum
2
12 1( ) ( )
O Ot
O O O Otdt
M H
M H H H
2
1O 1 2( ) ( )
t
O Otdt H M H
The total angular impulse on a particle of mass mabout a fixed point O equals the correspondingchange in angular momentum about that point.
Principle of Conservation of Angular Momentumif , then 0O H 1 2( ) =( )O OH Hor0O M
0
Engineering Mechanics Dynamics -- IAA41
9
6
( ) ( )
6(10 )74075(10 )
59 200 m/s
O A O B
A A B B
A AB
B
B
H H
mr mr
rr
Sample 3/25 Molynia OrbitA comet is in the highly eccentric orbit shown in the figure. Its speed ate the mostdistant 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.
Engineering Mechanics Dynamics -- IAA42
0 0OM H
so OH constant
min 6371 390 6761r km
max 2(13520) 6761 20279r km
For OH constant6371(33880) 11720 20279B A
11300 /B km h 19540 /A km h
B B
A
33880 /km h11720 /km h
A
ro
maxr minr
2(13520) km
The central attractive force F on an earthsatellite can have no moment about the centerO of the earth. For the particular elliptical orbitwith major and minor axes as shown, a satellitewill have a velocity of 33880 km/h at the perigeealtitude of 390 km. Determine the velocity of thesatellite at point B and at apogee A. The radiusof the earth is 6371 km.
Engineering Mechanics Dynamics -- IAA43
0H ;02 ( ) 2 (2 ) (2 ) 0mr r m r r 0 / 4
2 2 2 200 0
1 12( [ ] ) 2( [2 ] ) (3 / 4)2 2 4
T m r m r mr
2 2 2 20 0
3/ / 3 / 44
n T T mr mr
problem 03/228
The two spheres of equal mass m are able to slide along the horizontal rotatingrod. If they are initially latched in position a distance r from the rotating axis withthe assembly rotating freely with an angular velocity ω0 , determine the newangular velocity ω after the spheres are released and finally assume positionsat the ends of the rod at a radial distance of 2r. Also find the fraction n of theinitial kinetic energy of the system which is lost. Neglect the small mass of therod and shaft.
Engineering Mechanics Dynamics -- IAA44
From
0 0
2 2cos ( )
cos
M Hdmgl ml mldt
gl
so at
d d
2
0 0cos
2 | g dl
290
2sin g g
l l
90 2l gl
By work-energy21 2
2V T mgl m gl
mg
lT
O
The simple pendulum of mass m and length lis released from rest at θ = 0. Using only theprinciple of angular impulse and momentum,determine the expression for in terms of θand the velocity υ of the pendulum at θ = 90°.Compare this approach with a solution by thework-energy principle.
Engineering Mechanics Dynamics -- IAA45
Direct Central Impact
Engineering Mechanics Dynamics -- IAA46
0
0
0
0
' '1 1 0 0 1
2 0 1 1 00
' '2 2 0 2 0
2 0 2 0 20
' '2 1
1 2
[ ( )][ ( )]
( )( )
relative velocity of separationrelative velocity of approach
t
rtt
d
t
rtt
d
F dt memF dt
F dt memF dt
e
Coefficient of Restitution
Engineering Mechanics Dynamics -- IAA47
Engineering Mechanics Dynamics -- IAA48
Oblique Central Impact
' '1 1 2 2 1 1 2 2
'1 1 1 1
'2 2 2 2
' '2 1
1 2
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )( ) ( )
n n n n
t t
t t
n n n
n n n
m m m m
m mm m
VeV
Engineering Mechanics Dynamics -- IAA49
Sample 3/29
' ' '2 1 1
1 2'1
'1 1
'1 1
' ' 2 ' 2 2 21 1
'' 1 11
'1
( ) ( ) 0 ( )0.5( ) ( ) 50sin30 0
( ) 12.5 ft/sec
( ) ( )
( ) ( ) 50cos30 43.3 ft/sec
( ) ( ) 12.5 43.345.1 ft/sec
( ) 12.5tan tan( ) 43.3
n n n
n n
n
t t
t t
n t
n
t
e
m m
16.10
A ball is projected onto the heavy plate with a velocity of 50 ft/sec at the30° angular shown. If the effective coefficient of restitution is 0.5, computethe rebound velocity υ′ and its angle θ′.
