law of orbits - lsu...a cubical block of something, 10.0 cm on a side, floats at the interface...

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Page 1: Law of Orbits - LSU...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
Page 2: Law of Orbits - LSU...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

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

Page 3: Law of Orbits - LSU...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

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…

Page 4: Law of Orbits - LSU...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

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

Page 5: Law of Orbits - LSU...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

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.

Page 6: Law of Orbits - LSU...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

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

Page 7: Law of Orbits - LSU...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

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

Page 8: Law of Orbits - LSU...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

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

Page 9: Law of Orbits - LSU...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

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( )

Page 10: Law of Orbits - LSU...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

PreviousExamProblem

ptop Atop = pbottomAbottom

What we know: Ftop = Fbottom

ptop πR22( ) = ptop + ρwatergh( ) πR12( )

ptop π R1 + h cot60( )2( ) = ptop + ρwatergh( ) πR12( )

Page 11: Law of Orbits - LSU...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

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

Page 12: Law of Orbits - LSU...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

Chapter 15 : Oscillations

Mo2onsthatRepeatthemselvesinawellprescribedway(i.e.period)

Swingaballbackandforthwithcertainfrequencyf

xKinemaIcs:posi2on:x(t)velocity:v(t)accelera2on:a(t)

Dynamics:drivingforceenergy

Movingamassmbackandforthonafric2onlesswithcertainangularfrequencyω

Page 13: Law of Orbits - LSU...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

OscillaIonscharacterisIcsFrequencyf(#ofoscilla2ons/second)‐‐Units:hertz=1Hz=1oscilla2on/s=1s‐1

PeriodT(2meforoneoscilla2on‐cycle)=1/f

2xm 1period=1cycle

Don’tforgettoputyourcalculatorin“RADIAN”mode

Page 14: Law of Orbits - LSU...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

“SimpleHarmonicMoIon(SHM)”‐‐KinemaIcs

PosiIon

Velocity

AcceleraIon

InSHM,theaccelera2onispropor2onaltothedisplacementbutoppositesign.Thepropor2onalityisthesquareoftheangularfrequency.

Page 15: Law of Orbits - LSU...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

RelaIonshipsforSHM

Howtodeterminexmaxandφ:INITIALCONDITIONS

Knowingx(0)andv(0),

Page 16: Law of Orbits - LSU...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

SHMandHooke’sForceLaw(again…)‐‐Dynamics

“InSHM,theaccelera2onispropor2onaltothedisplacementbutoppositesign.Thepropor2onalityisthesquareoftheangularfrequency”.

Block‐springislinearSHO:

Largermass‐>longerperiod

NOTE:Independentofamplitude

Page 17: Law of Orbits - LSU...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

Amass(m)isahachedtooneendofaspring(k)andisfreetoslideonafric2onlesshorizontalsurface.Themassispulledbackfromitsequilibriumposi2onandreleasedfromrest.Whatistheaccelera2onofthemassattheequilibriumposi2on(x=0)?

1.  Atitsmaximumvalue

2.  Itdependsonthespringconstant3.  Itdependsonthemass

4.  Zero

Question 15-1

Page 18: Law of Orbits - LSU...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

Example Ablockwhosemassmis680gisfastenedtoaspringwhoseconstantkis65N/m.Theblockispulledadistancex=11cmfromitsequilibriumposiIonatx=0onafricIonlesssurfaceandreleasedfromrestatt=0.

(a)Whataretheangularfrequency,frequency,andperiodoftheresulIngmoIon?

(b)WhatisthephaseangleandamplitudeoftheoscillaIon?

(c)WhatisthemaximumspeedandacceleraIon?