:

Newton’s laws
Work and energy
Momentum & angular momentum
Newton’s laws application
Work and energy
Conservation of energy
Momentum and angular momentum
Conservation of
Momentum and angular 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

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

1.1 Newton’s first law
A body acted on by no net force moves with constant velocity and zero acceleration inertial force
1.2 Newton’s second law

F
 m
 a
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.


F

 d p
 dt d ( m


) dt

F
 d ( m

) dt
 m d v
 dt
 v
 dm dt
If m is a constant, then:

F
 m d

(net force)
 m
 a dt
Caution: This is a vector equation used for a instant.
We will use it in component form.
In the rectangular coordinate axis:
F
 x
F
 y
F
 z ma ma ma z x y
In the natural coordinate axis:
F n
 ma n
 m
 2
F t
 ma t
 m d

 dt m
eˆ t 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

Forces are the interactions between two or more objects .

2.1 Fundamental Forces:
Einstain :
S.L.Glashow
physicists unified field theory:gravity and electromagnetic interaction weak and electromagnetic interaction
Grand Unification Theory

2.2 the Common forces in mechanics
1) Gravitation

F
 
G m
1 m
2 r
2
2) elasticity r
F
  kx
k: force constant
x: elongation
F
3) Frictional force
Static friction
Kinetic friction
0
 f s
  s
N f k
  k
N o x
F

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: For m: x:Nsin

=ma y:Ncos

=mg m
N y
F
M mg
 x a=g.tg

For (M+m):
F=(M+m)a
=(M+m)g.tg


F s 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: N=ma perpendicular:

N=mg
So: a=g/
 。
If a

,then N

, then

N>mg, will the m go up?

 m
R
F s m mg
N
Discussion:What must be the
 if the block does not slide down the cylinder?
Draw a free body diagram of m.
horizontal: N=mR

2 perpendicular:

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=F min
.
Solution: so : F
 horizontal ： Fcos perpendicular ： Fsin

- f k
=0

+ N - mg=0
 mg f k
= µN cos
   sin

F=F min
： dF d


0 , d
2
F d

2

0
So: tg

= µ 。 that is: when L=h/sin

=2.92m
F=F min f k
N m mg
F

L 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: mg
 f
 ma ,
 mg
 kv
 m dv dt
Suppose: v t
 mg k f
 kv f o y mg

 mg
 kv
 m dv dt
0
 v v dv
 v t
  k m

0 t dt
Suppose: v
 v t
( 1
 e
 kt m ) v t
 mg k f final velocity: v t
 mg k
 c
Can you describe the man’s motion?
mg o y

Example:
 the picture.
v

0
 v
0
 j r

0

0 i
Solution: x : ma x
 f
0 t , a x
 dv x dt , so : dv x
 f
0 t dt m initial

0 v x dv x condition : t

0 , v x

0
 
0 t f
0 t m dt , so : v x
 f
0 t
2
2 m y m
 v x
 dx dt
 dx
 f
0 t
2 dt
2 m v
0
F=f
0 t i
O x

initial conditions : t

0 , x

0

0 x dx
 
0 t f
0 t
2 dt , so : x

2 m f
0 t
3
6 m y : y
 v
0 t so path eqation : x
 f
0
6 mv
0
3 y
3 y m v
0
O
F=f
0 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

2 Rsin
 perpendicular ： Ncos

=mg
R=0.1m
N cos

=g/ (

2 R)

=80.9
0
 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 b) N can not balance mg, so…

=4.0rev/s c) When

=1.0rev/s=2
 rad/s, cos

=2.5, so the bead stay at the bottom of the loop
N
 mg
R=0.1m

Example: A small bead at rest slide down a frictionless bowl of radius R from point A. Find N, a n
, a t at this position.
Solution: f n
=ma n
, f t
=ma t normal ： N-mgcos

= ma n
= m tangential ： mgsin

= ma t
= m
 d
R

2 dt so ： a t
=gsin
 mg
  m d
 dt
 m d
 d

 d
 dt
  m

A d
 d

 mg
N
 o
R
  m

R d
 d


0
 m
 d
   


2 mgR
 d

1
2 m

2
 mgR cos


4. Noninertial Frame of reference
B: mass m is accelerating.
A: mass m is at rest.
Which one is right?
B
A m k a
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.

a
 mB
 a
 mA
 a

AB

 a '
 a
 
F

 m a mB
Transposition:

F : real force
F
 i

F
: inertial force
 m
 a
 m assume :
 a '

F i
 a
  m
 a
: acceleration of
A relative to B
 m
 a '

 m a
Then:

F


F i
 m a

'

F i
A
  m a

F
 m k

 k x
B 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 m(g+a)sin a

=ma


=(g+a)sin
 m a

 m
 ma mg
N 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 ) x : N sin
 
Ma m
N
( for a ma y
N '
M mg a '

for m
ma
  mg


: ma sin
 
N
 ma
 mg cos
 x

Tracing problem
Plane: x=x
0
+vt, y=h
Missile: dY/dX=(y-Y)/(x-X)
So: dY/dt=k(y-Y) dX/dt=k(x-X)
And: k 2 [(y-Y) 2 +(x-X) 2 ]=u 2 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 y v u
O
Y(0)=0, X(0)=0 h x