Engineering Mechanics Dynamics -- IAA50
' '1 1 2 2 1 1 2 2
' '1 2
' ' ' '2 1 2 1
1 2' '1 2
' '1 1 1 1 1 1
( ) ( ) ( ) ( )
5.20 0 ( ) ( )
( ) ( ) ( ) ( ) 0.6 ( ) ( ) 5.20 0
( ) 1.039 m/s ( ) 4.16 m/s
( ) ( ) ( ) (
n n n n
n n
n n n n
n n
n n
t t t
m m m m
e
m m
' '
2 2 2 2 2 2
) 3 m/s
( ) ( ) ( ) ( ) 0t
t t t tm m
Sample 3/30
Spherical particle 1 has a velocity υ1 = 6 m/s in thedirection shown and collides with spherical particle2 of equal mass and diameter and initially at rest. Ifthe coefficient of restitution for these conditionsis , determine the resulting motion of eachparticle following impact. Also calculate thepercentage loss of energy due to the impact.
0.6e
Engineering Mechanics Dynamics -- IAA51
2gh , ' '2gh
' ' 1100 0.7242100
heh
'
2100 11002100
47.6%
mgh mghnmgh
As a check of the basketball before the start of a game, the refereereleases the ball from the overhead position shown, and the ballrebounds to about waist level. Determine the coefficient ofrestitution and the percentage n of the original energy lost duringthe impact.
Engineering Mechanics Dynamics -- IAA52
Central-Force Motion
202
2
20
mGmm r rr
r r
mr h
F r
Engineering Mechanics Dynamics -- IAA53
Orbital Mechanics
02
1 cos GmCr h
Engineering Mechanics Dynamics -- IAA54
D’Alembert’s Principle and Inertia Force
inertia forcemr
F ma 0F mr 0
D’Alembert’s Principle using inertia force to treat dynamics by statics
Engineering Mechanics Dynamics -- IAA55
fig_04_001
i i
c
mm
F f rF r
Equation of motion
c i im m r rMass center
Chap. 4 Kinetics of Systems of Particles
Principle of motion of the masscenter the resultant of theexternal forces on any system ofmasses equal the total masstimes the mass centeracceleration.
Engineering Mechanics Dynamics -- IAA56
Linear Momentum
( )
( )
i i
i c i
i c i i
c
mm
dm mdt
m
G rr ρ
r ρ
r
Engineering Mechanics Dynamics -- IAA57
fig_04_003
Kinetic Energy
12
1 ( 2 )21 12 2
i c i
i c i
Ti i i
T T Ti c c i c i i i i
T Tc c i i i
T m
m m m
m m
r r ρr r ρ
r r
r r r ρ ρ ρ
r r ρ ρ
Engineering Mechanics Dynamics -- IAA58
Angular Momentum about a Fixed Point
( )
( )O i i i
O i i i i i
i i
O
m
m mO
H r r
H r r r rr F
M
Angular Momentum about c.g.
( )( )( )
G i i i
i i c i
i i c i i i
i i i
mmm mm
H ρ rρ r ρρ r ρ ρρ ρ
( )
G i i C i i i i
i i
G
m mH ρ r ρ ρ rρ FM
Engineering Mechanics Dynamics -- IAA59
p_04_020
1 2 0 1 1 2 2 0(0) ( ) ( ) ( )2 2l lm l m m m s l x m s x m s
ls
2x 1x2m 1m
0mAB
C
With respect to C, i im x constant
Simplify and get
But they meet when
1 1 2 2
0 1 2
m x m xsm m m
2 1x x l so 1 2 1 2
0 1 2
( )m m x m lsm m m
problem 04/22
The man of mass m1 and the woman of mass m2 arestanding on opposite ends of the platform of mass m0 whichmoves with negligible friction and is initially at rest with s = 0.The man and woman begin to approach each other. Derivean expression for the displacement s of the platform whenthe two meet in terms if the displacement x1 of the relative tothe platform.
Engineering Mechanics Dynamics -- IAA60
sp_04_03_01
Sample 4/4
A shell with a mass of 20 kg is fired frompoint O, with a velocity u = 300 m/s in thevertical 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 Ais observed to rise vertically a distance of 500m above P, and fragment B is seen to have ahorizontal velocity vB and eventually lands atpoint Q. When recovered, the masses of thefragments A, B, and C are found to be 5, 9,and 6 kg, respectively. Calculate the velocitywhich fragment C has immediately after theexplosion. Neglect atmosphere resistance.
