Chapter 6 - Santa Rosa Junior Collegesrjcstaff.santarosa.edu/~lwillia2/20/20ch6_f14.pdf© 2010...

© 2010 Pearson Education, Inc.

PowerPoint® Lectures for College Physics: A Strategic Approach, Second Edition

Chapter 6

Circular Motion, Orbits, and Gravity

Circular Motion

Milky Way Galaxy

Orbital Speed of Solar System: 220 km/s Orbital Period: 225 Million Years

Mercury: 48 km/s

Venus: 35 km/s

Earth: 30 km/s

Mars: 24 km/s

Jupiter: 13 km/s

Neptune: 5 km/s

Equatorial Rotational Period: 27.5 days

Differential Solar Rotation Entangled Magentic Field Lines makes Sun Spots

464m/s

Precession causes the position of the North Pole to change over a period of 26,000 years.

Orbital Speed of Earth: ~ 30 km/s

Although the Moon is always lit from the Sun, we see

different amounts of the lit portion from

Earth depending on

where the Moon is

located in its month-long

orbit. Orbital Speed of Moon: ~ 1 km/s

155mph~70m/s

RADAR: RAdio Detecting And Ranging

200mph~90m/s

Electrons in Bohr Orbit: 2 million m/s

(Speed of Light: 200 million m/s)

Maximal Kerr Black Hole

Rotational Speed = Speed of Light

Uniform Circular Motion

The velocity vector is tangent to the path The change in velocity vector is due to the change in direction. The centripetal acceleration changes the direction of motion:

∆=∆

vat

Angular Position: Rotation Angle

∆θ = rotation angle

∆s = arc length

r = radius

sr

θ ∆∆ =

The rotation angle is the ratio of arc length to radius of curvature. For a given angle, the greater the radius, the greater the arc length.

: angular velocity (rad/s)v: tangential velocity (m/s)ω

vt rθω ∆

= =∆

v rω=

Angular & Tangential Velocity

is the same for every point & v varies with r. ω

Period & Frequenccy of Rotation

What is the period of the Earth? Moon?

Centripetal Acceleration

• The acceleration is always perpendicular to the path of the motion

• The acceleration always points toward the center of the circle of motion

• This acceleration is called the centripetal acceleration

• Magnitude is given by:

2

Cvar

=


Circular Motion: centripetal acceleration

There is an acceleration because the velocity is changing direction.

Slide 3-43

Space Station Rotation

A space station of diameter 80 m is turning about its axis at a constant rate. If the acceleration of the outer rim of the station is 2.5 m/s2, what is the period of revolution of the space station? a. 22 s b. 19 s c. 25 s d. 28 s e. 40 s

2 2, Cv ra vr

π= =

Τ

Centripetal and Centrifugal Center SEEKING and Center FLEEING

Factual = Centripetal Force Ffictitious = Centrifugal Force

In Physics, we use ONLY

CentriPETAL acceleration NOT

CentriFUGAL acceleration!

The Earth rotates once per day around its axis as shown. Assuming the Earth is a sphere, is the rotational speed at Santa

Rosa greater or less than the speed at the equator?

366 m/s

464 m/s

Is the centripetal acceleration greater at the Equator or at Santa

Rosa?

2

cvar

= 20.034 /m s

20.027 /m s

The Earth rotates once per day around its axis. Assuming the Earth is a sphere with radius 6.38 x 106m, find the tangential speed of a person at the equator and at 38 degrees latitude (Santa Rosa!) and their centripetal accelerations.

At the equator, r = 6.38 x 106m: 62 2 (6.38 10 ) 464 /

86,400r x mv m st sπ π

= = =∆

( )222

6

464 /.034 /

6.38 10= = =c

m sva m sr x m

At Santa Rosa, r = 6.38 x 106m cos38: 62 2 (6.38 10 )cos38 366 /

86,400r x mv m st sπ π

= = =∆

( )222

6

366 /.027 /

cos38 6.38 10 cos38= = =c

m sva m sr x m

ca

ca

What is the total acceleration acting on a person in Santa Rosa?

cag

The vector sum.

Is your apparent weight as measured on a spring scale more at the Equator or at Santa Rosa?

cag

Since you are standing on the Earth (and not in the can) the centrifugal force tends

to throw you off the Earth. You weigh less where the centripetal force is greatest because

that is also where the centrifugal force is greatest – the force that tends to throw you

out of a rotating reference frame.

Centripetal & Centrifugal Force Depends on Your Reference Frame

Inside Observer (rotating reference frame) feels Centrifugal Force pushing them against the can.

Outside Observer (non-rotating frame) sees Centripetal Force pulling can in a circle.

Center-seeking

Center-fleeing

Centrifugal Force is Fictitious? The centrifugal force is a real effect. Objects in a rotating frame feel a centrifugal force acting on them, trying to push them out. This is due to your inertia – the fact that your mass does not want to go in a circle. The centrifugal force is called ‘fictitious’ because it isn’t due to any real force – it is only due to the fact that you are rotating. The centripetal force is ‘real’ because it is due to something acting on you like a string or a car.

Artificial Gravity How fast would the space station segments A and B have to rotate in

order to produce an artificial gravity of 1 g?

56 / ~ 115Av m s mph=

104 / ~ 210Bv m s mph=


Important: Inside vs Outside the Rotating Frame


Horizontal (Flat) Curve

The force of static friction supplies the centripetal force

The maximum speed at which the car can negotiate the curve is

Note, this does not depend on the mass of the car

v grµ=

2

= =∑ cmvF f

r

0= − =∑ yF N mgµ=f mg


Example Problem A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?

Slide 6-28


A highway curve has a radius of 0.14 km and is unbanked. A car weighing 12 kN goes around the curve at a speed of 24 m/s without slipping. What is the magnitude of the horizontal force of the road on the car? What is μ? Draw FBD. a. 12 kN b. 17 kN c. 13 kN d. 5.0 kN e. 49 kN

Horizontal (Flat) Curve

Banked Curve These are designed with friction

equaling zero - there is a component of the normal force that supplies the centripetal force that keeps the car moving in a circle.

2

tan vrg

θ =

2

sinθ= =∑ rmvF n

rcos 0θ= − =∑ yF n mg

Dividing:

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? a. 0.68 kN b. 0.64 kN c. 0.72 kN d. 0.76 kN e. 0.52 kN

Banked Curve


Vertical Circle has Non-Uniform Speed

Where is the speed Max? Min? Where is the Tension Max? Min?


Example Problem: Loop-the-Loop A roller coaster car goes through a vertical loop at a constant speed. For positions A to E, rank order the:

• centripetal acceleration

• normal force

• apparent weight

Slide 6-32


A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant?

QuickCheck 8.10

Slide 8-80

Humps in the Road: Outside the Vertical Loop


A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant?

QuickCheck 8.10

Now the centripetal acceleration points down.

Slide 8-81

Humps in the Road: Outside the Vertical Loop


A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.

QuickCheck 8.11

Slide 8-82

Loop d’ Loops: Inside the Vertical Loop


Humps in the Road Outside the Vertical Loop

A roller-coaster car has a mass of 500 kg when fully loaded with passengers. The car passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.0 m/s. What is the force of the track on the car at the top of the hill? a. 7.0 kN up b. 7.0 kN down c. 2.8 kN down d. 2.8 kN up e. 5.6 kN down


:

n

mg

Humps in the Road

: 0=Take n

=v gr

What is the maximum speed the car can have as it passes this highest point without losing contact with the road?

Max speed without losing contact MEANS:

2mvmgr

=Therefore:

Maximum Speed for Vertical Circular Motion

Maximum Speed to not loose contact with road only depends on R! ROOT GRRRRRRRR


What is the maximum speed the vehicle can have at B and still remain on the track?


A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.

QuickCheck 8.11

The track is above the car, so the normal force of the track pushes down.

Slide 8-83



What is the minimum speed so that the car barely make it around the loop the riders are upside down and feel weightless ? R = 10.0m

Loop d’ Loops: Inside the Vertical Loop Minimum Speed to get to the Top.


A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip?

a. 3.2 kN

b. 8.1 kN

c. 4.9 kN

d. 1.7 kN

e. 5.3 kN



Universal Law of Gravity

d M

m

211

26.67 10 NmG xkg

−=

2

GmMFd

=


Gravity: Inverse Square Law

2

GmMFd

=


Inside the Earth the Gravitational Force is Linear.

Acceleration decreases as you fall to the center (where your speed is the greatest) and then the acceleration increases

but in the opposite direction, slowing you down to a stop at the other end…but then you would fall back in again,

bouncing back and forth forever!

Gravitational Force INSIDE the Earth


Finding little g

Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g?

2you E

E

Gm MF

R= youF m a=

2you E

youE

Gm Mm a

R=

Response to the Force

2E

E

GMaR

=Independent of your mass! This is why a rock and feather fall with the same acceleration!

Source of the Force

This is your WEIGHT!


Finding little g

2E

E

GMaR

=

Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g?

2you E

E

Gm MF

R= youF m a=

( )( )( )

11 2 2 24

26

6.673 10 / 5.98 10

6.38 10

−

=x Nm kg x kg

ax m

29.81 /a m s=

Source of the Force Reaction to the Force

= g!

