Advanced Engineering Dynamics 2010 Part 5 pdf
Advanced Engineering Dynamics 2010 Part 5 pdf
...
Therefore
(5. 5)
Gm sin0 d0
dV
=
-
2r
Using the cosine rule
-
2Rx
cos0
(5. 6)
y
=
x2
+
R~
and differentiating gives
(5. 7)
2
2r dr
=
2Rx
sin0 d0
Substituting equation
(5. 7)
into ...
m2(r2&
+
2wi2)
or
(5. 15)
o
=
1
~(m,r~w)
=
z
1
g(m2r:w)
rl
Substituting equation
(5. 12)
into equation
(5. 14)
gives
lF121
=
p2s
+
CLj:
(5. 16)
Now...
Advanced Engineering Dynamics 2010 Part 1 pdf
...
Dispersive waves
55
55
55
58
59
61
64
65
67
72
75
76
80
83
85
85
85
88
90
93
100
103
106
107
107
109
1
I8
1 25
1 25
1 25
128
130
132
133
136
138
141
1 45
149
2
Newtonian ...
The symmetrical body
Forced precession
Epilogue
5
Dynamics
of
Vehicles
5. 1
5. 2
5. 3
5. 4
5. 5
5. 6
5. 7
5. 8
5. 9
5. 10
5. 1
1
5. 12
Introd...
Advanced Engineering Dynamics 2010 Part 12 pdf
...
0 .5
-0 .50 0
0
-0.866
-0.779
0
0
0 0
I
222
Robot
arm
dynamics
and
The velocity of
0
is
dr,
dri-l
dt dt
+
-
(d,
+
a;)
+
0,
x
(dj
+
ai)
at
Vi=-=-
(8 .51 )
(8 .52 ) ... revolute joint the second term on the right hand side is zero.
The acceleration of
0;
is
(8 .54 )
(8 .55 )
For a revolute joint the second and fourth terms on the right hand side are ze...
Advanced Engineering Dynamics 2010 Part 2 doc
...
Thus equation (1 .53 ) can be written
Newton
's
laws
for
a group
of
particles
15
1.13
Newton's
laws
for
a
group
of
particles
Consider a group of
n
particles, three ... zero the moment of
momentum is conserved.
1. 15
Energy for a group of particles
Integrating equation (1. 45) with respect to displacement yields
(1 .55 )
The
first
term on the left hand...
Advanced Engineering Dynamics 2010 Part 3 ppt
...
general form for the Lagrangian of a particle is
map
aP
m
e
2
at
at
2
at
r= +- (5) ~p). (5) ~p>+m ((R~p)
-
v
=T
-
u,
-
u,
-
v
with
5)
=
oxi
+
o,,j
+
o,k
and
p
= ... time integral will be integrated by
Parts
because
6u
=
0
at
t,
and
t2.
The second term in equation (3.18) is
Integrating by parts gives
(3.19)
50
Hamilton
S
principle
Thus
the .....
Advanced Engineering Dynamics 2010 Part 4 pot
...
constant
(4 .55 )
Since for a rigid body there can be no internal energy losses and because there is no exter-
nal work being done the kinetic energy will be constant
1
1
2 2
(4 .56 )
We can ...
(4.47)
gives
(4 .50 )
(4 .5
1)
where
Z
is
the moment of inertia about the axis of rotation.
With
(e)
=
(1
m n)T
and noting that
[I]
is symmetrical, equation
(4 .50 )
expan...
Advanced Engineering Dynamics 2010 Part 6 docx
...
(5. 50)
shows that
Equation
(5. 47)
may
now be written as
(5. 49)
(5. 50)
(5. 5
1)
(5. 52)
which holds for
0
S
e
<
1.
Figure 5. 12 shows plots of
0
versus a non-dimensional time for ...
J(e
+
1)
-
{(e
-
l)tan(6/2)
Plots of equation (5. 55) , including equation (5. 54), for different values of
e
are shown in Fig.
5. 13.
5. 6
Effects
of
oblateness
I...
Advanced Engineering Dynamics 2010 Part 8 docx
...
u’)
Also
shown on Fig.
6. 15
is a plot of
-
X/X-
=
(1
-
e-‘)
(6 .55 )
and this shows a reasonably close resemblance to the plot of
(6 .54 ).
Equation
(6 .55 )
is of
the same form
as ... eventually to
(6.49)
(6 .50 )
21
213
+p+1
2p
+
1
I= (!)
3v
h
[iln(71
-
)
-
,3
arctan(
T)
+'$](6 .51 )
Now
Impact
of
a rigid mass
on
an elastic bar
1 45
&...
Advanced Engineering Dynamics 2010 Part 9 potx
... one-dimensional wave propagation
HELICAL
SPRING
DATA
Nmber
of
turm
11 .5
Coilradius
R
50 .0
mm
Wmradius
r
12 .5
mm
Mlterial,
steel
EN
498
Galcdatedjvm
jidl theory
Helix angle
s" ... spring
167
will separate into
two
groups of three.
So,
substituting equations (6. 157 ), (6. 158 ) and
(6. 159 ) into equations (6.164), (6.168) and (6.169) leads to
m
R
GJ(...
Advanced Engineering Dynamics 2010 Part 11 pps
... shall only be con-
cerned with the overall dynamics and not with the detail. This is a vast subject area
of
which
dynamics is a substantial and vital part.
8.2
Typical arrangements
8.2.1
CARTESIAN ...
SUCCESSIVE ROTATIONS ABOUT FIXED AXES
8
Robot
Arm
Dynamics
8.1
Introduction
In
this chapter we examine the way in which three-dimensional dynamics is applied to
a
sys-...
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