Embed Size (px)
Citation preview
Midterm 2 review
H ill b t d• How you will be seated• What you need to bring• Preparation of formula sheet• Exam Strategy• Review
PHY 231 1
Exam• Prepare formula sheet: one 8.5x11 sheet – both sides if needed• Bring sheet , pens or pencils, and a calculator to this room during the
normal lecture time. Also bring your student ID.• You will be seated in a specific row. A seating chart will be displayed p g p y
on the screen.• You will be given a personalized exam with attached scratch paper.• We will collect exam bubble sheet and scratch paper when you handWe will collect exam, bubble sheet, and scratch paper when you hand
in the exam. We will also check your ID. • We will insist on the time limit. Please do not force us to take the exam
from youfrom you. • Don’t forget to do the corrections set.
formula sheet
H f ll h d f l h t d d it t l• Hopefully you have prepared a formula sheet and used it to solve review problems.
• If you have not, you can select formulae from the first pages of each f l t d dd f l t if fi d it f l fof my lectures and add formulae or notes if you find it useful for
solving problems.
Physics 231 Lecture 11
• Main points of today’s lecture:• Examples of energy and power• Examples of energy and power• Impulses: forces that last only a
short time• Moment m• Momentum
• Impulse-Momentum theorem( )F
ΔΔΔ
vp m=
• Momentum conservation
( )ifmm vvvptF −=Δ=Δ=Δ
ffftot ,2,1, ppp +≡
• Momentum and external forcesitotii ,,2,1 ppp ≡+=
t ppF −=Δ itotftotext t ,, ppF −=Δ
Physics 231 Lecture 12
• Impulses• Momentum conservation tot,f 1,f 2,f 1,i 2,i tot ,i≡ + = + ≡p p p p p p
Main points of this lect.
• Isolated systems• Totally inelastic collisions ( ) ( )f 1 1 2 2 1 2v m m m m= + +v v
M i i t f t d ’ l t
Physics 231 Lecture 13
• Main points of today’s lecture:• Elastic collisions in one
dimension:( ) 2mmm −( )
( )10
120
122
2021
210
21
211
2
2
vmm
mvmmmmv
vmm
mvmmmmv
f
f
++
+−=
++
+−=
• Multiple impulses and rocket propulsion.
2121 mmmm ++
propellant f 0F t m v m vΔ = Δ ⋅ − Δ ⋅
( )
( )
propellant f 0
propellant f 0mF v v Thrustt
mTh
Δ= − = −Δ
Δ
• Center of Mass ( )0 f
mThrust v vt
Δ= −Δ
tot cm i ii
M X m x= tot cm i ii
M Y m y=i i
tot cm tot i ii
M V P m v= =
• Main points of today’s lecture:
Physics 231 Lecture 14
• Multiple impulses and rocket propulsion.
propellant f 0F t m v m vΔ = Δ ⋅ − Δ ⋅
( )
( )
propellant f 0mF v v Thrustt
mThrust v v
Δ= − = −Δ
Δ= −
• Things that oscillate • Mass and Spring
( )0 fThrust v vt
= −Δ
• Circular motionT 1/ f ; f 1/ T; 2 fω π= = =
2for mass and spring k / m ω =• Simple harmonic motion
0
0
x A cos( t )v A sin( t )
ω θω ω θ
= += − +
p g
x 02
x 0
0
v A sin( t )
a A cos( t ) and A are constants
ω ω θω ω θ
θ
+
= − +
• Main points of today’s lecture:
Physics 231 Lecture 15
• Main points of today’s lecture:• Simple harmonic motion
• Mass and Spring• Pendulum• Circular motionT 1/ f ; f 1/ T; 2 fω π= = =
x A cos( t)ω=
; ;2for mass and spring k / m ω =
x2
x
v A sin( t)
a A cos( t)A i t t
ω ωω ω
= −
= −A is a constant
( )2for pendulum g / L
tω
θ θ=
( )max= cos tθ θ ω
• Main new points of today’s
Physics 231 Lecture 16p y
lecture:• Description of circular motion in
cylindrical coordinates (r,θ):y ( , )
( ) ( )
2 2 1 yr x y ; tanx
x r cos ; y r sin
θ
θ θ
− = + =
= =• First equations for circular
motion:
( ) ( )x r cos ; y r sinθ θ= =
θ θΔ Δt 0; lim
t tif is in radians
θ θω ω
θω θ
Δ →Δ Δ= =Δ Δ
Δ Δ
• Centripetal acceleration:
tv r; s r ;ω θ= Δ = Δ
v2 vt
rvac =
vt
ac
• Main points of today’s lecture:Physics 231 Lecture 17
• Centripetal acceleration:
va2
c =vt
• Rotational motion definitions:limω ωα αΔ Δ
rac ac
• Rotational kinematics equations:θΔ
t 0
t t
, limt t
s r; v r;a r
α α
θ ω α
Δ →= =Δ Δ
Δ = Δ = =
( ) ( )ωωθωωω
αωωωθ
+=Δ+=
+==Δ
21
21
00
0
t
tt
• Rolling motion:θΔ==Δ rsx
θαωωαωθ Δ+=+=Δ 2 21 2
022
0 tt
α=ω= ra ;rv
M i i f d ’ l
Physics 231 Lecture 17
• Main points of today’s lecture:• Rotational motion definitions:
θωθω Δ=Δ= lim
tt
tt
t
t
ωαωα
ωω
ΔΔ=
ΔΔ=
Δ=
Δ=
→Δ
→Δ
lim,
lim,
0
0
• Rotational kinematics equations:αωωωθ +==Δ tt
rarvrstt
tt αωθ ==Δ=ΔΔΔ
;;
( ) ( )ωωθωωω
αωωωθ
+=Δ+=
+==Δ
121
21
00
0
t
tt
• Rolling motion:θΔ==Δ rsx
θαωωαωθ Δ+=+=Δ 2 21 2
022
0 tt
α=ω= ra ;rv
• Main points of today’s lecture:
Physics 231 Lecture 18
• Main points of today s lecture:• Rotational motion definitions:
t 0, limω ωα α Δ →Δ Δ= =Δ Δ
• Rotational kinematics equations:
t 0
t t
,t t
s r; v r;a rθ ω α
Δ →Δ ΔΔ = Δ = =
q
( ) ( )ωωθωωω
αωωωθ
+=Δ+=
+==Δ
21
21
00
0
t
tt
• Rolling motion:θαωωαωθ Δ+=+=Δ 2
21 2
022
0 tt
• Torqueα=ω=
θΔ==Δra ;rv
rsx
Fd I ; d is the lever armτ α= =q• Equilibrium
;0 ;0 == iiF τ
M i i f d ’ l
Physics 231 Lecture 19
• Main points of today’s lecture:• Torque
armlevertheisd;Fdτ
• Newton’s second law for rotations:
armlever theisd;Fd=τ
• Equilibrium:
ατ I=
0;0 == F τ
• Center of mass:0 ;0 == iiF τ
iirm
=
ii
iii
cm mr
Physics 231 Lecture 20
• Main points of today’s lecture:• Analogies between rotational and
translational motiontranslational motion• Solutions involving:
I
ω
ατ
IKE
I
rot =
=
2
21
• Rolling motion:
ωIL =2
Rolling motion:
22
21
21 ωIMvKEtotal +=
M i i f d ’ l
Physics 231 Lecture 21
• Main points of today’s lecture:• Angular momentum:
L Iω=• Newton’s law of universal
gravitation:GMmF GMm
• Kepler’s laws and the relation
2rGMmF =
rGMmPE −=
• Kepler s laws and the relation between the orbital period and orbital radius.
32
2 4 rGM
T
= π
• Main points of today’s lecture:
Physics 231 Lecture 23
• Gravitation potential energy
1 2grav
GM MPEr
= −
• Tensile stress and strainLYF Δ=Δ
12r
• Bulk stress and strain:0LA
VBPF ΔΔΔ
• Pressure in fluids:F
VBP
A=Δ=
• Buoyancy
ghPPAFP topbot ρ+== ;
gVWF displacedfdisplacedB ρ==
Physics 231 Lecture 24
• Main points of today’s lecture:• Buoyancyy y
• Bernouli’s equation21P ++
gVWF displacedfdisplacedB ρ==
0200
2
21
2
gyvP
gyvP fff
ρρ
ρρ
++=
++
• Viscous flow
• Poisseuille’s law
2
dAvF η=
( )L
PPRtV
ηπ
821
4 −=ΔΔ
Physics 231 Lecture 25• Main points of today’s lecture:p y• Bernoulli’s equation
2
21 gyvP fff ρρ ++
• Viscous flow and Poisseuille’sl
0200 2
1 gyvP ρρ ++=
law
dAvF η=
( )L
PPRtV
ηπ
821
4 −=ΔΔ
• Transverse and longitudinal waves
• traveling waves λfvwave =• Wave speed for a string
μFv =
h.mass/lengt theis and tension string theis Here
μ
μF
• Main points of today’s lecture:Physics 231 Lecture 26
• Viscous flow and Poisseuille’slaw
dAvF η=
( )L
PPRtV
ηπ
821
4 −=ΔΔ
• Transverse and longitudinal waves
• traveling waves λfvwave =
d η
g• Wave speed for a string
μFv =
h.mass/lengt theis and tension string theis Here
μ
μF
• Superposition and interference of waves; Reflection of waves.
• Sound waves:
I
• Sound intensity:• I=P/A, I0=1x10-12 W/m2
=
010log10
IIβ
Problem solving strategy
L k th h th d d th i t bl fi t• Look through the exam and do the easiest problems first.• On each problem, you can use the following “fool proof” procedure to
solve the problem:– Read the problem carefully.– If necessary, draw a picture.– Identify the physics principles that are involved in the problem.y p y p p p– Identify the known values given in the problem and unknowns
values you are expect to find.– Choose the appropriate equations that can be used to get theChoose the appropriate equations that can be used to get the
unknown you are suppose to find.– Count the number of equations and verify that the number of
equations is at least equal to the number of unknown quantities thatequations is at least equal to the number of unknown quantities that are in the equations.
– Solve.
Review
I h l t d 11 bl th t ill t t th i i l th t• I have selected 11 problems that illustrate the principles that are involved for problem sets 1-4.
• If I finish these problems, I will accept requests for additional bl ti llproblems as time allows.
• An answer key will be posted on the Physics 231 web site after the lecture for section 3
PHY 231 23
PHY 231 24
PHY 231 25
PHY 231 26
PHY 231 27
PHY 231 28
PHY 231 29
PHY 231 30
PHY 231 31
PHY 231 32
PHY 231 33
PHY 231 34
PHY 231 35
PHY 231 36
PHY 231 37
PHY 231 38
