Download Slides for Chapters 1, 2, 3, 4 and Review

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General Physics I
(aka PHYS 2013)
P ROF. VANCHURIN
( AKA V ITALY )
University of Minnesota, Duluth
(aka UMD)
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O UTLINE
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S ECTION 1.1: T HE N ATURE OF P HYSICS
I
Mathematics
I is the language of science: physics, astronomy, chemistry,
engineering, geology, etc.
I very abstract math ideas find their way into science (e.g.
complex numbers, differential geometry)
I It was conjectured that all of the mathematics has a
physical realization somewhere (e.g. multiverse theories)
I
Physics
I serves as a bridge (or foundation) to other sciences.
I one can never prove anything in physics, but math is also
not as pure as seems (Godel’s theorems)
I it is however remarkable how successful the language of
math is in describing the world around us.
I
Think of physics as a toolbox of ideas which can be used in
building scientific models in chemistry, biology, geology,
astronomy, engineering, etc.
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S ECTION 1.2: S OLVING P HYSICS P ROBLEMS
I
Concepts
I in most disciplines the more material you can memorized
the better your final grade will be. Not the case in physics.
I concepts is what you have to understand (actually there is
only a single concept and almost everything follows)
I learning how to learn physics concepts will be your first
and perhaps most difficult task.
I
Problems
I the only way to evaluate if you really understand concepts
is to solve problems. understanding solutions not enough.
I you might have hard time solving problems at first, but it is
essential to learn how to solve problems on your own.
I keep track of how many problems you solved by yourself
in each chapter (without anyone helping or googling)
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S ECTION 1.3: S TANDARDS AND U NITS
Units
I Physics is an experimental science and experiments
involves measurements (e.g. time, length, mass).
I To express the results of measurements we use units. Units
of length, units of time units of mass.
I There are some standard units just because historically we
decided so (e.g. seconds, meters, kilograms).
I One can always convert from one system of units to
another given a conversion dictionary.
I Time: measured in seconds, milliseconds, microseconds,...
I
I What is time and arrow of time is a deep philosophical question. We might discuss it a bit in the
last day of classes in context of the second law of thermodynamics.
I
Length: measured in meters, millimeters, micrometers, ....
I Despite of the fact that length and time appear to us very differently, there is a very deep
connection (symmetry) between them. We might discuss is briefly when we discuss gravitation.
I
Mass: measured in units of gram, milligram, microgram, ...
I Mass is also something very familiar to us in everyday life, but also has very deep properties
connecting it so length and time. We might mention it briefly in connection to black-holes.
I
Other units can be formed from seconds, meters and kilograms
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S ECTION 1.4: C ONVERTING U NITS
I
Dimension.
I Any physical quantity expressed in units of TIME, LENGTH or MASS is said to have dimensions of
time, length or mass respectively. More generally one can have physical quantities which have
mixed dimensions. For example if d has units of LENGTH and t has units of TIME, then quantity
v=
d
t
(1)
has units of LENGTH/TIME.
I Evidently Eq. (1) has quantities with the same dimension (i.e. LENGTH/TIME) on both sides of the
equation. This must be true for any equation that you write. Checking that the quantities on both
sides of equation have the same dimension is a quick, but very important test that you could do
whenever you setup a new equation. If the dimension is not the same than you are doing
something wrong.
Conversion.
I Sometimes you will need to convert from one system of units to another. This can always be done
with the help of conversion dictionary. For the case of conversion from standard system of units to
British system of units the dictionary is:
1 in
=
2.54 cm
(2)
1 pound
=
4.448221615260 Newtons
(3)
I If we are given a quantity in units of speed, then we can convert it from one system of units to
another.
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S ECTION 1.5: U NCERTAINTY AND S IGNIFICANT
F IGURES
Experimental Measurements.
I
Measurements are always uncertain, but it was always hoped
that by designing a better and better experiment we can improve
the uncertainty without limits. It turned out not to be the case.
I
There is a famous uncertainty principle of quantum mechanics,
but you will only learn it next year in (PHYS 2021) if you decide
to take it.
I
From our point of view uncertainty is nothing but uncertainty in
measurements. This (as well as significant figures) will be
discussed in your lab course.
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S ECTION 1.6: E STIMATES AND O RDER OF
M AGNITUDE
Theoretical Estimates.
Similarly to uncertainties in experimental measurements,
theoretical predictions are never exact. We always make
simplifying assumption and thus the best we can hope for is an
estimate for the physical quantities to be measured.
I A useful tool in such estimates is known as order-of-magnitude
estimate (also know as outcome of “back-of-the-envelope
calculations”).
I
I
Such estimates are often done using the so-called dimensional
analysis - i.e. just use the known quantities to form a quantity
with the dimension of the quantity that you are looking for.
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S ECTION 1.7: V ECTORS AND V ECTOR A DDITION
Vectors.
I Some physical quantities are describe by a single real number. We call these
quantities - scalar quantities or scalars.
I Other quantities also have a direction associated with them and thus are describe
by three real numbers.
~ = (Ax , Ay , Az ).
A
(4)
We call these quantities - vector quantities or vectors. There are also tensors etc.
I When dealing with vectors it is often useful to draw a picture:
I Vectors are nothing but straight arrows drawn from one point to another.
Zero vector is just a vector of zero length - a point.
I Length of vectors is the magnitude of vectors. The longer the arrow the
bigger the magnitude.
I It is assumed that vectors can be parallel transported around. If you attach
~ then the vector A
~ +~
beginning of vector ~
B to end of another vector A
B is a
~ to end of vector ~
straight arrow from begging of vector A
B.
Coordinates.
I The space around us does not have axis and labels, but we can imagine that
these x, y and z axis or the coordinate system to be there.
I This makes it possible to talk about position of, for example, point particles
using their coordinates - real numbers.
I Since one needs three real numbers to specify position it is a vector. Similarly,
velocity, acceleration and force are all vectors.
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S ECTION 1.8: C OMPONENTS OF V ECTORS
I Symmetry. You might complain that there is arbitrariness in how one chooses
coordinate system or what components of the vector are and you would be right.
It turns out that the physically observable quantities do not depend on the
choice of coordinate systems and thus one can choose it to be whatever is more
convenient. Moreover, this symmetry is an extremely deep property which gives
rise to conservation laws that we will learn in this course.
I Magnitude. The length of vector or magnitude is a scalar quantity
~ =A
~ = |A|
A
(5)
or in components
(Ax , Ay , Az ) =
q
A2x + A2y + A2z .
(6)
I Direction. One can also find direction of vector using trigonometric identities.
I Addition. Two vectors can be added together to get a new vector
~ =A
~ +~
C
B
(7)
an in component form
(Cx , Cy , Cz ) = (Ax , Ay , Az ) + (Bx , By , Bz ) = (Ax + Bx , Ay + By , Az + Bz ).
(8)
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S ECTION 1.9: U NIT V ECTORS
I Unit vectors is a vector (denoted with a hat) that has magnitude one.
|û| =
q
u2x + u2y + u2z = 1.
(9)
There are three unit vectors are so important that there are special letters
reserved to denote these vectors
î
=
(1, 0, 0)
ĵ
=
(0, 1, 0)
k̂
=
(0, 0, 1).
(10)
I Multiplication / division by scalar. Any vector can be multiplied by a scalar to
obtain another vector,
~ = CB.
~
A
(11)
In components from
(Ax , Ay , Az ) = C(Bx , By , Bz ) = (CBx , CBy , CBz ).
(12)
I Components. Any vector can be written in components in two ways:
~ = (Ax , Ay , Az ) = Ax î + Ay ĵ + Az k̂
A
(13)
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S ECTION 1.10: P RODUCT OF V ECTORS
I Scalar (or dot) product. Dot product is a multiplication between two vectors
which produces a scalar:
~ ·~
~ =C
A
B=~
B·A
(14)
In components
~ ·~
A
B = (Ax , Ay , Az ) · (Bx , By , Bz ) = Ax Bx + Ay By + Az Bz = AB cos φ.
(15)
One can also derive multiplication table for unit vectors.
I Vector (or cross) product. Cross product is a multiplication between two vectors
which produces a vector:
~
~ ×~
~ =C
A
B = −~
B×A
(16)
In components
~
~ ~
A×
B = (Ax , Ay , Az )×(Bx , By , Bz ) ≡ (Ay Bz −Az By , Az Bx −Ax Bz , Ax By −Ay Bx ) = C.
(17)
One can also derive multiplication table for unit vectors.
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S ECTION 2.1: D ISPLACEMENT, T IME , AVG . V ELOCITY
I
Point particles.
I
I
I
I
Objects (car, ball, stone) often modeled as point particles.
Position is described by a vector in some coordinate system
1D coordinate system for motion along a straight line.
Position.
I
As time progresses position vector changes with time
~r(t) = (x(t), y(t), z(t)).
I
I
(18)
For motion in 1D only one component is relevant: x(t).
Average velocity.
I
In 1D average velocity is defined as
vavg ≡
∆x
x(t2 ) − x(t1 )
=
.
∆t
t2 − t1
(19)
where ∆x ≡ x(t2 ) − x(t1 ) and ∆t ≡ t2 − t1 .
I
Question: Can traveled distance be larger than displacement?
Can it be smaller? Can it be the same?
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S ECTION 2.2: I NSTANTANEOUS V ELOCITY
I
Instantaneous velocity.
I
Instantaneous velocity (or just velocity) is defined as
v(t) ≡ lim vavg = lim
∆t→0
I
I
∆x(t)
dx(t)
=
.
∆t
dt
(20)
Given x as a function of t one can always find instantaneous
velocity at every moment of time by simple differentiation.
Graphical representation.
I
I
I
∆t→0
average velocity is a slope of a line joining the coordinates
of initial (x1 , t1 ) and final (x2 , t2 ) points points on x(t) graph
instantaneous velocity if a slope of the tangent line to x(t) at
a given time.
Question: Can instantaneous velocity be larger than average?
Can it be smaller? Can it be the same?
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S ECTION 2.3: A CCELERATION
I
Average acceleration.
I
Given instantaneous velocity as a function of time v(t), one
can calculate average acceleration:
aavg ≡
I
I
∆v
v(t2 ) − v(t1 )
=
.
∆t
t2 − t1
This is analogous to how the average velocity was defined.
Instantaneous acceleration.
I
Instantaneous acceleration (i.e. an acceleration at a given
moment of time) is defined by taking a limit,
a(t) ≡ lim aavg = lim
∆t→0
I
∆t→0
dv(t)
∆v
=
.
∆t
dt
(22)
Graphical representation.
I
I
I
(21)
avg. velocity is slope of line joining (v(t1 ), t1 ) and (v(t2 ), t2 )
instantaneous velocity if a slope of the tangent line to v(t) .
Question: Can instantaneous acceleration be larger than
average? Can it be smaller? Can it be the same?
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S ECTION 2.4: M OTION WITH C ONST. A CCELERATION
Constant Acceleration.
I Motion with constant acceleration is a motion for which
instantaneous acceleration is a constant function
a(t) = ax .
I
I
I
(23)
Velocity for such motion changes as
vx (t) = v0x + ax t
(24)
1
x(t) = x0 + v0x t + ax t2 .
2
(25)
Position changes as
Note that ax , v0x , x0 are some fixed numbers representing
acceleration, velocity and position at time t = 0, but t is a
variable which can take any non-negative value.
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S ECTION 2.4: M OTION WITH C ONST. A CCELERATION
Useful relations.
I
One can obtain other useful relations from Eqs. (24) and
(25) by expressing t in one equation and plugging it into
another
vx (t)2 = v20x + 2ax (x(t) − x0 ) .
I
Another useful relation is obtained from Eqs. (24) and (25)
by expressing ax in one equation and plugging it into
another
1
(vx (t) + v0x ) t.
(27)
2
Question: What would change in equations (23), (24), (25), (26)
and (27) if the initial time is at t0 6= 0?
x(t) − x0 =
I
(26)
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S ECTION 2.4: M OTION WITH C ONST. A CCELERATION
Example 2.4. A motorcyclist heading east through a small town
accelerates at a constant acceleration 4.0 m/s2 after he leaves city
limits. At time t = 0 he is 5.0 m east of the city-limits while he moves
east at 15 m/s. (a) Find position and velocity at t = 2.0 s. (b) Where
is he when his speed is 25 m/s?
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S ECTION 2.5: F REE FALLING B ODIES
Free falling.
I
Free fall is a very deep concept in physics. All it means
that in 4-dimensional space-time free falling object move
along straight lines (geodesics).
I
For now we are only interested in motion very close to
surface of the Earth and such motions can be
approximated as motions with constant acceleration
ax = g = 9.80 m/s2 .
(28)
I
Thus all of the concepts and equations considered in the
previous section apply.
I
Note that g is taken to be positive and thus it makes sense
to choose x-axis to point vertically and downward. If you
choose the x-axis to point upward then ax = −9.80 m/s2 .
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S ECTION 2.5: F REE FALLING B ODIES
Example 2.6. One-euro coin is dropped from the Leaning Tower of
Pisa and falls free from rest. What are its position and velocity after
1.0 s?
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S ECTION 2.5: F REE FALLING B ODIES
Example 2.7. You throw a ball vertically upward from the roof of a tall building. The ball
leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the
ball is then in free fall. On its way back, it just misses the railing. Find (a) the ball’s position
and velocity 1.00 s and 4.00 s after leaving your hand; (b) the ball’s velocity when it is 5.00 m
above the railing; (c) the maximum height reached; (d) the ball’s acceleration when it is at its
maximum height.
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S ECTION 2.6: V ELOCITY /P OSITION BY I NTEGRATION .
Velocity by Integration.
From definition of acceleration
dvx (t)
dt
(29)
dvx = ax (t)dt
(30)
ax (t) =
and thus in differential from
or in integral form
Z
vx (t2 )
t2
Z
dvx =
vx (t1 )
ax (t)dt.
(31)
t1
Therefore for t1 = 0 and t2 = T we have
T
Z
ax (t)dt.
vx (T) = vx (0) +
(32)
0
If we replace
T→t
t→τ
vx (0) → v0x
(33)
then we get
Z
vx (t) = v0x +
t
ax (τ )dτ
(34)
0
and in the case of constant acceleration
vx (t) = v0x + ax t.
(35)
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S ECTION 2.6: V ELOCITY /P OSITION BY I NTEGRATION .
Position by Integration.
From definition of velocity
dx(t)
dt
(36)
dx = vx (t)dt
(37)
vx (t) =
and thus in differential from
or in integral form
Z
x(t2 )
t2
Z
dx =
x(t1 )
vx (t)dt.
(38)
t1
Therefore for t1 = 0 and t2 = T we have
T
Z
x(T) = x(0) +
vx (t)dt.
(39)
0
If we replace
T→t
t→τ
x(0) → x0
(40)
then we get
t
Z
x(t) = x0 +
vx (τ )dτ
(41)
0
and in the case of constant acceleration
Z t
x(t) = x0 +
(v0x + ax τ )dτ
0
or
x(t) = x0 + v0x t +
1 2
ax t .
2
(42)
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S ECTION 2.6: V ELOCITY /P OSITION BY I NTEGRATION
Example 2.9. Sally is driving along a straight highway in her 1965 Mustang. At t = 0, when
she is moving at 10 m/s in the positive x-direction, she passes a signpost at x = 50 m. Her
x-acceleration as a function of time is
ax (t) = 2.0 m/s2 − 0.10 m/s3 t.
(43)
(a) Find her x-velocity vx (t) and x(t) as functions of time. (b) When is her x-velocity greatest?
(c) What is that maximum x-velocity? (d) Where is the car when it reaches that maximum
x-velocity?
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S ECTION 3.1: P OSITION AND V ELOCITY V ECTORS
Extra dimensions
I We now generalize motion to 2D and 3D, but one can extend it further
I Extra dim. (six of them) are also needed to formulate string theory
I These extra dim. are usually very small and can often be neglected
Position
I Vector ~r specifies position of an object in three dimensions
~r = (x, y, z) = xî + yĵ + zk̂
(44)
I If the object is in motion, then the position changes with time
~r(t) = (x(t), y(t), z(t)) = x(t)î + y(t)ĵ + z(t)k̂.
(45)
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S ECTION 3.1: P OSITION AND V ELOCITY V ECTORS
Velocity
I Average velocity vector defined as
~vavg ≡
~r(t2 ) −~r(t1 )
∆~r
=
,
∆t
t2 − t1
(46)
I Instantaneous velocity vector defined as
~v(t) ≡ lim ~vavg = lim
∆t→0
∆t→0
d~r(t)
∆~r
=
.
∆t
dt
(47)
dx(t)
dx(t)
dx(t)
î +
ĵ +
k̂
dt
dt
dt
(48)
or in components as
d~r(t)
=
dt
dx(t) dy(t) dz(t)
,
,
dt
dt
dt
and if we denote ~v ≡ vx , vy , vz then
~v =
vx
=
vy
=
vz
=
=
dx(t)
dt
dy(t)
dt
dz(t)
.
dt
(49)
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S ECTION 3.1: P OSITION AND V ELOCITY V ECTORS
Example 3.1. A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars
lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane.
The rover, which we represent as a point, has x- and y-coordinates that vary with time:
x(t)
=
y(t)
=
2.0 m − 0.25 m/s2 t2
(1.0 m/s) t + 0.025 m/s3 t3
z(t)
=
0m
(50)
(a) Find the rover’s coordinates and distance from the lander at t = 2.0 s. (b) Find the rover’s
displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. (c) Find a
general expression for the rover’s instantaneous velocity vector ~v. Express ~v at t = 2.0 s in
components from and in terms of magnitude and direction.
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S ECTION 3.2: T HE A CCELERATION V ECTOR
Acceleration
I Average accelaration vector defined as
~aavg ≡
~v(t2 ) − ~v(t1 )
∆~v
=
,
∆t
t2 − t1
(51)
I Instantaneous velocity vector defined as
~a(t) ≡ lim ~aavg = lim
∆t→0
∆t→0
d~v(t)
∆~v
=
.
∆t
dt
or in components as
dvy (t)
d~v(t)
dvx (t) dvy (t) dvz (t)
dvx (t)
dvz (t)
~a =
=
,
,
=
î +
ĵ +
k̂
dt
dt
dt
dt
dt
dt
dt
and if we denote ~a ≡ ax , ay , az then
ax
=
ay
=
az
=
dvx (t)
d2 x(t)
=
dt
dt2
dvy (t)
d2 y(t)
=
dt
dt2
dvz (t)
d2 z(t)
=
.
dt
dt2
(52)
(53)
(54)
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S ECTION 3.2: T HE A CCELERATION V ECTOR
Example 3.2. Let’s return to motion of the Mars from the previous
section. (a) Find the components of the average acceleration for the
interval t = 0.0 s to t = 2.0 s. (b) Find the instantaneous
acceleration at t = 2.0 s.
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S ECTION 3.2: T HE A CCELERATION V ECTOR
Parallel and Perpendicular Components.
I A useful example of a moving coordinate system in 2D is when one of the axis
point in the direction of the velocity vector ~v.
I Then we can decompose the acceleration vector into component parallel ~a|| and
perpendicular ~a⊥ to the direction of motion.
I Magnitude of velocity vector changes as
dv(t)
= v̂(t) · ~a|| (t)
dt
(55)
I Direction of the velocity vector changes as
~a⊥ (t)
dv̂(t)
=
.
dt
v(t)
I Parallel component is responsible for changes in magnitude of the velocity
vector (i.e. speed), but not the direction
 a||

speed is increasing
+ v ~v
~a|| = 0
speed is not changing

 a||
speed is decreasing.
− v ~v
(56)
(57)
I Perpendicular component is responsible for changes in direction of velocity,but
not in magnitude.
I If there is only a perpendicular component then the object is in circular motion.
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S ECTION 3.3: P ROJECTILE M OTION
Projectile motion
I Projectile motion is completely determined by gravitationally acceleration and
air resistance starting from initial condition determined by position and velocity.
I The fact that you have to specify an even number of initial data is a consequence of the
fact that (differential) equations of motion are of the second order in time.
I Projectile motion is always confined to a 2D plane determined by two vectors:
gravitational acceleration vector and initial velocity vector.
I It is convenient to choose a coordinate system so that one of the axis is vertical,
the other one is horizontal (and there is no motion in z-direction.)
I For y-axis pointing upward and x-axis horizontally
~a = (0, −g) = −gĵ
(58)
and thus
~v(t)
=
~r(t)
=
v0x î + (v0y − gt)ĵ
1
(x0 + v0x t) î + y0 + v0y t − gt2 ĵ
2
(59)
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S ECTION 3.3: P ROJECTILE M OTION
Example 3.6. A motorcycle stunt rider rides off the edge of a cliff.
Just at the edge his velocity is horizontal with magnitude 9.0 m/s.
Find the motorcycle’s position, distance from the edge of the cliff, and
velocity 0.50 s after it leaves the edge of the cliff.
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S ECTION 3.4: M OTION IN A C IRCLE
Uniform motion
I Uniform motion is when the direction of velocity vector changes, but magnitude
(or speed) does not change, i.e.
dv
=0
dt
dv̂
6= 0,
dt
and
(60)
I This implies that
d~v
=0
dt
but if the magnitude of acceleration also does not change,
~v ·
(61)
a(t) = a⊥ (t) = const.
then
0=
d
d
d 2
a =
(~a · ~a) =
dt
dt
dt
d~v d~v
·
dt dt
(62)
=2
d2~v d~v
·
dt2 dt
(63)
I By combining Eqs. (61) and (63) in 2D we must conclude that
~v ∝
d2~v
.
dt2
(64)
I If the proportionality constant is negative, then solutions of the above equation
are sines and cosines (you will see this differential equation over and over in
physics courses).
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R EVIEW
S ECTION 3.4: M OTION IN A C IRCLE
Uniform motion
I We have shown that
~v(t) ∝
d2~v(t)
.
dt2
(65)
I With little more of calculus (that we shall skip) one can show the such motion
gives rise to circular motion described by equation
R=
q
(x(t) − X)2 + (y(t) − Y)2
(66)
where (X, Y) is at the center of the circle and R is its radius.
I By choosing the origin of coordinates at the center of circle, i.e. making
(X, Y) = (0, 0)
(67)
q
x(t)2 + y(t)2 = R
(68)
we can simplify Eq. (66) to get
r=
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S ECTION 3.4: M OTION IN A C IRCLE
Radial (or centripetal) acceleration.
By differentiating Eq. (66) with respect to time once we get
0=
dx(t)
dy(t)
x(t) +
y(t)
dt
dt
(69)
or
~r · ~v = 0
(70)
and by differentiating it twice we get
0=
or
dx(t)
dt
2
+
d2 x(t)
dt2
x(t) +
dy(t)
dt
2
+
d2 y(t)
dt2
y(t)
v2 = −~a ·~r = −~a⊥ ·~r.
(71)
(72)
But since the two vectors ~a⊥ and ~r point in opposite directions we have
v2 = a⊥ R
(73)
v2
.
R
(74)
or
a⊥ =
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R EVIEW
S ECTION 3.4: M OTION IN A C IRCLE
Periodic motion.
I Since the motion is circular the object must come to were it started at some finite
time. This time is called period, T, and the motion is called also periodic.
I It is useful to write an exact solution for periodic motion around origin
t
t
~r(t) = R sin 2π + φ , R cos 2π + φ
(75)
T
T
where R, T and φ are some constants.
I Note that as time goes from t to t + T the arguments of the sin and cos functions
change by 2πand thus the position vector does not change
~r(t) = ~r(t + T).
(76)
I Evidently, an object in circular motion is traveling a distance 2πR with a constant
speed v and thus the period must be
T=
2πR
v
v=
and
2πR
.
T
(77)
I By combining it with
v2
R
(78)
4π 2 R
.
