physics 350 chapter 5 circular motion and the law of gravity
TRANSCRIPT
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Physics 350Physics 350Chapter 5Circular Motion and the Law of Gravity
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Centripetal AccelerationCentripetal Acceleration
Consider a car moving in a circle with constant velocity
Even though the car is moving with constant speed, it has an acceleration
The centripetal acceleration is due to the change in the direction of the velocity
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Centripetal AccelerationCentripetal Acceleration
Using vectors, rearrange to determine the change in velocity (direction)
The vector change is directed towards the center of motion
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Pictorial “Derivation” of Centripetal Acceleration
v1
v2
a = v/t
In uniform circular motion the acceleration is constant, directed towards the center. The velocity has constant magnitude, and is tangent to the path.
Top view:
a
a a
a
a
a = v2/r (r is radius of curve)
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Centripetal AccelerationCentripetal Acceleration
The magnitude of the centripetal acceleration is given by
◦This direction is toward the center of the circle
2
c
va
r
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Centripetal AccelerationCentripetal Acceleration
An object can have a centripetal acceleration only if some external force acts on it.◦In the case in the
figure, the force is the tension in the string
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Centripetal AccelerationCentripetal Acceleration
For the car moving on a flat circular track, the force is the friction between the car and the track.
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Centripetal AccelerationCentripetal AccelerationForces that act inward are
considered to be centripetal forces◦Examples:
Tension in example above Gravity on a satellite orbiting the Earth Force of friction
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Centripetal AccelerationCentripetal AccelerationApplying Newton’s Second Law
along the radial direction we can determine the net centripetal force Fc
Fc = mac = m (v2 / r)
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Centripetal AccelerationCentripetal Acceleration
If the centripetal force were removed, the object would leave its circular path and move in a straight line tangent to the circle◦Merry go round
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Curves, Centrifugal, Centripetal Forces
Going around a curve smushes you against window◦Understand this as inertia: you want to
go straight your body wants tokeep going straight
but the car is acceleratingtowards the center of the curve
Car acceleration is v2/r you think you’re being accelerated by v2/r relative to the car
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Centripetal, Centrifugal Forces, continued
The car is accelerated toward the center of the curve by a centripetal (center seeking) force
In your reference frame of the car, you experience a “fake”, or fictitious centrifugal “force”◦ Not a real force, just inertia relative to car’s
acceleration
Centripetal Forceon car velocity of car
(and the way you’d rather go)
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Centripetal ForcesCentripetal Forces Fictitious Force - Centrifugal
Driving in a car around a curve feels like you are applying a force to the car outward
Not really a force, the force one feels is the car applying a force on you from the frictional force it applies to the road
Inertia keeps our bodies wanting to move forward, the car applies a force to push it inwards
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Rotating Drum RideVertical drum rotates, you’re pressed
against wall◦ Friction force against wall matches gravity◦ Seem to stick to wall, feel very heavy
The forces real and perceived
Real Forces:
Friction; upCentripetal; inwardsGravity (weight); down
Perceived Forces:
Centrifugal; outwardsGravity (weight); down
Perceived weight; down and out
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Centripetal AccelerationCentripetal AccelerationGravitron
◦Accelerating upwards?Climbing car
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Works in vertical direction too…
Roller coaster loops:◦ Loop accelerates you downward (at top) with
acceleration greater than gravity◦ You are “pulled” into the floor, train stays on
track it’s actually the train being pulled into you!
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Vertical Circular Motion
Consider the forces acting on a motorcycle performing a loop-to-loop:
Can you think of other objects that undergo similar motions?
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Old-Fashioned SwingsThe angle of the ropes tells
us where the forces are:Ropes and gravity pull on
swingers If no vertical motions (level
swing), vertical forces cancel
Only thing left is horizontal component pointing toward center: centripetal force
Centripetal force is just mv2/r (F = ma; a = v2/r)
gravity (mg)
swing ropes: what you feel from your seat
resultant: centripetal
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What about our circular motions on Earth?Earth revolves on its axis once per dayEarth moves in (roughly) a circle about
the sunWhat are the accelerations produced
by these motions, and why don’t we feel them?
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Earth RotationVelocity at equator: 2r / (86,400 sec) = 463
m/sv2/r = 0.034 m/s2
◦ ~300 times weaker than gravity, which is 9.8 m/s2
Makes you feel lighter by 0.3% than if not rotating
No rotation at north pole no reduction in gIf you weigh 150 pounds at north pole, you’ll
weigh 149.5 pounds at the equator◦ actually, effect is even more pronounced than this
(by another half-pound) owing to stronger gravity at pole: earth’s oblate shape is the reason for this
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ConcepTest 5.1ConcepTest 5.1 TetherballTetherball
In the game of In the game of
tetherball, the struck tetherball, the struck
ball whirls around a ball whirls around a
pole. In what direction pole. In what direction
does the does the net forcenet force on on
the ball point?the ball point?
1) toward the top of the poletoward the top of the pole2) toward the groundtoward the ground3) along the horizontal component of along the horizontal component of
the tension forcethe tension force4) along the vertical component of the along the vertical component of the
tension forcetension force5) tangential to the circletangential to the circle
W
T
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The vertical component of the vertical component of the
tensiontension balances the weightweight.
