Physics 350 Chapter 5 Circular Motion and the Law of Gravity.

Physics 350Physics 350Chapter 5Circular Motion and the Law of Gravity

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


Using vectors, rearrange to determine the change in velocity (direction)

The vector change is directed towards the center of motion

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)


The magnitude of the centripetal acceleration is given by

◦This direction is toward the center of the circle

2

c

va

r


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


For the car moving on a flat circular track, the force is the friction between the car and the track.

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

Centripetal AccelerationCentripetal AccelerationApplying Newton’s Second Law

along the radial direction we can determine the net centripetal force Fc

Fc = mac = m (v2 / r)


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

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

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)

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

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

Centripetal AccelerationCentripetal AccelerationGravitron

◦Accelerating upwards?Climbing car

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!

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?

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

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?

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

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

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


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


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.)

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.

Planetary Motion and Newtonian Gravitation

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.

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

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

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

Kepler’s LawsKepler’s LawsKepler’s Third Law

◦They do have the same K!

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

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

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.)

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

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.”

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

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?

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)

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

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

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

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

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

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

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).

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.

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

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

Kepler’s LawsKepler’s Laws

Example:

Physics 350 Chapter 5 Circular Motion and the Law of Gravity.

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Transcript of Physics 350 Chapter 5 Circular Motion and the Law of Gravity.