Transcript Newtons Laws and Its Application

Chapter 4 & 5
Dynamics:
Newton's Laws and Its
Application
“The discovery of the laws of dynamics,
or the laws of motion, was a dramatic
moment in the history of science... after
Newton there was complete
understanding.
—— R. Feynman
Isaac Newton
(1643-1727)
“Nature and Nature's laws
Mathematician and
lay hid in night;
physicist, the greatest
God said, Let Newton be!
scientist of all time.
And all was light.”
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“If I have been able to see a little farther than
other man, it’s because I have stood on the
shoulder of giants.”
—— Isaac Newton
Work and energy
Newton’s laws
Momentum
Angular momentum
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Force
(action capable of accelerating an object)
F
m
G
F
G
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Newton’s First Law of Motion
Aristotle (384-322 B.C. )
Force is necessary to keep a body in motion
Galileo Galilei (1564-1642)
Idealized experiment with no friction
No forces → constant speed in a straight line
Net force
F2
F1
F
0
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Newton’s first law of motion:
Every body continues in its state of rest or of
uniform speed in a straight line as long as no net
force acts on it.
Inertia: tendency to keep one’s state of motion
Inertial reference frames
Noninertial reference frames
How to distinguish?
By checking to see if Newton’s first law holds.
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Mass
Mass is a measure of the inertia of a body.
More mass ←→ harder to change its velocity
Unit of mass: kilogram (kg)
◆
Mass and weight
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Newton’s Second Law of Motion
No forces → constant speed in a straight line
F
0
velocity will change
Newton’s second law of motion:
The acceleration of an object is directly proportional
to the net force acting on it and is inversely
proportional to its mass.
The direction of the acceleration is in the direction
of the net force acting on the object.
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F

a
m
 F  ma
Component form in rectangular coordinates
F
x
 max ,
F
y
 may ,
F
z
 maz.
Unit of force is Newton (N)
1N=1kg·m/s2
Limitations:
Macro world, low speed motion, inertial frames
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Newton’s Third Law of Motion
Where do forces come from?
Action ←→ reaction
F1
F2
Newton’s third law of motion:
Whenever an object exerts a force on a second
object, the second object exerts an equal and
opposite force on the first.
◆ Action
force and reaction force are acting
on different objects!
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Weight and normal force
Weight: the magnitude of gravity
FG  mg
g = 9.80m/s2
Direction: down toward the center of Earth
Contact force:
∥ Friction
⊥ Normal force
FN
FG
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Weight loss
Example1: Someone reads his weight 600N in an
elevator when it accelerates at 0.2g to the ground,
what is the real weight?
Solution:
FN =600N
FG  FN  m  0.2 g
a=g
 FN  0.8mg
a=0.2g
FN
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FG  mg  FN  750 N
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What about this acceleration?
FG
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Friction
Why there is a friction? (P105, Figure 5-1)
Kinetic friction
F   k N
k : coefficient of kinetic friction
Static friction
F   s N
 s : coefficient of static friction
 k and  s depend on the surfaces:
materiel, smoothness, lubrication, …
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How to solve problems
1. Identify all the bodies
2. Draw a free-body diagram, show all the forces
3. Choose a convenient x-y coordinate system
4. Component equations of Newton’s second law
5. Solve all the equations
▲ Be careful about limitations of the formulas!
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Tensions in rope
Example2: An object (m=10kg) is hanged on a rope
as the figure. What are the tensions in the rope?
(Ignoring the mass of rope)
60°30°
Solution: 1) free-body diagram
2) coordinate system
3) component equations:
FT1
m
y
o
x
FT2
mg

3
mg
 FT 1 cos 60  FT 2 cos 30  0  FT 1 
2

1
FT 1 sin 60  FT 2 sin 30  mg  FT 2  mg

2
FT 1 =?
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*Mass of rope & Catenary
Consider a rope with uniformly distributed mass,
what is the shape of a hanging rope?
x
y  cosh( )  C
a
Catenary
Search and read literatures if you are interested
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Pulling a box
Example3: Someone is pulling a box, where µ=0.75,
h=1.5m. If he wants to move with constant velocity
and exert a minimum force, L=?
Solution:
x : F cos   F  0
F
y : F sin   FN  mg  0

  FN 
y
o
 mg
F
cos    sin 
dF
 0  tan   
d
L=h/sin =2.5m
L
Fµ
FN
m
µ
mg
F
x
h

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Broken Atwood’s machine
Example4: Two masses connected by a rope and a pulley
(Atwood’s machine). The connection part of m2 is broken
and m2 is moving with constant acceleration a0 relative to
the rope. What are the accelerations of m1 and m2 relative
to the ground? (Ignoring the mass of rope and pulley)
Solution:
T
T
m1: m1g- T = m1a1
m2：m2g- T= m2 a2
m1
m2 ao
a2G  a2 R  aRG
m1g
m2g
 a2  a0  a1 or a1  a2  a0
a1 m1
m2 a2
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m1: m1g- T = m1a1
m2：m2 g- T = m2 a2
a2G  a2 R  aRG
or：a0 = a1 + a2
Solve these equations:
m 2 a 0  ( m1  m 2 ) g
a1 
m1  m 2
m1a 0  ( m 2  m1 ) g
a2 
m1  m 2
( 2 g  a 0 ) m1m 2
T
m1  m 2
a1 m1
T
m2
a2
T
m1
m2 ao
m1g
m2g
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Uniform circular motion
Radial acceleration
Net force
aR  v 2 / r
2
F

ma

mv
/r

R
Centripetal/radial force
v
aR
F
r o
Not some new kind of force
It can be tension, weight, friction or the net
force of these kind of forces.
Determined by aR
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Conical pendulum
Example5: conical pendulum.
a) What is the direction of acceleration?
b) Calculate the speed and period.
Solution: a) direction of acceleration
2
F
sin


mv
/r
b) horizontal: T
vertical: FT cos   mg  0
 r  L sin  
v
Lg sin 2 
,
cos 
L
FT
m
r O
aR
G
2 L sin 
T
v
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Rotating hoop
Example6: A thin circular horizontal hoop of mass
m and radius R rotates at frequency f. Determine
the tension within the hoop.
T
Solution: Consider a tiny section
of the hoop with an angle d,
d
m
it has a mass dm  d
2
speed v  2 f  R
R
T
v2
d
d
 2T 
Centripetal force F  dm  2T sin
2
2
R
Tension:
T  2 mRf 2
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Highway curves
Where does centripetal force come from?
Unbanked curves
v
Centripetal force: friction
F
Banked curves
Centripetal force:
N
Net force of N and G
F
(and friction)
G
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Skating in a bowl
Example7: Someone is skating in a bowl shape track
with constant , determine the height h. (no friction)
horizontal: Nsin =m2Rsin
Solution:

R
vertical:

N
mg
h
Ncos =mg
g
 cos  2
 R
g
h  R  R cos  R  2

2
  g / R  h  0?
Limitation of model
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Nonuniform circular motion
2
a

v
/r
Radial acceleration
R
dv
Tangential acceleration atan 
dt
Total acceleration
a
Ftan
FR
2
a tan
 a R2
atan
Fnet
aR

a
Components of the net force:
2
v
dv
FR  maR  m , Ftan  matan  m
r
dt
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Example8: An object (mass m) slides down along a
bowl from position A with no friction, find the radial
and tangential acceleration of the object.
2
v
Solution: Radial: N  mg cos   maR  m
R
dv
Tangential: mg sin   matan  m
dt
So: atan = gsin
o
A
N 

mg
R
Conservation
1 2
mv  mgRcos of energy
2
v2
 aR   2 gcos
R
What is the normal force N ?
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Velocity-dependent force
Question: An object falls from rest, under the
action of gravity and the air friction F =- v, what
is the speed at time t, and when t→∞?
dv
Solution:
mg   v  ma  m
dt
Separation of variables

dv

  dt
v  mg / 
m



t
mg

mg
m
ln(v  )   t  C  v 
 C1e

m



t
mg
m
Using initial conditions v 
(1  e ) t→∞?

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Example9: An object moves along a semi-circular
wall on smooth horizontal plane. Known: v0 and µ,
determine its speed v at time t.
Solution:
v2
Radial force N  m
R
dv
Tangential force   N  m
dt
2
dv

v dv
  2 
dt
  
v
R
R dt
1 
v  t  0   v0
 t  C,
v R
v0 R
Speed: v 
R  v0 t


µ
v0
v

.
o
R
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*Shooting problem
A cannon ball is fired with v0 at angle , and the
air friction F =- v, how to describe the motion,
and what is the range?
F
dv x
m
 Fx   v x
y
dt
dv y
o
x
m
 Fy  mg   v y  mg
G
dt
vx  vx 0 ebt


g  bt g

v y   v y 0  b  e  b




where b 
m
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vx 0

 bt
vx  vx 0 ebt
x

1

e



b



g  bt g   1

 y   v y 0  g  1  ebt   gt
v y   v y 0  b  e  b




b
b
b
vx 0  vy 0  300m/s; a) b  0.05s1
b) b  0.3s 1
How about F =-’ v2 ?
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