Conservative Forces
A force F( x, y, z ) is conservative if there exists a single-valued scalar function V ( x, y, z ) such
that
F = V
(1)
x
= y .
z
(2)
where
The work done on a particle by a conservative force while the particle moves from point A t
Static equilibrium position
The system is at static equilibrium position
Find : 1 2
Lagrange Method
Maple Code
2l
1
l
l
1
> restart;
> Eq1:=sin(t1)+sin(t2)-1;
Eq1 := sin( t1 ) + sin( t2 ) 1
> Eq2:=3*tan(t1)-tan(t2);
m
Eq2 := 3 tan ( t1 ) tan ( t2 )
m
2
l
System of Particles
z
z
mk
m1
rc
r1
mk
rk
mn
k
rc
rn
y
y
x
x
(a)
(b)
Consider a system of n particles such as in Figure (a) above for n = 3 . Let mk be the
mass of the k-th particle, k = 1,2,., n , and let m is the total mass of the particles. Let rk
be t
Orbital Dynamics
Keplers Laws of Planetary Motion
I. The orbit of each planet is an ellipse with the sun at one focus
II. The radius vector from the sun to a planet sweeps over equal of area in equal time
III. The square of the periods of the planets are
Rocket Problem
In vacuum with no gravity
m = m0 bt
b = e Ae ve
Impuls and Momentum
pe Ae t = (m bt )(v + v ) + bt (v ve ) mv
b mass rate of burned fuel ejected
Devide by t and set t 0
ve exit velocity of gases relative to the rocket
pe Ae t = mv bve t
Ex
Vector Operations
Let A = ( A1
A3 ) and B = (B1
T
A2
B2
B3 ) be two vectors in three-dimensional space.
T
e3
A
Dot Product
B
The dot product A B is a scalar
e2
O
A B = AT B = AB cos
(1)
e1
where is the angle between A and B . The graphical meaning of (1)
Relative Motion
Let XYZ be a stationary coordinate system and let xyz be a moving coordinates as
shown in the figure. Let be the angular velocity of xyz .
Relative Position
r =R+
Relative Velocity
& &
&
r = R + ( )r +
Relative Acceleration
& = R + ( )r +
Coordinate Systems
Cylindrical Coordinates
Spherical Coordinates
e r e e z
e r e e
z
z
P
r
ez
z
e
P
r
er
e
O
y
y
er
e
r
Q
x
x
Q
The vectors e z , r, OQ, and e are in the same plan
r = re r + ze z
&
= e z
&
&
&
v = re r + re + ze z
(
r = re r
&
&
= e + e
Mass-Spring-Damper System
Mass-Spring System
k
m
F sin t
x
Equation of motion:
m& + kx = F sin t
x
(1)
Initial conditions:
x(0 ) = x0
(2)
Solution:
v(0 ) = v0
x(t ) = A sin nt + B cos nt +
mF
sin t ,
2
n = k m
2
n
(3)
Mass-Spring-Damper System
k
m
c
x
Eq
Laws of Dynamics
Newtons Second Law
m& = F
r
(1)
Work and Energy
Integration of (1) in the e t direction gives
B
B
A
y
A
r
F dr = m&dr
&dr =
r
(2)
O
( )
et
r
1 d
& &
(r r )dt = 1 d v 2
2 dt
2
(3)
z
B
en
v2
A
W = TB T A
x
v1
s
1
T mv 2
2
where W F dr = m
Eulers Equations for Rigid Body
Newtons second law
F = ma
G
where aG is the acceleration of the center of gravity G .
Rotation about Fixed Point O
is a rigid body with fixed points of rotation O with respect to an inertial coordinate
system.
x y z is a