- •Preface
- •Textbook Layout and Design
- •Preliminaries
- •See, Do, Teach
- •Other Conditions for Learning
- •Your Brain and Learning
- •The Method of Three Passes
- •Mathematics
- •Summary
- •Homework for Week 0
- •Summary
- •1.1: Introduction: A Bit of History and Philosophy
- •1.2: Dynamics
- •1.3: Coordinates
- •1.5: Forces
- •1.5.1: The Forces of Nature
- •1.5.2: Force Rules
- •Example 1.6.1: Spring and Mass in Static Force Equilibrium
- •1.7: Simple Motion in One Dimension
- •Example 1.7.1: A Mass Falling from Height H
- •Example 1.7.2: A Constant Force in One Dimension
- •1.7.1: Solving Problems with More Than One Object
- •Example 1.7.4: Braking for Bikes, or Just Breaking Bikes?
- •1.8: Motion in Two Dimensions
- •Example 1.8.1: Trajectory of a Cannonball
- •1.8.2: The Inclined Plane
- •Example 1.8.2: The Inclined Plane
- •1.9: Circular Motion
- •1.9.1: Tangential Velocity
- •1.9.2: Centripetal Acceleration
- •Example 1.9.1: Ball on a String
- •Example 1.9.2: Tether Ball/Conic Pendulum
- •1.9.3: Tangential Acceleration
- •Homework for Week 1
- •Summary
- •2.1: Friction
- •Example 2.1.1: Inclined Plane of Length L with Friction
- •Example 2.1.3: Find The Minimum No-Skid Braking Distance for a Car
- •Example 2.1.4: Car Rounding a Banked Curve with Friction
- •2.2: Drag Forces
- •2.2.1: Stokes, or Laminar Drag
- •2.2.2: Rayleigh, or Turbulent Drag
- •2.2.3: Terminal velocity
- •Example 2.2.1: Falling From a Plane and Surviving
- •2.2.4: Advanced: Solution to Equations of Motion for Turbulent Drag
- •Example 2.2.3: Dropping the Ram
- •2.3.1: Time
- •2.3.2: Space
- •2.4.1: Identifying Inertial Frames
- •Example 2.4.1: Weight in an Elevator
- •Example 2.4.2: Pendulum in a Boxcar
- •2.4.2: Advanced: General Relativity and Accelerating Frames
- •2.5: Just For Fun: Hurricanes
- •Homework for Week 2
- •Week 3: Work and Energy
- •Summary
- •3.1: Work and Kinetic Energy
- •3.1.1: Units of Work and Energy
- •3.1.2: Kinetic Energy
- •3.2: The Work-Kinetic Energy Theorem
- •3.2.1: Derivation I: Rectangle Approximation Summation
- •3.2.2: Derivation II: Calculus-y (Chain Rule) Derivation
- •Example 3.2.1: Pulling a Block
- •Example 3.2.2: Range of a Spring Gun
- •3.3: Conservative Forces: Potential Energy
- •3.3.1: Force from Potential Energy
- •3.3.2: Potential Energy Function for Near-Earth Gravity
- •3.3.3: Springs
- •3.4: Conservation of Mechanical Energy
- •3.4.1: Force, Potential Energy, and Total Mechanical Energy
- •Example 3.4.1: Falling Ball Reprise
- •Example 3.4.2: Block Sliding Down Frictionless Incline Reprise
- •Example 3.4.3: A Simple Pendulum
- •Example 3.4.4: Looping the Loop
- •3.5: Generalized Work-Mechanical Energy Theorem
- •Example 3.5.1: Block Sliding Down a Rough Incline
- •Example 3.5.2: A Spring and Rough Incline
- •3.5.1: Heat and Conservation of Energy
- •3.6: Power
- •Example 3.6.1: Rocket Power
- •3.7: Equilibrium
- •3.7.1: Energy Diagrams: Turning Points and Forbidden Regions
- •Homework for Week 3
- •Summary
- •4.1: Systems of Particles
- •Example 4.1.1: Center of Mass of a Few Discrete Particles
- •4.1.2: Coarse Graining: Continuous Mass Distributions
- •Example 4.1.2: Center of Mass of a Continuous Rod
- •Example 4.1.3: Center of mass of a circular wedge
- •4.2: Momentum
- •4.2.1: The Law of Conservation of Momentum
- •4.3: Impulse
- •Example 4.3.1: Average Force Driving a Golf Ball
- •Example 4.3.2: Force, Impulse and Momentum for Windshield and Bug
- •4.3.1: The Impulse Approximation
- •4.3.2: Impulse, Fluids, and Pressure
- •4.4: Center of Mass Reference Frame
- •4.5: Collisions
- •4.5.1: Momentum Conservation in the Impulse Approximation
- •4.5.2: Elastic Collisions
- •4.5.3: Fully Inelastic Collisions
- •4.5.4: Partially Inelastic Collisions
- •4.6: 1-D Elastic Collisions
- •4.6.1: The Relative Velocity Approach
- •4.6.2: 1D Elastic Collision in the Center of Mass Frame
- •4.7: Elastic Collisions in 2-3 Dimensions
- •4.8: Inelastic Collisions
- •Example 4.8.1: One-dimensional Fully Inelastic Collision (only)
- •Example 4.8.2: Ballistic Pendulum
- •Example 4.8.3: Partially Inelastic Collision
- •4.9: Kinetic Energy in the CM Frame
- •Homework for Week 4
- •Summary
- •5.1: Rotational Coordinates in One Dimension
- •5.2.1: The r-dependence of Torque
- •5.2.2: Summing the Moment of Inertia
- •5.3: The Moment of Inertia
- •Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End
- •5.3.1: Moment of Inertia of a General Rigid Body
- •Example 5.3.2: Moment of Inertia of a Ring
- •Example 5.3.3: Moment of Inertia of a Disk
- •5.3.2: Table of Useful Moments of Inertia
- •5.4: Torque as a Cross Product
- •Example 5.4.1: Rolling the Spool
- •5.5: Torque and the Center of Gravity
- •Example 5.5.1: The Angular Acceleration of a Hanging Rod
- •Example 5.6.1: A Disk Rolling Down an Incline
- •5.7: Rotational Work and Energy
- •5.7.1: Work Done on a Rigid Object
- •5.7.2: The Rolling Constraint and Work
- •Example 5.7.2: Unrolling Spool
- •Example 5.7.3: A Rolling Ball Loops-the-Loop
- •5.8: The Parallel Axis Theorem
- •Example 5.8.1: Moon Around Earth, Earth Around Sun
- •Example 5.8.2: Moment of Inertia of a Hoop Pivoted on One Side
- •5.9: Perpendicular Axis Theorem
- •Example 5.9.1: Moment of Inertia of Hoop for Planar Axis
- •Homework for Week 5
- •Summary
- •6.1: Vector Torque
- •6.2: Total Torque
- •6.2.1: The Law of Conservation of Angular Momentum
- •Example 6.3.1: Angular Momentum of a Point Mass Moving in a Circle
- •Example 6.3.2: Angular Momentum of a Rod Swinging in a Circle
- •Example 6.3.3: Angular Momentum of a Rotating Disk
- •Example 6.3.4: Angular Momentum of Rod Sweeping out Cone
- •6.4: Angular Momentum Conservation
- •Example 6.4.1: The Spinning Professor
- •6.4.1: Radial Forces and Angular Momentum Conservation
- •Example 6.4.2: Mass Orbits On a String
- •6.5: Collisions
- •Example 6.5.1: Fully Inelastic Collision of Ball of Putty with a Free Rod
- •Example 6.5.2: Fully Inelastic Collision of Ball of Putty with Pivoted Rod
- •6.5.1: More General Collisions
- •Example 6.6.1: Rotating Your Tires
- •6.7: Precession of a Top
- •Homework for Week 6
- •Week 7: Statics
- •Statics Summary
- •7.1: Conditions for Static Equilibrium
- •7.2: Static Equilibrium Problems
- •Example 7.2.1: Balancing a See-Saw
- •Example 7.2.2: Two Saw Horses
- •Example 7.2.3: Hanging a Tavern Sign
- •7.2.1: Equilibrium with a Vector Torque
- •Example 7.2.4: Building a Deck
- •7.3: Tipping
- •Example 7.3.1: Tipping Versus Slipping
- •Example 7.3.2: Tipping While Pushing
- •7.4: Force Couples
- •Example 7.4.1: Rolling the Cylinder Over a Step
- •Homework for Week 7
- •Week 8: Fluids
- •Fluids Summary
- •8.1: General Fluid Properties
- •8.1.1: Pressure
- •8.1.2: Density
- •8.1.3: Compressibility
- •8.1.5: Properties Summary
- •Static Fluids
- •8.1.8: Variation of Pressure in Incompressible Fluids
- •Example 8.1.1: Barometers
- •Example 8.1.2: Variation of Oceanic Pressure with Depth
- •8.1.9: Variation of Pressure in Compressible Fluids
- •Example 8.1.3: Variation of Atmospheric Pressure with Height
- •Example 8.2.1: A Hydraulic Lift
- •8.3: Fluid Displacement and Buoyancy
- •Example 8.3.1: Testing the Crown I
- •Example 8.3.2: Testing the Crown II
- •8.4: Fluid Flow
- •8.4.1: Conservation of Flow
- •Example 8.4.1: Emptying the Iced Tea
- •8.4.3: Fluid Viscosity and Resistance
- •8.4.4: A Brief Note on Turbulence
- •8.5: The Human Circulatory System
- •Example 8.5.1: Atherosclerotic Plaque Partially Occludes a Blood Vessel
- •Example 8.5.2: Aneurisms
- •Homework for Week 8
- •Week 9: Oscillations
- •Oscillation Summary
- •9.1: The Simple Harmonic Oscillator
- •9.1.1: The Archetypical Simple Harmonic Oscillator: A Mass on a Spring
- •9.1.2: The Simple Harmonic Oscillator Solution
- •9.1.3: Plotting the Solution: Relations Involving
- •9.1.4: The Energy of a Mass on a Spring
- •9.2: The Pendulum
- •9.2.1: The Physical Pendulum
- •9.3: Damped Oscillation
- •9.3.1: Properties of the Damped Oscillator
- •Example 9.3.1: Car Shock Absorbers
- •9.4: Damped, Driven Oscillation: Resonance
- •9.4.1: Harmonic Driving Forces
- •9.4.2: Solution to Damped, Driven, Simple Harmonic Oscillator
- •9.5: Elastic Properties of Materials
- •9.5.1: Simple Models for Molecular Bonds
- •9.5.2: The Force Constant
- •9.5.3: A Microscopic Picture of a Solid
- •9.5.4: Shear Forces and the Shear Modulus
- •9.5.5: Deformation and Fracture
- •9.6: Human Bone
- •Example 9.6.1: Scaling of Bones with Animal Size
- •Homework for Week 9
- •Week 10: The Wave Equation
- •Wave Summary
- •10.1: Waves
- •10.2: Waves on a String
- •10.3: Solutions to the Wave Equation
- •10.3.1: An Important Property of Waves: Superposition
- •10.3.2: Arbitrary Waveforms Propagating to the Left or Right
- •10.3.3: Harmonic Waveforms Propagating to the Left or Right
- •10.3.4: Stationary Waves
- •10.5: Energy
- •Homework for Week 10
- •Week 11: Sound
- •Sound Summary
- •11.1: Sound Waves in a Fluid
- •11.2: Sound Wave Solutions
- •11.3: Sound Wave Intensity
- •11.3.1: Sound Displacement and Intensity In Terms of Pressure
- •11.3.2: Sound Pressure and Decibels
- •11.4: Doppler Shift
- •11.4.1: Moving Source
- •11.4.2: Moving Receiver
- •11.4.3: Moving Source and Moving Receiver
- •11.5: Standing Waves in Pipes
- •11.5.1: Pipe Closed at Both Ends
- •11.5.2: Pipe Closed at One End
- •11.5.3: Pipe Open at Both Ends
- •11.6: Beats
- •11.7: Interference and Sound Waves
- •Homework for Week 11
- •Week 12: Gravity
- •Gravity Summary
- •12.1: Cosmological Models
- •12.2.1: Ellipses and Conic Sections
- •12.4: The Gravitational Field
- •12.4.1: Spheres, Shells, General Mass Distributions
- •12.5: Gravitational Potential Energy
- •12.6: Energy Diagrams and Orbits
- •12.7: Escape Velocity, Escape Energy
- •Example 12.7.1: How to Cause an Extinction Event
- •Homework for Week 12
350 |
Week 8: Fluids |
Finally, we take the limit z → 0 and identify the definition of the derivative to get:
|
|
dP |
= ρg |
(710) |
|||
|
|
|
dz |
||||
|
|
|
|
|
|
|
|
Identical arguments but without any horizontal external force followed by x → 0 and |
y → 0 lead |
||||||
to: |
|
|
|
dP |
|
|
|
|
dP |
|
= |
= 0 |
(711) |
||
|
|
|
|
||||
|
dx |
|
dy |
|
as well – P does not vary with x or y as already noted136.