Engineering Mechanics Dynamics -- IAA61
sp_04_03_01
2 2
/ 300(4/5)/9.81 24.5s
[(300)(4/5)] 2940m2 2(9.81)
2 2(9.81)(500) 99.0m/s/ 4000/ 24.5 163.5m/s
z
z
A A
B
t u g
uhg
ghs t
1 2[ ] A A B B C Cm m m m G G v v v v
2 2 2
320(300)( ) 5(99.0 ) 9(163.5)( cos45 sin45 ) 65
6 2560 1040 495427 173.4 82.5 m/s
(427) (173.4) (82.5) 468m/s
C
C
C
C
i k i j v
v i j kv i j k
Sample 4/4
Engineering Mechanics Dynamics -- IAA62
Steady Mass Flow
1 1 1 2 2 2A A m
2 1 2 1( ) ( ) ( )m m m G v v v v
m F v
Engineering Mechanics Dynamics -- IAA63
( )
106(680 1000/3.6) 4(680)45400 N 45.4 kN
0
45.4 32 4.6(9.81)sin 0
sin 0.296 17.22
a f
x x
T m u m u
F ma
problem 04/35
T
N
4.6(9.81) kNmg
1000 km/h
x32 kNR
The jet aircraft has a mass of 4.6 Mg and a drag (airresistance) of 32 kN at a speed of 1000 km/h at a particularaltitude. The aircraft consumes air at the rate of 106 kg/sthrough 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 tothe exhaust nozzle, determine the maximum angle ofelevation at which the jet can fly with a constant speed of1000 km/h at the particular altitude in question.
Engineering Mechanics Dynamics -- IAA64
problem 04/38
Density of salt water, 31030 kg/m
Resistance R equals net trust T
where ( )T m u Nozzle velocity 20.082/ 41.8 m/s(0.050)
4
u Q A
10001030(0.082) 84.5 kg/s, 70 19.44 m/s3600
84.5(41.8 19.44) 1885 N
m Q
R T
The jet water ski has reached its maximum velocity of 70 km/hwhen operating in salt water. The water intake is in thehorizontal tunnel in the bottom of the hull, so the water entersthe intake at the velocity of 70 km/h relative to the ski. Themotorized pump discharge water from the horizontal exhaustnozzle of 50-mm diameter at the rate of 0.082 m3/s. Calculatethe resistance R of the water to the hull at the operating speed.
64
Engineering Mechanics Dynamics -- IAA
With reversers in place
so
F mu
sin30
(50 0.65)(650)sin30 50(55.6 0)16460 2780 19240 N
R g aT m u m
19240 0.63830100
n
problem 04/45
Without reversers
(50 0.65)650 50(55.6)32900 2780 30100 N
g aT m u m
RT x
650 m/s
650 m/s
30
30
200/3.6 55.6 m/s
The jet-engine thrust reverser to reduce an aircraft speed of 200km/h after landing employs folding vanes which deflect theexhaust gases in the direction indicated. If the engine isconsuming 59 kg of air and 0.65 kg of fuel per second, calculatethe braking thrust as a fraction n of the engine thrust without thedeflector vanes. The exhaust gases have a velocity of 650 m/srelative to the nozzle.
65
Engineering Mechanics Dynamics -- IAA66
Variable Mass system
0( )R m muF R m
F m mu
Engineering Mechanics Dynamics -- IAA67
mu pA mg R m
Rocket Propulsion
Engineering Mechanics Dynamics -- IAA68
Sample 4/12
00 0
0
,
ln
m t
m
T mg m T mu mu mu mg mdm dmd u gdt d u g dtm mmu gtm
0
0max 0
( ) /
ln ( )
b b
bb
t m m mm gu m mm m
let mb:mass of rocket when burnout occurs
Solution I (F=ma solution)
A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuelis exhausted. The relative nozzle velocity of the exhaust gas has a constant value u atatmospheric pressure throughout the flight. If the residual mass of the rocket structureand machinery is mb when burnout occurs, determine the expression for the maximumvelocity reached by the rocket. Neglect atmospheric resistance and the variation ofgravity with altitude.
Engineering Mechanics Dynamics -- IAA69
Sample 4/12
Solution II (Variable-Mass solution)
,F mg F m mu mg m mumu mu T T mg m
same as Solution I
A rocket of initial total mass m0 is fired vertically with constant acceleration until the fuelis exhausted. The relative nozzle velocity of the exhaust gas has a constant value u atatmospheric pressure throughout the flight. If the residual mass of the rocket structureand machinery is mb when burnout occurs, determine the expression for the maximumvelocity reached by the rocket. Neglect atmospheric resistance and the variation ofgravity with altitude.
Engineering Mechanics Dynamics -- IAA70
fig_05_001
Chap.5 Planar Kinematics
Engineering Mechanics Dynamics -- IAA71
fig_05_004
( )
rr r
r r rω
ω ω ω
displacement velocity acceleration
Fixed Axis RotationZ
Y
X
Engineering Mechanics Dynamics -- IAA72
sp_05_03_01
2
2
2
[ ] 2 (0.4 0.3 ) 0.6 0.8 m/ s
[ ( )] 2 (0.6 0.8 ) 1.6 1.2 m/ s
[ ] 4 (0.4 0.3 ) 1.2 1.6 m/ s
[ ] 2.8 0.4 m/ s
n n
t t
n t
v v k i j i j
a a k i j i j
a a k i j i j
a a a a i j
r
r
r
2 rad/s k
24 rad/s k
2 20.6 0.8 1m/s
2 22.8 0.4 2.83 m/sa
Sample 5/3The right-angle bar rotates clockwise with an angular acceleration
Write the vector expressions for the velocity and acceleration of point A when
Engineering Mechanics Dynamics -- IAA73
p_05_002
2
2
2
( 6k 45 j)
270i mm/s
4k 45 j 6 (45 j)
180i 1620 j mm/s
A A
A A A
r
a r r
(a)
(b)
2
2
2
6k ( 30i 45 j)
270i 180 j mm/s
4k ( 30i 45 j) 6 ( 30i 45 j)
900i 1740 j mm/s
B B
B B B
r
a r r
24rad/s rad/s, 6
Determine the velocity and acceleration of (a) point A and (b) point B with
Engineering Mechanics Dynamics -- IAA74
sp_05_04_01
,where , ,
, and
O O
O O O
s rr a r
s a s
2
sin ( sin )
(1 cos ) (1 cos )
(1 cos ) sin
(1 cos ) sin
O
O O
O
x s r rx r
x
a r
2
cos (1 cos )
sin sin
sin cos
sin cos
O
O O
O
y r r ry r
y
a r
20x y r and
Sample 5/4
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.
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.
Engineering Mechanics Dynamics -- IAA75
Sample 5/7
3 0 0 10 3 1.7320.1732 0.1 0
4 1.732 m/s
A
i j kv i i j i
i j
2 24 (1.732) 19 4.36 m/sA
/ 0A O A O O v v v v rω
0
-10 rad/ s0.2( cos30 sin30 ) 0.1732 0.1 m3 m/ s
O
kr i j i jv i
ω=
Calculate the velocity of point A on the wheel without slipping for the instant represented.
sp_05_07_01
Engineering Mechanics Dynamics -- IAA76
Sample 5/11
[ / ]
/ 3/ 0.300 10 rad / sO
r
OC
2 2(0.300) (0.200) 2(0.300)(0.200)cos1200.436 m
AC
[ ]
0.436(10) 4.36 m/sA
r
AC
Locate the instantaneous center of zero velocity and use it to find the velocity of point A for the position indicated.
Engineering Mechanics Dynamics -- IAA77
sp_05_08_01
Sample 5/8
1 2 3
1 1 2 2 3 3
1 2
1 2
2
1 2
0 100 (175 50 ) 2 75100 50
175 1503 6,7 7
ω ωω ωω
ω ω
r r r rr r r r
k j k i j k iω ω ω
r
r1
r2
r3
Engineering Mechanics Dynamics -- IAA78
sp_05_08_01
Sample 5/14
1 2 3
1 1 2 2 3 3
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
( )( )( )
r r r rr r r rr r r
r rr r
ω ω ωω ω ωω ω ωω ω ω
r
r1
r2
r3
1 23 3100 ( ) ( 100 ) ( 175 50 )7 7
6 6( [( ) ( 175 50 )] 0 2 (2 [ 75 ])7 7
r k j k k j k i j
)k k i j k k i
21 0.1050 rad/s 2
2 4.34 rad/s
Engineering Mechanics Dynamics -- IAA79
sp_05_09_01
Sample 5/9
/A B A B v v v
1
[ ]5 1500(2 ) 65.4 f / sec
12 605 14
sin sin60sin 0.309 18.02
B
r
t
//
65.4 67.3 ft /secsin78.0 sin72.0
65.4 34.4 ft /secsin30 sin72.0
AA
A BA B
/ 34.4[ / ] 29.5 rad/sec14/12
A BABr
AB
/G B G B v v v
/ /4 (34.4) 9.83 ft / sec
1464.1ft /sec
G B AB A B
G
GBGBAB
Engineering Mechanics Dynamics -- IAA80
sp_05_09_01
Sample 5/9
r
r1r2
y
x1 2
1 1 2 2
1 1 1 1 1
2 2 2 2 2
( )( )
r r rr r rr r r
r r
ω ωω ω ωω ω ω
Engineering Mechanics Dynamics -- IAA81
sp_05_15_01
/ /( ) ( )A B A B n A B ta a a a 2[ ]na r
2
2
5 1500[2 ]( )12 60
10280 ft / sec
Ba
2[ ]na r
2/
2
14( ) (29.5)12
1015 ft / sec
A B na
/10280cos60 1015cos18.02 ( ) sin18.02A A B ta
/0 10280sin60 1015cos18.02 ( ) cos18.02A B t
2/( ) 9030 ft /secA B t 23310 ft /secA
[ / ]ta r 2 2
/ 9030/(14/12) ft /sec 7740 rad/secA B
23310 ft / secA 2/( ) 9030 ft /secA B t
Sample 5/15
Engineering Mechanics Dynamics -- IAA82
fig_05_005
General Motion: Rotation + Translation
fig_05_006
Engineering Mechanics Dynamics -- IAA83
fig_05_007
Instantaneous Center
Engineering Mechanics Dynamics -- IAA84
fig_05_011
Body-Fixed Coordinates in Rotation
A B
A B
A B
B
r rr rr r
r
ρω ρ+ ρω ω ρ ω ρ ω ρ+ω ρ+ ρ
ω ω ρ ω ρ+ 2ω ρ ρ
Ar
Coriolis Acceleration
2A B
r r ω ω ρ ω ρ ρ+ ω ρ
Coriolis acceleration
Engineering Mechanics Dynamics -- IAA85
sp_05_16_01
A v ρ ρ4 6 5 24 5 in./secA v k i i j i
2 2(24) (5) 24.5 in./secA
( ) 2A a ρ ρ ρ ρ2( ) 4 (4 6 ) 4 24 961 in./sec k k i k j ρ
210 6 60 in./sec k i j ρ22 2(4 ) 5 40 in./sec k i j ρ
281 in./sec iρ2(81 96) (40 60) 15 20 in./secA a i j i j
2 2 2(15) (20) 25 in./secAa
Sample 5/16The motion of slider A is separately controlled, and at this instant, r = 6 in., =5 in./sec, and =81 in./sec . Determine the absolute velocity and acceleration of A for this position.
r r
y
x
2
Engineering Mechanics Dynamics -- IAA86
( )
rr r
r r rω
ω ω ω
displacement velocity acceleration
With and without Body-Fixed Coordinate
Let
5 , 2 , 310
-20 30
r i k kr r jr r r
= i + j
Engineering Mechanics Dynamics -- IAA87
p_05_161
( sin j cos k)
22 k ( sin j cos k)
2 sin i
cora
(west)
The track provides the necessary westwardacceleration so that the velocity vector is properlyrotated and reduced in magnitude.
B y
z
A
For 500 km/h
(a) Equator,
(b) North pole,
0 0cora
5 250090 2(7.292 10 ) 0.0203 m/s3.6cora
problem 05/163
A vehicle A travels with constant speed v along a north-south track. Determine the Coriolisacceleration aCor as a function of the latitude θ at (a) the equator and (b) the north pole.
Engineering Mechanics Dynamics -- IAA88
Chap. 6 Dynamics of Planar Rigid BodyEquation 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.
G
G G
m
F r
M H
Engineering Mechanics Dynamics -- IAA89
Equation of Motion in 2D
2
G i i i
i i i
mm
dm
I
H ρ ρρ ω ρ
ω
ω
Angular momentum
c
G
mI
F rM ω
Engineering Mechanics Dynamics -- IAA90
Moment about a Fixed Point
0,if P is fixed
P G C
G C
P P
P
P O
mI mI m
I I
M H ρ rω ρ rω ρ r
rω ω
Engineering Mechanics Dynamics -- IAA91
when tipping impends
F mg mx
(0.4) (0.6) 0 GM mg mg I
23
ANAF
G
50(9.81)N0.8m
1.2m
Determine the value of the force P whichwould cause the cabinet to begin to tip.What coefficient μs of static friction isnecessary to ensure that tipping occurswithout slipping?
AN mg
23
x g g
As a whole :
1( ) (50 10) 60 40 P m m x x x g
Engineering Mechanics Dynamics -- IAA92
Center of Percussion
or
C
G G
O O
mM I
M I
F rω
The resultant force at the center of percussion is zero.
ok
ok Radius of gyration about point O
Engineering Mechanics Dynamics -- IAA93
Center of Percussion
-
012
23
X
Y
O
F R mxmg R my
F h Iy
x
h
XR
YR
hmg
F
Engineering Mechanics Dynamics -- IAA94
2
2
120(9.81)(0.2) 20(0.4)3
36.8rad/s
O OM I
20.2 36.8 7.36m/s
x r
xG0.2m 0.2m
t
20(9.81)N
2F
O 20(9.81) 2 20( 7.36)12 49.04
24.5N
x
A B
F mxF
F mg
F F F
problem 06/34
The 20-kg uniform steel plate is freely hinged about the z-axis asshown. Calculate the force supported by each of the bearings at Aand B an instant after the plate is released from rest in thehorizontal y-z plane.
Engineering Mechanics Dynamics -- IAA95
mgy
O
r GO
mg
O
r GO
22
/ 2
( )2
/ 2
O O
y
M I mgr mr
g rgF my mg O mrr
O mg
(a)
(b) 2 21( )2
2 / 32( )3
/ 3
O O
y
M I mgr mr mr
g rgF my mg O mrr
O mg
problem 06/38
Determine the angular acceleration and the force on the bearingat O for (a) the narrow ring of mass m and (b) the flat circulardisk of mass m immediately after each is released from rest inthe vertical plane with OC horizontal.
Engineering Mechanics Dynamics -- IAA96
Sample 6/5
A metal hoop with a radius r = 6 in. isreleased from rest on the 20° incline. If thecoefficients of static and kinetic friction areμs = 0.15 and μk = 0.12, determine theangular acceleration α of the hoop and thetime for the hoop to move a distance of 10 ftdown the incline.
Engineering Mechanics Dynamics -- IAA97
Sample 6/5[ ]x xF ma sin20mg F ma
[ 0]y yF ma cos20 0N mg
[ ]GM I 2Fr mr
0.1710F mg cos20 0.940N mg mg
max[ ]sF N max 0.15(0.940 ) 0.1410F mg mg
max[ ]kF N 0.12(0.940 ) 0.1128F mg mg
x[ ]xF ma sin20 0.1128mg mg ma 20.229(32.2) 7.38 ft/seca
[ ]GM I 20.1128 ( )mg r mr 20.1128(32.2) 7.26 rad/sec6 /12
21[ ]2
x at2 2(10) 1.646 sec
7.38xt
Assume pure rolling 4 equations for 4 unknowns a r
and
Check if the assumption valid. The friction force be bounded by N
So it is slipping Solve again the 4 unknowns . kf N
Engineering Mechanics Dynamics -- IAA
Sample 6/7
98
The slender bar AB weighs 60 lb and moves in the vertical plane, with its endsconstrained to follow the smooth horizontal and vertical guides. If the 30-lb force isapplied at A with the bar initially at rest in the position for which θ = 30°, calculatethe resulting angular acceleration of the bar and the forces in the small end rollersat A and B.
Engineering Mechanics Dynamics -- IAA99
Sample 6/72cos30 2 cos30 1.732 ft/secxa a 2sin30 2 sin30 1.0 ft/secya a
[ ]CM I m d 21 6030(4cos30 ) 60(2sin30 ) (4 )12 32.2
60 60(1.732 )(2cos30 ) (1.0 )(2sin30 )32.2 32.2
4.39 9.9424.42 rad/sec
[ ]y yF ma[ ]x xF ma
6060 (1.0)(4.42)32.2
A
6030 (1.732)(4.42)32.2
B
68.2 lbA
15.74 lbB
Engineering Mechanics Dynamics -- IAA100
mg
r
x
G
A
N
0 0OM I I M
0s Hence no friction force and
sinx x AF ma mg mr sinAgr
problem 06/82
Determine the angular acceleration of each of the two wheelsas they roll without slopping down the inclines. For wheel Ainvestigate the case where the mass of the rim and spokes isnegligible and the mass of the bar is concentrated along itscenterline. For wheel B assume that the thickness of the rim isnegligible compared with its radius so that all of the mass isconcentrated in the rim. Also specify the minimum coefficient ofstatic friction μs required to prevent each wheel from slopping.
Engineering Mechanics Dynamics -- IAA101
NC
F
mg
r
x
G
B
2sin 2
sin2
C C B
B
M I mgr mrgr
1 sin / cos2
1 tan2
s
s
F mg mgN
2 sin2gM I Fr mrr
1 sin2
F mg
problem 06/82
Determine the angular acceleration of each of the two wheelsas they roll without slopping down the inclines. For wheel Ainvestigate the case where the mass of the rim and spokes isnegligible and the mass of the bar is concentrated along itscenterline. For wheel B assume that the thickness of the rim isnegligible compared with its radius so that all of the mass isconcentrated in the rim. Also specify the minimum coefficient ofstatic friction μs required to prevent each wheel from slopping.
Engineering Mechanics Dynamics -- IAA102
b
mg
C
45
Ar I
Ama/( )G A tm ar
b
G
T
2
2
12 6 2 2
34
1 3( )6 42
2 2 (12)(9.81) 20.8 N8 8
A
G
M I madmgb b bmb m
gb
M Ib gT mb
b
T mg
problem 06/84
The uniform 12-kg square panel is suspended from point Cby the two wires at A and B. If the wire at B suddenlybreaks, calculate the tension T in the wire at A an instantafter the break occurs.
Engineering Mechanics Dynamics -- IAA103
Kinetic Energy
212
T m
212 OT I
2 21 12 2 CT m I
Engineering Mechanics Dynamics -- IAA104
mg
A
x
mg
B
x
2 2
sin1 12 2
U TU mgx
T m I
2
2 2 2 2
1 021 1 ( )2 2
T m
T m mr mr
2
2
1sin 2 sin2
sin sin
A
B
mgx m gx
mgx m gx
case A:
case A:
case B:
case B:
problem 06/116
Engineering Mechanics Dynamics -- IAA105
1 1 2 2T V T
2 2 21 1 ( )2 2 12 2l lmg x ml m x
22 2
( )2
6 2 2
lg x
l lx x
max 0.211 2 2 2
( 0.211 )2 1.861
0.211 (0.211 )6 2 2
x
lg l gll l l
20d
dx
0.789x l 0.211x lset obtain or
0.789x lThe solutionwould yield the sameonly then the motion is CCW.
max
For the pivoted slender rod of length l, determine the distance xfor which the angular velocity will be a maximum as the barpasses the vertical position after being released in thehorizontal position shown. State the corresponding angularvelocity.