The horizontal component of horizontal component of
tensiontension provides the centripetal centripetal
forceforce that points toward the
center of the circle.
1) toward the top of the poletoward the top of the pole2) toward the groundtoward the ground3) along the horizontal component of along the horizontal component of
the tension forcethe tension force4) along the vertical component of the along the vertical component of the
tension forcetension force5) tangential to the circletangential to the circle
ConcepTest 5.1ConcepTest 5.1 TetherballTetherball
In the game of In the game of
tetherball, the struck tetherball, the struck
ball whirls around a ball whirls around a
pole. In what direction pole. In what direction
does the does the net forcenet force on on
the ball point?the ball point?
W T
W
T
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Centripetal AccelerationCentripetal Acceleration
The tangential component of the acceleration is due to changing speed
The centripetal component of the acceleration is due to changing direction
Total acceleration can be found from these components 2
C2t aaa
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Centripetal AccelerationCentripetal Acceleration
Example A 1,000kg car rounds a curve on a
flat road of radius 50.0m at a speed of 50.0km/hr (14.0m/s). Will the car make the turn if:
a) the pavement is dry and the coefficient of static friction is 0.800?
b) the pavement is icy and the coefficient of static friction is 0.200?
(Note: use max static friction here for the extreme case of the tires almost slipping.)
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Centripetal AccelerationCentripetal AccelerationExample: An engineer wishes to design a curved
exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. He does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path.
a) Show that curve must be banked at tan θ = v2/rg.
b) Find the angle at which the curve needs to be banked for a 50m radius and a speed of 13.4 m/s.
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Planetary Motion and Newtonian Gravitation
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Kepler’s LawsKepler’s Laws1. All planets move in elliptical orbits
with the Sun at one of the focal points.
2. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.
3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun.
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Kepler’s LawsKepler’s LawsKepler’s First Law“All planets move in
elliptical orbits with the Sun at one of the focal points.”
◦Any object bound to another by an inverse square law will move in an elliptical path
◦Second focus is empty
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Kepler’s LawsKepler’s Laws Kepler’s Second Law “A line drawn from the
Sun to any planet sweeps out equal areas in equal time intervals.”
Objects near the Sun will need to cover more distance per time
◦ Faster velocity
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Kepler’s LawsKepler’s LawsKepler’s Third Law“The square of the orbital period of
any planet is proportional to cube of the average distance from the Sun to the planet”◦Orbital Period = time it takes a planet to
make one full orbit around the sun◦For orbit around the Sun, T2/r3 = K = KS =
2.97x10-19 s2/m3
◦K is independent of the mass of the planet Therefore, all planets should have the same K
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Kepler’s LawsKepler’s LawsKepler’s Third Law
◦They do have the same K!
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Kepler’s LawsKepler’s LawsPlanetary Data relative to Earth
Kepler's 3rd Law
T in years, a in astronomical units; then T2 = a3
Planet Period TDist. a fr.
SunT2 a3
Mercury 0.241 0.387 0.05808 0.05796
Venus 0.616 0.723 0.37946 0.37793
Earth 1 1 1 1
Mars 1.88 1.524 3.5344 3.5396
Jupiter 11.9 5.203 141.61 140.85
Saturn 29.5 9.539 870.25 867.98
Uranus 84 19.191 7056 7068
Neptune 165 30.071 27225 27192
Pluto 248 39.457 61504 61429
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Newtonian MechanicsKepler’s Laws described the
kinematics of the motion of the planets but didn’t answer why the planets move the way they do
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The Universal Law of Gravity• Any two bodies are attracting each
other through gravitation, with a force proportional to the product of their masses and inversely proportional to the square of their distance:
F = G m1m2
r2
(G is the Universal constant of gravity.)
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Newtonian GravitationNewtonian GravitationNewton’s Universal Law of Gravitation
Fg = G ————
where G is the gravitational constant, G = 6.673 x 10-11 m3 kg-1 s-2
Example of the inverse-square law
m1m2
r2
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Newtonian GravitationNewtonian Gravitation
Applying Newton’s third law to two masses:
Action-Reaction Pair
F21 = -F12“Every pair of particles exerts
on one another a mutual gravitational force of
attraction.”
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Newtonian GravitationNewtonian GravitationThe gravitational force exerted
by a uniform sphere on a particle outside the sphere is the same as the force exerted if the entire mass of the sphere were concentrated at its center
This is called Gauss’s Law.◦Applies to electric fields also
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Why was the Law of Gravitation not obvious (except to Newton).How big are gravitational forces between ordinary objects?
Conclusion:• G is very small, so…need huge masses to get perceptible forces
212
2112
r
mmG|F|
1 Newton is about the force needed to support 100 grams of mass on the EarthAbout the weight of a small apple
9.1x10-31 kgelectron
5x10-11 meterorbit radius
4x10-47 N.1.7x10-27 kg
proton
m1 m2 r12 F12
1 kg a liter of soda
1 kg sandwich 1 meter 6.67x10-11 N.
100 kga person
100 kg another person 1 meter 6.67x10-7 N.
106 kga ship
100 meters 0.67 N.106 kganother ship still hard to detect
Does gravitation play a role in atomic physics & chemistry?
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Circular Orbits(a pretty good approximation for all the planets because the eccentricities are much less than 1.)
Centripetal acceleration
2
rv
ar
Newton’s second law
2
2
rGMm
rmv
There is a subtle approximation here: we are approximating the center of mass position by the position of the sun. This is a good approximation.
(velocity)
(acceleration)
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Circular Orbits
The planetary mass m cancels out. The speed is then
r
GMv
Time = distance / speedi.e., Period = circumference / speed
or ,
GM
rT
GMr
rv
rT
322 4
22
Kepler’s third law: T 2 r 3
Period of revolution
2
2
rGMm
rmv
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Generalization to elliptical orbits(and the true center of mass!)
)( mMGa4
T32
2
where a is the semi-major axis of the ellipse
The calculation of elliptical orbits is difficult mathematics.
GMa4 32
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Finding the Value of G Henry Cavendish first
measured G directly (1798) Two masses m are fixed at
the ends of a light horizontal rod (torsion pendulum)
Two large masses M were placed near the small ones
The angle of rotation was measured
Results were fitted into
Newton’s Law G=6.67x10-11 N.m2/kg2
G versus g:• G is the universal gravitational constant, the same everywhere• g = ag is the acceleration due to gravity. It varies by location.• g = 9.80 m/s2 at the surface of the Earth
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Superposition: The net force on a point mass when there are many others nearby is the vector sum of the forces taken one pair at a time
All gravitational effects are between pairs of masses. No known effects depend directly on 3 or more masses.
1413121
11 ,,,i
,i on FFF F F
m1
m4
m3
m2
21F
12r
41F
31F
14r
13r
r
Gm- a g
symmetry by 0 g m F
2i1,
igi,1,i
m at1m on 11
1
Example:
m1
m4
m3m2 r
m5
r
'r
'r
m2 = m3
m4 = m5
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Newtonian GravitationNewtonian GravitationAcceleration due to gravity
◦Determined experimentally◦g value varies with altitude
ag = GME / r2
Altitude Altitude (km)(km)
aagg (m/s (m/s22)) Altitude ExampleAltitude Example
00 9.839.83 Mean Earth SurfaceMean Earth Surface
8.88.8 9.809.80 Mt. EverestMt. Everest
36.636.6 9.719.71 Highest manned Highest manned balloonballoon
400400 8.708.70 Space shuttle orbitSpace shuttle orbit
35,70035,700 0.2250.225 Comm. SatelliteComm. Satellite
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Gravitational “field” transmits the force
Concepts for g-fields:
◦ No contact needed: “action at a distance”.
◦ Acceleration Field created by gravitational mass transmits the force as a distortion of space that another (inertial) mass responds to.
12212
21112 r̂
r
mmG)r(gm F 2
A piece of mass m1 placed somewhere creates a “gravitational field” that has values described by some function g1(r) everywhere in space. Another piece of mass m2 feels a force proportional to g1(r) and in the same direction, also proportional to m2.
The field g1(r) is the gravitational force per unit mass created by mass m1, present at all points whether or not there is a test mass m2 located there The gravitational field vectors point in the direction of the acceleration a particle would experience if placed in the field at each point.
Field lines help to visualize strength and direction. close together strong field, direction force on test mass
2ˆg GM
m r
Fg r
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Mass of the EarthMass of the Earth
We can use the equation: g = G*Mearth/Rearth
2
to solve for Mearth since we know ◦g = 9.8 m/s2 (from our lab
experiment), ◦G = 6.67 x 10-11 Nt-m2/kg2 (from
precise gravity force experiments), and◦Rearth = 6,400 km (since we know the
circumference of the earth = 25,000 miles).
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Mass of the EarthMass of the Earthg = G*Mearth/Rearth
2 or
Mearth = g*Rearth2/G
= 9.8 m/s2 * (6.4 x 106 m)2 / 6.67 x 10-11 Nt-m2/kg2
= 6.0 x 1024 kg .
Actual value = 5.9742 x 1024kgThis value is certainly large as we expect
the mass of the earth to be large.
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You weigh yourself on a scale inside
an airplane that is flying with
constant speed at an altitude of
20,000 feet. How does your
measured weight in the airplane
compare with your weight as
measured on the surface of the
Earth?
1) greater than
2) less than
3) same
ConcepTest ConcepTest Fly Me AwayFly Me Away
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You weigh yourself on a scale inside
an airplane that is flying with
constant speed at an altitude of
20,000 feet. How does your
measured weight in the airplane
compare with your weight as
measured on the surface of the
Earth?
1) greater than
2) less than
3) same
At a high altitude, you are farther away from the
center of Earth. Therefore, the gravitational
force in the airplane will be less than the force
that you would experience on the surface of the
Earth.
ConcepTest ConcepTest Fly Me AwayFly Me Away
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Kepler’s LawsKepler’s Laws
Example: