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Classic Mechanics : Dynamics
Dynamics
Newton’s laws
Newton’s
application
laws
Work and energy
Momentum & angular
momentum
Conservation
Momentum
Work Conservation
of Momentum
and
of
and
and
angular
energy
angular
energy
momentum
momentum
"Nature and Nature's laws lay
hid in night; God said, Let Newton
be! and all was light."
Newton, Sir Isaac (1642-1727),
mathematician and physicist, one
of the foremost scientific intellects
of all time.
a profound impact:
astronomy, physics, and
mathematics
achievements:
• reflecting telescope
• three laws of motion;
• the law of universal gravitation
• the invention of calculus
Chapter4-5 Newton’s Laws of Motion
Key terms:
dynamics
force mass
superposition
net force
Newton’s law of motion
Inertia
equilibrium
action-reaction pair
elasticity
tension
friction force
gravitational interaction
free-body diagram
Chapter4-5 Newton’s Laws of Motion
1. Newton’s law of motion
1.1 Newton’s first law
A body acted on by no net force moves with
constant velocity and zero acceleration
inertial
force


F  ma
1.2 Newton’s second law
If a net external force acts on a body, the body
accelerates. The direction of acceleration is the
same as the direction of the net force. The net
force vector is equal to the mass of the body
times the acceleration of the body.


 dp
d ( m )
F

dt
dt


 d ( m
)
dv  dm
F
m
v
dt
dt
dt
If m is a constant, then:


d
F m
dt

(net force)  ma
Caution: This is a vector equation used for a instant.
We will use it in component form.
In the rectangular
coordinate axis:
Fx  ma x
F y  ma y
Fz  ma z
In the natural
coordinate axis:
2
Fn  ma n  m

d
Ft  ma t  m
dt
eˆ t
m
eˆ n
1.3 Newton’s third law
Whenever two bodies interact, the two forces that
they exert on each other are always equal in
magnitude and opposite indirection.
1.4 Application area of Newton’s law
low speed
macroscopic
practicality
inertial frame
2. What is a force?
Forces are the interactions between two or more objects.
2.1 Fundamental Forces:
Einstain :
unified field theory:gravity and
electromagnetic interaction
S.L.Glashow
weak and electromagnetic interaction
physicists
Grand Unification Theory
2.2 the Common forces in mechanics
1) Gravitation

m1m 2
F  G 2 rˆ
r
2) elasticity
F  kx
o
k: force constant
x: elongation
x
F
3) Frictional force
Static friction
Kinetic friction
0  fs  s N
fk  k N
F
3. Applications of Newton’s law
Problem-solving strategy
* Identify the body
 Examine the force, draw a free body diagram
 Construct a coordinate
* Write the Newton’s law in component form
* Calculate the equation
Example: A wedge with mass M rests on a frictionless
horizontal table top. A block with mass m is placed on the
wedge, and a horizontal force F is applied to the wedge.
What must be the magnitude of F if the block is to remain at
a constant height above the table top?
Solution:
m
F
M
x:Nsin=ma
y:Ncos=mg
For m:
mg
N
a=g.tg 
y
For (M+m):
x

F=(M+m)a
=(M+m)g.tg 
Fs
m
a
m
N
mg
Discussion:What must be the magnitude of a if the
block does not slide down the wedge?
Draw a free body diagram of m.
horizontal:
perpendicular:
So:
N=ma
N=mg
a=g/ 。
If a,then N , then N>mg, will the m go up?

Fs
m
m
R
N
mg
Discussion:What must be the  if the block does not
slide down the cylinder?
Draw a free body diagram of m.
horizontal:
perpendicular:
N=mR2
N=mg
so :  
g
R
Example: A man pulls a box by a rope with constant
speed along a straight line, known:k=0.6, h=1.5m. Find
how long the rope is when F=Fmin.
Solution:
horizontal：
Fcos - fk=0
perpendicular： Fsin + N - mg=0
fk = µN
mg
so : F 
cos   sin
F=Fmin：
dF
d 2F
 0,
0
2
d
d
So: tg = µ。
that is: when L=h/sin=2.92m
F=Fmin
N
fk
m
mg
L
F

h
Example: A parachute man drop into air, the resisting
force is approximately proportional to the man’s speed v,
find the velocity in any instantaneous time and the final
velocity?
Solution: Draw a free body diagram of the man,
establish Newton’s law in y direction:
f  kv
mg  f  ma ,
dv
 mg  kv  m
dt
Suppose:
vt 
f
mg
k
o
y
mg
dv
 mg  kv  m
dt

v
0
Suppose:
mg
vt 
k
dv
k t
   dt
v  vt
m 0
v  vt ( 1  e

kt
m
)
mg
final velocity: v 
c
t
k
f
o
y
Can you describe the man’s motion?
mg
Example: Try to find the path of the particle according to




