Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body ...

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Lecture 23: MOI & Torque

Transcript of Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body ...

Page 1: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Lecture 23: MOI & Torque

Page 2: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Kinetic energy of Rotation

• K = sum of ½ m v2 for all parts of the body Moment of inertia I K = ½ I ω2

Page 3: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Example: Two objects connected with a massless rod I

• Mass of 1 kg at -2m, of 4kg at 1m I = 1kg (-2m)^2 + 4kg (1m)^2

= 8 kg m^2

• Different I around different axis!

• Example: rotate around midpoint:

I’ = 11.25 kg m^2

Page 4: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Post-lecture Exercise (10.1 – 10.6)  

• Two masses of 2 kg are connected by a massless 1m rod and rotated around their center of mass with a period of 2s. Calculate the rotational kinetic energy of this configuration.

• K =  9.87 J

• Use Eqs. (10-33) & (10-34): I = 2kg(-0.5m)^2+ 2kg(0.5m)^2 = 1 kg m^2

ω = 2π/T = π Hz, so K = ½ I ω ^2 = ½ π^2 J

Page 5: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Determining MOI

• Integral

• Table

• Parallel-axis theorem

Page 6: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Torque

• Torque is force times lever arm

• Lever arm is distance to rotation axis along a direction perpendicular to the force

• Later: τ = r x F

• | τ | = |r| |F| sin φ

Page 7: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Pre-lecture Exercise (10.7 – 10.10)

• In the simulation Torque how large does the red force have to be (if the red position is negative 1m and all other quantities at their initial values) such that the sum of the torques produced by the blue and the red forces is zero, i.e. that there is no net force, and hence no net angular acceleration, and hence no rotational motion of the bar? 

• Fred = 10N• Torque = force times lever arm. If the lever arm is half as long,

we need twice as much force: f= 10 N

Page 8: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Newton II for Rotation

Page 9: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Work and Rotational Kinetic Energy

• W = ∫ τ dθ

• P = dW/dt = τω

Page 10: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Lecture 24: General Rotations

Page 11: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Worksheet: Torque and angular acceleration

Page 12: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Rotation plus Translation

• Rolling is a combination of motion of the COM and rotation about the COM

• Point of contact remains stationary, while point on top of wheel moves with twice the velocity of the COM.

Page 13: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Rolling as pure Rotation

• Rolling can also be viewed as a pure rotation around the point of contact with floor

• Need parallel axis theorem to calculate correct MOI

• Two parts represent contributions from rotation and translation to KE

Page 14: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Forces of rolling

• Friction is static, since point of contact is stationary.

• Rolling down a ramp:– Can calculate is in linear coordinates: static

friction and mg sin β determine linear acc.– Torque due to static friction acting at radius R

determines angular acc.: torque = I α

Page 15: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Torque as a Vector

Page 16: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Pre-lecture Exercise (11.1 – 11.6)

• What is the direction of the torque produced by a force pointing in the SW direction and at a point 2m directly below the origin?

• a. SW b. Down

• c. NE d. NW• e. 45 degrees upward from West • f. None of the above.

• Answer: torque = r x F, where r is in –k, and F in –i – j direction, so –i+j or NW or d)

Page 17: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down due to

a force exerted on its edge. At the instant shown, the torque on the disc is pointing in …

• -z direction

• -y direction

• +y direction

• +z direction

z

x

yF

Page 18: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Reminder: Vector Product

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What is A x A ?

• Zero

• A

• -A

• A2

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Lecture 25: Angular Momentum

Page 21: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Angular Momentum

Page 22: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down due to

a force exerted on its edge. The angular momentum of the bug is pointing in …

• -z direction

• -y direction

• +y direction

• +z direction

z

x

yF

Page 23: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Pre-lecture Exercise (11.7-11.10) • What is the direction of the angular momentum (around

the origin) of a particle located 2m directly above the origin (i.e. 2m along the z axis) with a momentum vector in the SW (i.e. –i–j) direction?

• a. SW b. Down• c. NE d. NW• e. 45 degrees upward from West • f. None of the above.• Answer: l = r x p, where r is in k, and p in –i – j

direction, so i-j or SE so f)

Page 24: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Angular Momentum of Rigid Body about fixed axis

• L = sum over angular momenta of parts

• Use l = r p = m r v = m r (r ω)

L = I ω

Page 25: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Dynamics Transliteration

• Mass Moment of inertia

• Force Torque

• Momentum Angular Momentum

• Newton II

• Momentum Conservation Angular Momentum Conservation

Page 26: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Angular Momentum Conservation

• Angular momentum is conserved if no external net torque is present

• Demos: turntable + weights, turntable and bikewheel, bikewheel spinning on rope

Page 27: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Post-lecture Exercise (11.7 – 11.10)

• In the sample problem on page 285, what are the magnitude and direction of the net angular momentum L about point O of the two-particle system if the velocity of particle two is reversed (180 degrees direction change)? The direction is going to be either out of the page (positive L) or into the page (negative L).

• Answer: The only change is the sign of the vector l2, so L = (10 +8) kg m/s2= 18 kg m/s2

Page 28: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Lecture 26: Rotational Energy

Page 29: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Pre-lecture Exercise (11.10 – 11.12) • By what factor does the spinning volunteer’s

period of rotation (p. 291) change if he is able to reduce is moment of inertia by a factor of 1.5? (Hint: your answer should be smaller than one if the period is reduced, and bigger than one if the period gets longer.)

• Answer: I ω = const, so if I goes down by 1.5, ω goes up by 1.5, so the period goes down by 1.5, or f = 1/1.5 =2/3 = 0.6667.

Page 30: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Precession of a Gyroscope

• Demo: Little Gyroscope

• Demo: Bike Wheel on stick– Rate of precession becomes larger as wheel

slows down

Page 31: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Gravitation

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History

• Ptolemy

• Copernicus

• Brahe

• Galilei

• Kepler

• Newton

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From Galileo to Newton - the Birth of Modern Science

1609 1687

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Precursor: Nicolas Copernicus (1473–1543)

• Rediscovers the heliocentric model of Aristarchus

• Planets on circles needs 48(!!) epicycles to explain

different speeds of planets• Not more accurate than Ptolemy

Major Work : De Revolutionibus Orbium Celestium

(published posthumously)

Page 35: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Geocentric vs Heliocentric: How do we know?

• Is the Earth or the Sun the center of the solar system?

• How do we decide between these two theories?

• Invoke the scientific methods: – both theories make (different) predictions– Compare to observations– Decide which theory explains data

Page 36: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Phases of Venus

Heliocentric

Geocentric

Page 37: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Sunspots

• MPEG video from Galileo Project (June 2 – July 8, 1613)

Page 38: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Galileo and his Contemporaries

• Elizabeth I. (1533-1603) – Queen of England• Tycho Brahe (1546-1601) – Danish Astronomer• Francis Bacon (1561-1626) – English Philosopher• Shakespeare (1564- 1616) – Poet & Playwright• Galileo Galilei (1564-1642) – Italian PAM• Johannes Kepler (1571-1630) – German PAM• Rene Descartes (1596 - 1650) – French PPM• Christiaan Huygens (1629-1695) – Dutch PAM• Isaac Newton (1643-1727) – English PM • Louis XIV (1638-1715) – French “Sun King”

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Tycho Brahe Johannes Kepler Galileo Galilei

Observations Phenomenology/Theory Experiment

Data Predictions test predictions

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Johannes Kepler–The Phenomenologist

• Key question:

How are things happening?

Major Works:• Harmonices Mundi (1619)• Rudolphian Tables (1612)• Astronomia Nova• Dioptrice

Johannes Kepler (1571–1630)

Page 41: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Kepler’s First Law

The orbits of the planets are ellipses, with the Sun at one focus

Page 42: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Ellipses

a = “semimajor axis”; e = “eccentricity”

Page 43: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Lecture 27: Gravitation

• Piazza Convocation Short Lab!

• Starry Monday tonight 7pm, here (Sci 238)

Page 44: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Kepler’s Second LawAn imaginary line connecting the Sun to any planet sweeps out equal areas of the ellipse in equal times

Page 45: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Why is it warmer in the summer than in the winter in the USA?

• Because the Earth is closer to the Sun

• Because the Sun is higher in the sky in the summer

• None of the above

Page 46: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Axis Tilt – earth as gyroscope• The Earth’s rotation axis is tilted 23½ degrees

with respect to the plane of its orbit around the sun (the ecliptic)

• It is fixed in space sometimes we look “down” onto the ecliptic, sometimes “up” to it

Path around sun

Rotation axis

Page 47: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The Seasons• Change of seasons

is a result of the tilt of the Earth’s rotation axis with respect to the plane of the ecliptic

• Sun, moon, planets run along the ecliptic

Page 48: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Animation

• TeacherTube video

Page 49: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Position of Ecliptic on the Celestial Sphere• Earth axis is tilted w.r.t. ecliptic by 23 ½ degrees

• Equivalent: ecliptic is tilted by 23 ½ degrees w.r.t. equator! Sun appears to be sometime above (e.g. summer

solstice), sometimes below, and sometimes on the celestial equator

Page 50: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The vernal equinox happens when the sun enters the zodiacal sign of Aries, but is actually located in the constellation of Pisces.

Page 51: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Precession of the Equinoxes

Precession period

about 26,000 years

“The dawning of the age of

Aquarius”

Page 52: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Kepler’s Third LawThe square of a planet’s orbital period is proportional to the cube of its orbital semi-major axis:

P 2 a3

a P

Planet Orbital Semi-Major Axis Orbital Period Eccentricity P2/a3

Mercury 0.387 0.241 0.206 1.002Venus 0.723 0.615 0.007 1.001Earth 1.000 1.000 0.017 1.000Mars 1.524 1.881 0.093 1.000Jupiter 5.203 11.86 0.048 0.999Saturn 9.539 29.46 0.056 1.000Uranus 19.19 84.01 0.046 0.999Neptune 30.06 164.8 0.010 1.000Pluto 39.53 248.6 0.248 1.001

(A.U.) (Earth years)

Page 53: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

“Strange” motion of the Planets

Planets usually move from W to E relative to the stars, but sometimes strangely turn around in a loop, the so called retrograde motion.

Page 54: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The heliocentric Explanation of retrograde planetary motion

Page 55: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The New Physics & Astronomy in a Nutshell: Newton’s Principia

• Newton’s key question:

Why are things happening?

• Invented calculus and physics while on vacation from college

• His three Laws of Motion, together with the Law of Universal Gravitation, explain all of Kepler’s Laws (and more!)

• Principia (1687) [Full title: Philosophiae naturalis

principia mathematica] has his famous three laws on page 19 of 443. Isaac Newton (1642–1727)

Page 56: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Newton’s Synthesis: Unify sub- and super-lunar phenomena!

• Gravity on earth: a = g = 9.8 m/s2

– Due to force of earth on object a earth radius R away

• Effect on Moon: a = v2 /r– From length of month, distance to moon: 384,000 km

= 60 R (known to Greeks)– Acceleration is a = 0.00272 m/s2 = g/3600

• Conclusion: Force falls off like distance squared!

Page 57: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Law of Universal Gravitation

Force = G Mearth Mman / |r|2

Vector: Fman,earth = - G Mearth Mman rto man, from earth/ |rm,e|3

2

MEarth

Mman

|r|

Page 58: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Which of the following depends on the inertial mass of an object (as

opposed to its gravitational mass)?

• The time it takes on object to fall from a certain height

• The weight of an object on a bathroom-type spring scale

• The acceleration given to the object by a compressed spring

• The weight of the object on an ordinary balance

Page 59: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Orbital Motion

Page 60: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Cannon “Thought Experiment”

• http://www.phys.virginia.edu/classes/109N/more_stuff/Applets/newt/newtmtn.html

Page 61: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Suppose Earth had no atmosphere, and a ball were fired from the top of Mt. Everest in a direction tangent to the ground. If the initial speed were

high enough to cause the ball to travel in a circular trajectory around Earth, the ball’s acceleration

would be…

•Much less than g (b/c the ball doesn’t fall to the ground)•Be approximately g•Depend on the ball’s speed•None of the above

Page 62: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Lecture 28: Rest of Gravity

Page 63: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of

centripetal force acting on B to that acting on A?

• 1/8

• ¼

• ½

• 1

Page 64: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Principle of Superposition

• Gravitational forces can be added together as vectors, of course

• Newton’s shell theorem– A uniform spherical shell of matter attracts a

particle that is outside of the shell as if all the shell’s mass were concentrated at its center

Page 65: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Gravitation near the surface

• F=G Mm/r2

• F =ma a =GM/r2

• Approximations: – Earth is not uniform, not a sphere– Is rotating

Page 66: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Applications

• From the distance r between two bodies and the gravitational acceleration a of one of the bodies, we can compute the mass M of the other

F = ma = G Mm/r2 (m cancels out)

– From the weight of objects (i.e., the force of gravity) near the surface of the Earth, and known radius of Earth RE = 6.4103 km, we find ME = 61024 kg

– Your weight on another planet is F = m GM/r2

• E.g., on the Moon your weight would be 1/6 of what it is on Earth

Page 67: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Applications (cont’d)

• The mass of the Sun can be deduced from the orbital velocity of the planets: MS = rOrbitvOrbit

2/G = 21030 kg – actually, Sun and planets orbit their common center of mass

• Orbital mechanics. A body in an elliptical orbit cannot escape the mass it's orbiting unless something increases its velocity to a certain value called the escape velocity– Escape velocity from Earth's surface is about 25,000 mph (7

mi/sec)

Page 68: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Gravity Inside the Earth

• A uniform shell of matter exerts no net gravitational force on a particle located inside of it

Page 69: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Gravitational Potential Energy

• U = -GM/r

• Proof by calculating work (integral)

• Force from potential energy

• Escape speed

Page 70: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Einstein & Gravity

• General relativity

• Equivalence principle

• Curvature of space

Page 71: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

General Relativity ?! That’s easy!

(Actually, it took Prof. Einstein 10 years to come up with that!)

Rμν -1/2 gμν R = 8πG/c4 Tμν

What does that mean?

Page 72: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The Idea behind General Relativity

Page 73: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

More General

• General Relativity is more general in the sense that we drop the restriction that an observer not be accelerated

• The claim is that you cannot decide whether you are in a gravitational field, or just an accelerated observer

• The Einstein field equations describe the geometric properties of spacetime

Page 74: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Do bowling balls fall faster than apples?

Page 75: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

No!

Galileo: In the absence of air, all objects experience the same acceleration (change in motion) near Earth’s surface

http://www.youtube.com/watch?v=5C5_dOEyAfk

Page 76: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Equivalence Principle

A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body. For Newton's equation of motion in a gravitational field, written out in full, it is:

(Inertial mass) (Acceleration) = (Intensity of the gravitational field) (Gravitational mass).

It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body. — Albert Einstein

Page 77: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Meaning

• We cannot decide whether we live in an accelerated reference frame, or near a strong gravitational field.

Page 78: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

The Idea behind General Relativity

– We view space and time as a whole, we call it four-dimensional space-time.

• It has an unusual geometry

– Space-time is warped by the presence of masses like

the sun, so “Mass tells space how to bend”

– Objects (like planets) travel in “straight” lines through this curved space (we see this as orbits), so

“Space tells matter how to move”

Page 79: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Planetary Orbits

Sun Planet’s orbit

Page 80: Lecture 23: MOI & Torque. Kinetic energy of Rotation K = sum of ½ m v 2 for all parts of the body  Moment of inertia I  K = ½ I ω 2.

Effects of General Relativity

Bending of starlight by the Sun's gravitational field (and other gravitational lensing effects)