Velocity is a relative
quantity
Disentangling Coordinates
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
2
Reference Frames
y
• Consider two frames of reference the O-frame (label
events according to t,x,y,z) and the O'-frame (label
Vt'
events according to t',x',y',z') moving at a constant
velocity V, with respect to each other at let the origins
O
coincide at t= t' = 0. In the Galilean transformations
the O and O' frames are related as follows:
z'
z
t = t!
Δt = t 2 − t1 = Δt ! = t 2! − t1!
x = x ! + Vt !
Δx = x2 − x1 = x2! − x1! + V (t 2! − t1! ) = Δx! + VΔt !
Time is
absolute!
y = y!
z = z!
Δy = Δy !
Δz = Δz !
y'
V
Event
O: (t,x,y,z)
O': (t'.x',y',z')
x'
x
O'
x
x'
Classical velocity
addition formula!
• Galilean Velocity Transformation:
Δx Δ x '
Δt ' Δx '
vx =
=
+V
=
+ V = v!x + V
Δt Δt
Δt Δt '
Δy Δy ' Δy '
vy =
=
=
= v!y
Δt Δt Δt '
Δz Δz ' Δ z '
vz =
=
=
= v!z
Δt Δt Δt '
R. Field 9/6/2012
University
FloridaLecture
PHY2053,
Fallof2013,
PHY 2053
6 – Newton’s Laws
Page 8
3
Postulates of Classical Physics
y
• First Postulate of Classical Physics (“Relativity
Principle”):
y'
V
The basic laws of physics are identical in all systems of
reference (frames) which move with uniform
(unaccelerated) velocity with respect to one another. The
laws of physics are invariant under a change of inertial
frame. The laws of physics have the same form in all
inertial frames. It is impossible to detect uniform motion.
Vt'
Event
O: (t,x,y,z)
O': (t'.x',y',z')
x'
x
O
O'
x
z
x'
z'
• Second Postulate of Classical Physics (Galilean Transformation):
The O and O' frame are related by the Galilean Transformation.
t = t!
x = x! + Vt !
y = y!
Δt = Δt !
Δx = Δx ! + VΔt !
Δy = Δy !
z = z!
Δz = Δz !
R. Field 9/6/2012
University
FloridaLecture
PHY2053,
Fallof2013,
PHY 2053
6 – Newton’s Laws
Classical velocity
addition formula!
v x = v!x + V
v y = v!y
v z = v!z
Page 9
4
H-ITT Problem #1, 3 min
Relative Velocity
5
•
1)
2)
3)
4)
5)
see
H-ITT Problem #1, 3 min
Car A is driving 45 mph south. Car B is driving 60 mph
at 45º west of south. What is the magnitude and
direction of the velocity of car A as seen by a
passenger in car B? Direction is expressed in angle
with respect to East.
vBA ≃ 115 mph, direction ≃ 50º
vBA ≃ 115 mph, direction ≃ 25º
vBA ≃ 100 mph, direction ≃ 50º
vBA = 100 mph, direction ≃ 25º
vBA = 85 mph, direction ≃ 50
✖
✖
✖
✖
✖
No valid solution is listed. See following slides.
PHY2053, Lecture 6, Newton’s Laws
6
The problem is asking us to compute the magnitude
and direction of a vector. We know how to calculate
these quantities if we know the x and y components
of a vector:
q
~ = ax x̂ + ay ŷ
A
~ =
|A|
a2x + a2y ; tan ✓ =
This means that breaking the calculation up by x
and y components will simplify solving the problem.
Broken up into x and y components, the two
velocity vectors are:
ay
ax
~vA = [ 45 ŷ] mph
~vB = [( 60 sin ↵)x̂ + ( 60 cos ↵)ŷ] mph
What vector calculation do we want to perform? We know the
velocities of both cars with respect to the ground. It then makes
sense to take the reference frame fixed to the ground to be the
reference frame “from” which we are transforming (O). We want to
know what an observer in car “B” would see. So, the reference
frame of car “B” should be the reference frame “into” which we are
transforming (O’). We defined the velocity vector V as the velocity
with which reference frame O’ [car B] is moving when observed
from reference frame O[the ground], in other words,
. The relevant velocity transformation formula is:
~
V = ~vB
PHY2053, Lecture 6, Newton’s Laws
0
~vA
= ~vA
~
V
h
~ , compute
Given ~vA and V
0
~vA
.
i
7
Having broken down the velocities of car A and car B, we now do the calculation of va’ by
components:
0
~ = ~vA ~vB
~vA
= ~vA V
=
[( 45 ŷ)
[( 60 sin ↵)x̂ + ( 60 cos ↵)ŷ]] mph
=
[ 45 ŷ + 60 sin ↵ x̂ + 60 cos ↵ ŷ] mph
=
[42.43 x̂ + (42.43
=
[42.43 x̂
45) ŷ] mph
2.57 ŷ] mph.
The magnitude of the velocity with which an observer in car B sees car A moving (vA’) is
0
vA
=
=
=
=
0
|~vA
|
hq
i
2
2
vA,x + vA,y
hp
i
42.432 + 2.572 mph
42.51 mph
The direction angle can be computed from
0
vA,y
tan ✓ = 0
vA,x
!
✓ = tan
1
= tan
1
PHY2053, Lecture 6, Newton’s Laws
⇣
0
vA,y
0
vA,x
⌘
= tan
( 6.06) =
1
⇣
2.57 mph
42.43 mph
6.05 rad =
⌘
3.47
8
PHY2053, LECTURE 6:
Newton’s Laws
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Newton’s First Law
Law 1: Objects in motion tend
to stay in motion, and objects at
rest tend to stay at rest unless an
outside force acts upon them.
Note - “motion” in this
translation implies
“motion with a constant velocity
along a straight line”
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Experiment
Inertial ball
Newton’s Second Law
Law 2: The rate of change of the
LAW
II:Mutationem
The alteration
of
Lex
II:
motus
momentum of a body is directly
motion
is ever proportional
to
proportionalem
esse
vi
motrici
proportional to the net force
the
motive force
impress’d,
and
impressae,
fierithe
secundum
acting on it,etand
direction of
is
maderectam
in the direction
of the
lineam
qua
vis
illa
the change in momentum takes
right
line in which that force is
imprimitur.
place in the direction of the net
impress’d. (translation by Motte, 1792)
force
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Net force
● Force is a vector quantity
“Net Force”:
● For systems with constant mass, the rate of change of
momentum is
, so 2nd Newton’s law becomes:
● Mass [inertial] m is a proportionality constant
between force and acceleration → the property of an
object to resist velocity changes
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Four Fundamental Forces
● Electromagnetic Force
● electricity, magnetism, atoms, light
● Gravitational Force
● planets, galaxies
● Weak Force
● nuclear decay - keeps the stars “burning”
● Strong Force
● proton, neutron
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Normal Force
●
●
●
●
Not a fundamental force
Due to atomic structure of every-day objects
Objects resist other objects trying to enter their volume
Counter-examples: radiation, (exotic) particles
● Model with a perfectly rigid, impenetrable surface
● Surface exactly counters normal component of contact
force exerted on it by an object
● Example: You are sitting / standing / not falling
through the floor to the center of the Earth right now
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
Normal Force Example: Incline
The angle of the frictionless incline is α = 30°. Mass m1
slides down the incline, starting from rest. What is the
speed of the mass after it slid 10 meters downhill?
[use g = 10 m/s2]
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws