ページ3:

Mechanical
energy
E-K+U
Conservation of
.
:
energy
AK+AU+Uint = 0
六、 Momentum, Collision
Momentum
P = m V
•合外力 = 0 ⇒
E
v
• Conservative force (127).
弹、万(重)、静电
K
Nonconservative force (非保守)(路径有关
作功力学能不守恆)
*
戸单位同
PP (ΣP-m₁₁ + m₁₁ = mx + mv.)
平均力
AP
At
Impulse
Collision
J-Fat
impulse
F-t
graph
=
ΣF dt
- momentum
theorem p
Elastic collision
F、K皆守恆,
Ven 不变
Inelastic collision 户守恆,K损失, Vs = Ven
Center of mass motion :
ΣFort-Macm
tRotation
ballistic pendulum
→ K→ C
MB V₁ = (MB +Mw) V₂
± (m₂+Mw) = (m + mwjgh V₁ = √mgh
Angular displacement
:(角位移, 日 )
S = r
I rad
radian 12€
radius
弧长 S = re => 10 = $
(rad)
Angular velocity
(角速度Ò)
40 V
W =
3
At
r
dt
(rad/s)
方向:右手 rule
'
counterclockwise (逆时针) ⇒ 正值
· Angular frequency (speed) :)
W = = = 2πf
Angular acceleration
α =
AW
At
(角頻率/速率,W)
(角加速度,a)
w
speeding up
slowing down
D

ページ4:

V = V₁ + at
S = Vot + ± at²
V₁ = √₁² + zas
linear and angular kinematics :
• linear speed:
2πr
V =
= wr
T
• tangential acceleration :
W = W₁ + αt
a = ra
0 = W₁t + at
W = Wi + 2x0
V'
4π'r
=
=
w'r
centripetal (radial) acceleration: a₁ = and = \--
=>
Fc = mac
= m⋅
= mw'r
energy
Rotational kinetic
moment
of inertia(转动惯量、惯性矩): I = mr = [mirk
•
K=Iw= mrw=
Parallel - axis theorem ( 16 ) I₁ = Icm +
Torque (Moment, =)
.
= Fx = Iα
Angular
JP
F
Md"
linear acceleration
ā
/arad
v = wr
Sede
The who
The w
momentum : ( L )
- FxP=Fxmv
=
Iŵ
合外力矩=0
⇒
- Rolling
rotation +
K = MVCM² + ICM W
VCM = wr
[守恆 ([-定)
translating
=
ex Rotational collision:
I,w, I, w, (I + I) w
→[守恆
work
'
power
W = S°` t do = {^"^[w do + ±1w* - ±1w;*
P = zw
· Equilibrium
移动 衡
ΣF = 0
靜力平衡
(static equilibrium)
转动平衡
Σt = 0
Center of Gravity
mm1=14
mm 1-24
mam 134
mm 14%
mam 154
mm 16%
The cylinders with higher moment of inertia roll down a slope
with a smaller acceleration, as more of their potential energy
needs to be converted into the rotational kinetic energy.

ページ5:

+ Elasticity
Stress (5)
6 =
A
Strain (应变):
force
per
unit
area
deformation due to the
stress
AL
Ɛ =
Lo
Elastic moduli:(弹性模数)
elastic moduli
stress
=
(Hooke's law)
(Pa = N/m²)
strain
(40
Tensile / Compressive
(linear)
• Young's Modulus
Y
stress
•Bulk Modulus
F./A
AL/L.
B =
• Shear Modulus
AP
AV/V₁
#*+
Pressure
Sh=
Shear strin
Tan
- Shear
F/A
Volume
Area A
- V₂+ AV
(AV <0)
Elasticity & Plasticity
.
(弹性、塑性)
Elastic limit or yield point
deformation
Plastic
Proportional
limit
tt 191 tk på - stress oc strain (Hooke's law) $ 2
弹性板限– 超过迸塑性行为
Elastic
behavior
0<1%
Plastic
behavior
Permanent
set
Fracture
point
屈服点开始永久形变
断裂点
Strain
30%
477
mg
= g = GM
(G= 6.67 x 10" N m/h)
=
W-GRIP
RE
+ Gravitation
gravitational
F₂ =
force
GMm
r
=
gravitational potential energy
U-GMM
r
circular satellite orbit
F₁ = GMmm x = m 4x2 = F₁
→ Circular orbit
V=
GM
√ R
(第一宇宙迷度)
>
•
E K+Umm ) + (- GMm) =
GMM
→ escape speed
V
2GM
VR
(第二宇宙建度)