Engineering Mechanics Dynamics -- IAA106
2
11 2
t
tdt
F G
G F G
Linear Momentum Angular Momentum
O OH I
2
11 2( ) ( )
O Ot
O O Ot
M H
H M dt H
Engineering Mechanics Dynamics -- IAA107
Sample 6/16The uniform rectangular block of dimensions shown is sliding to the lefton the horizontal surface with a velocity v1 when it strikes the small stepat O. Assume negligible rebound at the step and compute the minimumvalue of v1 which will permit the block to pivot freely about O and justreach the standing position A with no velocity. Compute the percentageenergy loss n for b = c.
Engineering Mechanics Dynamics -- IAA108
Sample 6/16[ ]O OH I
2 2 2 2 2 22 2 2
1( ) { ( ) [( ) ( ) ]} ( )12 2 2 3O
c b mH m b c m b c
1 2[( ) ( ) ]O OH H 2 21 2( )
2 3b mm b c 1
2 2 23
2( )b
b c
2 2 3 3[ ]T V T V 2 2
22
1 0 0 [ ]2 2 2 2O
b c bI mg
2 2 2 2 212 231 ( )[ ] ( )
2 3 22( )bm mgb c b c b
b c
22 2
1 22( (1 )( )3g c b c b
b
2 2 22 2 2 21 2 2 222 2 2 22 11 2
1 13 32 2 1 11 3 2( )
4 12
OO
m IE k b c bn b cE b c cm
b
62.5%n b c
Engineering Mechanics Dynamics -- IAA109
problem 06/188
The homogeneous sphere of mass m and radius r is projected along theincline 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 tof the period of slopping. In addition, state the velocity v of the masscenter G and the angular velocity ω at the end of the period of slipping.
Engineering Mechanics Dynamics -- IAA110
problem 06/188
0
0
0
0
0
( ) 0 cos
( )
( cos sin ) ( )
ty y y
tx x x
k
F dt m N mg
F dt m
mg mg t m
00
2
( )
2( cos )5
tG
k
M dt I
mg r t mr
r
7 cos 2sink
2 tan7k
0
0
0
2(7 cos 2sin )
57 2tan
57 2 tan
k
k
k
k
k
tg
r r
mgG
r
N kN
y
x(1)
(2)
We desire the time t when (3)Solution of Eqs. (1)-(3):
For slipping to cease,
or
Engineering Mechanics Dynamics -- IAA111
Chap.7 3D Kinematics and KineticsTranslation Rotation
v r
( ) a r r
Engineering Mechanics Dynamics -- IAA
Sample 7/1The 0.8-m arm OA for a remote-control mechanism ispivoted about the horizontal x-axis of the clevis, andthe entire assembly rotates about the z-axis with aconstant speed N = 60 rev/min. Simultaneously, thearm is being raised at the constant rate rad/s/ forthe position where , determine (a) the angularvelocity of OA, (b) the angular acceleration of OA, (c)the velocity of point A, and (d) the acceleration ofpoint A. If, in addition to the motion described, thevertical shaft and point O had a linear motion, say, inthe z-direction, would that motion change the angularvelocity or angular acceleration of OA?
Inertial coordinates X-Y-ZBody fixed coordinates x-y-zA: point of interestB: origin of body-fixed coordinates, often
the mass center AΩ: angular velocity of the rigid angular
velocity of x-y-z about X-Y-Z
or 2
A B
A B
r rr r
Engineering Mechanics Dynamics -- IAA116
Sample 7/3,7/4
Crank CB rotates about the horizontal axiswith an angular velocity ω1 = 6 rad/s which isconstant for a short interval of motion whichincludes the position shown. The link AB hasa ball-and-socket fitting on each end andconnects crank DA with CB. For the instantshown, determine the angular velocity ω2 ofcrank DA and the angular velocity ωn of linkAB.
Engineering Mechanics Dynamics -- IAA117
Sample 7/3,7/4/
2[ ] 50 , 100(6) 600 mm/sA B n A B
A Br
v v rv j v i i
250 600
50 100 100x y zn n n
i j kj i
2
6
2
0 2
y z
x z
x y
n n
n n
n n
2 6 rad/s
/[ 0] 50 100 100 0x y zn A B n n n r
2 2 22 2 4 5 2 5 rad/s3n
4 8 10rad/s rad/s rad/s3 3 3x y zn n n
2( 2 4 5 ) rad/s3n i j k
Engineering Mechanics Dynamics -- IAA
Angular velocity
1 2 3
1 1 2 2 3 3T
2 2
2 3
1 2 3
1T
2 2T
3 3
2 3
0
0
, , 0
0
0
r = r r rr = r r r
r
r r rr
r
r
4 equations for 4 unknows; andif and are general vectors,
then
6 equations for 6 unknows; and
1r
2r
r3r
118
Engineering Mechanics Dynamics -- IAA
Angular Acceleration
1 2 3
1 1 2 2 3 3
1 1 1 1
2 2 2 2
3 3 3 3
T2 2
2 3
0
0
0
r = r r rr = r r r
r = r
r
r
r
4 equations for 4 unknows; and
1r
2r
r3r
119
Engineering Mechanics Dynamics -- IAA120
Angular Momentum
[ ( )]G dm H
[ ( )]O dm H r r
(a)
(b)
x-y-z Body-fixed coordinates at C.G.ω: angular velocity of the rigid body
3 3 3 1
G G
xx xy xz
yx yy yz
zx zy zz
xx x xy y xz z
yx x yy y yz z
zx x zy y zz z
I I II I II I I
I I II I I
I I I
H I
Engineering Mechanics Dynamics -- IAA
Inertia Matrix
xx x yy y zz zI I I H i j k
Principal Axes
0 00 00 0
xx
yy
zz
II
I
2 2
2 2
2 2
( )
( )
( )
xx xy
yy xz
zz yz
I y z dm I xydm
I z x dm I xzdm
I x y dm I yzdm
xx xy xz
yx yy yz
zx zy zz
I I II I II I I
121
Engineering Mechanics Dynamics -- IAA122
problem 07/64
b
b
x
y
a ay
z
z
Introduce axes0, ,
2 2x y z
2 21 10, (2 )12 3x y y yI I m a ma
0, 0y z x zI I 2 2 2 21 1[(2 ) (2 ) ] ( )
12 3z zI m a b m a b
x y z Eq. 7/11 applied to gives
2 2 21 1( ) [ ( )]3 32 2
y y y z z zI I
ma m a b
H j k
j k1cos45 sin45 ( )2
j j k j k
1sin45 cos45 ( )2
k j k j k
2 2 21 3[ (2 ) ] 20 ( 0.04 0.06 )6 6
( 0.4 0.6 ) Nms
m b a b
H j k j k
j k21 1 (20 ) ( 0.4 0.6 ) 6.0 59.2 J
2 2TT H k j k
But
The rectangular plate, with amass of 3 kg and a uniform smallthickness, is welded at the 45° angle tothe vertical shaft, which rotates with theangular velocity of 20π rad/s. Determinethe angular momentum H of the plateabout O and find the kinetic energy of theplate.
Engineering Mechanics Dynamics -- IAA123
problem 07/64
b
b
x
y
a ay
z
z
z
45 y
20 rad/sz
xM
2
2
22 2
1 2 0 0120 0
10 0 0 2 012
0 0 10 0 4 412
G G
G G
x x'G y y
z z
m aI
I m bI
m a b
H I
H I
I
about x-y-z
about x′ -y′ -z′
1 0 0 0 00 cos sin 0 sin0 sin cos cos
'
G G G G
Α
H A H A I A I
Note that G G I A I A
Kinetic energy T T1 1T2 2G G
I I
Engineering Mechanics Dynamics -- IAA
Euler’s Equation
F G
M H
( )
( )
xyz
x y z
ddtH H H
HM H
i j k H
( )
( )
( )
x y z z y
y z x x z
z x y y x
H H H
H H H
H H H
M i
j
k
x x y z z y
y y z x x z
z z x y y x
M H H H
M H H H
M H H H
( )
( )
( )
x xx x yy zz y z
y yy y zz xx z x
z zz z xx yy x y
M I I I
M I I I
M I I I
124
Engineering Mechanics Dynamics -- IAA
problem 07/82
z
45 y
20 rad/sz
xM
Eq. 7/23 2x yz zM I
/ 22 2
/ 2
3 3 2 22
2
2 3(0.2)( ) 0.02 kgm3 6 62 2 2 2
b
yz bI yzdm y dl y dy
b b mb
20.02(20 ) 79.0 NmxM
on plate, 79.0 Nmx M i
but acting on shaft, 79.0 Nm M i
The plate has a mass of 3 kg and iswelded to the fixed vertical shaft, whichrotates at the constant speed of 20πrad/s. Compute the moment M appliedto the shaft by the plate due to dynamicsimbalance.