Transcript Document

v  v0  a t
a  const
 v  v0 
v 
t

0
0

 2 
2
1
x  x0  v0 t  2 a t
v  v  2a  x  x0 
2
2
0
Newton’s First Law of Motion
Every body continues in its state of rest or uniform
speed in a straight line unless acted on by a nonzero
net force.
Preceding defines INERTIA
Preceding defines MASS
(m  W)
Best when observed in the absence of friction.
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.
a=
F
m
F = m a
dp
(F=
)
dt
Newton’s Third Law of Motion
Whenever one object exerts a force on a second
object, the second exerts an equal and opposite
force on the first.
The Bane of Galileo: Friction
Friction is everywhere! There is little wonder
why it played such a prominent part in
MECHANICS for the ancients (e.g., Aristotle).
Galileo recognized friction as separate from
motion, so the equations of kinematics could be
discovered (using geometry).
Friction - treated as a force (though not a vector)
- always opposite to the motion
- sometimes related to the motion,
sometimes not so related
Dynamics of Circular Motion
Galileo sez: Circular motion is not natural;
straight line motion is natural
What is required for an object to move in a circle?
v2
 FR  maR  m r
v2
aR 
r
(v constant, a constant,
v changing, a changing)
v
R 
r
FR  m  R2 r
angular velocity
What happens to an object when the
centripetal force quits?
A 0.150-kg ball on the end of a 1.10-m cord is swing
in a vertical circle. What is the minimum speed that
the ball must have to continue moving in a circle?
FT  W  FR
v2
mg  m
r
v  gr
 9.80 m/s 2 1.10 m
 3.28 m/s
FTB  W  FRB
FTB  W  FRB
Banked Curves - Rotations and Inclines
A 1000-kg car rounds a curve on a flat road of radius
50 m at a speed of 50 kph (14 m/s). Will the car make
the turn if the pavement is icy and s = 0.25?
v2
FR  m
r
14 m/s 

 1000 kg
 3920 N
50 m
2
Ffr   s FN
 s m g
 0.24  1000 kg  9.80 m/s 2
 2450 N
Ffr  FR  the car will slide
An airplane traveling at 520 kph attempts to turn around.
By banking at an angle of 38, how long will this
maneuver take?
Need to find radius of curve,
then time =  radius / speed
Fy  W  0
FN cos   mg  0
mg
FN 
cos 
Fx  FR  0
2
v
FN sin   m  0
r
v2
FN sin   m
r
mg
v2
sin   m
cos 
r
v2
g tan  
r
v2
r
g tan 

144 m/s 
1
2
2
9.80 m/s 2  tan 38
 2700 m
circumference
t
v
 r  2700 m


v
144 m/s
 59 s
Nonuniform Circular Motion
Easiest to describe in terms
of the circle.
atan 
dv
(  a )
dt
v2
aR 
r
a  aR2  a2
(polar coordinates)
a  aR  a
Terminal Velocity
For velocity-dependent friction force (e.g., air
resistance), the frictional force can equal the
motive force.
F W  F
D
 ma
for example, consider FD  v 2
a  0  v  terminal velocity
Newton’s Law of Universal
Gravitation
In 1687, very controversial law of mechanics
action at a distance - no direct contact
(first of the force fields)
universal - applies to both terrestrial and
celestial motion
verification required experiments, calculus,
and history
Law of Universal Gravitation
Every particle in the universe attracts every other
particle with a force that is proportional to the product
of their masses and inversely proportional to the square
of the distance between them. This force acts along the
line joining the two particles.
m1 m2
F G 2
r
G  6.67  10
11
Nm
kg
2
2
What is the force of attraction between a 50-kg
person and a 75-kg person sitting 50 cm apart?
m1 m2
F G 2
r
50 kg  75 kg
 6.67 E  11 N m /kg
(0.50 m)2
 1.0E  6 N
2
2
m1 m 2
ˆ
F12  G
r
12
2
r12
rˆ12  rˆ21
F1  F12  F13  F14 
  F1i
i1
Weighing the Earth
Mm
FG  G 2  mg
rE
M
g G 2
rE
g rE2
M
(1798)
G
9.80 m/s 2  (6.38E 6 m) 2

6.67 E  11 N m 2 /s 2
 5.98E 24 kg
What is g on the top of Mt Everest, 8848 m high?
M
M
g G 2 G
2
r
 rE  h 
 6.67 E  11 N m /kg
2
2
 9.77 m/s 2 ( -0.3%)
5.98E24 kg
 6380  8.8 km 
2
What is g on the moon?
mM = 7.35E22 kg rm = 1.74E6 m
M
g G 2
r
 6.67 E  11 N m /kg
2
 1.62 m/s
2
2
 1.0 


 6.05 
7.35E 22 kg
1.74E6 m 
2
What is g in the Shuttle?
h = 240 km
ME
M
g G 2 G
2
r
 rE  h 
 6.67 E  11 N m /kg
2
 9.1 m/s
2
 -7% 
2
5.98E24 kg
 6380 + 240 km 
2
What is the effect of the Earth’s rotation
on the value of g?
W  FR  ma
v2
g  g 
rE
v 2 (464 m/s) 2
g  
rE
6380 km
 0.0337 m/s
Why is the Space Shuttle weightless?
W  FR  0
Mm
v2
G 2 m
r
r
GM
GM
v

r
rE  h
2 r
T
v
Kepler’s Laws and Gravitation
Kepler found three laws of
planetary motion in Brahe’s data.
He was looking for celestial music.
1) Planetary orbits are ellipses
2) Equal areas in equal time
3) T2  s3
Kepler’s Third Law from Newton’s Laws
consider circular orbits and Newton's laws
FG  FR  0
M m m v2
G 2 
0
r
r
G
Mm
2 mr

4

r2
T2
4 2 3
T 
r
GM
2
v

T 2  r3
2 r
T
Gravitational Field
Force fields are simpler than contact forces.
Project action at a distance.
Interaction only for pairs of objects.
Force depends on the “charge” of an object.
F
GM
g    2 rˆ
m
r
Forces in nature :
gravitation
electromagnetic
weak nuclear
strong nuclear
-
mass
electric charge
electric charge
baryon
Electric and magnetic forces combined by Faraday
and Maxwell around 1840.
Electromagnetic and weak nuclear combined in 1967.
Who is next?
Principle of Equivalence
FG  FR  0
M mG mR v
G 2 
r
r
2
Conceptually, the gravitational mass and the inertial
mass are different. But, numerically, they are equal
to high precision. What about the laws of mechanics?