the picture.
v0  v0 j
r0  0i
Solution:
x : ma x  f 0 t , a x  dv x dt ,
f0 t
so : dv x 
dt
m
initial condition : t  0 , v x  0

vx
0
dv x  
dx
vx 
dt
t
0
f0 t
f0 t 2
dt , so : v x 
m
2m
f0 t 2
 dx 
dt
2m
y
m
v0
O
F=f0 t i
x
initial conditions : t  0 , x  0

x
0
dx  
t
0
f0 t 2
f0 t 3
dt , so : x 
2m
6m
y : y  v0 t
f0
3
so path eqation : x 
y
6 mv 03
y
m
v0
O
F=f0 t i
x
Example: A small bead can slide without friction on a
Circular hoop that is in a vertical plane and has a radius of
R=0.1m. The hoop rotates at a constant rate of =4.0rev/s
about a vertical diameter.
a) Find the angle  at which the bead is in vertical equilibrium.
b) Is it possible for the bead to “ride” at the same elevation as
the center of the loop?
c) What will happen if the hoop rotates at 1.00rev/s
p. 161, 5-103
=4.0rev/s
Solution: a) For the bead
normal：
Nsin=m2Rsin
perpendicular： Ncos=mg
cos=g/ (2R)
=80.90
R=0.1m
N

mg
Example: A small bead can slide without friction on a
Circular hoop that is in a vertical plane and has a radius of
R=0.01m. The hoop rotates at a constant rate of =4.0rev/s
about a vertical diameter.
a) Find the angle  at which the bead is in vertical equilibrium.
b) Is it possible for the bead to “ride” at the same elevation as
the center of the loop?
c) What will happen if the hoop rotates at 1.00rev/s
=4.0rev/s
b) N can not balance mg, so…
c) When =1.0rev/s=2rad/s,
cos=2.5, so the bead stay at
the bottom of the loop
R=0.1m
N

mg
Example: A small bead at rest slide down a frictionless
bowl of radius R from point A. Find N, an , at at this position.
Solution:
fn=man , ft=mat
normal：N-mgcos = man = m
2
R
d
tangential：mgsin = mat = m
dt
so： at=gsin
A
o
N 
R

mg
 d
d d
d
d
 m
m

  m
mgsin  m
R d
d dt
d
dt


0
md 



2
mgRs ind
1
m 2  mgRcos
2
4. Noninertial Frame of reference
A
B: mass m is accelerating.
A: mass m is at rest.
m k
B
a
Which one is right?
The bus is not a inertial frame of reference.
A frame of reference in which Newton’s first law
is valid is called an inertial frame of reference.
Any frame of reference will also be inertial if it
moves relative to earth with constant velocity.
Newton’s laws of motion become valid in non-inertial
system by applying a inertial force on the object.









amB  amA  a AB  a' a
F  mamB  ma'  ma



Transposition: F  ma  ma'



assume : Fi   ma
F : real force

Fi : inertial force
Then:

a : acceleration of
A relative to B


Fi  ma
 

F  Fi  ma'
A
B


F  kx
m k
a
Example: A wedge rests on the floor of a elevator. A block
with mass m is placed on the wedge. There is no
friction between the block and the wedge. The elevator
is accelerating upward. The block slides along the
wedge. Try to find the acceleration of the block with
respect to the elevator.
Solution: The elevator is the frame of reference, there are
three forces acts on the block.
a
N
m(g+a)sin=ma
a=(g+a)sin
m
m
a


mamg
a
Example: A wedge with mass M rests on a frictionless
Horizontal table top. A block with mass m is placed on
The wedge. There is on friction between the block and
the wedge. The system is released from rest.
a) Calculate the acceleration of the wedge and the
Horizontal and vertical components of the acceleration of
the block.
b) Do your answer to part (a) reduce to the correct
results When M is very large?
c) As seen by a stationary observer, what is the shape of
the trajectory of the block?
(see page 182, 5-108)
m
( for
N
( for
ma
a
N'
y
M
x
mg
M ) x : N sin  Ma
a'

m)
// : ma cos   mg sin   ma '
: ma sin  N  mg cos 
Tracing problem
Plane: x=x0+vt, y=h
Missile: dY/dX=(y-Y)/(x-X)
y
v
So: dY/dt=k(y-Y)
dX/dt=k(x-X)
k2[(y-Y)2+(x-X)2]=u2
And:
k=u/[(y-Y)2+(x-X)2]-1/2
So: Y(n+1)=Y(n)+k(y-Y)t
X(n+1)=X(n)+k(x-X)t
h
u
O
Y(0)=0, X(0)=0
x