Calculate the magnitude of the force of gravity between the Earth and the Moon. The distance between the Earth and Moon centers is 3.84x108m

2EMGmMF

d=

( )( )( )( )

11 2 2 24 22

28

6.673 10 / 5.98 10 7.35 10

3.84 10

−

=EM

x Nm kg x kg x kgF

x m

202.01 10EMF x N=

Earth-Moon Gravity

Calculate the acceleration of the Earth due to the Earth-Moon gravitational interaction.

EME

E

Fam

=

20

24

2.01 105.98 10

x Nx kg

=

5 23.33 10 /Ea x m s−=

Earth-Moon Gravity

EMM

M

Fam

=

20

22

2.01 107.35 10

x Nx kg

=

3 22.73 10 /Ma x m s−=

Calculate the acceleration of the Moon due to the Earth-Moon gravitational interaction.

Earth-Moon Gravity

The acceleration of gravity at the Moon due to the Earth is:

3 22.73 10 /Ma x m s−=

5 23.33 10 /Ea x m s−=

The acceleration of gravity at the Earth due to the moon is:

Why the difference?

FORCE is the same. Acceleration is NOT!!!

BECAUSE MASSES ARE DIFFERENT!

Earth-Moon Gravity

Force is not Acceleration!

The forces are equal but the accelerations are not!

Earth on Moon Moon on EarthF F= −


Checking Understanding: Gravity on Other Worlds A 60 kg person stands on each of the following planets. Rank order her weight on the three bodies, from highest to lowest.

A. A > B > C B. B > A > C C. B > C > A D. C > B > A E. C > A > B

Slide 6-38


Answer A 60 kg person stands on each of the following planets. Rank order her weight on the three bodies, from highest to lowest.

A. A > B > C B. B > A > C C. B > C > A D. C > B > A E. C > A > B

Slide 6-39

Recall: Curvature of Earth

If you threw the ball at 8000 m/s off the surface of the Earth (and there were no buildings or mountains in the way)

how far would it travel in the vertical direction in 1 second?

Curvature of the Earth: Every 8000 m, the Earth curves by 5 meters!

The ball will achieve orbit.

2 2 21 ~ 5 / (1 ) 52

y gt m s s m∆ = ⋅ =

Orbital Velocity If you can throw a ball at 8000m/s, the Earth curves away

from it so that the ball continually falls in free fall around the Earth – it is in orbit around the Earth!

Above the atmosphere

Ignoring air resistance.

Projectile Motion/Orbital Motion Projectile Motion is Orbital motion that hits the Earth!

Orbital Motion| & Escape Velocity 8km/s: Circular orbit

Between 8 & 11.2 km/s: Elliptical orbit 11.2 km/s: Escape Earth

42.5 km/s: Escape Solar System!

Orbit Question Find the orbital speed of a satellite 200 km above the Earth. Assume a circular orbit.

2var

=2

s Em M GFr

= sF m a=

2

2s E

sm M G vF m

r r= = E

E

M GvR h

=+

24 11 2 2

6

(5.97 10 )(6.67 10 / )6.58 10

x kg x Nm kgvx m

−

=

37.78 10 /v x m s=

24 65.97 10 , 6.38 10E EM x kg R x m= =

Notice that this is less 8km/s!

What is this?

Orbit Question

2 rvπ

Τ =

What is the period of a satellite orbiting 200 km above the Earth? Assume a circular orbit.

2 rv π=

Τ

5314 88minsΤ = =

24 65.97 10 , 6.38 10E EM x kg R x m= =

6

3

2 (6.58 10 )7.78 10 /

x mx m s

π=

2

2GMm vF m

rr= =

2(2 / )rmr

π Τ=

22 34 r

GMπ

Τ =

Kepler’s 3rd!

If you don’t know the velocity:

Period increases with r!

g and v Above the Earth’s Surface If an object is some

distance h above the Earth’s surface, r becomes RE + h

( )2E

E

GMgR h

=+

The tangential speed of an object is its orbital speed and is given by the centripetal acceleration, g:

2v gr=

2

E

vR h+

( )E

E

GMvR h

=+

( )2E

E

GMR h

=+

2GMmF

r=

Orbital speed decreases with increasing altitude!


Additional Questions A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities increases?

A. Speed B. Angular speed C. Period D. Centripetal acceleration E. Gravitational force of the earth

Slide 6-43


Answer A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities increases?

A. Speed B. Angular speed C. Period D. Centripetal acceleration E. Gravitational force of the earth

Slide 6-44

Chapter 6 - Santa Rosa Junior Collegesrjcstaff.santarosa.edu/~lwillia2/20/20ch6_f14.pdf© 2010...

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