T2
(79)
a⊥ =
we get
a⊥ =
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R EVIEW
S ECTION 3.4: M OTION IN A C IRCLE
Example 3.12. Passengers on a carnival ride move at a constant speed in a horizontal circle of
radius 5.0 m, making a complete circle in 4.0 s. What is their acceleration?
Step 1: What is the coordinate system? Let choose a (moving) coordinate system with ||
direction in the direction of motion and ⊥ direction in the radial direction.
Step 2: What is given? Period
T = 4.0 s
and radius
R = 5.0 m
Step 3: What do we have to find? Then
a⊥ =
4π 2 (5.0 m)
= 12 m/s2 = 1.3 g
(4.0 s)2
and
a|| = 0.
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R EVIEW
S ECTION 3.5: R EFERENCE FRAME .
Reference frame.
I We have already mentioned how the choice of coordinate system is important
for calculations, but should not matter for physically observable quantities.
I We have also discussed (in context of 2D motions) how it is sometime useful to
work with respect to a moving coordinate system.
I More generally, let object A move with respect to a coordinate system (or
reference frame) of object B, described by position vector
~rA/B (t)
(80)
and let object B move with respect to a coordinate system (or reference frame) of
object C, described by position vector
~rB/C (t)
(81)
then we say that to describe object A with respect to a coordinate system (or
reference frame) of object C, we use the following rule
~rA/C (t) = ~rA/B (t) +~rB/C (t).
(82)
I By taking a time derivative of Eq. (82) we get a rule for adding velocities
d~rA/C (t)
dt
~vA/C (t)
=
=
d~rA/B (t)
d~rB/C (t)
+
dt
dt
~vA/B (t) + ~vB/C (t).
(83)
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S ECTION 3.5: R EFERENCE FRAME .
Example 3.14-15. There is a 100 km/h wind from west to east. (a) If an airplane’s compass
indicates that it is headed due north, and its airspeed indicator shows that it is moving through
the air at 240 km/h, what is the velocity of the airplane relative to earth? (b) What direction
should the pilot head with speed 240 km/h to travel due north?
Step 1: Choose coordinate system. Let x-axis to point east and y-axis to point north.
Step 2: What is given?
~vA/E = 100 km/h î
Step 3: What do we have to find? (a) The airspeed indicator tells us that
~vP/A = 240 km/h ĵ
and thus
~vP/E = ~vP/A + ~vA/E = 100 km/h î + 240 km/h ĵ
(b) In general velocity of airplane relative to air is
~vP/A = x î + y ĵ
with
vP/A =
q
x2 + y2 = 240 km/h
and thus
~vP/E = ~vP/A + ~vA/E = x î + y ĵ + 100 km/h î = (x + 100 km/h)î + y ĵ = y ĵ.
Thus we have two equations with two unknowns with solution
~vP/A = x î + y ĵ = −100 km/h î + 218 km/h ĵ.
R EVIEW
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S ECTION 4.1: F ORCES AND I NTERACTIONS
Fundamental forces.
I
There are four types of fundamental forces:
1) electromagnetic,
2) weak,
3) strong and
4) gravitational.
I
The electromagnetic and weak (and to some extend strong)
forces had been successfully unified into electroweak (and to
some extend GUT) theory.
I
The situation is much worse with regards to gravitational force
which is manifests itself not through exchange of participles like
other forces, but through curvature of space-time.
I
The string theory does describe a way of how to think about
gravity (perturbatively), but it is too naive to expect that we will
know the final answer any time soon.
I
In our everyday experience the (microscopic) forces manifest
themselves trough (macroscopic) forces.
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S ECTION 4.1: F ORCES AND I NTERACTIONS
Superposition of forces.
There might be a number of different forces action on a given object,
and the total force (or net force) is given by
X
~Fi = ~F1 + ~F2 + ~F3 + ...
~ =
(84)
R
i
where the usual vector addition is used, i.e.
!
X
(Rx , Ry , Rz ) =
Fix ,
X
i
Strength of the net force is then
v
!2
u
q
u X
R = R2x + R2y + R2z = t
Fix +
i
Fiy ,
i
X
Fiz
(85)
.
i
!2
X
i
Fix
!2
+
X
i
Fix
. (86)
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R EVIEW
S ECTION 4.1: F ORCES AND I NTERACTIONS
Example 4.1. Three professional wrestlers are fighting over a champion’s
belt. The forces are on horizontal plane and have magnitudes and directions:
F1
=
250 N and θ1 = 127◦
F2
=
50 N and θ2 = 0◦
F3
=
120 N and θ3 = 270◦ .
Find the components of the net force on the belt and it magnitude and
direction.
(87)
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R EVIEW
S ECTION 4.2: N EWTON ’ S F IRST L AW
First Law.
I
A body acted on by no net force, i.e.
X
~Fi = 0
(88)
i
has a constant velocity (which may be zero) and zero acceleration.
I
(This tendency for a body to continue its motion is know as
inertia and is extremely important concept in theory of general
relativity.)
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S ECTION 4.2: N EWTON ’ S F IRST L AW
Example 4.2. In the classic 1950 science-fiction film Rocketship X-M,
a space-ship is moving in the vacuum of the outerspace, far from any
star or planet, when it engine dies. As a result, the spaceship slows
down and stops. What does Newton’s first law say about this scene?
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S ECTION 4.2: N EWTON ’ S F IRST L AW
Example 4.3. You are driving a Maserati GranTurismo S on a
straight testing track at a constant speed of 250 km/h. You pass a
1971 Volkswagen Beetle doing a constant speed 75 km/h. On which
car is the net force greater?
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S ECTION 4.2: N EWTON ’ S F IRST L AW
Inertial frames.
I
It is important to note that the Newton’s first law is not obeyed
in all reference frames (e.g. inside of an accelerating train).
I
Those frames of references where it is obeyed is called inertial
frame of reference (e.g. inside a train moving with constant
velocity).
I
Surface of Earth is not exactly an inertial reference frame (why?)
but it is pretty close to being inertial.
I
For inertial reference frames one can easily go from one frame to
another using
~rA/C (t) = ~rA/B (t) +~rB/C (t).
(89)
I
and
~vA/C (t) = ~vA/B (t) + ~vB/C (t).
which makes such frames particularly useful.
(90)
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S ECTION 4.2: N EWTON ’ S F IRST L AW
Examples. In which of the following situations is there zero net force on the
body?
I
An airplane flying due north as a steady 120 m/s and at a
constant altitude?
I
A car driving straight up a hill with a 3o slope at a s constant 90
km/h
I
A hawk circling at a a constant 20 km/h at a constant height of
15 m above an open field?
I
A box with slick, firctionless surface in the back truck as the
truck accelerates on a a level road at 5 m/s2 .
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S ECTION 4.2: N EWTON ’ S S ECOND L AW
Second Law.
I
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 mass of
the body times the acceleration vector of the body equals to the net force
vector, i.e.
X
~Fi = m~a
(91)
i
or
P~
Fi
~a = i .
(92)
m
I As a vector equation in 3D it is equivalent to three equations
X
Fix = max
i
X
Fiy
= may
Fiz
= maz .
i
X
i
(93)
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S ECTION 4.2: N EWTON ’ S S ECOND L AW
Units.
I
With the help of second law we can now relate the units of
force to units of mass and acceleration.
I
Since each equation must have the same units on both
sides we see that
N = kg · m/s2 .
(94)
I
If a 1 kg object moves with acceleration 1 m/s2 then there
must be a net fore of 1 N applied to it.
I
In British system of units
1 lb = 4.448 N.
(95)
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R EVIEW
S ECTION 4.2: N EWTON ’ S S ECOND L AW
Example 4.4. A worker applies a constant horizontal force with
magnitude 20 N to a box with mass m = 40 kg resting on a level floor
with negligible friction. What is the acceleration of the box?
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R EVIEW
S ECTION 4.3: M ASS AND W EIGHT
Weight.
I
Is a gravitational force acting on an object close to the surface of
the Earth (to be precisely at the see level)
~ = m~g.
w
I
(96)
It is a vector, but one often refers to the magnitude of the weight
force as weight
w = mg
(97)
or even to mass itself just because one, i.e. m, is simply related to
the other, i.e. w.
I
One should however be careful when weight is measured above
or below the sea level (e.g. on the airplane), as the weight force
can vary.
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S ECTION 4.3: M ASS AND W EIGHT
Example 4.7. A 2.49 × 104 N Rolls-Royce Phantom traveling in the
+x direction makes an emergency stop; the x-component of the net
force acting on it is −1.83 × 104 N. What is its acceleration?
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S ECTION 4.4: N EWTON ’ S T HIRD L AW
Third Law. If a body A exerts a force on body B (an “action”), then
body B exerts a force on body A (a “reaction”). These two forces have
the same magnitude, but are opposite in direction. These two forces
act on different bodies.
~FA on B = −~FB on A .
(98)
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S ECTION 4.4: N EWTON ’ S T HIRD L AW
Example 4.8. After your sport car breaks down, you start to push it
to the nearest repair shop. While the car is starting to move, how does
the force exert on the car compare to the force the car exerts on you?
How do these forces compare when you are pushing the car along at a
constant speed?
R EVIEW
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S ECTION 4.4: N EWTON ’ S T HIRD L AW
Example 4.9. An apple sits at rest on a table, in equilibrium (i.e.
static). What forces act on the apple? What is the reaction force to
each of the forces acting on the apple? What are the action-reaction
pairs?
R EVIEW
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S ECTION 4.4: N EWTON ’ S T HIRD L AW
Example 4.10. A stonemason drags a marble block across a floor by
pulling on a (weightless) rope attached to the block. The block is not
necessarily in equilibrium. How are the various forces related? What
are the action-reaction pairs?
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S ECTION 4.4: N EWTON ’ S T HIRD L AW
Example. You are driving a car on a country road when a mosquito
splatters on the windshield. Which has the greater magnitude: the
force that the car exerts on the mosquito or the force that the mosquito
exerted on the car? Or are the magnitudes the same? If they are the
different, how can you reconcile this fact with Newton’s third law? If
they are equal, why is mosquito splattered and damaged while the car
is undamaged?
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R EVIEW
S ECTION 4.5: F REE - BODY DIAGRAM
Free-body diagram. A useful tool in solving problems on Newton’s
laws is to draw free-body diagram, where a chosen body appears by
itself without of its surroundings, and with vectors drawn to show the
magnitude and directions of all the forces that act on the body.
The tricky part here is to include all of the forces that act on a
chosen body, but not to include any other forces that act on
other bodies.
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S ECTION 4.5: F REE - BODY DIAGRAM
Free-body diagram.
R EVIEW
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C HAPTER 0: M ATHEMATICS
I
Algebra
I
I
I
I
Trigonometry
I
I
I
I
I
Solving linear equations
Solving quadratic equations
Solving system of two equations with two unknowns
Hypotenuse from two sides
Side from hypotenuse and another side
Angle from hypothenuse and one side
Angle from two sides, etc.
Calculus
I
I
I
I
Differentiation of standard functions
Product rule, quotient rule, chain rule.
Indefinite integrals (i.e. antiderivitives)
Definite integrals over interval
R EVIEW
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C HAPTER 1: U NITS , P HYSICAL Q UANTITIES AND
V ECTORS
I
Units
I
I
I
I
Physical quantities
I
I
SI system of units
British system of units
Conversion of units
Scalars, Vectors, etc.
Vector algebra
I
I
I
I
I
I
I
I
I
Vector representations (graph., comp., unit vec.)
Magnitude and direction of a 2D vector from components
Components of a 2D vector from magnitude and direction
Vector addition/subtraction (graph., comp., unit vec.)
Vector multiplication by scalar (graph., comp., unit vec.)
Dot product between 2 vectors (graph., comp., unit vec.).
Finding angle between 2 vectors from dot product
Cross product between 2 vectors (graph., comp., unit vec.)
Determine direction of cross product using right hand rule.
R EVIEW
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C HAPTER 2: M OTION A LONG A S TRAIGHT L INE
I
Definition of relevant physical quantities
I Position and displacement
I Average velocity and instantaneous velocity
I Average acceleration and instantaneous acceleration
I
Graphing of physical quantities vs. time
I Graphs a-vs-t, v-vs-t and x-vs-t (obtain one from another)
I Calculate quantities from graphs of a-vs-t, v-vs-t or x-vs-t
I
Differentiation and integration
I Calculate functions a(t), v(t) or x(t) (calc. one from another)
I Calculate quantities from functions of a(t), v(t) or x(t)
I
1D kinematic problem (e.g. free fall)
I Step 1: Choose 1D coordinate system (x-axis, origin)
I Step 2: Determine initial conditions (e.g. position, velocity)
I Step 3: Determine final conditions (e.g. position, velocity)
I With many stages of motion consider each stage separately
I Final cond. for one stage are initial cond. for next stage
R EVIEW
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C HAPTER 3: M OTION IN T WO OR T HREE DIMENSIONS
I
2D kinematic problem (e.g. projectile motion)
I Step 1: Choose 2D coordinate system (x-axis, y-axis, origin)
I Step 2: Determine initial conditions (e.g. position, velocity)
I Step 3: Determine final conditions (e.g. position, velocity)
I With many phases solve for each phase separately.
I Final cond. for one phase are initial cond. for next phase
I
Hints
I Intermediate answers can be algebraic, not numeric.
I Every equation must have the same units on both sides.
I Solve system if there are as many equations as unknowns.
I Remember that quadratic equation may have two solutions.
I
Uniform circular motion
I Acceleration is perp. to velocity and points towards center
I Relations between: acceleration, velocity, radius, period.
I
Relative motion
I Find relative quantities by vector addition/subtraction
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C HAPTER 4: N EWTON ’ S L AWS OF M OTION
I
Newton’s First Law
I Distinguishing inertial and non-inertial reference frames.
I Apply the first law to object moving with zero acceleration
(which may or may not have zero velocity)
I
Newton’s Second Law
I Determine magnitudes of all forces applied to object, the
net force and acceleration along a given direction.
I Apply the second law to object moving with non-zero
acceleration (which may or may not be have const. speed).
I
Newton’s Third Law
I Determine all action-reaction pairs of objects.
I Apply the third law to every pair of objects.
I
Forces problems
I Step 1: Choose a coordinate system (1D or 2D).
I Step 2: Draw a free-body diagram for each object.
I Step 3: Apply Newton’s Laws (1st, 2nd and/or 3rd)
R EVIEW