In order to find P (z) from this di erential expression (which applies, recall, to any confined fluid in static equilibrium in a gravitational field) we have to integrate it. This integral is very simple if the fluid is incompressible because in that case ρ is a constant. The integral isn’t that di cult if ρ is not a constant as implied by the equation we wrote above for the bulk compressibility. We will therefore first do incompressible fluids, then compressible ones.
8.1.8: Variation of Pressure in Incompressible Fluids
In the case of incompressible fluids, ρ is a constant and does not vary with pressure and/or depth. Therefore we can easily multiple dP/dz = ρg above by dz on both sides and integrate to find P :
dP |
= |
ρg dz |
|
Z |
|
Z |
|
dP |
= |
ρg dz |
|
P (z) |
= |
ρgz + P0 |
(712) |
where P0 is the constant of integration for both integrals, and practically speaking is the pressure in the fluid at zero depth (wherever that might be in the coordinate system chosen).
Example 8.1.1: Barometers
Mercury barometers were originally invented by Evangelista Torricelli137 a natural philosopher who acted as Galileo’s secretary for the last three months of Galileo’s life under house arrest. The invention was inspired by Torricelli’s attempt to solve an important engineering problem. The pump makers of the Grand Duke of Tuscany had built powerful pumps intended to raise water twelve or more meters, but discovered that no matter how powerful the pump, water stubbornly refused to rise more than ten meters into a pipe evacuated at the top.
Torricelli demonstrated that a shorter glass tube filled with mercury, when inverted into a dish of mercury, would fall back into a column with a height of roughly 0.76 meters with a vacuum on top, and soon thereafter discovered that the height of the column fluctuated with the pressure of the outside air pressing down on the mercury in the dish, correctly concluding that water would behave exactly the same way138. Torricelli made a number of other important 17th century discoveries, correctly describing the causes of wind and discovering “Torricelli’s Law” (an aspect of the Bernoulli Equation we will note below).
In honor of Torricelli, a unit of pressure was named after him. The torr is the pressure required to push the mercury in Torricelli’s barometer up one millimeter. Because mercury barometers
136Physics and math majors and other students of multivariate calculus will recognize that I should probably be using
partial derivatives here and establishing that ~ P = ρ~g, where in free space we should instead have had ~ P = 0 P
→
constant.
137Wikipedia: http://www.wikipedia.org/wiki/Evangelista Torricelli. ,
138You, too, get to solve Torricelli’s problem as one of your homework problems, but armed with a lot better
understanding.
Week 8: Fluids |
351 |
were at one time nearly ubiquitous as the most precise way to measure the pressure of the air, a specific height of the mercury column was the original definition of the standard atmosphere. For better or worse, Torricelli’s original observation defined one standard atmosphere to be exactly “760 millimeters of mercury” (which is a lot to write or say) or as we would now say, “760 torr”139.
Mercury barometers are now more or less banned, certainly from the workplace, because mercury is a potentially toxic heavy metal. In actual fact, liquid mercury is not biologically active and hence is not particularly toxic. Mercury vapor is toxic, but the amount of mercury vapor emitted by the exposed surface of a mercury barometer at room temperature is well below the levels considered to be a risk to human health by OSHA unless the barometer is kept in a small, hot, poorly ventilated room with someone who works there over years. This isn’t all that common a situation, but with all toxic metals we are probably better safe than sorry140.
At this point mercury barometers are rapidly disappearing everywhere but from the hands of collectors. Their manufacture is banned in the U.S., Canada, Europe, and many other nations. We had a lovely one (probably more than one, but I recall one) in the Duke Physics Department up until sometime in the 90’s141, but it was – sanely enough – removed and retired during a renovation that also cleaned up most if not all of the asbestos in the building. Ah, my toxic youth...
Still, at one time they were extremely common – most ships had one, many households had one, businesses and government agencies had them – knowing the pressure of the air is an important factor in weather prediction. Let’s see how they work(ed).
P = 0
H
P = P 0
Figure 104: A simple fluid barometer consists of a tube with a vacuum at the top filled with fluid supported by the air pressure outside.
A simple mercury barometer is shown in figure 104. It consists of a tube that is completely
139Not to be outdone, one standard atmosphere (or atmospheric pressures in weather reporting) in the U.S. is often given as 29.92 barleycorn-derived inches of mercury instead of millimeters. Sigh.
140The single biggest risk associated with uncontained liquid mercury (in a barometer or otherwise) is that you can
easily spill it, and once spilled it is fairly likely to sooner or later make its way into either mercury vapor or methyl mercury, both of which are biologically active and highly toxic. Liquid mercury itself you could drink a glass of and it would pretty much pass straight through you with minimal absorption and little to no damage if you – um – “collected” it carefully and disposed of it properly on the other side.
141I used to work in the small, cramped space with poor circulation where it was located from time to time but never very long at a time and besides, the room was cold. But if I seem “mad as a hatter” – mercury nitrate was used in the making of hats and the vapor used to poison the hatters vapor used to poison hat makers – it probably isn’t from this...
352 |
Week 8: Fluids |
filled with mercury. Mercury has a specific gravity of 13.534 at a typical room temperature, hence a density of 13534 kg/m3). The filled tube is then inverted into a small reservoir of mercury (although other designs of the bottom are possible, some with smaller exposed surface area of the mercury). The mercury falls (pulled down by gravity) out of the tube, leaving behind a vacuum at the top.
We can easily compute the expected height of the mercury column if P0 is the pressure on the exposed surface of the mercury in the reservoir. In that case
P = P0 + ρgz |
(713) |
as usual for an incompressible fluid. Applying this formula to both the top and the bottom, |
|
P (0) = P0 |
(714) |
and |
|
P (H) = P0 − ρgH = 0 |
(715) |
(recall that the upper surface is above the lower one, z = −H). From this last equation: |
|
P0 = ρgH |
(716) |
and one can easily convert the measured height H of mercury above the top surface of mercury in the reservoir into P0, the air pressure on the top of the reservoir.
At one standard atmosphere, we can easily determine what a mercury barometer at room temperature will read (the height H of its column of mercury above the level of mercury in the reservoir):
P0 = 13534 |
kg |
× 9.80665 |
m |
× H = 101325Pa |
(717) |
|
|
||||
m3 |
sec2 |
Note well, we have used the precise SI value of g in this expression, and the density of mercury at “room temperature” around 20◦C or 293◦K. Dividing, we find the expected height of mercury in a barometer at room temperature and one standard atmosphere is H = 0.763 meters or 763 torr
Note that this is not exactly the 760 torr we expect to read for a standard atmosphere. This is because for high precision work one cannot just use any old temperature (because mercury has a significant thermal expansion coe cient and was then and continues to be used today in mercury thermometers as a consequence). The unit definition is based on using the density of mercury at 0◦C or 273.16◦K, which has a specific gravity (according to NIST, the National Institute of Standards in the US) of 13.595. Then the precise connection between SI units and torr follows from:
P0 = 13595 |
kg |
× 9.80665 |
m |
× H = 101325Pa |
(718) |
|
|
||||
m3 |
sec2 |
Dividing we find the value of H expected at one standard atmosphere:
Hatm = 0.76000 = 760.00 millimeters |
(719) |
Note well the precision, indicative of the fact that the SI units for a standard atmosphere follow from their definition in torr, not the other way around.
