- •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
Week 5: Torque and Rotation in One Dimension |
243 |
5.3: The Moment of Inertia
We begin with a specific example to help smooth the way.
Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End
y |
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pivot |
dm = λ dx |
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x |
L |
x |
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dx |
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Figure 63: A solid rod of length L with a mass M uniformly pivoted about one end. One can think of such a rod as being the “massless rod” of the previous section with an infinite number of masses mi uniformly distributed along its length, that sum to the total mass M .
In figure 63 above a massive rod pivoted about one end is drawn. We would like to determine how this particular rod will rotationally accelerate when we (for example) attach a force to it and apply a torque. We therefore must characterize this rod as having a specific mass M , a specific length L, and we need to say something about the way the mass is distributed, because the rod could be made of aluminum (not very dense) at one end and tungsten (very dense indeed) at the other and still “look” the same. We will assume that this rod is uniformly distributed, and that it is very thin and symmetrical in cross-section – shaped like a piece of wire or perhaps a wooden dowel rod.
In a process that should be familiar to you from last week and from the previous section, we know that the moment of inertia of a sum of discrete point masses hung on a “massless” rod (that only serves to assemble them into a rigid structure) is just:
X |
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Itot = miri2 |
(477) |
i |
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the sum of the moments of inertia of the point masses.
We can clearly approximate the moment of inertia of the continuous rod by dividing it up into N
pieces, each of length x = L/N and mass |
M = M/N , and treating each small piece as a “point |
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mass” located at xi = i x: |
N |
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µ |
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N |
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I |
M |
i L |
= |
M x2 |
(478) |
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rod ≈ i=1 |
N |
N |
i=1 |
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¶ |
i |
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X |
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X |
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As before, the limit of this sum as N → ∞ is by definition the integral, and in this limit the sum exactly represents the moment of inertia of the rod.
We can easily evaluate this. To do so, we chant our ritual expression: “The mass of the chunk is the mass per unit length of the chunk times the length of the chunk”, or dm = λdx = ML dx, so:
Irod = Z0 |
x2dm = L Z0 |
x2dx = |
3 |
(479) |
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L |
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M |
L |
M L2 |
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5.3.1: Moment of Inertia of a General Rigid Body
This specific result can easily be generalized. If we consider a blob-shaped distribution of mass, the di erential moment of inertia of a tiny chunk of the mass in the distribution about some fixed axis
244 |
Week 5: Torque and Rotation in One Dimension |
pivot axis
dm
r
Figure 64: A “blob-shaped” chunk of mass, perhaps a piece of modelling clay, constrained to rotate about an axis through the blob, perhaps a straight piece of nearly massless coat-hanger wire.
of rotation is clearly:
dI = r2dm |
(480) |
By now you should be getting the idea that summing up all of the little chunks that make up the
object is just integrating:
Z
Iblob = r2dm (481)
blob
where it is quite one thing to write down this formal expression, quite another to be able to actually do the integral over all of the chunks of mass that make up an object.
It isn’t too di cult to do this integral for certain simple distributions of mass, and we will need a certain “stock repertoire” of moments of inertia in order to solve problems. Which ones you should learn to do depends on the level of the course – math/physics majors should learn to integrate over spheres (and maybe engineers as well), but everybody else can probably get away learning to evaluate the moment of inertia of a disk. In practice, for any really complicated mass distribution (like the blob of clay pictured above) one would either measure the moment of inertia or use a computer to actually break the mass up into a very large number of discrete (but small/point-like) chunks and do the sum.
First let’s do an example that is even simpler than the rod.
Example 5.3.2: Moment of Inertia of a Ring
z
M
dθ R
ds = Rd θ
Figure 65: A ring of mass M and radius R in the x-y plane rotates freely about the z-axis.
We would like to find the moment of inertia of the ring of uniformly distributed mass M and radius R portrayed in figure 65 above. A di erential chunk of the ring has length ds = R dθ. It’s mass is thus (say the ritual words!):
dm = λds = |
M |
R dθ = |
M |
dθ |
(482) |
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2πR |
2π |
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Week 5: Torque and Rotation in One Dimension |
245 |
and its moment of inertia is very simple:
Iring = Z |
r2dm = Z0 |
2π R2 |
dθ = M R2 |
(483) |
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2π M |
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In fact, we could have guessed this. the axis of rotation, so its moment of and has no “vector” character) is just
All of the mass M in the ring is at the same distance R from inertia (which only depends on the mass times the distance M R2 just like a point mass at that distance.
Because it is so important, we will do the moment of inertia of a disk next. The disk will be many things to us – a massive pulley, a wheel or tire, a yo-yo, a weight on a grandfather clock (physical) pendulum. Here it is.
Example 5.3.3: Moment of Inertia of a Disk
dA = rdθdr
rdθ
M |
dr |
|
dθ
R r
Figure 66: A disk of mass M and radius R is pivoted to spin freely around an axis through its center.
In figure 66 a disk of uniformly distributed mass M and radius R is drawn. We would like to find its moment of inertia. Consider the small chunk of disk that is shaded of area dA. In plane polar coordinates (the only ones we could sanely hope to integrate over) the di erential area of this chunk is just its di erential height dr times the width of the arc at radius r subtended by the angle dθ, r dθ. The area is thus dA = r drdθ.
This little chunk was selected because the mass dm in it moves in a circle of radius r around the pivot axis. We need to find dm in units we can integrate to cover the disk. We use our litany to set:
dm = σdA = |
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M |
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r drdθ |
(484) |
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πR2 |
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and then write down: |
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dI = r2dm = |
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M |
r3 drdθ |
(485) |
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πR2 |
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We integrate both sides to get (note that the integrals are independent one dimensional integrals that precisely cover the disk):
Idisk = |
πR2 |
à |
0 |
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r3 dr! |
0 |
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dθ |
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M |
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Z |
R |
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µZ |
2π |
¶ |
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= |
πR2 |
µ |
4 |
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¶ |
(2π) |
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M |
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R4 |
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= |
1 |
M R2 |
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(486) |
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2 |
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This is a very important and useful result, so keep it in mind.