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Circular Motion
Circular motion is very common and very important in
our everyday life. Satellites, the moon, the solar system
and stars in galaxies all rotate in “circular” orbits. The
term circular here is being used loosely since even
repetitive closed motion is generally not a perfect
circle.
At any given instant a moving object not moving in a
straight line is moving along the arc of a (maybe
changing) circle.
So if we understand motion
in a circle we can understand
more complicated trajectories.
Remember at any instant the velocity is along the path
of motion but the acceleration can be in any direction.
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Physics 214
Spring 2009
1
Circular motion
F = ma = mv2/r
a = v2/r
If the velocity of an object changes direction then the
object experiences an acceleration and a force is
required.
These are called “centripetal” acceleration and force
and are directed toward the center of the circle.
This is the effect you feel rounding a corner in a car
Week 4
Physics 214
Spring 2009
2
Circular motion
F = ma = mv2/r
a = v2/r
Where does this come from?
a = Δv/Δt = vxv/r
Look at the Δv vector: it’s proportional to v.
(and is directed toward the center of the circle.)
Also, the rate that Δv is changing is proportional to the
rate that the angle between v1 and v2 is changing (that
is, the rate that vector v is swinging around.) This rate of
change is just v/r (the angle is in RADIANS)
Week 4
Physics 214
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3
Questions Chapter 5
Q6 A ball on the end of a string
is whirled with constant speed in
a counterclockwise horizontal
circle. At point A in the circle, the
string breaks. Which of the
curves sketched below most
accurately represents the path
that the ball will take right after
the string breaks (as seen from
above)? Explain.
.
4
•
3
2
A
1
A=1, B=2, C=3 D=4
Path number 3
Week 4
Physics 214
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4
Balance of forces
We need to understand the forces that
are acting along both horizontal and
vertical directions.
In the case shown the tension or force
exerted by the string has components
which balance the weight in the vertical
direction and provide the centripetal
force horizontally.
Th = mv2/r
Tv = W = mg
Week 4
Physics 214
Spring 2009
5
Cars
Ff
Ff
Rear view
When a car turns a corner it is friction
between the tires and the road which
provides the centripetal force.
If the road is banked then the normal force
also provides some centripetal force.
For each banked track there is a velocity for
which no friction is required.
Note Ff adjusts itself (up to break point)
And it is the HORIZONTAL component of Ff
which may help supply the needed Fcentrip.
Note: Nv must balance W.
Question: if v = 0, where does Ff point?
A. Upslope
Week 4
View from Above
Ff
W = mg
B. Downslope
Physics 214
Spring 2009
6
Vertical circles
N
v
W = mg
mg – N = mv2/r
If v = 0 then N = mg
As v increases N
becomes smaller When
v2/r = g
The car becomes
weightless. Same as
weightless-training
airplane, for astronauts
Week 4
g
We choose
+ always
toward the
center of
the circle
If car were
running on
“capture rails”
that could also
hold it down, big
v would cause N
to point
Physics 214
downward
Spring 2009
Ferris wheel
At the bottom
N - mg = mv2/r
At the top
mg – N = mv2/r
With seatbelt, at top if v2 is
big enough, seatbelt would hold
you down, your “weight” would
7
be upside down!
Gravitation and the planets
Astronomy began as soon as man was able to observe
the sky. Records exist going back several thousand
years. In particular, of the yearly variation of the stars
in the sky and the motion of observable objects such
as planets.
People observed the “fixed” North Star and, for
example, the rising of Sirius signaling the flooding of
the Nile.
Copernicus was the first person to advocate a suncentered solar system. Followed by Galileo who used
the first telescopes (moons of Jupiter like small solar
system.)
Tycho Brahe was the most famous naked eye
astronomer.
Kepler, his assistant used the data to draw quantitative
Week
4
Physics 214
8
conclusions.
Spring 2009
Keplers Laws
1) Orbits are ellipses. For 0 “eccentricity” Æ circle
2) The radius vector sweeps out
equal areas in equal times
3) T2 proportional to r3
T is the period. For the earth, T
is one year and r is the average radius. Earth’s
eccentricity is small, but does affect climate cyclically.
For circular motion with constant velocity v
the circumference of a circle is 2πr and the
Period T = 2πr/v = distance (arc length) / speed
Week 4
Physics 214
Spring 2009
9
Keplers Laws
1) Orbits are ellipses. For 0 “eccentricity” Æ circle
2) For the Earth, ellipse is rather
close to a circle
3) T2 proportional to r3
T = 1 year, is the period for the earth..
For circular motion with constant velocity v
the circumference of a circle is 2πr and the
Period T = 2πr/v = distance (arc length) / speed
So v = 2πr/T = circumf,/period = 2π 94M mi/1 yr
Week 4
Physics 214
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Newton and Gravitation
Newton developed the Law of
Universal Gravitation
force between ANY two objects is
proportional to M1m2/r2 .
