1) Law of Orbits - all planets move in elliptical orbits with the Sun at one focus
2) Law of equal areas - A line that connects a planet to the Sun sweeps out equal areas in equal time.
3) The Law of Periods : The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit
€
⇒ rsat3 =
GM4π 2
T 2
⇒ T 2 =4π 2
GM
rsat
3
ANGULAR MOMENTUM IS CONSERVED
€
pat h = p0 + ρgh
Pressure applied to a confined fluid increases the pressure throughout by same amount
Pascal’s Principle
€
pout = pin
Fout = Fin
Aout
Ain
Demo…
BuoyancyandArchimedes’PrincipleBuoyant Force - is equal to the weight of fluid displaced by the object
- is directed UPWARDs
- Does not depend on the shape of the object ONLY volume.
- Applies to partially or completely immersed object
€
FB = mFgUPWARD
A stone drops
Wood will rise
A “bag” of water stays put
€
FB −mg( ) < 0 a→ downward
€
FB −mg( ) > 0 a→ upward €
FB −mg( ) = 0 a = 0
BuoyancyandArchimedes’PrincipleBuoyant Force - equal to the weight of fluid displaced by the object
- is directed UPWARDs - Does not depend on shape of object - ONLY volume. - Applies to partially or completely submerged object
€
FB = mFgUPWARD
Which has larger buoyant force?
If the volumes are the same, they displace the
same mass of fluid so the buoyant forces are the
same
The answer is 2. What matters is the mass of the displaced fluid.
Sink or float? A 200-ton ship is in a tight-fitting lock so that the mass of fluid left in the lock is much less than the mass of the ship. Does it float?
1. No. The ship touches the bottom since it weighs more than the water.
2. Yes, as long as the water gets up to the ship’s waterline.
Youareonadistantplanetwheretheaccelera2onduetogravityishalfthatonEarth.Wouldyoufloatmoreeasilyinwateronthatplanet?
1. Yes,youwillfloathigher
2. Floa2ngwouldnotbechanged.3. No,youwillfloatdeeper.
Clicker Question
FBD of system:
Weight (downward) = mobject”g” Buoyant force (upward) = mdisplaced”g”
FB
W
Archimedes’Principle:ApparentweightinafluidWeigh in air, then weigh in water. From this, and knowing ρwater you can get object’s ρ.
€
Weightapparent =Weightactual − FB
Is it gold???
€
Weightapparent =Weightactual − FB= ρcrownVg − ρwaterVg
€
Wactual
Wactual −Wapparent
=ρcrownVgρwaterVg
=ρcrownρwater
Specific gravity
€
14.7 ⋅ kg14.7 ⋅ kg−13.4 ⋅ kg
=11.3 ???
Specific gravities: Gold (Au) 19.3 Lead (Pb) 11.3
Archemedes’ : Is the King’s crown gold??? (Hiero III 306-215 BC)
Because of buoyant force, apparent weight in water is less than actual weight €
ma
€
mg
€
FB
=
€
13.4 =14.7 − FB
ρwaterVg =1.3N V =1.3×10−4m3
ρcrown =14.7Vg
=11300kgm3
22.422400Os
11.411400Pb
8.98960Cu
7.87800Fe
2.72700Al
11000water
0.0011.21air
Specificgravityρ(kg/m3)Material
SpecificGravity
€
specific gravity =ρ
ρWATER
Rather than using the large SI unit it is sometimes convenient to use specific gravity
HeliumBlimpsLength 192 feet Width 50 feet Height 59.5 feet Volume 202,700 cubic feet (5740 m3) Maximum Speed 50 mph Cruise Speed 30 mph Powerplant: Two 210 hp fuel-injected, air-cooled piston engines
What is the maximum load weight of blimp (WL) in order to fly?
At static equilibrium
€
F = 0 : ∑ WHe +WL = FB
WL = FB −WHe
€
WL = mairg −mHeg
“Maximum Gross Weight 12,840 pounds”
€
= (5740 ⋅m 3 )(9.8 ⋅m /s 2 ) 1.21− 0.179 ⋅ kg /m 3( )= 58 ⋅ kN =13,000 ⋅ lbs
€
ρHe = 0.179 ⋅ kg /m 3 & ρair = 1.21⋅ kg /m 3
€
WHe
€
WL€
FB
The “fluid” blimp is in is:
air
€
= ρairVship g − ρHeVshipg
=Vshipg ρair − ρHe( )
PreviousExamProblem
€
ptop Atop = pbottomAbottom
What we know: Ftop = Fbottom
€
ptop πR22( ) = ptop + ρwatergh( ) πR12( )
€
ptop π R1 + h cot60( )2( ) = ptop + ρwatergh( ) πR12( )
A cubical block of something, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below the interface (see the cross section view in the figure). The density of the oil is 790 kg/m3. (a) What is the gauge pressure at the upper face of the block? (b) What is the gauge pressure of the lower face of the block? (c) What is the mass and density of the block?
Δptop− surface = ρoilgy0a)
Δplower− surface = ρoilghoil + ρwaterg(L − hy0 )b)
mblockg = Fbuoyant = FB (oil) + FB (water)
ρblockL3g = ρoilL
2y0g + ρwaterL2 (L − y0 )g
ρblock =y0 ρoil − ρwater( )
L+ ρwater
c)
Let the top be at a distance y0 from the top surface Let the length of a side be L Let h be distance from the top of the block and the interface Let hoil be the position of the interface.
ρblock =1.5 790 −1000( )
10+1000 = 968.5km / m3
PreviousExamProblem
Chapter 15 : Oscillations
Mo2onsthatRepeatthemselvesinawellprescribedway(i.e.period)
Swingaballbackandforthwithcertainfrequencyf
xKinemaIcs:posi2on:x(t)velocity:v(t)accelera2on:a(t)
Dynamics:drivingforceenergy
Movingamassmbackandforthonafric2onlesswithcertainangularfrequencyω
OscillaIonscharacterisIcsFrequencyf(#ofoscilla2ons/second)‐‐Units:hertz=1Hz=1oscilla2on/s=1s‐1
PeriodT(2meforoneoscilla2on‐cycle)=1/f
2xm 1period=1cycle
Don’tforgettoputyourcalculatorin“RADIAN”mode
“SimpleHarmonicMoIon(SHM)”‐‐KinemaIcs
PosiIon
Velocity
AcceleraIon
InSHM,theaccelera2onispropor2onaltothedisplacementbutoppositesign.Thepropor2onalityisthesquareoftheangularfrequency.
RelaIonshipsforSHM
Howtodeterminexmaxandφ:INITIALCONDITIONS
Knowingx(0)andv(0),
SHMandHooke’sForceLaw(again…)‐‐Dynamics
“InSHM,theaccelera2onispropor2onaltothedisplacementbutoppositesign.Thepropor2onalityisthesquareoftheangularfrequency”.
Block‐springislinearSHO:
Largermass‐>longerperiod
NOTE:Independentofamplitude
Amass(m)isahachedtooneendofaspring(k)andisfreetoslideonafric2onlesshorizontalsurface.Themassispulledbackfromitsequilibriumposi2onandreleasedfromrest.Whatistheaccelera2onofthemassattheequilibriumposi2on(x=0)?
1. Atitsmaximumvalue
2. Itdependsonthespringconstant3. Itdependsonthemass
4. Zero
Question 15-1
Example Ablockwhosemassmis680gisfastenedtoaspringwhoseconstantkis65N/m.Theblockispulledadistancex=11cmfromitsequilibriumposiIonatx=0onafricIonlesssurfaceandreleasedfromrestatt=0.
(a)Whataretheangularfrequency,frequency,andperiodoftheresulIngmoIon?
(b)WhatisthephaseangleandamplitudeoftheoscillaIon?
(c)WhatisthemaximumspeedandacceleraIon?