ページ6:

+ Periodic Motion
SHM
restoring
force
displacement
toward equilibrium position
Restoring force F
0
Fx=-kx
/
ax
Hooke's law =
circle motion projection SHM :
x = A cos o
VX
= - Aw sin o
O = wt
ax
=
Aw cos 0
-w³x
.
W
=
厝
simple pendulum
Fo= mg sino
= 3.
energy
:
1mg s in @
A
L
m
=>
2π
k
=>
T = 2π
EK+Um² + ± kx² = ±² k A² constant
Mechanical waves
Displacement a
W= 2πf=
21
十三、
types
。 Transverse
wave
(横)
Longitudinal
wave
(縱)
Surface
Wave
periodic
Wave
(sinusoidal
wave)
.
v = λf
power
amplitude (A)
period (T)
frequency (f)
phase (4)
• wavelength (A)
y (x-t) = A cos (kxwt)
→ wave number
2
w= kv
入
厶质点运动:
对七偏微
ay
Uy
=
Aw sin (kx-wt)
at
ay
ay
=
- Aw cos (kx-wt)
厶波位移:
对x偏微
+ 31/31 ==
ay (x-t)
= wave equation :
ay
·- - Aksin (kx-wt)
ay
-Ak cos (kx-wt)
18x+
speed
on
string
:
V =
ať"
=
V
ay (x-t)
ax"

ページ7:

· P(x,t) = Fg (x. t) Vy (x. t) = √μF w`A`sin" (dx-wt)
wave
intensity I=
boundary conditions.
P
4πr
r
r fixed end - FR K
P = √μF WA
Pav ==√μF WA
M
反射
free end
同相
Standing
:
wave
二反向波干涉
y(x,t)
=
y, (x7) + y (x. t) = - A cos (kx+wt) + Acos (4x-wt) =
2A sin kx sin wt
(kx = nπ. n = 0.1.2...)
nλ
nv
2L
(n-1.2.3.
n'λ
端昇L
f
n'v
(n. 1.3.5.
4L
夺数
• (1%) / 1st / (2nd harmonic )
谐
fundamental forgang
Sound
wave.
Displaced particles
overtone
y (x. +) = Aws (kx-wt)
P(x) = Bk A sinckx-wt) P
antinode
node
Resonance carve graph of amplitude A
ves driving frequency f. Peaks occur at
somal-male frequencies of the pipe
--M...
A sound wave
displaces the left
and the
end of the cylinder right end by
BKA
(Ap = - B AV)
by yy.)-
K
speed
V =
P
√(solid liquid)
RT
'
V=
M
(gas)
P =
=
A
pv
√B WA² = P = Pi
20v 2√√B
intensity =7
Beats : (拍音)
Two sound waves
with slightly
different frequencies
↳ decibal scale : B = (10dB) log=
(I₁ = 10th W/m²)
二频率相近波干涉 周期性強弱交替
Waves in
Waves out of
phase with
each other
phase with each
other
(a)
AAA
ww w v v w w wwww w w VV w w w
-Time
0.25 s
0.50 s
0.75 s
1.00 s
fu... = f. - f₂
www www
(b)
The two waves interfere
constructively when they are in phase
Beat
and destructively when they are a half-cycle out of phase. The
resultant wave rises and falls in intensity, forming beats
the Doppler effect :
^ in front =
A bahind
=
=
=
listener
source
✓ V₁ V - Vs
fi
f
V+V
V+Vs
f,
A behind
=
V+V₁
V+V
to
=
= ₤ V+V₁
V+V₁
Time
①V₁
V₂
0.
Ar
x+ Ar
The change in volume of the disturbed
cylinder of fluid is Sty-y₁)

ページ8:

+ Electric charge & field
Triboelectric: (摩擦起电)
Plastic (-)
Charge by Induction:
Glass (+)
Silk (-)
fur (+)
(感应起电)
le.c. = 1.6 × 10-1 C.
Low
F
k
8.8.1
1
|qq|
4πE. r²
k = = = 9 × 10" (Nm/) E. - 8.854× 10th (c/pm)
+
Electric force :
Coulomb's
Electric field :
E = 121
r'
> F = & E
Electric
dipoles
V/m = = = №/c
(电偶极)
diople
moment
P = q d
(C·m)
torque
7
== P× E
P >>
U = -P-E
+ Gauss's law.
electric flux :
vector ->
outward
=
E· A
= EA cos &
Gauss's law :
₁ = E dA
=
q
E.
E= 8.8510 (F/M)
spherical surface = EA = (
=>>
E
Ein
=
q
4πr
乞。
q
4πE. r
。
Gaussian surfaces
-2R and M
F(R)
4
Inside the sphere, the
electric field is zero
Oute
the square of the radial distance
E = -4
ER

ページ9:

L
• uniformly sphere :
(r<R) p=
(TCR 3)
.
Q end = Vout = (478 3/3) (*)
E = EA = E (4x+7)=
=
=>>
E =
Qr
E. R
4RE.R
↳ line charge : E
wo sheet of charge :
= EA = E (2πrl)- Qul =
E
入
2πe, r
E = E (2A) =
Qeal
⇒
E - 26.
L
→ parallel plate :
= 0
E.
Conductor
charge 分布于表面
空腔內側生感应电荷
E₁₁ = 0
field at surface:
E =
(σ = 1)
十六
Electric potential
Electric potential energ
-Spherical insulator
EUR 0
Awe R
E=
Or
4me R
Gaussian
surface
Eα
E r
Gaussian
surface
surface
(a) Solid conductor with charge c
(b) The same conductor with an internal cavity
"
Arbitrary
Gaussian
surface A
Cavity
conductor
The charge qetesides emirely on the face of
the conductor. The sinution is clectic,
E-0 within the conduct
Became at all points within the conductor
the electie field at all pets on the G
face me be rero
(c) As isold charge placed in the cavity
For Eto be ao at all posts on the Gaussian
surface the face of the cavity
have
W--AU (保守力)
(a) q and go have the same sign.
(b) and q have opposite signs.
K + U = 定
U = f = q V
Electric potential :
V =
U
q
(J=C+V)
KQ
=
(Volt = 5/c)
r
AV = V₁ - V₁₂ = -SĒJ
potential gradient
:
U
®
U>O
Asr-0, U+x,
Asr, U→0
U
As 0, U
Asr, U-0.
+
+
E是V 的负梯度
(负梯度:V/n=-= % )
ax
Fe
力:Fe
kQ g
E=
=
场:E =
kQ
Fe- & E
AUF AT
EU =
r
有土
AV = Ed
有
Ve
v =
位:V = RQ
Ve-V
r
19
4760
E=0
v
19
ATE R
19
4mer

ページ10:

• solenoid:
匝数
B-II
Bout = 0
Ampere's law
-4a-3a-2a
Pl
a 2a 3a 4a
L
• toroidal solenoid:
B = NI
2πr
special
=
case:
=>
B = L
circle path around long straight conductor
(1) ₤ B. d = μI
(逆时针⇒+)
(2)沒闭合导线: fB.17 =。
+
Electromagnetic induction
induced electromotive force.
Faraday's law
E -N
doR
It
Lenz's law :
B
B
di
changing
=>
induced emf (E)
=
induced current
安右手
产生抵抗。改变方之感应电流
motional electromotive force.
导体切割磁力线
(a) Isolated moving rod
(b) Rod connected to stationary conductor
X
B
B
×
×
× ×
EN
FEAT
× X x X
X
x
E = lv B » 1- - B
X
×
x
bo
x
x
X
F-qE
1=
F-1/B
X
Maxwell's equation
。 Gauss's law for Ĕ :
• Gauss's law for B :
B· JA = 0
© Ampere's law § 3. dl = µ₁Ime = pubic+ io)] = ƒ(ic+ €4
Faraday's law ₤E-11--4-
= E
Path for
Ampere's law
Bulging surface
Plane
surface
0
Path 2
Path 3
El = lv B
Displacement
current

ページ11:

十七་
Magnetic
.
No magnetic monopoles !
The oth pole is actually
The cath's mugott
offringeographic
封闭曲线
Magnetic field (B): (Tecla - N/A-m)
magnetic
field lines =
Magnetic force :
.
FB TE I (BLF)
F
=
x
④右于开掌
→圆周运动: F= &B = m - F
current car
mv
1818
carrying
Magnetic flux (P₂) :
conductor: FIX B
.
。
(曲线只看超終
=
B. A
=
(Wb Tm=N/A)
close surface
f
Current
.
loop
Magnetic
moment
: =IA
U--μ· B
十八
current segment
Magnetic field
moving charge:
'
law of
11x7
M11x
Biot - Savart
4ㄦ r
4π r
A 1
4%
llo - 4π × 10-7 (Wb/Aim - T-m/A · N-s/c)
L
↳o long.
.
wire
M.I
B = [1
parallel conductors F = LUB = =
L
coil
B =
M. Ia'
2(x²+α)
=
Vxi
4π
r
M.I.I
2πr
同向吸
A

ページ12:

1. When an object is executing simple harmonic motion, the acceleration is a
maximum when the displacement of the object is a maximum.
O 。
○ ×
19.When an object is executing simple harmonic motion, the velocity is a
maximum when the displacement of the object is a maximum.
C °
O
x
6.0 m and mass M 90 kg rests
15m and equidistant from the
A uniform plank of length L.
on sawhorses separated by D
center of the plank. Cousin Throckmorton wants to stand on the
right-hand end of the plank. If the plank is to remain at rest, how
massive can Throckmorton be?
质心
8-150-
tge
XCM =
Mco)+m()
M+m
D
=
2M+m)
2
6m 15 (90+m)
m-30
16 If we double only the mass of a vibrating ideal mass-and-
spring system, the mechanical energy of the system increases by
a factor of 2.
○ °
O x
E=AA
(X) A proton is placed in a uniform electric field. The direction of the
acceleration of this proton due to this field is opposite to the electric field.
--+
dipole
moment
electric field : +→-
2. A spring is mounted horizontally, with its left end fixed. A spring balance attached to the free end
and pulled toward the right indicates that the stretching force is proportional to the displacement, and
a force of 8.0 N causes a displacement of 0.040 m. We replace the spring balance with a 0.50-kg
glider, pull it 0.030 m to the right along a frictionless air track, and release it from rest. (a) Find the
force constant k of the spring (b) Find the angular frequency of the resulting oscillation (c) Find the
magnitude of the maximum acceleration of the glider. (d) Find the magnitude of the maximum
velocity of the glider. (e) Find the kinetic energy of the glider at x-0.01 m
(a)
=
N = 980 (W)
8 = 0.04 k
k. 200
(b)
W =
20
(ra)
Sir Lancelot, who weighs 800 N, is assaulting a castle by climbing
a uniform ladder that is 5.0 m long and weighs 180 N (Fig. 11.9)
The bottom of the ladder rests on a ledge and leans across the moat
in equilibrium against a frictionless, vertical castle wall. The ladder
makes an angle of 53.1" with the horizontal. Lancelot pauses one-
third of the way up the ladder. (a) Find the normal and friction forces
on the base of the ladder. (b) Find the minimum coefficient of static
friction needed to pervent slipping at the base. (c) Find the magni
tude and direction of the contact force on the base of the ladder.
f
(a)
(b)
·x: f- N'=o
J: N-180-800=0
Σ14N' =
800 + 1.5.180
N = 267.5 (N) - f
f=μN-980 μ-267.5
M = 0.27
(c)
√780" + 267.5² = 1·15.85
are sin()74.25"
A block of mass M attached to a horizontal spring with force con-
stant k is moving in SHM with amplitude A₁. As the block passes
through its equilibrium position, a lump of putty of mass mis
dropped from a small height and sticks to it. (a) Find the new
amplitude and period of the motion. (b) Repeat part (a) if the putty
is dropped onto the block when it is at one end of its path.
Two equal postive chap 20 C are located a
desert on a hard chap 0-40Cx 040-07
wwwwwww
wwwwwwww.
(a) T = 2π
M+m
k
文
(0)
a--w`x = -20°-0-03 - -12 ("/")
(d) VmWA =
20 x 0.03 0.6 (~/s)
(e) K = A²±²² = 200 (005-001) -0.08 (1)
E=m=A
MV, +0 = M +m) V₂
1 V₁ = WA₁ = √ A
E₁ = E₁₂
与抵场
Freak
0.30m
650m
20.6712 (N)
Q-40C
0.30ml
050
-20 CY
m
A₁ =
√M+n
A₁
A
(b)
unchange