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MECHANICS
Lecture notes for Phys 111
Dr. Vitaly A. Shneidman
Department of Physics, New Jersey Institute of Technology, Newark, NJ 07102
(Dated: April 22, 2017)
Abstract
These notes are intended as an addition to the lectures given in class. They are NOT designed
to replace the actual lectures. Some of the notes will contain less information then in the actual
lecture, and some will have extra info. Not all formulas which will be needed for exams are
contained in these notes. Also, these notes will NOT contain any up to date organizational or
administrative information (changes in schedule, assignments, etc.) but only physics. If you notice
any typos - let me know at [email protected]. For convenience, I will keep all notes in a single file each time you can print out only the added part. Make sure the file is indeed updated, there is a
date indicating the latest modification. There is also a Table of Contents, which is automatically
updated. For convenience, the file with notes will be both in postscript and pdf formats. A few
other things:
Graphics: Some of the graphics is deliberately unfinished, so that we have what to do in class.
Advanced topics: these will not be represented on the exams. Read them only if you are really
interested in the material.
Computer: Mostly, the use of a computer will not be required in the lecture part of this course.
If I need it (e.g., for graphics), I will use Mathematica. You do not have to know this program, but
if you are interested I will be glad to explain how it works.
1
file: notes111.tex
Contents
I. Introduction
2
A. Physics and other sciences
2
B. Point mass
2
C. Units
2
1. Standard units
2
2. Conversion of units
3
D. Units and dimensional analysis
4
II. 1-dimensional motion
6
A. v = const
6
B. Variable velocity
7
1. a = const
13
C. Free fall
15
III. Projectile motion
19
A. Introduction: Object dropped from a plane
19
B. General
20
C. Examples
21
IV. Vectors
24
A. System of coordinates
25
B. Operations with vectors
26
1. Single vector
26
2. Two vectors: addition
27
3. Two vectors: dot product
28
4. Two vectors: vector product
29
C. Preview. Forces as vectors. Elements of static equilibrium
V. 2D motion
31
37
A. Introduction: Derivatives of a vector
37
B. General
37
C. ~a = const
38
0
D. ~a = ~g (projectile motion)
39
E. Uniform circular motion
39
1. Preliminaries
39
2. Acceleration
41
3. An alternative derivation
41
F. Advanced: Classical (Galileo’s) Relativity
VI. Newton’s Laws
42
43
A. Force
43
1. Units
43
2. Vector nature
43
B. The Laws
44
1. Gravitational force
46
2. FBD, normal force
47
C. Statics
47
D. Dynamics: Examples
49
VII. Newton’s Laws: applications to friction and to circular motion
A. Force of friction
55
55
1. Example: block on inclined plane
55
B. Fast way to solve quasi-one-dimensional problems
61
C. Centripetal force
63
1. Conic pendulum
64
2. Satellite
65
3. Turning car/friction/incline
66
D. (Advanced) ”Forces of inertia”
67
VIII. Work
68
A. Scalar (dot) product - reminder
68
B. Units
68
C. Definitions
68
D. 1D motion and examples
71
1
IX. Kinetic energy
72
A. Definition and units
72
B. Relation to work
73
1. Constant force
73
2. Variable force
74
C. Power
74
X. Potential energy
76
A. Some remarkable forces with path-independent work
76
B. Relation to force
77
XI. Conservation of energy
77
A. Conservative plus non-conservative forces
80
B. Advanced: Typical potential energy curves
82
C. Advanced: Fictitious ”centrifugal energy”
84
D. Advanced: Mathematical meaning of energy conservation
85
XII. Momentum
86
A. Definition
86
B. 2nd Law in terms of momentum
86
XIII. Center of mass (CM)
88
A. Definition
88
B. Relation to total momentum
89
C. 2nd Law for CM
90
D. Advanced: Energy and CM
90
XIV. Collisions
91
A. Inelastic
91
1. Perfectly inelastic
91
2. Explosion
94
B. Elastic
94
1. Advanced: 1D collision, m 6= M
94
2. Advanced: 2D elastic collision of two identical masses
96
2
XV. Kinematics of rotation
98
A. Radian measure of an angle
98
B. Angular velocity
99
C. Connection with linear velocity and centripetal acceleration
99
D. Dynamics of centripetal motion (see also sec. VII C)
100
E. Angular acceleration
102
F. Connection with tangential acceleration
102
G. Rotation with α = const
103
XVI. Kinetic Energy of Rotation and Rotational Inertia
104
A. The formula K = 1/2 Iω 2
104
B. Rotational Inertia: Examples
105
1. Collection of point masses
105
2. Hoop
105
3. Rod
106
4. Disk
106
5. Advanced: Solid and hollow spheres
107
C. Parallel axis theorem
108
1. Distributed bodies plus point masses
110
2. Advanced: Combinations of distributed bodies.
110
D. Conservation of energy, including rotation
112
E. ”Bucket falling into a well”
112
1. Advanced: Atwood machine
113
2. Rolling
114
XVII. Torque
116
A. Definition
116
B. 2nd Law for rotation
117
C. Application of τ = Iα. Examples.
118
1. Revolving door
119
2. Rotating rod
120
3. Rotating rod with a point mass m at the end.
120
4. ”Bucket falling into a well” revisited.
121
3
5. Advanced: Atwood machine revisited.
121
6. Rolling down incline revisited.
123
D. Torque as a vector
125
1. Cross product
125
2. Vector torque
125
XVIII. Angular momentum L
125
A. Single point mass
125
B. System of particles
126
C. Rotating symmetric solid
126
1. Angular velocity as a vector
126
~
D. 2nd Law for rotation in terms of L
127
XIX. Conservation of angular momentum
A. Examples
128
128
1. Free particle
128
2. Student on a rotating platform
129
3. Chewing gum on a disk
130
4. Measuring speed of a bullet
131
5. Rotating star (white dwarf)
132
XX. Equilibrium
133
A. General conditions of equilibrium
133
B. Center of gravity
133
C. Examples
133
1. Seesaw
133
2. Beams
135
3. Ladder against a wall
137
XXI. Fluids
139
A. hydrostatics
139
1. Density
139
2. Pressure
139
4
3. Archimedes principle
142
B. Hydrodynamics
144
1. Continuity equation and the Bernoulli principle
XXII. Gravitation
145
147
A. Solar system
147
B. Kepler’s Laws
147
1. 1st law
148
2. 2nd law
149
3. 3rd law
149
C. The Law of Gravitation
150
1. Gravitational acceleration
150
2. Satellite
151
D. Energy
153
1. Escape velocity and Black Holes
E. Advanced: Deviations from Kepler’s and Newton’s laws
5
153
155
Dr. Vitaly A. Shneidman, Phys 111, Lecture
I.
INTRODUCTION
A.
Physics and other sciences
in class
B.
Point mass
The art physics is the art of idealization. One of the central concepts in mechanics is a
”particle” or ”point mass”
i.e. a body the size or structure of which are irrelevant in a given problem. Examples:
electron, planet, etc.
C.
1.
Units
Standard units
In SI system the basic units are:
m (meter), kg (kilogram) and s (second)
Everything else in mechanics is derived. Examples of derived units (may or may not have a
special name): m/s, m/s2 (no name), kg · m/s2 (Newton), kg · m2 /s2 (Joule), etc.
Major variables, their typical notations and units:
variable
units
name
speed, v
m/s
-
acceleration (a or g)
m/s2
-
force (F, f, N, T )
N=kg m/s2 Newton
energy (K, U, E)
J=kg m2 /s2
Joule
2
2.
Conversion of units
Standard path: all units are converted to SI. E.g., length:
1 in = 0.0254 m , 1 f t = 0.3048 m , 1 mi ≃ 1609 m
Examples:
mi
1609 m
m
= 70
≃ 31.3
h
3600 s
s
2
3 cm2 = 3 10−2 m = 3 10−4 m2
70
1 mm3 = 1(10−3 m)3 = 10−9 m3
Non-standard:
Earth is 150 · 106 km from Sun; find speed in km/s if 1 year ≃ 365 days
2π
150 · 106 km
≃ 30 km/s
365 · 24 · 3600 s
Express µm = 10−6 m in Å = 10−10 m and in nm = 10−9 m
1 µm = 10−6 m = 10−6 · 1010 Å = 104 Å
1 µm = 10−6 m = 10−6 · 109 nm = 103 nm
A certain load of grain dries up in the storage, losing 2 kg in 4 weeks. Find the drying rate in
µg/s:
103 g
103 · 106 µg
109 µg
µg
2 kg
= 0.5
= 0.5
= 0.5
≃ 827
5
4 week
7 · 24 · 3600 s
7 · 24 · 3600 s
6.06 · 10 s
s
A certain laboratory pipette releases droplets about 1.2 mm in diameter. How many drops of
water are in a 1.0 L bottle? (Ans. 106 )
On average, a human heart pumps 45 cm3 of blood per beat, while the beat rate is about 60
beats per min. How many gallons of blood does the heart pump in 1 year? ( 1 gallon= 3800 cm3 ).
Vol =
45 cm3 60 beat
45 cm3 · 60
·
· 1 year =
60 · 24 · 365 min
beat
min
min
Vol ≃ 1.42 · 109 cm3 = 1.42 · 109 cm3 /(3800 cm3 /gal ≃ 0.37 · 106 gal
3
D.
Units and dimensional analysis
Verification of units is useful to check the math. More interesting, however, is the possibility to get some insight into a new problem before math is done, or before it is even
possible. E.g., suppose we do not know the formula for displacement in accelerated motion
with v0 = 0. Let us guess, having at our disposal only the mass m, [kg], acceleration a, [m/s2 ]
and time t, [s]. Since all functions exp, sin, etc. can have only a dimensionless argument,
sin t − wrong!
sin
t
− correct
T
Look for a power law
x ∼ aα tβ mγ or [m] = [m/s2]α [s]β [kg]γ
To get correct dimensions,
α=1, β=2, γ=0
(cannot get the coefficient, which is 1/2 but otherwise ok). Important: not too many
variables, otherwise could have multiple solutions, which is as good as none: e.g., if v0 6= 0
is present can construct a dimensionless at/v0 and any function can be expected.
Problems.
Galileo discovered that the period of small oscillations of a pendulum is independent of its amplitude. Use this to find the dependence of the period T on the length of the pendulum l, gravitational
acceleration g and, possibly, mass m. Namely, look for
T ∼ l α g β mγ
and find α, β and γ.
Solution (detail in class):
T =L
α
L
T2
β
Mγ
1
1
γ = 0, β = − , α = , T ∼
2
2
s
l
g
Example: The force created by a spring stretched by x meters is given by F = −kx (”Hook’s
law”) where the spring constant k is measured in N/m and N = kg · m/s2 . Find the dependence
4
of the period of oscillations T of a body of mass m attached to this spring on the values of k, m
and the amplitude A.
Solution: look for
T ∼ Aα kβ mγ
L 1 β γ
α
T =L M 2 ·
M
T
L
1
1
α = 0, β = − , γ = , T ∼
2
2
5
r
m
k
Dr. Vitaly A. Shneidman, Phys 111, Lecture
II.
1-DIMENSIONAL MOTION
Position x(t)
Displacement:
∆x = x (t2 ) − x (t1 )
Distance:
D ≥ |∆x| ≥ 0
Velocity:
∆x
, ∆t = t2 − t1
∆t
with a small ∆t (later, we distinguish between average velocity with a finite ∆t and instanv=
taneous with ∆t → 0).
Speed:
s=
A.
D
≥ |v| ≥ 0
∆t
v = const
See fig. 1
x
5
v
2.5
2
4
1.5
3
1
2
0.5
1
0.5
1
1.5
2
t
0.5
-0.5
-1
-1
-1.5
-2
1
1.5
2
t
FIG. 1: Velocity (left) and position (right) plots for motion with constant velocity: Positive (red)
or negative (blue). Note that area under the velocity line (positive or negative) corresponds to the
change in position: E.g. (red) 2 × 2 = 4.5 − 0.5, or (blue) 2 × (−1) = −2 − 0.
Displacement
∆x = v ∆t
6
(1)
Distance
D = |∆x|
(2)
s = D/∆t = |v|
(3)
Speed
Example. A motorcycle with VM = 60 m/s is catching up with a car with VC = 40 m/s,
originally D = 200 m ahead. When and where will they meet? Give the graphic solution.
VM t = VC t + D , thus, t =
D
= 10 s
VM − VC
Above is the meeting time; note ”relativity” - in a reference frame moving with VM the motorcycle
is stationary, while the car approaches it with speed of VM − VC . The meeting point - the distance
from the original position of the motorcycle - is
x = VM t = D
VM
= 600 m
VM − VC
Graphically, the solution is intersection in t, x plane, or when the area between red and blue
horizontal lines equals D in the t, v plane.
B.
Variable velocity
Starting example:
7
Example (trap!):
S1 = 2 km/h; , S2 = 4 km/h . Sa − ?
D = 2AB , t1 = AB/S1 , t2 = AB/S2
Sa =
D
2AB
2S1 S2
=
=
6= 3 km/h
t1 + t2
AB/S1 + AB/S2
S1 + S2
For explicit x(t) plots for v 6= const see fig. 4.
Average velocity:
vav =
8
∆x
∆t
(4)
x
25
20
15
10
5
1
2
3
4
5
6
t
FIG. 2: A sample position vs. time plot (blue curve), and determination of the average velocities
- slopes (positive or negative) of straight solid lines. Slope of dashed line (which is tangent to x(t)
curve) is the instantaneouus velocity at t = 2 .
Average speed
sav =
D
≥ |vav |
∆t
(5)
Distance:
D ≥ |∆x|
(6)
Instantaneous velocity
v = lim vav =
∆t→0
9
dx
dt
(7)
v
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
t
v
1.0
0.8
0.6
0.4
0.2
t
FIG. 3: Determination of displacement for a variable v(t). During the ith small interval of duration
∆t the velocity is replaced by a constant vi shown by a horizontal red segment. Corresponding
P
displacement is ∆xi ≈ vi · ∆t (the red rectangular box). The total displacement ∆x =
∆xi
is then approximated by the area under the v(t) curve. The error -the total arera of the small
triangles- becomes small for small ∆t (compare left and right figures) and vanishes in the strict
limit ∆t → 0, when the sum becomes an integral.
Displacement ∆x - area under the v(t) curve or Advanced
Z t
∆x(t) =
v (t′ ) dt′
t1
Acceleration
aav =
10
∆v
∆t
(8)
v
3
x
4
3
2
2
1
1
0.5
1
1.5
t
2
0.5
-1
1
1.5
2
t
-1
-2
-2
FIG. 4: Velocity (left) and position (right) plots for motion with constant acceleration: Positive
(red) or negative (blue). Again, area under the velocity line (positive or negative) corresponds to
the change in position. E.g. (red) (2.5 + 0.5) × 2/2 = 4 − 1 or (blue) (−2) × 2/2 = −2 − 0.
a = lim aav
∆t→0
dv
d2 x
=
= 2
dt
dt
Example: Given (t in seconds, x in meters)
x(t) = −
t4 t3
+ + t2 − t + 1
4
3
Find: a) vav between t1 = 1 s and t2 = 3 s; b) v(t) , a(t)
vav =
−17/4 − 13/12
8
x(t2 ) − x(t1 )
=
=−
t2 − t1
3−1
3
b) from
d n
t = ntn−1 : v(t) = −t3 + t2 + 2t − 1 , a(t) = −3t2 + 2t + 2
dt
11
(9)
12
1.
a = const
Notations: Start from t = 0, thus ∆t = t; v(0) ≡ v0 .
∆v = at
(10)
or
v = v0 + at
Displacement - area of the trapezoid in fig. 4 (can be negative!):
∆x =
v0 + v
t = v0t + at2 /2
2
(11)
A useful alternative: use t = (v − v0 ) /a:
∆x =
v0 + v v − v0
v 2 − v02
=
2
a
2a
(12)
(A more elegant derivation follows from conservation of energy,... later)
SUMMARY: if
a = const ,
(13)
v = v0 + at
1
x = x0 + v0t + at2
2
2
v − v02
v0 + v
t=
x − x0 =
2
2a
(14)
13
(15)
(16)
Examples: meeting problems (car VC = 40 m/s, aC = 0 and motorcycle VM = 0, aM =
2 m/s2 )
1
XC = VC t , XM = aM t2
2
1
1
VC t = aM t2 , t = 2VC /aM = 40 s , Xmeet = t · VC = 1600 m = aM t2
2
2
14
C.
Free fall
Reminder:
if a = const then
(17)
v = v0 + at
1
x = x0 + v0t + at2
2
2
v0 + v
v − v02
x − x0 =
t=
2
2a
(18)
Free fall:
a → −g , x → y , x0 → y0 (or, H)
15
(19)
(20)
a = −g = −9.8 m/s2
(21)
v = v0 − gt
1
y = y0 + v0 t − gt2
2
2
v − v2
y − y0 = 0
2g
(22)
(23)
(24)
Example: max height:
v = 0 , ymax − y0 = v02 /2g
(25)
Example: the Tower of Piza (v0 = 0 , y0 = H ≃ 55 m). Find t, v upon impact. Explore
v0 = ±8 m/s.
0 = H + 0t − gt2 /2 , t =
0 − H = −v 2 /2g , v =
16
s
p
2H
g
2gH
Lab.
Let w be the width of the flag and t1 and t2 the small time intervals to cross each of the
gates. Then
v1 =
w
w
, v2 =
t1
t2
and from
x − x0 =
after adjusting notations
a=
v 2 − v02
2a
v22 − v12
2(x − x0 )
(the theoretical value is g sin θ = g · H/L, neglecting friction).
HW
The rocket-driven sled Sonic Wind No. 2, ..., runs on a straight, level track of length
1030 m . Starting from rest, it can reach a speed of 222 m/s in a time 0.865 s .
Compute the acceleration, assuming that it is constant: from v = v0 + at with v0 = 0
a=
v
t
What is the distance covered in a time of 0.865 s ?
1
x = at2 = . . .
2
17
A jet fighter pilot wishes to accelerate from rest at 5.00 g to reach Mach-3 (three times
the speed of sound) as quickly as possible. Experimental tests reveal that he will black out
if this acceleration lasts for more than 5.00 s . Use 331 m/s for the speed of sound.
t = v/a = v/(5g) ∼ 1000/50 > 5 s
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 15 s before
the package strikes the ground, how high above the ground was the package when it was
released if air resistance is negligible?
1
0 = y0 + v0 t − gt2 , y0 = . . .
2
A car accelerates from 6 m/s to 17 m/s at a rate of 3 m/s2 . How far does the car travel
while accelerating? No time given or asked!!!, thus
v 2 − v02
= ...
2a
x − x0 =
A airplane that is flying level needs to accelerate from a speed of 200 m/s to a speed of
240 m/s while it flies a distance of 1.20 km. What must be the acceleration of the plane? same as above, a = . . .
A ball rolls across a floor with an acceleration of a = −0.1m/s2 in a direction opposite
to its velocity. The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the
floor. What was the initial speed of the ball? - same as above with a < 0
v 2 − v02 = 2ax , v0 = . . . > v
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second
interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second
interval. How does the distance traveled by the truck compare to that of the car?
No acceleration given, thus
x=
v0 + v
t
2
with v0 = 0 for the car (and v = 20) and v = 0 for the truck (and for the truck v0 = 20).
18
III.
A.
PROJECTILE MOTION
Introduction: Object dropped from a plane
Given: y0 = H = 100 m and ~v0 = V î, with V = 200 m/s (horizontal). Find: (a) vertical
displacement after t1 = 2 s, (b) horizontal distance L to hit the ground, (c) the speed v upon
impact and the angle and (d) position of the object relative to the plain. Solution:
Vertical motion (horizontal velocity does not matter!):
1
y(t) = H − gt2
2
(a) vertical displacement at t1 :
1
H − y(t1 ) = gt2 = 0.5 · 9.8 · 22 ≈ 20 m
2
(b) Time to fall, y(t) = 0
t=
From x-direction
L = Vt = V
p
p
2H/g
p
2H/g = 200 2 · 100/9.8 = . . .
(c) Speed upon impact. From y-direction
vy2 = 2gH
From x-direction
vx = const = V
v 2 = vx2 + vy2 = v02 + 2gH
19
(the last formula is also valid for non-horizontal launch with v0 being the full initial speed).
Angle of impact with horizontal:
p
tan θ = vy /vx = − 2gH/V
(d) since vx = const = V the object is right under the plane (!)
B.
General
: Select x-axis horizontal, y-axis vertical.
ax = 0 , ay = −g
(26)
From eq. (44) written in components one has
vx = v0,x = const , vy = v0,y − gt
(27)
v0,x = v0 cos θ , v0,y = v0 sin θ
(28)
with
Displacement:
x = x0 + v0,x t
(29)
1
y = y0 + v0,y t − gt2
2
(30)
Also,
y − y0 =
2
v0y
− vy2
v2 − v2
= 0
2g
2g
Max elevation:
2
v0,y
ymax − y0 =
2g
v0,x v0,y
xmax =
g
20
(31)
(32)
Range:
R = 2xmax = 2
v0 cos θ v0 sin θ
g
v02
R = sin (2θ)
g
(33)
Note maximum for θ = π/4.
Trajectory: (use x0 = y0 = 0). Exclude time, e.g. t = x/v0,x . Then
v0,y 1 x2
g
y=x
− g 2 = x tan θ − 2
x2
2
v0,x 2 v0,x
2v0 cos θ
(34)
This is a parabola - see Fig. 5.
y
5
4
3
2
1
2
4
6
8
10
x
FIG. 5: Projectile motion for different values of the initial angle θ with a fixed value of initial speed
v0 (close to 10 m/s). Maximum range is achieved for θ = π/4.
C.
Examples
Problem. A coastguard cannon is placed on a cliff y0 = 60 m above the sea level. Three
shells are fired at angles θ = 0 , ±30o with horizontal, each with an initial speed v0 = 80 m/s.
Find the following:
1. the horizontal distance x from the cliff to the point where each projectile hits the water
2. the speed upon impact
21
Solution: θ = 0 exactly as for the plane. Consider θ = +30o :
vx = v0 cos θ , v0y = v0 sin θ
Note y = 0 at the end. Find time from vertical motion only
1
0 = y0 + v0y t − gt2
2
(a quadratic equation - select a positive root). Then, the horizontal distance
x = vx t
Speed upon impact (just as for the plane)
q
v = v02 + 2gy0 > v0
Angle θ does not affect the final speed.
Lab.
22
1
y = y0 − gt2
2
x = v0 t
First, y = 0, x - measured. Then,
t=
Second,
p
2y0 /g , v0 =
x
= ...
t
y = y0 − H , x = D
and
t1 =
p
2(y0 − H)/g , D = v0 t1 = . . .
23
IV.
VECTORS
A vector is characterized by the following three properties:
• has a magnitude
• has direction (Equivalently, has several components in a selected system of coordinates).
• obeys certain addition rules (”Tail-to-Head” or, equivalently ”rule of parallelogram”).
This is in contrast to a scalar, which has only magnitude and which is not changed when a
system of coordinates is rotated.
How do we know which physical quantity is a vector, which is a scalar and which is
neither? From experiment (of course). Examples of scalars are mass, time, kinetic energy.
Examples of vectors are the displacement, velocity and force.
Tail-to-Head addition rule.
24
A.
System of coordinates
Unit vectors ~i, ~j
25
B.
Operations with vectors
1.
Single vector
0.5
ay
0.4
a
0.3
0.2
Θ
0.1
ax
0.2
0.4
0.6
0.8
Consider a vector ~a with components ax and ay (let’s talk 2D for a while). There is an
associated scalar, namely the magnitude (or length) given by the Pythagorean theorem
q
a ≡ |~a| = a2x + a2y
(35)
Note that for a different system of coordinates with axes x′ , y ′ the components ax′ and ay′
can be very different, but the length in eq. (35) , obviously, will not change, which just
means that it is a scalar.
Primary example: position vector (note two equivalent forms of notation)
~r = (x, y) = x~i + y~j
with |~i| = |~j| = 1.
Polar coordinates:
r=
p
x2 + y 2 , θ = arctan(y/x)
x = r cos θ , y = r sin θ
Note: arctan might require adding 180o - always check with a picture!
Another operation allowed on a single vector is multiplication by a scalar. Note that the
physical dimension (”units”) of the resulting vector can be different from the original, as in
F~ = m~a.
26
2.
Two vectors: addition
3
2.5
2
~
C
1.5
1
0.5
~
A
-2
-1.5
~
B
-1
-0.5
0.5
1
~ = A+
~ B.
~ Note the use of rule of parallelogram (equivalently, tail-toFIG. 6: Adding two vectors: C
~ = (−2, 1), B
~ = (1, 2)
head addition rule). Alternatively, vectors can be added by components: A
~ = (−2 + 1, 1 + 2) = (−1, 3).
and C
For two vectors, ~a and ~b one can define their sum ~c = ~a + ~b with components
cx = ax + bx , cy = ay + by
(36)
The magnitude of ~c then follows from eq. (35). Note that physical dimensions of ~a and ~b
must be identical.
Note: for most problems (except rotation!) it is allowed to carry a vector parallel to
itself. Thus, we usually assume that every vector starts at the origin, (0, 0).
~ = −2î + ĵ is followed by B
~ = î + 2ĵ. Find magnitude and
Example Displacement A
direction of the resultant displacement.
~ =A
~+B
~ = −î + 3ĵ
C
27
C=
3.
√
−1
Cx
= √ = −...
C
10
o
θ = . . . > 90
10 , cos θ =
Two vectors: dot product
If ~a and ~b make an angle φ with each other, their scalar (dot) product is defined as
~a · ~b = ab cos (φ)
or in components
~a · ~b = ax bx + ay by
(37)
A different system of coordinates can be used, with different individual components but
with the same result. For two orthogonal vectors ~a · ~b = 0. Preview. The main application
of the scalar product is the concept of work ∆W = F~ · ∆~r, with ∆~r being the displacement.
Force which is perpendicular to displacement does not work!
Example. See Fig. 6.
~·B
~ = (−2)1 + 1 · 2 = 0
A
(thus angle is 90o ).
~ and C
~ in Fig. 6.
Example Find angle between 2 vectors B
General:
cos θ =
In Fig. 6:
B=
√
12 + 22 =
cos θ =
√
~a · ~b
ab
5, C =
p
√
(−1)2 + 32 = 10
1
(−1) · 1 + 3 · 2
√
= √ , θ = 45o
5 2
2
Example. ~a = î + 2ĵ − k̂. Find the angle β it makes with the y-axis. Note: |ĵ| = 1, thus
cos β =
ay
2
~a · ĵ
=
=√
2
a
a
1 + 22 + 12
~ B
~
Advanced Example: Prove the Pythagorean theorem C 2 = A2 + B 2 . Proof. Let A,
~ the hypotenuse - see Fig. 7.
represent the legs of a right triangle and C
28
~
C
~
B
~
A
FIG. 7:
One has
~ =B
~ −A
~
C
or
~ ·A
~ = A2 + B 2
C 2 = B 2 + A2 − 2B
~ ·A
~ = 0 for perpendicular vectors.
since B
~ ·A
~ 6= 0:
Example. The cosine theorem. The same thing, only now B
~ ·A
~ = A2 + B 2 − 2AB cos θ
C 2 = B 2 + A2 − 2B
~ and B.
~
θ being the angle between A
4.
Two vectors: vector product
At this point we must proceed to the 3D space. Important here is the correct system of
coordinates, as in Fig. 8. You can rotate the system of coordinates any way you like, but
you cannot reflect it in a mirror (which would switch right and left hands). If ~a and ~b make
an angle φ ≤ 180o with each other, their vector (cross) product ~c = ~a × ~b has a magnitude
c = ab sin(φ). The direction is defined as perpendicular to both ~a and ~b using the following
rule: curl the fingers of the right hand from ~a to ~b in the shortest direction (i.e., the angle
must be smaller than 180o). Then the thumb points in the ~c direction. Check with Fig. 9.
Changing the order changes the sign, ~b × ~a = −~a × ~b. In particular, ~a × ~a = ~0. More
generally, the cross product is zero for any two parallel vectors.
29
y
x
z
z
x
y
FIG. 8: The correct, ”right-hand” systems of coordinates. Checkpoint - curl fingers of the RIGHT
hand from x (red) to y (green), then the thumb should point into the z direction (blue). (Note
that axes labeling of the figures is outside of the boxes, not necessarily near the corresponding axis;
also, for the figure on the right the origin of coordinates is at the far end of the box, if it is hard
to see in your printout).
FIG. 9: Example of a cross product ~c (blue) = ~a (red) × ~b (green). (If you have no colors, ~c is
vertical in the example, ~a is along the front edge to lower right, ~b is diagonal).
Suppose now a system of coordinates is introduced with unit vectors î, ĵ and k̂ pointing
in the x, y and z directions, respectively. First of all, if î, ĵ, k̂ are written ”in a ring”, the
cross product of any two of them equals the third one in clockwise direction, i.e.
î × ĵ = k̂ , ĵ × k̂ = î , k̂ × î = ĵ
, etc. (check this for Fig. 8 !).
Example. Fig. 6:
~ = −2î + ĵ , B
~ = î + 2ĵ
A
30
~ ×B
~ = (−2î + ĵ) × (î + 2ĵ) = (−2) · 2î × ĵ + ĵ × î =
A
= −4k̂ − k̂ = −5k̂
~×B
~ goes into the page, as
(Note: in Fig. 6 k̂ goes out of the page; the cross product A
indicated by ”-”.)
More generally, the cross product is now expressed as a 3-by-3 determinant
î ĵ k̂ ay az
ax az
ax ay ~
~a × b = ax ay az = î − ĵ + k̂ by bz bx bz bx by bx by bz (38)
The two-by-two determinants can be easily expanded. In practice, there will be many zeros,
so calculations are not too hard.
Preview. Vector product is most relevant to rotation.
Example. See Fig. 6.
~ ×B
~ = k̂((−2)2 − 1 · 1) = −5k̂
A
C.
Preview. Forces as vectors. Elements of static equilibrium
Besides displacement ~r, and velocity ~v , forces represent another example of a vector. Note
that they are measured in different units, N (newtons), i.e. each component of the force
Fx , Fy , Fz is measured in N. How do we know that force is a vector? From experiments
on static equilibrium which demonstrate that forces indeed add up following the standard
vector addition rule of parallelogram (or, that they add up by components, which is the
same thing).
Consider your Lab experiment
For equilibrium one expects
F~1 + F~2 + F~3 = 0
or
In the example:
F~3 = − F~1 + F~2
F1 = 0.9 N , θ1 = 180o ; F2 = 0.75 N , θ2 = 315o = −45o
31
FIG. 10: The force table (left) and an example of its schematic representation (right)
Thus
F1x = −0.9 , F1y = 0 ; F2x = 0.53 , F2y = −0.53
with
F3x = 0.37 , F3y = 0.53
and
θ3 = tan−1
0.53
≃ 55o
0.37
(check with picture that this is about correct; the angle 55o + 180o = 235o has the same
tan). Similarly, the magnitude
F3 =
√
.372 + .532 = 0.646 (N)
32
Example: Equilibrium on incline.
~ + m~g = 0
T~ + N
The x-components, down the incline:
mg sin θ , −T , and 0 for N
thus
−T + mg sin θ = 0 , and T = mg sin θ
The y-components, normal to incline:
−mg cos θ , 0 for T , and N
thus
N − mg cos θ = 0 , and N = mg cos θ
33
Example: ”bird in the middle of a cord”
From symmetry
|T~1 | = |T~2 | ≡ T
x - automatic; from y:
T sin α + T sin α − mg = 0 , T = mg/(2 sin α)
Note the limits!
34
Example: ”weight between two cords”
T~1 + T~2 + m~g = 0
x : −T1 + T2 cos α + 0 = 0
y : 0 + T2 sin α − mg = 0
Thus,
T2 = mg/ sin α
T1 = T2 cos α = mg cot α
Example: ”weight between two cords (general)”
T~1 + T~2 + m~g = 0
−T1 cos α + T2 sin β = 0
35
T1 sin α + T2 cos β − mg = 0
Then use T2 = T1 cos α/ sin β...
36
Dr. Vitaly A. Shneidman, Phys 111, Lecture
V.
2D MOTION
A.
Introduction: Derivatives of a vector
~r(t) = (x(t) , y(t)) = x(t)~i + y(t)~j
d
dx
dx dy
dy
= ~i + ~j
~r =
,
dt
dt dt
dt
dt
d2
d2 x~ d2 y ~
d2 x d2 y
=
~
r
=
,
i+ 2j
dt2
dt2 dt2
dt2
dt
B.
General
Position:
~r = ~r(t)
(39)
∆~r
∆t
(40)
Average velocity:
~vav =
(see Fig. 11).
Instantaneous velocity:
~v = lim ~vav =
∆t→0
d~r
dt
(41)
Average acceleration:
~aav =
37
∆~v
∆t
(42)
Instantaneous acceleration:
~a = lim ~aav
∆t→0
d~v
d2~r
=
= 2
dt
dt
(43)
y
100
90
80
70
60
50
40
30
10
20
30
40
50
60
70
80
x
FIG. 11: Position of a particle ~r(t) (blue line), finite displacement ∆~r (black dashed line) and the
average velocity ~v = ∆~r/∆t (red dashed in the same direction). The the instantaneous velocity at
a given point is tangent to the trajectory.
C.
~a = const
∆~v = ~a · ∆t
or with t0 = 0
~v = ~v 0 + ~a · t
(44)
1
~r = ~r0 + ~v0 · t + ~a · t2
2
(45)
Displacement:
38
(The above can be proven either by integration or by writing eq. (44) in components and
using known 1D results).
Example. Describe the x and y motion for ~a = −î + 2ĵ (in m/s2 ) and ~v0 = 10î (in m/s). At
t = 0 the particle is at the origin. Solution
1
1
x(t) = 10t − t2 , y(t) = 2t2
2
2
√
Can eliminate t via (the simpler) y(t) from which we get t = y and
√
x = 10 y − y/2
D.
~a = ~g (projectile motion)
see previous lecture
HW
A car comes to a bridge during a storm and finds the bridge washed out. The driver
must get to the other side, so he decides to try leaping it with his car. The side the car is on
is H=21.1 m above the river, whereas the opposite side is a mere h=2.4 m above the river.
The river itself is a raging torrent L=61.0 m wide.
s
2(H − h)
, vx = L/t = . . .
t=
g
q
vy = 0 − gt , v = vx2 + vy2
or h − H =
p
0 − vy2
, vy = − 2g(H − h) , v = . . .
2g
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 39 m/s
. The plane is in a 18 m/s wind blowing toward the southwest relative to the earth.
E.
1.
Uniform circular motion
Preliminaries
Radian measure of an angle:
l = rθ
39
FIG. 12: Angle of 1 rad ≈ 57.3o . For this angle the length of the circular arc exactly equals the
radius. The full angle, 360o , is 2π radians.
Consider motion around a circle with a constant speed v. The velocity ~v, however,
changes directions so that there is acceleration.
Period of revolution:
T = 2πr/v
(46)
with 1/T - ”frequency of revolution”. Angular velocity (in rad/s):
ω=
2π
v
=
T
r
40
(47)
2.
Acceleration
Consider counterclockwise rotation with representative position vectors ~r(t) and ~r(t+∆t)
(black) which are symmetric with respect to the vertical. Note that ~v (blue) is always
perpendicular to ~r. Thus, from geometry vectors ~v (t + ∆t) , ~v (t) and ∆~v (dashed blue) form
a triangle which is similar to the one formed by ~r (t + ∆t) , ~r(t) and ∆~r (dashed black). Or,
|∆~r|
|∆~v |
=
v
r
|∆~r|
v
|∆~v |
= lim
∆t→0 ∆t
r ∆t→0 ∆t
a = lim
Or
a = v 2/r = ω 2 r
3.
(48)
An alternative derivation
We can use derivatives with the major relation
d
d
sin(ωt) = ω cos(ωt) ,
cos(ωt) = −ω sin(ωt) ,
dt
dt
One has
~r(t) = (x, y) = (r cos ωt, r sin ωt)
~v (t) =
d~r
= (−rω sin ωt, rω cos ωt)
dt
41
(49)
~a =
d~v
= −rω 2 cos ωt, −rω 2 sin ωt = −ω 2~r
dt
(50)
which gives not only magnitude but also the direction of acceleration opposite to ~r, i.e.
towards the center.
FIG. 13: Position (black), velocities (blue) and acceleration (red) vectors for a uniform circular
motion in counter-clockwise direction.
F.
Advanced: Classical (Galileo’s) Relativity
New reference frame:
~ =R
~ 0 + V~ t , V~ = const
R
(51)
~ = ~r − V~ t − R
~0
~r′ = ~r − R
(52)
~v ′ = ~v − V~
(53)
New coordinates, etc.:
Example: Projectile with vx 6= 0 and new ref. frame with V~ = (vx , 0).
42
Dr. Vitaly A. Shneidman, Phys 111, 4th Lecture
VI.
A.
1.
NEWTON’S LAWS
Force
Units
”Newton of force”
N = kg
2.
m
s2
(54)
Vector nature
From experiment, action of two independent forces F~1 and F~2 is equivalent
to action of the resultant
F~ = F~1 + F~2
43
(55)
F~1 + F~2 + F~3 = 0
~
~
~
F3 = − F1 + F2
(see the Force Table experiment in the lecture on Vectors).
B.
The Laws
1. If F~ = 0 (no net force) then ~v = const
2.
F~ = m~a
(56)
F~12 = −F~21
(57)
3.
44
Notes: the 1st Law is not a trivial consequence of the 2nd one for F~ = 0,
but rather it identifies inertial reference frames where a free body moves
with a constant velocity. In the 3rd Law forces F~12 and F~21 are applied to
different bodies; both forces, however, are of the same physical nature (e.g.
gravitational attraction).
Examples. Force from equilibrium or from acceleration (without specifying
the nature of force).
• resultant of two forces (in class)
• A force F acts on a object with mass m which accelerates from rest to
speed v during time t. Find F . Solution:
From v − v0 = at and v0 = 0, find a = v/t; then, F = ma.
• A car with mass M increases speed from v1 to v2 over distance ∆x. Find
45
the force. Solution. From ∆x = 12 (v1 + v2) t find t. Then, a = (v2 −v1)/t
and F = Ma. (Alternatively, can use ∆x = (v22 − v12)/(2a) to find a
directly.)
1.
Gravitational force
F~g = m~g
(58)
Weight - the force an object excerts on a support. Will conside with m~g
for a stationary object. (Note: the textbook uses the latter as a definiton
with weight always equal m~g ; other books will use the former definition of
”weight”, or ”apparent weight” as force on a support. That can be different
from m~g in case of acceleration, with the extreme example of ”weightlessness”, as will be discussed in class).
46
2.
FBD, normal force
~ acts on the body, not on the surface. According
Note: the normal force N
~ acts on the surface, but that force is
to the 3rd Law a reaction force −N
of no interest in most cases, unless the acceleration of the surface is to be
evaluated. Also, the normal force which acts on the body may or may not
compensate other forces which act on the same body -in the above diagram
~ + m~g 6= 0 there will be acceleration, as in an
, the force of gravity. If N
elevator.
C.
Statics
In equilibrium
F~ ≡
X
F~i = 0
i
(all forces are applied to the same body!).
Example:
47
(59)
~ + m~g = 0
T~ + N
x) : −T + mg sin θ = 0 , T = mg sin θ
y) : N − mg cos θ = 0
Example:
From symmetry
|T~1 | = |T~2 | ≡ T
x - automatic; from y:
T sin α + T sin α − mg = 0 , T = mg/(2 sin α)
Note the limits!
Example:
48
T~1 + T~2 + m~g = 0
−T1 cos α + T2 sin β = 0
T1 sin α + T2 cos β − mg = 0
Then use T2 = T1 cos α/ sin β...
(see other examples in the lecture on Vectors).
D.
Dynamics: Examples
The following examples will be discusses in class:
• finding force between two accelerating blocks
• apparent weight in an elevator
• Atwood machine
• block on inclined plane (no friction)
49
Two blocks: a horizontal force F~ is applied to a block with mass M (red)
~
which in turn pushes a block with mass m (blue); ignore friction. Find N
and ~a. (note that the 3rd Law is already used in the diagram).
F
N
-N
Solution (quick): first treat the 2 blocks as a single solid body. From 2nd
Law for this ”body”
~a = F~ /(M + m)
From 2nd Law for mass m:
~ = m~a
N
Solution (academic): write 2nd Law for each of the 2 block separately. For
M
For m, as above
~
~
F + −N = M~a
~ = m~a
N
which will lead to the same result (add the 2 equations together to find ~a).
50
Weight in an elevator: Find the ”apparent weight” of a person of mass m
if the elevator accelerates up with acceleration ~a.
Solution. Forces on the person: m~g (down, black, applied to center-of-mass)
~ (reaction of the floor, blue, applied to feet) - up. [A force −N
~ -not
and N
shown- acts on the floor and determines ”apparent weight”]. The 2nd Law
for the person
~ = m~a
m~g + N
From here
~ = m (~a − ~g )
N
or in projection on vertical axis
N = m(g + a)
Note: N > mg for ~a up and N < mg for ~a down; N = 0 for ~a = ~g .
51
Atwood machine:
✻
✻
~
T~1
T2
❄M~g
m~g
❄
FIG. 14: Atwood machine. Mass M (left) is almost balanced by a slightly smaller mass m. Pulley
has negligible rotational inertia, so that both strings have the same tension T~1 = T~2 = T
• 2nd Law(s) for each body; for M axis down, for m - up.
• constrains: a-same, tensions - same
From 2nd Law(s):
Mg − T = Ma , T − mg = ma
Add together
(M − m)g = (M + m)a , or
M −m
M +m
Note a ”crude” way to get the same result: take the net contribution of all
a=g
external forces acting on the system Mg − mg and divide by the full mass of
the system, M + m. What if need tensions (should be the same)?
T1 = Mg − ma < Mg , T2 = mg + ma > mg
Frictionless incline - Fig. 15.
52
y
~
N
~a
x
m~g
FIG. 15: A block on a frictionless inclined plane which makes an angle θ with horizontal.
~ in our case and the assumed acceleration ~a
• identify forces, m~g and N
(magnitude still to be found).
~ + m~g = m~a
• write the 2nd Law (vector form!) N
• select a ”clever” system of coordinates x, y.
• write down projections of the vector equation on the x, y axes, respectively (note, components of the force of gravity in such coordinates are
always mg sin θ downhill parallel to the incline and −mg cos θ in perpendicular direction):
x:
mg sin θ = ma
y : N − mg cos θ = 0
the x-equation will give acceleration
a = g sin θ
53
(which is already the solution); the y equation determines N .
• before plugging in numbers, a good idea is to check the limits. Indeed,
for θ = 0 (horizontal plane) a = 0 no acceleration, while for θ = π/2
one has a = g, as should be for a free fall.
A few practical remarks to succeed in such problems.
• The original diagram should be BIG and clear. If so, you will use it as
a FBD, otherwise you will have to re-draw it separately with an extra
possibility of mistake.
• In the picture be realistic when dealing with ”magic” angles of 30, 45, 60
and 90 degrees. Otherwise, a clear picture is more important than a
true-to-life angle.
• Vectors of forces should be more distinct than anything else in the picture; do not draw arrows for projection of forces - they can be confused
with real forces if there are many of them.
• the force of gravity in the picture should be immediately identified as
m~g (using an extra tautological definiton, such as F~g = m~g adds an
equation and confuses the picture).
• If only one body is of interest, do not draw any forces which act on other
~ which acts on the
bodies (in our case that would be, e.g. a force −N
inclined plane).
• select axes only after the diagram is completed and the 2’d Law is written in vector form. As a rule, in dynamic problems one axes is selected
in the direction of acceleration (if this direction can be guessed).
54
Dr. Vitaly A. Shneidman, Phys 111, 5th Lecture
VII.
NEWTON’S LAWS: APPLICATIONS TO FRICTION AND TO CIRCULAR
MOTION
A.
Force of friction

N

f

F
f
st
at
ic
kinetic
M
g
μsN
F
Force of friction on a moving body:
f = µN
(60)
Direction - against velocity; µ (or µk ) - kinetic friction coefficient.
Static friction:
fs ≤ µs N
with µs - static friction coefficient.
1.
Example: block on inclined plane
(see next page)
55
(61)
y
~
N
f~
◗
❦
◗
◗
◗
~a
x
m~g
FIG. 16: A block sliding down an inclined plane with friction. The force of friction f~ is opposite
to the direction of motion and equals µk N .
~ and the acceleration ~a (magnitude still to be found).
• identify forces, f~, m~g and N
• write the 2’d Law (vector form!)
~ + m~g = m~a
f~ + N
• select a ”clever” system of coordinates x, y.
• write down projections of the vector equation on the x, y axes, respectively:
x:
−f + mg sin θ = ma
y : N − mg cos θ = 0
• relate friction to normal force:
f = µk N
the above 3 equations can be solved (in class). Answer:
a = g (sin θ − µk cos θ)
56
• Note: the body will keep moving only for
µk < tan θ
The body will start moving only for
µs < tan θ
57
~ and Mg.
~
First, try without friction, i.e. f = 0 and do not need N
• for body M on the table
T = Ma
• for hanging body m:
mg − T = ma
Thus, adding the two equations together (to get rid of T ):
mg = Ma + ma , or a = g
m
m+M
With frition:
• for body M on the table
T − f = Ma (horizontal)
N − Mg = 0 (vertical)
• for hanging body m (same as before):
mg − T = ma
•
f = µN = µMg (new)
58
Thus,
mg − f = (M + m)a , a =
mg − µMg
m − µM
=g
>0
M +m
M +m
(if µM > m motion is impossible - friction too strong).
59
N
T
T
f
mg
Mg
Assume large enough M to move left. Hanging mass (axis down):
Mg − T = Ma
(62)
T − mg sin θ − f = ma
(63)
g(M − m sin θ) − f = (M + m)a
(64)
Mass on incline (x-axis uphill):
From above
How to find friction? - from N:
N − mg cos θ = 0
(65)
N = mg cos θ , f = µN
(66)
g(M − m sin θ − µm cos θ) = (M + m)a
(67)
a = ...
(68)
60
B.
Fast way to solve quasi-one-dimensional problems
Only external forces or their projections along the motion. No tension, no normal force.
Friction
f = µmg (horizontal) , f = µmg cos θ (inclined)
General
Example: Atwood machine
X
Fext = Mtot a
Mg − mg = (m + M)a , a = . . .
if need T
T − mg = ma , T = . . .
Example: external force (no friction)
F − mg sin θ = (M + m)a , a = . . .
Tension (if need) from
F − T = Ma , T = F − Ma
Example: 2 blocks with friction, large M
61
N
T
T
f
mg
Mg
select positive direction uphill
Mg − µmg cos θ − mg sin θ = (M + m)a , a = . . .
Mg − T = Ma , T = M(g − a)
Example: 3 blocks with friction, M > m
Mg − mg − µM1 g = (m + M1 + M)a , a = . . .
Tension T1 between M1 and M from
Mg − T1 = Ma
Tension T between M1 and m from
T − mg = ma
62
C.
Centripetal force
Newton’s laws are applicable to any motion, centripetal including. Thus, for centripetal
force of any physical origin
v2
Fc = mac ≡ m = mω 2 r
r
(69)
Centripetal force is always directed towards center, perpendicular to the velocity - see Fig.
17. Examples: tension of a string, gravitational force, friction.
FIG. 17: Position (black), velocities (blue) and centripetal force (red) vectors for a uniform circular
motion in counter-clockwise direction. The direction of force coincides with centripetal acceleration
(towards the center). The value of centripetal force at each point is determined by the vector sum
of actual physical forces, e.g. normal force and gravity in case of a Ferris Wheel, or tension plus
gravity in case of a conic pendulum.
Simple example: A particlle with mass around a circle with a diameter d = 2 m. It takes
t = 3 s to complete one revolution. Find the magnitude of the centripetal force F .
Solution: Since F = mac , need acceleration. Use
ac = v 2 /r with v = πd/t and ac = 2π 2
63
d
t2
Equivalently
ac = ω 2
d
with ω = 2π/t
2
Thus,
F = 2π 2
1.
md
= ...
t2
Conic pendulum
See Fig. 18, forces will be labeled in class.
FIG. 18: A conic pendulum rotating around a vertical axes (dashed line). Two forces gravity (blue)
and tension (black) create a centripetal acceleration (red) directed towards the center of rotation.
The angle with vertical is θ, length of the s tring is L and radius of revolution is r = L sin θ.
One has the 2nd Law:
T~ + m~g = m~ac
In projections:
x : T sin θ = mac = mω 2 r
y : T cos θ − mg = 0
64
Thus,
ω=
r
g
≃
L cos θ
r
g
L
(the approximation is valid for θ ≪ 1). The period of revolution
2π
≃ 2π
ω
s
L
g
Note that mass and angle (if small) do not matter.
2.
Satellite
F~g = m~ac
Fg = mg , ac = v 2 /r
Thus,
v2
r
√
v = gr
g=
about 8 km/s for Earth. Period of revolution
2π
p
65
r/g
3.
Turning car/friction/incline

N
θ
m
g
~ + f~ + m~g = m~a
friction : N
x-axis - towards the center (dashed):
f = ma = mv 2 /R
y-up:
N − mg = 0
and
f ≤ µs N = µs mg
Thus,
v 2 /R ≤ µs g
~ + m~g = m~a
incline : N
N sin θ = mv 2 /R
N cos θ − mg = 0
mg tan θ = mv 2 /R
tan θ =
66
v2
gR
D.
(Advanced) ”Forces of inertia”
F~ = m~a
Define
F~i = −m~a
F~ + F~i = 0
”statics”.
Deflection of a pendulum in an accelerating car. Left: from the point of inertial reference
frame. Right: from the point an accelerating reference frame associated with the car; note
a new ”field of gravity” −~a which combines with regular ~g .
Regular solution (φ - angle with vertical):
T~ + m~g = m~a
x : T sin φ = ma ,
y : T cos φ − mg = 0
Thus,
tan φ =
a
g
Solution in non-inertial reference frame: no acceleration; pendulum hangs in the direction
~ = ~g + (−~a). Thus, tan φ = a/g, the same.
of new ”full gravity” G
67
Dr. Vitaly A. Shneidman, Phys 111, 6th Lecture
VIII.
A.
WORK
Scalar (dot) product - reminder
see introduction on vectors
θ
~a · ~b = ab cos θ = ax bx + ay by + az bz
q
q
a = a2x + a2y + a2z , b = b2x + b2y + b2z
Example: ~a = î + 2ĵ + 3k̂, ~b = 2î − 4ĵ + k̂. Find θ.
cos θ =
B.
1 × 2 + 2 × (−4) + 3 × 1
~a · ~b
√
= ...
= √
ab
12 + 22 + 32 ×
22 + 42 + 12
Units
Joule (J):
1 J = 1 N · m = kg
C.
m2
s2
(70)
Definitions
Constant force:
W = F~ · ∆~r ≡ F ∆r cos θ ≡ Fx ∆x + Fy ∆y + Fz ∆z
68
(71)

F
Δr
Example: work by force of gravity ; x-”east”, y-”North”, z-up
Fx = 0 , Fy = 0 , Fz = −mg
Wg = −mg∆z
(72)
(and ∆x , ∆y do not matter!)
Let m = 3 kg, and the particle is moved from ~r1 = 2î − 4ĵ + k̂ to î + 2ĵ + 3k̂ (in meters).
Find Wg
Wg = −3 kg · 9.8
m
· (3 − 1) m ≃ −60 J
s2
Example:
F~ = ~i + 2~j , ~r1 = 3~i + 4~j , ~r2 = 6~i − 4~j
∆~r = 3~i − 8~j , W = 3 · 1 + (−8) · 2 = −13 (J)
Example:
F~ = 2~j , ~r1 = 3~i + 4~j + 5~k , ~r2 = 6~i − 4~j + ~k
W = (−4 − 4) · 2 = −16 (J)
(x and z components of displacement do not matter!)
Example: An M = 600 kg elevator is going up at a constant speed of v = 1 m/s for
t = 10 s. How much work is done by the motor?
Since v = const
Fy − Mg = 0 , thus Fy = Mg
69
∆y = vt , W = Mgvt = 600 · 9.8 · 1 · 10 = . . .
(Note: the work done by gravity Wg = −Mg∆y is negative of the same value)
Example: An M = 80 kg skydiver falls 200 m with a constant speed of 100 m/s. Find
the work done by viscous air friction.
Since v = const
Fy − Mg = 0 , thus Fy = Mg
(otherwise speed does not matter!)
∆y = −200 m , W = Mg ∆y = 80 · 9.8 · (−200) = − . . .
[Alternatively: the work done by gravity Wg = −Mg∆y = 80 · 9.8 · 200 > 0 (since goes
down), and W = −Wg < 0]
Example: An 10 kg projectile is displaced 200 m horizontally and 50 m vertically with
respect to its initial position. Find Wg .
Horizontal motion does not matter!
Wg = −Mgy = −10 · 9.8 · 50 ≃ −5 kJ
Variable force:
Let us break the path from ~r1 to ~r2 in small segments ∆~ri (blue), each with a force F~i ≃ const
(red). Then
Wi ≃ F~i · ∆~ri
and total work
W =
X
i
Wi →
70
Z
~r2
~r1
F~ · d~r
(73)
Note: forces which are perpendicular to displacement do not work, e.g. the centripetal
force, normal force.
D.
1D motion and examples
For motion in x-direction only
W =
Z
x2
Fx dx
(74)
x1
If Fx is given by a graph, work is the area under the curve (can be negative!).
Example: find work from graphs for 0 ≤ x ≤ 6 (left) and for 0 ≤ x ≤ 4 (middle and
right). In each case find the final speed of an m = 2 kg particle with v0 = 10 m/s (use the
KET - later)
F(x)
3
F(x)
3
2
2
1
1
1
2
3
6
x
1
-1
-1
-2
-2
-3
-3
left : W = 16 J ,
1
1
middle; W = 3 · 3 + [(6 − 3) + (6 − 4)](−1) = 2 J
2
2
1
1
right : W = · (3 + 1)2 + [1 + (−3)](4 − 2) = 2 J
2
2
(note: part of area can be negative)
1
1
W = mv 2 − mv02
2
2
v 2 = v02 + 2W/m = . . . > 0
71
2
4
5
6
x
Spring:
F = −kx
(75)
k - ”spring constant” (also known as ”Hook’s law”). Work done by the spring
1
1
1
1
Wsp = (F2 + F1 )(x2 − x1 ) = (−kx2 − kx1 )(x2 − x1 ) = − kx22 + kx21
2
2
2
2
1
1
Wsp = + kx21 − kx22
2
2
(76)
(If the spring is stretched from rest, the term with x21 will be absent and the work done by
the spring will be negative for any x2 6= 0.)
IX.
A.
KINETIC ENERGY
Definition and units
1
K = mv 2
2
or if many particles, the sum of individual energies.
Units: J (same as work).
Note: v 2 = vx2 + vy2 + vz2 , thus K = 12 m vx2 + vy2 + . . . .
72
(77)
B.
1.
Relation to work
Constant force
W = Fx ∆x + Fy ∆y = m [ax ∆x + ay ∆y]
According to kinematics
∆x =
and
ax ∆x =
2
2
vy2 − v0y
vx2 − v0x
, ∆y =
2ax
2ay
2
2
vy2 − v0y
vx2 − v0x
, ay ∆y =
2
2
Thus,
W = ∆K
(78)
which is the ”work-energy” theorem.
Examples: gravity and friction (in class).
(gravity). Max hight for vertical initial v:
v2
1
−mgh = 0 − mv 2 , h =
2
2g
(friction). The ”policeman problem”. Given: L, µ, find vo
Friction: f = µmg.
W = −f L = −µmgL < 0 (!)
1
∆K = 0 − mv02
2
−mv02 /2 = −µmgL
p
v = 2µgL
(79)
(80)
(81)
(82)
(friction+another force). A heavy crate is pushed from A to B (with AB = 3 m) by a
force P = 200 N at 60o to horizontal. Find the work Wf done by friction if KA = 300 J and
73
KB = 100 J.
KB − KA = Wf + WP , WP = P · AB · cos 60o
(83)
Wf = KB − KA − WP = 100 − 300 − 3 · 200 = −800 J
(Note: negative!)
2.
Variable force
Again, break the displacement path into a large number N of small segments ∆ri . For
each
Wi = ∆Ki ≡ Ki − Ki−1
Thus,
W =
N
X
i
Wi = (K1 −K0 )+(K2 −K1 )+. . .+(Ki−1 −Ki−2 )+(Ki −Ki−1 )+. . .+(KN −K0 ) ≡ ∆K
or
W = ∆K
(84)
which is the ”work-energy” theorem in a general form (also, can be applied to a system of
particles).
Example: spring.
1
k
m v 2 − v02 = − x2 − x20
2
2
C.
Power
P =
dW
= F~ · ~v
dt
Units: Watts. 1W = 1 J/s
Re-derivation of work-energy theorem:
d
dm 2 md
d~v
dW
K=
v =
(~v · ~v ) = m · ~v = F~ · ~v =
dt
dt 2
2 dt
dt
dt
74
Example: An M = 500 kg horse is running up an α = 30o slope with v = 4 m/s. Find P :
1
P = mg sin α · v ≃ 500 · 9.8 4 ≈ . . .
2
Example. For a fast bike the air resistance is ∼ v 2 . How much more power is needed to
double v? Ans.: 8 times more.
75
A
B
FIG. 19: Generally, the work of a force between points A and B depends on the actual path.
However, for some ”magic” (conservative) forces the work is path-independent. For such forces one
can introduce potential energy U and determine work along any path as W = UA − UB = −∆U .
Dr. Vitaly A. Shneidman, Phys 111, 7th Lecture
X.
POTENTIAL ENERGY
A.
Some remarkable forces with path-independent work
See Fig. 19 and caption.
Examples:
• Constant force
W =
X
F~ · ∆~ri = F~ ·
with potential energy
X
∆~ri = F~ · ∆r = F~ · ~rB − F~ · ~rA
U (~r) = −F~ · ~r
(85)
(86)
Example force of gravity with Fx = 0 , Fy = −mg and
Ug = mgh
(87)
• Elastic (spring) force
X xi + xi+1
kX 2
k
2
W ≃−
k
(xi+1 − xi ) = −
xi+1 − x2i = − x2f − x(88)
i
2
2
2
76
with potential energy
1
Usp(x) = kx2
2
(89)
• Other: full force of gravity F = −mg with r-dependent g, and any force which does
not depend on velocity but depends only on distance from a center.
• Non-conservative: kinetic friction (depends on velocity, since points against ~v)
B.
Relation to force
U (x) = −
XI.
Z
F (x) dx
(90)
F = −dU/dx
(91)
CONSERVATION OF ENERGY
Start with
∆K = W
If only conservative forces
W = −∆U
(92)
K + U = const ≡ E
(93)
thus
Examples:
77
• Maximum hight.
1
E = mgh + mv 2
2
1
0 + mv02 = mghmax + 0
2
v2
hmax = 0
2g
• Galileo’s tower. Find speed upon impact
1
mgh + 0 = 0 + mv 2
2
v 2 = 2gh
• Coastguard cannon (from recitation on projectiles). Find speed upon impact
1
1
mgH + mv02 = 0 + mv 2
2
2
v 2 = v02 + 2gH
(note that the angle does not matter!).
• A skier with no initial velocity slides down from a hill which is H = 18 m high and
then, without losing speed, up a smaller hill which is h = 10 m high. What is his
speed in m/s at the top of the smaller hill? Ignore friction.
1
(full initial energy) MgH + 0 = Mgh + Mv 2 (full final)
2
v 2 = 2g(H − h) > 0 , v = . . .
78
• In the Atwood machine the heavier body on left has mass M = 1.01 kg, while the
lighter body on right has mass m = 1 kg. The system is initially at rest, there is no
friction and the mass of the pulley is negligible.
Find the speed after M lowers by h = 50 cm; use only energy considerations.
1
(M on top) : E = Mgh + 0 + 0 = mgh + (M + m)v 2 (m on top)
2
M −m
v 2 = 2gh
, v = ...
M +m
• A spring with given k, m is stretched by X meters and released. Find vmax .
1
1
E = kx2 + mv 2
2
2
p
1
1
2
kX 2 + 0 = 0 + mvmax
, vmax = X k/m
2
2
• A block approaches the spring with speed v0 . Find the maximum compression
(full final energy) :
1 2
1
kxmax + 0 = 0 + mv02 (full initial energy)
2
2
p
|xmax | = v0 m/k
79
• find speed v at some given location x
first find kinetic energy at x
1
K(x) + U(x) = E = 0 + mv02
2
1
1
1
K(x) = mv02 − U(x) = mv02 − kx2 > 0
2
2
2
p
v(x) = 2K(x)/m , |x| ≤ |xmax |
A.
Conservative plus non-conservative forces
For conservative forces introduce potential energy U, and then define the total mechanical
energy
Emech = K + U
One has
W = −∆U + Wnon−cons
(94)
Then, from work-energy theorem
∆Emech = ∆ (K + U ) = Wnon−cons
(95)
Examples:
• Galileo’s Tower revisited (with friction/air resistance). Given: H = 55 m, m = 1 kg
and Wf = −100 J is lost to friction (i.e. the mechanical energy is lost, the thermal
energy of the mass m and the air is increased by +100 J). Find the speed upon impact.
1 2
0 + mv − (mgH + 0) = Wf < 0
2
1 2
mv = mgh − |Wf | , v = . . .
2
80
• Sliding crate. Given L, m, θ, µ find the speed v at the bottom.
h ≡ L sin θ
f = µmg cos θ
Wf = −f L
1
(0 + mv 2 ) − (mgh + 0) = Wf
2
2
v 2 = 2gh − |Wf |
m
• The same skier as before, but assume a small average friction force of 40 N and the
combined length of the slopes L = 200 m. The mass of the skier is M = 80 kg; find
the speed at the top of the smaller hill.
Wf = −f L < 0
1
(full final energy - full initial) (Mgh+ Mv 2 )−(MgH+0) = Wf (work by non-cons. force)
2
2Wf
v 2 = 2g(H − h) +
≥ 0, v = ...
M
(note: if v 2 < 0, motion is impossible and the skier will stop before reaching the 2nd top.)
Example: Problem which is too hard without energy
pendulum or one way rock-on-a-string?. Left: forces. Right: energy solution.
θ - angle with vertical. Then,
Ug = mgh = mgr (1 − cos θ) , Ugmax = 2mgr
1
E = Ug + mv 2 , E > 2mgr − rotation , 0 < E < 2mgr − oscillations
2
81
What if need T ? Lowest point
1
Ug = 0 , E = mv 2
2
T − mg = mv 2 /r = 2E/r
——————————————————————–
B.
Advanced: Typical potential energy curves
1
U = kx2
2
82
(96)
1
U = ky 2 + mgy
2
(97)
U = mgL(1 − cos θ)
(98)
83
p
K = E − U ≥ 0 , v = ± 2K/m
C.
(99)
Advanced: Fictitious ”centrifugal energy”
1
U = − mω 2 r2
2
84
(100)
1
U = Ug + Ucent = mgh − mω 2 r2 = const
2
ω2
h(r) = h(0) + r2
g
D.
(101)
(102)
Advanced: Mathematical meaning of energy conservation
Start from 2nd Law with F = F (x) (not v or t !)
mẍ = F (x) | × ẋ
(103)
d
ẋẍ = (ẋ)2 /2
Z dt
d
d
F (x) dx = − U(x)
ẋF (x) =
dt
dt
Thus
d
(K + U ) = 0
dt
K + U = const = E
85
(104)
(105)
Dr. Vitaly A. Shneidman, Phys 111, 8th Lecture
XII.
MOMENTUM
A.
Definition
One particle:
p~ = m~v
(106)
System:
~ =
P
X
p~i =
i
X
mi~vi
(107)
i
Units: kg · m/s (no special name).
B.
2nd Law in terms of momentum
Single particle:
m~a = m
d~v
d(m~v )
=
dt
dt
d~p
= F~
dt
∆~p =
R t2
t1
Z
t2
dt F~ (t) = F~av ∆t
t1
dt F~ (t) -impulse. ”It takes an impulse to change momentum”.
86
(108)
(109)
Example. A fast ball of mass m=0.2 kg moving with velocity ~v = −50î (in m/s) strikes
a vertical wall as shown. It rebounds with V~ = 30î m/s after a brief collision with the wall
lasting about ∆t = 0.04 s. Estimate the average force (in N) which acts on the ball during
the collision.
m
~
~
p~2 = mV , p~1 = m~v , thus ∆~p = m V − ~v = 0.2(30 + 50)î = 16î kg
s
∆p
Fav =
= 400 N
∆t
Example. A steel ball with mass m = 100 g falls down on a horizontal plate with v =
3 m/s and rebounds with V = 2 m/s up. (a) find the impulse from the plate; (b) estimate
Fav if the collision time is 1 ms.
Select the positive direction up. Then, p1 = −3 × 0.1 = −0.3 kg m/s and after the collision
p2 = mV = 0.2 kg m/s. (a) Impulse p2 − p1 = 0.5 kg · m/s. (note: |p1 | and |p2 | add up!).
(b) Fav = 0.5/0.001 = 500 N (large!).
Example. An M = 70 kg Olympic diver from an H = 10 m platform stops ∆t = 0.1 s
after hitting the surface. Find the average force on the diver during this time.
First, find velocity v when hitting the water (axis - up). From energy conservation while in
air:
p
m
1
Mv 2 = MgH , v = − 2gH ≃ −14
2
s
Then, find the impulse and average force while entering the water
∆p = M · 0 − Mv = 0 − 70(−14) = 980 kg
87
m
, Fav = 980/0.1 ≈ 104 N
s
System of two particles with NO external forces:
d~p1 /dt = F~12
d~p2 /dt = F~21
~
dP
= F~12 + F~21 = 0
dt
(110)
~ = const
P
(111)
i.e.
and the same for any number of particles.
XIII.
A.
CENTER OF MASS (CM)
Definition
X
X
~ = 1
R
mi~ri , M =
mi
M i
i
(112)
Example: CM of a two-particle system (”Sun-Earth”, r = 150 · 106 km, from center of Sun
on the x axis towards Earth; M ≃ 2 · 1030 kg , m ≈ 6 · 1024 kg)
m1 = M, ~r1 = (0, 0) ; m2 = m, ~r2 = (r, 0)
~ =
R
m
1
[(0, 0) + m(r, 0)] ≈
(r, 0)
M +m
M
(much closer to the bigger mass).
88
Example: CM of a triangle.
B.
Relation to total momentum
d ~
1 X
d
1 ~
V~cm = R
=
mi ~ri =
P
dt
M i
dt
M
~ = M V~CM
P
(113)
Example: Find V~CM if m = 1 kg (red), M = 2 kg (blue), ~v1 = (0, 3) m/s, ~v2 = (2, 0) m/s
P~ = (4, 3) kg m/s , V~CM = (4/3, 1) m/s
with
VCM = 5/3 m/s
89
C.
2nd Law for CM
Consider two particles:
d~p1 /dt = F~12 + F~1,ext
d~p2 /dt = F~21 + F~2,ext
d ~
P = F~1,ext + F~2,ext ≡ F~ext
dt
but
d
d ~
P = M V~cm = M~acm
dt
dt
thus
M~aCM = F~ext
(114)
Example: acrobat in the air - CM moves in a simple parabola.
If F~ext = 0
~CM = const
V
(115)
Examples: boat on a lake, astronaut.
D.
Advanced: Energy and CM
~vi = ~vi′ + V~CM
Note:
X
mi~vi′ = 0
i
1 X ′ ~ 2
mi ~vi + VCM
2 i
!
X
1X
1
2
2
E=
mi (vi′ ) +
mi~vi′ · V~CM + MVCM
2 i
2
i
E=
E=
1
1X
2
2
mi (vi′ ) + MVCM
2 i
2
2nd term - KE of the CM, 1st term - KE relative to CM. (will be very important for rotation).
90
XIV.
COLLISIONS
Very large forces F acting over very short times ∆t, with a finite impulse Fa ∆t. Impulse
of ”regular” forces (e.g. gravity) over such tiny time intervals is negligible, so the colliding
bodies are almost an insolated system with
~ ≃ const
P
before and after the collision. With energies it is not so simple, and several options are
possible.
A.
1.
Inelastic
~ = const , K 6= const
P
(116)
~2
V~1 = V
(117)
Perfectly inelastic
After collision:
91
FIG. 20: Example (1D). A chunk of black coal of mass m falls vertically into a vagon with mass
M , originally movin with velocity V . (Only) the horizontal component of momentum is conserved:
m · 0 + M · V = (M + m)vf inal
FIG. 21:
Example (2D). A car m (red) moving East with velocity v collides and hooks up with a
truck M moving North with velocity V . Find the magnitude and direction of the resulting
velocity V1 . Solution:
Direction:
tan θ =
Py
MV
V1y
=
=
V1x
Px
mv
Magnitude:
(M + m)V~1 = m~v + M V~
V1x = mv/(M + m) , V1y = MV /(M + m) , V1 =
92
q
2
V1x
+ V1y2
Example. A bullet of mass m = 9 gram is fired horizontally with unknown speed v at a
wooden block of mass M = 4.5 kg resting on a frictionless table. The bullet hits the block
and becomes completely embedded within it. After that, the block, with the bullet in it, is
traveling at speed V = 2 m/s. (a) Calculate the initial speed of the bullet in m/s; (b) find
the average force on the bullet if the collision lasted ∆t = 10−4 s; (c) estimate the energy
loss
mv + M · 0 = (m + M)V , v = V
∆pbul = m(V − v) ≈ 9 kg
m
, Fbul
s
M +m
m
≃ 1000
m
s
∆pbul
=
≃ 9 · 104 N
∆t
1
1
∆K = (M + m)V 2 − mv 2 = 0.5 · 4.5 · 22 − 0.5 · 0.009 · 10002 ≈ −8 kJ < 0
2
2
Note: in inelastic collision energy is always lost:
1 2 1
1
2
2
(M + m)V −
mv + Mv2 < 0
2
2 1 2
Example. The same block and bullet as above, but now the initial speed of the bullet
is known to be 700 m/s and it passes through the block, emerging on the other side with
vb = 300 m/s. Find the speed of the block V in m/s after the collision.
mv + M · 0 = mvb + MV , V = (v − vb )
m
= ...
M
Example. The same, but initially the block is also moving with vB = 5 m/s in the same
direction as the bullet. The bullet gets stuck. Find the final speed V of block+bullet.
mv + MvB = (m + M)V , V =
93
mv + MvB
= ...
m+M
2.
Explosion
A rocket with M, ~v brakes into m1 , V~1 and m2 , V~2 with m1 + m2 = M.
M~v = m1 V~1 + m2 V~2
Energy is increased.
B.
Elastic
~ = const , K = const
P
(118)
Example: 1D elastic collision of a body m , v with an identical stationary body M = m
mv + 0 = mV1 + mV2
1 2 1
1
mv = mV12 + mV22
2
2
2
or
v 2 = (V1 + V2 )2
v 2 = V12 + V22
Thus, V1 V2 = 0 or (exchange of velocities):
V1 = 0 , V2 = v
1.
Advanced: 1D collision, m 6= M
Collision of a body m , v with a stationary body M.
From conservation of momentum:
m (v − V1 ) = MV2
From conservation of energy:
M 2
m 2
v − V12 =
V
2
2 2
94
Divide 2nd equation by the 1st one (V2 6= 0 !) to get
v + V1 = V2
Use the above to replace V2 in equation for momentum to get
V1 =
m−M
v
m+M
V2 =
2m
v
M +m
and then
Note limits and special cases:
• m = M: V1 = 0, V2 = v (”total exchange”)
• M → ∞ (collision with a wall): V1 = −v (reflection)
• m (”tennis racket”) ≫ M (”tennis ball”): V2 = 2v
FIG. 22: Elastic collision of a moving missile particle m (red) with a stationary target M . (a)
m < M ; (b) m = M ; (c) m > M . In each case both energy and momentum are conserved.
95
2.
Advanced: 2D elastic collision of two identical masses
Collisions of 2 identical billiard balls (2nd originally not moving).
~ (before collision) = P
~ (after)
P
K (before collision) = K (after)
Momentum conservation gives
~v = V~1 + V~2
Energy conservation gives
v 2 /2 = V12 /2 + V22 /2
or
V~1 · V~2 = 0
which is a 90o angle for any V2 6= 0.
FIG. 23: Off-center elastic collision of 2 identical billiard balls (one stationary). Both energy and
momentum are conserved. Note that the angle between final velocities is always 90o .
Advanced Collisions of elementary particles
p2
K=
2m
only for v ≪ c. General
E=
Two limits:
p
m2 c4 + c2 p2
m = 0 , E = cp
96
(119)
(”photon”, v = c) and
m 6= 0 , v ≪ c : E ≈ mc2 +
97
p2
2m
Dr. Vitaly A. Shneidman, Phys 111 (Honors), Lectures on Mechanics
XV.
A.
KINEMATICS OF ROTATION
Radian measure of an angle
see Fig. 24. Arc length
l = rθ
(120)
if θ is measured in radians.
FIG. 24: Angle of 1 rad ≈ 57.3o . For this angle the length of the circular arc exactly equals the
radius. The full angle, 360o , is 2π radians.
98
B.
Angular velocity
Notations: ω (omega). Units: rad/s
Definition:
ω=
dθ
∆θ
≈
dt
∆t
(121)
Conversion from revolution frequency (ω = const):
Example: find ω for 45 rev/min
45
C.
rev
2π rad
rad
= 45
≈ 4.7
min
60 s
s
Connection with linear velocity and centripetal acceleration
v=
v
dl
d(rθ)
=
= ωr , ω =
dt
dt
r
ac = v 2/r = ω 2 r
(122)
(123)
Advanced. ~ω as a vector:
Direction - along the axis of revolution (right-hand rule). Relation to linear
velocity
~v = ~ω × ~r
99
D.
Dynamics of centripetal motion (see also sec. VII C)
Examples. Left: During an air show a plane flying at speed v = 250 m/s
performs a loop-of-death a vertical circle with a radius R = 1800 m (see
figure - not to scale). The pilot has a mass M = 80 kg. Find the magnitude
of the normal forces N1 and N2 (apparent weights) acting on the pilot at the
lowest and at the highest points of the loop. Under which conditions the
pilot feels ”weightless”, i.e. N2 = 0? What is N1 in that case? Right, the
same, for a person on a Ferris Wheel (FW).
N2
v
m
g
N1
v
m
g
bottom: acceleration and axis - up
N1 − Mg = Mac = M
v2
v2
, N1 = Mg + M > Mg
R
R
top: acceleration and axis - down; normal force: down (pilot), up (FW)
v2
v2
pilot : N2 + Mg = Mac = M , N2 = M − Mg
R
R
v2
v2
FW : −N2 + Mg = Mac = M , N2 = −M + Mg
R
R
weightless : N2 = 0 ⇒ ac = g , N1 = 2Mg
100
Example. A skier slides down from a hill which is 30 m high and then,
without losing speed, up a hill which is 10 m high. The mass of the skier
is R = 80 kg. Near the lowest point the path has a radius R = 20 m (see
figure) while near the top of the smaller hill r = 10 m. In each case find the
apparent weight of the skier.
mgH
1
mgh+ mv2
2
h
H
Solution. The force diagrams are almost identical to that of the MGR, but
v’s are different at the lowest and elevated points and are found from energy
conservation:
Thus,
p
1
2
lowest : MgH = Mv ⇒ v = 2gH
2
p
1
2
top of h-hill : MgH = Mgh + Mv ⇒ v = 2g(H − h)
2
bottom: acceleration and axis - up
N1 − Mg = M
v2
2H
2gH
= Mg
, N1 = Mg + Mg
> Mg
R
R
R
top: acceleration and axis - down; N2 - up
2(H − h)
2(H − h)
v2
, N2 = Mg − Mg
−N2 + Mg = M = Mg
r
r
r
(if N2 < 0 will fly off the hill).
101
Example A car goes along a horizontal curved road with a small radius r
and speed v, and then along a similar road with bigger radius R and speed
V . Find the ratio of centripetal forces f /F . What should be µ for each road
so that the car does not skid?
R
V
F
v
f
r
v 2 R
V2
f
v2
⇒
=
f =M ,F =M
r
R
F
V
r
2
v
f − friction ≤ µMg ⇒
≤ µg
r
E.
Angular acceleration
Definition:
α=
dω
d2 θ
= 2
dt
dt
(124)
Units:
[α] = rad/s2
F.
Connection with tangential acceleration
aτ =
dv
dω
=r
= rα
dt
dt
102
(125)
(very important for rolling problems!)
G.
Rotation with α = const
Direct analogy with linear motion:
x → θ ,
v → ω ,
ω = ω0 + αt , θ − θ0 =
a → α
ω + ω0
t
2
ω 2 − ω02
1
θ = θ0 + ω0 t + αt2 , θ − θ0 =
2
2α
(126)
(127)
New: connection between θ (in rads) and N (in revs) and ω in rad/s and
frequency f = 1/T in rev/s:
θ = 2πN , ω = 2πf
(128)
Examples:
• a free spinning wheel makes N = 1000 revolutions in t = 10 seconds,
and stops. Find α. Solution: in formulas
ω = 0 , θ0 = 0 , θ = 2πN
2θ
ω0
t ⇒ ω0 =
2
t
2θ
4πN
ω0
α = − = − 2 = − 2 = ...
t
t
t
θ=
• Given ω(0) = 5 rad/s , t = 1 s , N = 100 rev (does not stop!). Find α.
2πN − ω(0)t
1
= ...
2πN = ω(0)t + αt2 , α = 2
2
t2
103
XVI.
A.
KINETIC ENERGY OF ROTATION AND ROTATIONAL INERTIA
The formula K = 1/2 Iω 2
For any point mass
1
Ki = mi vi2
2
(129)
For a solid rotating about an axis
vi = ωri
(130)
with ri being the distance from the axis and ω, the angular velocity being
the same for every point. Thus, the full kinetic energy is
K=
X
i
1 2X
1
Ki = ω
mi ri2 ≡ Iω 2
2
2
i
(131)
Here I, the rotational inertia, is the property of a body, independent of ω
(but sensitive to selection of the rotational axis):
I=
X
i
1
mi ri2 , K = Iω 2
2
Or, for continuous distribution of masses:
Z
I=
dl λr2
for a linear object with linear density λ (in kg/m); or
Z
I=
dS σr2
for a flat object (S-area) with planar density σ (in kg/m2 ), or
Z
I=
dV ρr2
for a 3D object (V -volume) with density ρ (in kg/m3 )
104
(132)
(133)
(134)
(135)
B.
1.
Rotational Inertia: Examples
Collection of point masses
Two identical masses m at x = ±a/2. Rotation in the xy plane about the
z-axis through the CM.
1
I = 2 · m(a/2)2 = ma2 /2 = Ma2 , M = m + m
4
(136)
FIG. 25: 3 identical point masses m on a massless rod of length L. Left: axis through CM.
ICM = m(L/2)2 + m · 0 + m(L/2)2 = mL2 /2. Right: axis through the end. Iend = m · 0 +
m(L/2)2 + mL2 = (5/4)mL2 > ICM (!).
2.
Hoop
Hoop of mass M, radius R in the xy plane, center at the origin. Rotation
in the xy plane about the z-axis.
Linear density
Ihoop
λ = M/(2πR)
Z 2πR
=
dl λR2 = MR2
0
(the same for hollow cylinder about the axis)
105
(137)
3.
Rod
Uniform rod of mass M between at x = ±l/2. Rotation in the xy plane
about the z-axis through the center of mass.
x
FIG. 26: Evaluating rotational inertia of a rod
Linear density
λ = M/l
Thus,
I=
Or,
Z
l/2
−l/2
2
dx λx = 2λ ·
Irod =
4.
Z
l/2
0
1
dx x2 = 2λ (l/2)3
3
Ml2
12
(138)
Disk
Uniform disk of mass M, radius R in the xy plane, center at the origin.
Rotation in the xy plane about the z-axis.
Planar density
Elementary area
σ = M/ πR2
dS = 2πr dr
106
FIG. 27: Evaluating rotational inertia of a disk
Idisk =
Z
R
0
1
dr 2πrσr2 = MR2
2
(139)
(the same for solid cylinder about the axis).
5.
Advanced: Solid and hollow spheres
Solid sphere: slice it into collection of thin disks of thickness dz and radius
√
r = R2 − z 2
Isph =
Z
R
−R
1
2
dz πr2 ρr2 = MR2
2
5
Hollow:
2
Ih.sph = MR2
3
107
(140)
FIG. 28: Evaluating rotational inertia of a solid sphere
C.
Parallel axis theorem
I = Icm + MD2
(141)
with Icm being rotational inertia about a parallel axis passing through the
center of mass and D - distance to that axis.
Proof:
Introduce
X
~ cm = 1
~ri mi
R
M i
~ cm
~ri′ = ~ri − R
with
X
~ri′ mi = 0
i
108
and
Icm =
X
2
mi (ri′ )
i
Now
I=
X
i
2
X
′
~ ·
~
mi ~ri + D = Icm + MD2 + 2D
~ri′ mi
i
where the last sum is zero, which completes the proof.
Example: rod M, L about the end
ICM = ML2 /12
with D = L/2:
Irod,end = ML2 /12 + M(L/2)2 = ML2 /3
109
1.
Distributed bodies plus point masses
FIG. 29: Modification of I by a point mass m added at distance a from center. Left: rod mass M ,
length L. Right: disk of mass M and radius R.
rod : I = Irod + ma2 = ML2 /12 + ma2 = L2(M/3 + m)/4 if a ≃ L/2
disk : I = Idisk + ma2 = MR2 /2 + ma2 = R2(M/2 + m) if a ≃ R
2.
Advanced: Combinations of distributed bodies.
Example:
Solid spheres
with radius a and mass M, rod length L; rotation axis through center.
110
1) negligible size of spheres (a ≪ L) and massless rod
I = 2 × M(L/2)2 = ML2 /2
2) non-negligible size of spheres
2
I = 2 × Ma2 + ML2 /2
5
3) rod of mass m - add mL2 /12
Example:
Solid spheres with radius a and mass M each, rods with length L each;
rotation axis through center, perpendicular to plane.
1) negligible size of spheres (a ≪ L) and massless rods
I = 4 × M(L/2)2 = ML2
Note: central mass does not contribute!
2) non-negligible size of spheres
2
I = 5 × Ma2 + ML2
5
111
Note: all 5 spheres contribute.
3) rods of mass m each - add 2 × mL2 /12
D.
Conservation of energy, including rotation
K + U = const
(142)
where K is the total kinetic energy (translational and rotational for all
bodies) and U is total potential energy.
E.
”Bucket falling into a well”
T
T
mg
Suppose mass m goes distance h down starting from rest. Find final velocity
v. Pulley is a disk M, R. Solution: ignore T (!), use energy only.
• energy conservation
1 2 1 2
mv + Iω = mgh
2
2
112
• constrains
v = ωR
Thus
v 2 = 2gh
m
m + I/R2
If I = MR2 /2 (disk)
v 2 = 2gh
1.
1
, v = ...
1 + M/(2m)
Advanced: Atwood machine
FIG. 30: Atwood machine. Mass M (left) is almost balanced by a slightly smaller mass m. Pulley
has rotational inertia I and radius R.
Suppose the larger mass goes distance h down starting from rest. Find
final velocity v.
• energy conservation
1
1
(M + m)v 2 + Iω 2 = (M − m)gh
2
2
• constrains
v = ωR
113
Thus
v 2 = 2gh
M −m
M + m + I/R2
Acceleration from
h = v 2/2a
which gives
a=g
M −m
M + m + I/R2
Note the limit: m = 0, I = 0 gives a = g (free fall).
2.
Rolling
1 2 1 I v 2
gH = gh + v +
2
2m r
v 2 = 2g(H − h)
114
1
1 + I/mr2
FIG. 31: Rolling down of a body with mass m, rotational inertia I and radius R. Potential energy
at the top, mgh equals the full kinetic energy at the bottom, mv 2 /2 + Iω 2 /2.
Suppose the body rolls vertical distance h = L sin θ starting from rest. Find
final velocity v.
• energy conservation
1 2 1 2
mv + Iω = mgh
2
2
• constrains
v = ωR
Thus
v 2 = 2gh
1
1 + I/ (mR2 )
1
1+1
1
disk : I = MR2 /2 , v 2 = 2gh
1 + 1/2
hoop : I = MR2 , v 2 = 2gh
The disk wins!
Advanced. Acceleration from
a = v 2 /2l , l = h/ sin θ
which gives
a = g sin θ
1
1 + I/ (mR2 )
Note the limit: I = 0 gives a = g sin θ.
115
Dr. Vitaly A. Shneidman, Phys 111 (Honors), Lectures on Mechanics
XVII.
A.
TORQUE
Definition
Consider a point mass m at a fixed distance r from the axis of rotation. Only motion in
tangential direction is possible. Let Ft be the tangential component of force. The torque is
defined as
τ = Ft r = F r sin φ = F d
(143)
with φ being the angle between the force and the radial direction. r sin φ = d is the ”lever
arm”. Counterclockwise torque is positive, and for several forces torques add up.
Units:
[τ ] = N · m (same as Joules).
116
2D convention:
(not to scale!)
positive direction is indicated
τ+ = T1 R , τ− = T2 r
τ = τ+ − τ−
In equilibrium:
τ+ = τ− , T1 = mg , T2 = Mg
B.
2nd Law for rotation
Start with a single point mass. Consider the tangential projection of the 2nd Law
Ft = mat
Now multiply both sides by r and use at = αr with α the angular acceleration.
τ = mr 2 α
For a system of particles mi each at a distance ri and the same α
X
τ = Iα
117
(144)
C.
Application of τ = Iα. Examples.
Thhe rod-disk system is brought from ω0 = 0 to ω = 1000 rad/s in t = 10 s. Find T
(everything else is known, r ≪ R)
Solution.
α=
1
1
ω
, I = mrod r 2 + Mdisk R2
t
2
2
τ = Iα
118
1.
Revolving door
How long will it take to open a heavy, freely revolving door by 90 degrees starting from rest, if
a constant force F is applied at a distance r away from the hinges at an angle φ, as shown in the
figur? Make some reasonable approximations about parameters of the door, F and r. For I you
can use 1/3 M L2 (”rod”), with L being the horizontal dimension.
Solution: 1) find torque; 2) use 2nd law for rotation to find α; 3) use kinematics to estimate t
1.
τ = F r sin φ
2.
α = τ /I = 3F r sin φ/(M L2 )
if r = L
α = 3F/(M L) sin φ
3. To maximize α use φ = π/2. From
θ = 1/2 αt2
r
p
I
t = 2θ/α = π
FL
Using, e.g. M = 30 kg, F = 30 N , L = 1 m one gets t ∼ 1 s, which is reasonable.
119
2.
Rotating rod
α = τ /I = (1/2 Mgl)
1 2
Ml
3
3
= g/l
2
Linear acceleration of the end:
3
a = αl = g > g (!)
2
3.
Rotating rod with a point mass m at the end.
Will it go faster or slower?
Solution:
same as above, but
I → 1/3 Ml2 + ml2 , τ → 1/2 Mgl + mgl
Thus,
α=
3 g 1 + 2m/M
2 l 1 + 3m/M
Linear acceleration of the end:
3 1 + 2m/M
α= g
2 1 + 3m/M
(which is smaller than before, but still larger than g)
120
4.
”Bucket falling into a well” revisited.
T
T
mg
2nd Law for rotation:
α = τ /I = T R/I
2nd Law for linear motion:
ma = mg − T
Constrain:
α = a/R
Thus (divide 1st equation by R/I and add to the 2nd one, replacing α):
aI/R2 + ma = mg
or
a=g
5.
1
1 + I/ (mR2 )
Advanced: Atwood machine revisited.
Let T1 and T2 be tensions in left string (connected to larger mass M) and in the right
string, respectively.
• 2nd Law(s) for each body (and for the pulley with τ = (T1 − T2 ) R)
121
✻
✻
T~
T~1
2
M~g
❄
m~g
❄
FIG. 32: Atwood machine. Mass M (left) is almost balanced by a slightly smaller mass m. Pulley
has rotational inertia I and radius R.
• constrains a = αR
From 2nd Law(s):
Mg − T1 = Ma , T2 − mg = ma , (T1 − T2 ) = Iα/R
Add all 3 together to get (with constrains)
(M − m)g = (M + m)a + Iα/R = (M + m + I/R2 )a
which gives
a=g
M −m
M + m + I/R2
and α = a/R. What if need tension?
T1 = Mg − ma < Mg , T2 = mg + ma > mg
122
6.
Rolling down incline revisited.
Start with equilibrium due to additional force F up the incline
FIG. 33: Equilibrium of a body with mass m and radius R. Four forces act on the body: external
~ and f~ - static friction at the point of contact, up the plane; N
~ - normal reaction, perforce F
pendicular to the plain at the point of contact and m~g is applied to CM. Note that only F and
friction have torques with respect to CM (and only F and mg have torques with respect to point
of contact with the incline).
CM : F R = f r , F + f = mg sin θ
1
F = mg sin θ
2
or
1
contact point : F · 2R = mg · R sin θ , F = mg sin θ
2
123
f
N
mg
FIG. 34: Rolling down of a body with mass m, rotational inertia I and radius R. Three forces
~ - normal reaction,
act on the body: f~ - static friction at the point of contact, up the plane; N
perpendicular to the plain at the point of contact and m~g is applied to CM. Note that only friction
has a torque with respect to CM.
• 2nd Law(s) for linear and for rotational accelerations with torque τ = f R (f - static
friction)
• constrains a = αR
2nd Law (linear)
~ + m~g = m~a
f~ + N
or with x-axis down the incline
−f + mg sin θ = ma
2nd Law (rotation)
f R = Iα
or with constrain
f = aI/R2
Thus,
−aI/R2 + mg sin θ = ma
or
a = g sin θ
1
1 + I/mR2
the same as from energy conservation.
124
Advanced. Alternative solution: ”Rotation” about point of contact with I ′ = I + mR2
(”parallel axes theorem”). Now only gravity has torque τ = mgR sin θ. Thus
α = τ /I ′ =
g
1
sin θ
R
1 + I/mR2
which gives the same a = αR.
D.
1.
Torque as a vector
Cross product
see Introduction on vectors
2.
Vector torque
~τ = ~r × F~
XVIII.
A.
(145)
ANGULAR MOMENTUM L
Single point mass
~ = ~r × ~p
L
with ~p = m~v , the momentum.
125
(146)
~ for circular motion.
Example. Find L
Solution: Direction - along the axis of rotation (as ~ω !). Magnitude:
L = mvr sin 90o = mr 2 ω
or
~ = mr2 ~ω
L
B.
System of particles
~=
L
C.
1.
(147)
X
i
~ri × p~i
Rotating symmetric solid
Angular velocity as a vector
Direct ~ω allong the axis of rotation using the right-hand rule.
Example. Find ω
~ for the spinning Earth.
Solution: Direction - from South to North pole. Magnitude:
ω=
2π rad
rad
≃ ...
24 · 3600 s
s
126
(148)
If axis of rotation is also an axis of symmetry for the body
L=
X
mi ri2 ω = Iω
(149)
i
or
~ = I~ω
L
D.
~
2nd Law for rotation in terms of L
Start with
d~p
F~ =
dt
Then
~
dL
~τ =
dt
127
(150)
Dr. Vitaly A. Shneidman, Phys 111 (Honors), Lectures on Mechanics
XIX.
CONSERVATION OF ANGULAR MOMENTUM
Start with
~τ =
~
dL
dt
If ~τ = 0 (no net external torque)
~ = const
L
(151)
valid everywhere (from molecules and below, to stars and beyond).
For a closed mechanical system, thus
~ = const
E = const , P~ = const , L
For any closed system (with friction, inelastic collisions, break up of material, chemical
or nuclear reactions, etc.)
~ = const
E 6= const , P~ = const , L
A.
Examples
1.
Free particle
~ = ~r(t) × m~v
L(t)
with ~v = const and
~r(t) = ~r0 + ~v t
Thus, from ~v × ~v = 0
~ = (~r0 + ~vt) × m~v = ~r0 × m~v = L(0)
~
L(t)
128
2.
Student on a rotating platform
(in class demo) Let I be rotational inertia of student+platform, and
I ′ ≃ I + 2Mr 2
the rotational inertia of student+platform+extended arms with dumbbels (r is about the
length of an arm). Then,
L = Iω = I ′ ω ′
or
ω′ = ω
I
1
=ω
′
I
1 + 2Mr 2 /I
129
3.
Chewing gum on a disk
An m = 5 g object is dropped onto a uniform disc of rotational inertia I = 2 · 10−4 kg · m2
rotating freely at 33.3 revolutions per minute. The object adheres to the surface of the disc
at distance r = 5 cm from its center. What is the angular momentum? What is the final
angular velocity of the disk? How much energy is lost?
Solution.
L = Iω
Similarly to previous example, from conservation of angular momentum
L = Iω = I ′ ω ′
New rotational inertia
I ′ = I + mr 2
Thus
ω′ = ω
I
1
I
=
ω
=
ω
I′
I + mr 2
1 + mr 2 /I
Energy:
1
1
Ki = Iω 2 , Kf = I ′ (ω ′)2
2
2
K i − Kf = . . . > 0
130
4.
Measuring speed of a bullet
To measure the speed of a fast bullet a rod (mass Mrod ) with length L and with two wood
blocks with the masses M at each end, is used. The whole system can rotate in a horizontal
plane about a vertical axis through its center. The rod is at rest when a small bullet of mass
m and velocity v is fired into one of the blocks. The bullet remains stuck in the block after
it hits. Immediately after the collision, the whole system rotates with angular velocity ω.
Find v. ( use L = 2 meters, ω = 0.5 rad/s and m = 5 g , M = 1 kg , Mrod = 4 kg).
Solution. Let r = L/2 = 1 meter be the distance from the pivot. Angular momentum:
before collision
L=
L
mv
2
(due to bullet); after:
L = Iω
with
I = I0 + m(L/2)2
Here I0 is rotational inertia without the bullet:
I0 = Mrod L2 /12 + 2M(L/2)2
Due to conservation of angular momentum
L
mv = Iω
2
and
v = 2Iω/(mL)
131
5.
Rotating star (white dwarf )
A uniform spherical star collapses to 0.3% of its former radius. If the star initially rotates
with the frequency 1 rev/day what would the new rotation frequency be?
Solution - in class
132
XX.
A.
EQUILIBRIUM
General conditions of equilibrium
X
F~i = 0
(152)
X
~τi = 0
(153)
Theorem. In equilibrium, torque can be calculated about any point.
Proof. Let ~ri determine positions of particles in the system with respect to point O.
Selecting another point as a reference is equivalent to a shift of every ~ri by the same ~ro .
Then,
~τnew =
X
i
B.
(~ri + ~ro ) × F~i = ~τ + ~ro ×
X
F~i = ~τ
i
Center of gravity
Theorem. For a uniform field ~g the center of gravity coinsides with the COM.
Proof. Torque due to gravity is
τg =
X
i
C.
1.
~ri × mi~g =
X
i
Examples
Seesaw
133
mi~ri
!
~ CM × ~g
× ~g = M R
FIG. 35: Two twins, masses m and m (left) against their dad with mass M . Force of gravity on
the seesaw and reaction of the fulcrum are not shown since they produce no torque.
If 2d, d and D are distances from the fulcrum for each of the twins and the father,
mg · 2d + mgd = MgD
or
3md = MD
~
Note that used only torque condition of equilibrium. If need reaction from the fulcrum N
use the force condition
~ + (2m + M + Mseesaw ) ~g = 0
N
134
2.
Beams
FIG. 36: Horizontal beam of mass M and length L supported by a blue cord making angle θ with
horizontal. Only forces with non-zero torque about the pivot (tension T~ -blue- and gravity M~g
-black) are shown.
Cancellation of torques gives
T L sin θ = Mg
L
2
or
T = Mg/2 sin θ
Note: if you try to make the cord horizontal, it will snap (T → ∞). For θ → π/2 one has
T → Mg/2, as expected. The force condition will allow to find reaction from the pivot:
~ + M~g + T~ = 0
R
~ will be the force on the pivot).
(and −R
135
N
Mg
mg
FIG. 37: Horizontal beam of mass M and length L with additional mass m at the end. The rod
is pivoted at the left end, and there is a support at a distance d < L/2. Find the force N .
Consider torques about the pivot
N · d = Mg
L
+ mg · L , N = . . .
2
T
Mg
m
FIG. 38: A beam of mass M and length L with additional mass m at the end. The rod is pivoted
at the left end and makes an angle φ1 with vertical. There is a supporting cord which makes an
angle φ2 with vertical. Find the tension T .
T L sin φ3 = Mg
L
sin φ1 + mgL sin φ1 , φ3 = 180o − φ1 − φ2
2
136
A beam with M = 100 kg makes an angle 45o with horizontal, and is supported by a
cable which makes 30o with the beam. Find T .
T L sin 30o = Mg ·
3.
L
sin 45o
2
Ladder against a wall
FIG. 39: Blue ladder of mass M and length L making angle θ with horizontal. Forces: M~g (blue),
~ (vertical), friction f~ (horizontal).
wall reaction F~ (red), floor reaction N
137
Torques about the upper point:
π
L
− θ − f L sin θ − Mg sin
−θ =0
NL sin
2
2
2
π
But from force equilibrium (vertical) N = Mg and (on the verge) f = µN. Thus,
1
cos θ − µ sin θ = 0
2
138
XXI.
A.
1.
FLUIDS
hydrostatics
Density
ρ=
m
, [ρ] = kg/m3
V
(154)
E.g. for water
ρ ≃ 1000 kg/m3 = 1 g/cm3
Example. Recently, a thief stole 1, 000, 000 $ worth of gold from an armour
truck in NYC. Estimate the volume, if gold is worth about $25 per gram,
with density about 20 gram per cm3 .
m = 106/25 = 4 · 104 gram ; V =
2.
m
= 2 · 103 cm3 = 2 · 10−3 m3
ρ
Pressure
P = Fn /A
(155)
[P ] = P a = N/m2 , 1 atm ≃ 105 P a ≃ 760 mm Hg
(156)
139
”Barostatic equation”
P
dy
P+dP
dP = −ρg dy
For ρ = const (”incompressible”)
|∆P | = ρgy
(157)
Pascal’s principle: extra pressure δP is transferred to every point of the
fluid
δP = const
140
2.0
1.5
1.0
0.5
0.0
-5
-4
-3
-2
-1
0
1
2
Pressure under any of the tubes is the same and is determined only by the
level of the fluid.
Hydrolic lift (pistons not shown). Extra pressure δP = f /A1 (due to a
small force f applied to the narrow column) is transferred to the wider
column, creating a bigger force F = A2δP .
141
3.
Archimedes principle
The force on the upper surface is P A; the force on the lower is
(P + ρg∆h)A (where ρ is the density of fluid!). The resultant
FA = ρg∆V
where ∆V is the ”displaced volume”.
Example: find the density of a body (”crown”) from its weight in air ≃ mg =
ρc V g and ”weight” in fluid W .
W = mg − FA = ρc V g − ρV g = V g(ρc − ρ)
ρ
W
=1−
mg
ρc
mg
ρc = ρ
mg − W
mg = ρc V g ,
142
Example: Iceberg, ρice < ρ, V - volume of the iceberg, v - volume above
water
FA = ρ(V − v)g = Mg = ρiceV g
1−
v
v
ρ − ρice
= ρice /ρ ,
=
V
V
ρ
143
B.
Hydrodynamics
Laminar flow of ideal fluid (non-viscous and incompressible) around a
cylinder
144
1.
Continuity equation and the Bernoulli principle
ρ1 v1A1 = ρ2 v2A2 , ρ 6= const
(158)
v1A1 = v2A2 , ρ = const
(159)
Example. Crude oil is pumped through a pipe of radius R = 0.1 m. The
speed of the flow is v = 2 m/s. What is the volume flow speed, in L/s ?
L
dV
= πR2 v = π × 10−2 × 2 × 103
dt
s
Example. A doctor finds a partial blockade in a vessel. The cross-sectional
area available for blood flow near the blockage is Ab = 0.1 mm2. The flow
rate dV /dt in the region of the partial blockage is 2500 ml/hour. In a nearby
vessel that is clear, the area is Ac = 0.5 mm2. What is the blood flow rate,
and what is the velocity vc in m/s in the clear region?
dV /dt = 2500 ml/h − same!
dV /dt
10−6 m3
vc =
= 2500
= ...
Ac
3600 s × 0.5 · 10−6 m2
145
Bernoulli. Consider element of fluid bounded by flowlines and moving
right with m = ρV and small volume V = A∆x. The energy of the element
when it moves ∆x is changed due to work by pressure difference
W = −∆P · A∆x = −∆P V
(160)
(ignore graviity). The change in kinetic energy of the element
1 2
1
∆K = ∆
mv = V ∆(ρv 2)
2
2
(161)
(note: V = const (!) ). From
1 2
W = ∆K , ∆P + ∆
ρv
2
=0
1
P + ρv 2 = const , (no gravity)
2
1 2
P + ρgh + ρv = const , (with gravity)
2
The wing:
146
(162)
(163)
Dr. Vitaly A. Shneidman, Phys 111 (Honors), Lectures on Mechanics
XXII.
A.
GRAVITATION
Solar system
1 AU ≃ 150 · 106 km, about the average distance between Earth and Sun.
Mer - about 1/3 AU (0.39)
V - about 3/4 AU (0.73)
Mars - about 1.5 AU (1.53)
J - about 5 AU (5.2)
...
Solar radius - about 0.5% AU
B.
Kepler’s Laws
Cartesian:
x2 /a2 + y 2 /b2 = 1
(164)
x = r cos φ , y = r sin φ
(165)
r(φ) = a(1 − ǫ2 )/(1 + ǫ cos φ)
(166)
Polar:
ǫ > 1 - hyperbola, ǫ = 1 - parabola (a → ∞).
147
b
a
FIG. 40: The ellipse and Kepler’s Law’s:
a and b - major and minor semi-axes.
√
f = ± a2 − b2 - foci (Sun in one focus, the other one empty!)
ǫ = |f |/a - eccentricity
a = b - circle
1.
1st law
Path of a planet - ellipse, with Sun in the focus.
(Justification is hard, and requires energy, momentum and angular momentum conservation, plus explicit gravitational force - Newton’s law)
148
2.
2nd law
Sectorial areas during the same time intervals are same - see figure.
(Justification is easy, and requires ONLY angular momentum conservation in class)
dS =
3.
1
1
|~r × ~v dt| =
L · dt = const · dt
2
2m
3rd law
T 2 ∝ a3
(167)
and b does not matter!!!
(Justification is hard, and requires energy, momentum and angular momentum conservation, plus explicit gravitational force; will derive explicitly only
149
for a circle, a = b , ǫ = 0.)
C.
The Law of Gravitation
F =G
Mm
r2
(168)
or in vector form
Mm~r
F~ = −G 3
r
(169)
with G ≈ 6.7 · 10−11 N · m2 /kg 2 .
1.
Gravitational acceleration
M
g = F/m = G 2 = gs
r
2
R
r
(170)
with gs acceleration on the surface and M, R - mass and radius of the central
planet (or, of the Sun). In vector form
M~r
~g = −G 3 = −r̂gs
r
2
~r
R
, r̂ =
r
r
Example. Find gV if RV ≈ 0.4 RE and MV ≈ 0.056 ME
2
g
M
R
V
V
E
g ∝ M/R2 thus
≈ 0.4
=
gE
ME RV
150
(171)
2.
Satellite
v2
= g(r) = GM/r2
r
v=
p
GM/r
T = 2πr/v = 2πr
3/2
(172)
√
/ GM
T 2 = 4π 2 /GM r3
(173)
the 3rd law. (r is distance from center !)
Example. Find v for earth around sun, and check if T = 1 year.
v=
p
6.7 · 10−11 2 · 1030/(150 · 109) ≈ 30
T =
2πr
≈ 1 year
v
151
km
s
Low orbit:
2
R
v /r = g(r) = gs
r
2
for r ≃ R
v≃
about 8 km/s for surface of Earth.
p
gs R
152
(174)
D.
Energy
U=
Z
r
∞
′
′
F (r )dr = −GMm
U = −G
Z
∞
r
dr′
(r′)2
Mm
r
(175)
(176)
(177)
Probe m in combined field of Sun and planet
U = −G
1.
ME m
MS m
−G
rpS
rpE
Escape velocity and Black Holes
E =K +U
U (∞) = 0, K(∞) = E > 0 for escape. Thus
1 2
Mm
2
mv − G
≥ 0 , vesc
/2 = GM/R
2
r
p
p
vesc = 2GM/R = 2gsR ≈ 11 km/s
for the surface of Earth.
153
(178)
Example. A missile m is launched vertically up from the surface with
v0 = 12 vesc . Find h
Mm
1 2
1
2
)
E = mv0 + −G
= m(v02 − vesc
2
R
2
Mm
Mm R
1 2
R
= −G
= − mvesc
R+h
R R+h
2
R+h
R
3 2
2
2
vesc
= vesc
− v02 = vesc
R+h
4
3
h
4
R
= , 1 + = , h = R/3
R+h 4
R 3
E = −G
vesc = c ≃ 3 · 105 km/s - ”Black Hole”. Radius of a Black Hole:
RB =
2GM
c2
For Sun:
RB ≈ 9 km
154
E.
Advanced: Deviations from Kepler’s and Newton’s laws
0.5
0.5
-0.5
-0.5
in class
0.3
0.2
0.1
-0.4
-0.3
-0.2
0.1
-0.1
0.2
-0.1
-0.2
-0.3
155
0.3