Curiously, this value is invariably given in both textbooks and even the wikipedia article on atmospheric pressure as the average atmospheric pressure at sea level, which it almost certainly is not – a spatiotemporal averaging of sea level pressure would have been utterly impossible during Torricelli’s time (and would be di cult today!) and if it was done, could not possibly have worked out to be exactly 760.00 millimeters of mercury at 273.16◦K.
Week 8: Fluids |
353 |
Example 8.1.2: Variation of Oceanic Pressure with Depth
The pressure on the surface of the ocean is, approximately, by definition, one atmosphere. Water is a highly incompressible fluid with ρw = 1000 kilograms per cubic meter142. g ≈ 10 meters/second2. Thus:
or |
P (z) = P0 + ρwgz = ¡105 + 104z¢ |
Pa |
(720) |
|
P (z) = (1.0 + 0.1z) bar = (1000 + 100z) mbar |
(721) |
Every ten meters of depth (either way) increases water pressure by (approximately) one atmosphere!
Wow, that was easy. This is a very important rule of thumb and is actually fairly easy to remember! How about compressible fluids?
8.1.9: Variation of Pressure in Compressible Fluids
Compressible fluids, as noted, have a density which varies with pressure. Recall our equation for the compressibility:
P = −B |
V |
(722) |
V |
If one increases the pressure, one therefore decreases occupied volume of any given chunk of mass, and hence increases the density. However, to predict precisely how the density will depend on pressure requires more than just this – it requires a model relating pressure, volume and mass.
Just such a model for a compressible gas is provided (for example) by the Ideal Gas Law143 :
P V = N kbT = nRT |
(723) |
where N is the number of molecules in the volume V , kb is Boltzmann’s constant144 |
n is the number |
of moles of gas in the volume V , R is the ideal gas constant145 and T is the temperature in degrees Kelvin (or Absolute)146 . If we assume constant temperature, and convert N to the mass of the gas by multiplying by the molar mass and dividing by Avogadro’s Number147 6 × 1023.
(Aside: If you’ve never taken chemistry a lot of this is going to sound like Martian to you. Sorry about that. As always, consider visting the e.g. Wikipedia pages linked above to learn enough about these topics to get by for the moment, or just keep reading as the details of all of this won’t turn out to be very important...)
When we do this, we get the following formula for the density of an ideal gas:
ρ = |
M |
P |
(724) |
|
RT |
||||
|
|
|
where M is the molar mass148 , the number of kilograms of the gas per mole. Note well that this result is idealized – that’s why they call it the Ideal Gas Law! – and that no real gases are “ideal” for all pressures and temperatures because sooner or later they all become liquids or solids due to molecular interactions. However, the gases that make up “air” are all reasonably ideal at temperatures in the ballpark of room temperature, and in any event it is worth seeing how the pressure of an ideal gas varies with z to get an idea of how air pressure will vary with height. Nature will probably be somewhat di erent than this prediction, but we ought to be able to make a qualitatively accurate model that is also moderately quantitatively predictive as well.
142Good number to remember. In fact, great number to remember.
143Wikipedia: http://www.wikipedia.org/wiki/Ideal Gas Law.
144Wikipedia: http://www.wikipedia.org/wiki/Boltzmann’s Constant. ,
145Wikipedia: http://www.wikipedia.org/wiki/Gas Constant. ,
146Wikipedia: http://www.wikipedia.org/wiki/Temperature.
147Wikipedia: http://www.wikipedia.org/wiki/Avogadro’s Number.
148Wikipedia: http://www.wikipedia.org/wiki/Molar Mass.