Transcript Dynamics

Dynamics
Kinematic Vectors

Kinematic quantities refer to
the motion only.
• Position
• Velocity
• Acceleration

  dr
v r 
dt
vi  xi 
  
a  v  r
dxi
dt
ai  vi  xi
x2

Kinematic vectors of velocity
and acceleration are
coplanar.

a
 1

r1
v1

r2

a
 2
v2
x1
Force and Mass

vA

aA


vB
mA
mB
Two particles can influence
each other.
• Masses related to
accelerations

aB

a
mA
 B
mB a A


p  mv
 
pF

Momentum is defined be
mass and velocity.
• Derivative is force
• Defines equations of motion
Forces on One Particle

There is a net force on each
particle in an object.

This corresponds to the particle’s
acceleration.
m

r

2
2

d
r
d
(
m
r


 )
F  m a  m

2
dt
dt 2
Forces on a System

For N particles in a system
the forces add.
• Particles of constant mass
M   m
 1

 FINT (  )  0

Some forces are internal
and some are external to
the system.
• Internal forces cancel

FINT (  )
N

FEXT (  )
m

r

N 
N d 2 (m r )

 
Ftotal   F  
dt 2
 1
 1

N 
N 
N d 2 (m r )

 
Ftotal   FEXT (  )   FINT (  )  
dt 2
 1
 1
 1

2


M N d (m r )
d2 1 N
Fnet 

M
m
r

 
M  1 dt 2
dt 2 M  1
Center of Mass

The weighted average of
the positions of the particles
is the center of mass.

Fnet
 
 N

m
r

    
d 2   1
M 2

dt
M




M



m
r

N

The system acts like a
single particle.
• Force at center of mass
• Translational change at
center of mass

rCM 
 
 1
M

rCM

Fnet



drCM
d
d 
Fnet  M
 MvCM  pCM
dt
dt
dt
Sliding Inclined Plane


F1
• Two frictionless surfaces
iˆ
mg sin q
M
mg cos q
q


Fg (b )  mg
 


m( A  a )  Fg (b )  F1
mA cos q  ma  mg sin q
mAsin q  0  F1  mg cosq
The block and inclined plane
are both free to move.

The coordinates should point
along the surface.
• Normal force is the force of
constraint
• The motion will be along the
surface
• Acceleration from plane and
block relative to plane
Motion of Plane
mA cos q  ma  mg sin q

mAsin q  F1  mg cosq
The inclined plane has two
forces from constraints.
• Upward from table
• Reaction from block

F2
M
q

 F1


Fg ( p )  Mg
 
 
MA  Fg (b )  F2  F1
MA   F1 sin q
0  Mg  F2  F1 cosq
iˆ

The system of linear
equations are solved for the
accelerations.
A
 g sin q cos q
sin 2q  M m


cos 2q

a  g sin q 1  2
 sin q  M m 
Newtonian Mechanics

The result compares to the
simple problem of a fixed
plane.

There were four unknowns in
the problem.
• Two accelerations
• Two forces of constraint


F2
g sin q
M

 F1

F1
q


Fg ( p )  Mg


Fg (b )  mg
The constraint forces can be
eliminated by using work.
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