The (very small) constant of
proportionality was measured
by Cavendish more than 100 years
later G = 6.67 x 10-11 N.m2/kg2.
Since at the earths surface
mg = m(GMe/r2) the experiment
measured the mass of the earth
Week 4
Physics 214
Spring 2009
http://www.physics.purdue.
edu/class/applets/Newtons
Cannon/newtmtn.html
11
Newton and Gravitation
The force between ANY two objects
is ATTRACTIVE and = GM1m2/r2 .
N3 (third law) happens here too:
each M pulls the other,
equal and opposite.
For a spherical object, all its mass acts as if at the
CENTER.
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Planetary orbits
For a simple circular orbit
GmM/r2 = mv2/r where M is the
mass of the sun and m the mass
of the earth [or M is the mass of
the earth and m the mass of an
earth satellite.] Cancel m & solve
GM/r3=v2/r2 but
v/r = 2π/T
GM/r3=(2π/T)2
Week 4
T2/r3 = 4π2/GMs
Use average
radius for
elliptical orbits.
Works for other
attractors:
Earth, Jupiter.
Physics 214
Spring 2009
Use right Mass13
Satellite Orbits
For a geosynchronous orbit
period is 24 hours and height
above the earths surface is
~22,000 miles (add ~4000 miles to
get the distance to the CENTER)
T2/r3 = 4π2 /GMe
R3 =T2GMe /4π2
Take square root both sides
Week 4
Physics 214
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Geosynchronous
orbit must be
CIRCULAR, and will
be off-scale on the
above picture.
Also is equatorial.
14
Summary of Chapter 5
™Circular motion and centripetal
acceleration and force.
Fc = mv2/r
Ferris wheel, car around a corner or
over a hill. “Circular” pendulum.
™Gravitation and Planetary orbits
For a simple circular orbit
GmM/r2 = mv2/r where M is the mass
of the sun and m the mass of the
earth.
v2 = GM/r T = 2πr/v
T2 = 4π2r2/v2 = 4π2r3/GMs
At any given instant, FG is given by the
distance BETWEEN CENTERS OF MASS
Week 4
Physics 214
Spring 2009
Wrong r !!
use AVG.
T2/r3 = 4π2/GMs
15
Examples of circular motion
Vertical motion
N
Looking down on
stunt driver’s
cylindrical cage
N
v
v
W = mg
mg – N =
mv2/r
N
mg
N - mg = mv2/r
Week 4
N = mv2/r
Side view
Red force
arrows should
Ff
be equal, if
enough static
friction is
N
available. If μs is
too small, not
mg
enough Ff and
car will slide mg = Ff
downwards.
Physics 214
Spring 2009
v
T
mg
Swing object
on a string in a
vertical circle
mg + T = mv2/r top
T - mg = mv2/r bottom
16
Quiz question
.
Two objects each of mass 1 Metric ton are 2 meters
apart (between centers). What is the force between
them? G = 6.67x10-11 and one Metric ton = 1,000kg
A. 1.7x10-11 N
.
D. 1.7x10-8 N
Week 4
B. 1.7x10-17 N
C. 1.7x10-5 N
E. 1.7x106 N
Physics 214
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Moon and tides
anim0012.mov
Earth tides are dominantly due
to the gravitational force
X
E+M Centerof Mass is 80
exerted by the moon. Since
times closer to Earth since Earth
the Earth turns faster than the
is 80 times heavier than Moon
Moon orbits, the tidal bulges
Earth & Moon both orbit around
sweep around the Earth, with
their mutual CofM
high tides happening approx.
twice a day. Because of the
friction generated by tides the
Moon is gaining energy from
the Earth’s spin, and moving
away fromhttp://www.sfgate.com/getoutside/1996/jun/tides.html
the earth.
Week 4
whytides.gif
Physics 214
Spring 2009
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Moon and tides
The tidal bulge away from the moon
is where “centrifugal force” is bigger
than gravitational force.
The bulge towards the moon is
where the gravity outweighs the
centrifugal force.
This works because the entire earth
moves with the same angular
velocitiy around the CofM. The
outward bulge is where FG is less
and Fc (because of v2) is more, than
required for the orbit of the Earth’s
center. And vice versa for the
Moon-ward bulge.
Week 4
anim0012.mov
~250,000 mi
~384,000 km
X
E+M Centerof Mass is 80
times closer to Earth’s center,
since Earth is 80 times heavier
than Moon
Earth & Moon both orbit around
their mutual CofM
http://www.sfgate.com/getoutside/1996/jun/tides.html
whytides.gif
Physics 214
Spring 2009
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Moon and tides
The Moon itself has a tidal bulge, due to
ITS orbit. The Earth pulls on this bulge
and keeps the Moon “locked” with the
bulge facing the Earth. [This required the
Moon to lose some of its spin energy
early-on in the life of the solar system.]
The Earth spins faster than the Moon
orbits, so the Earth’s tidal bulges “torque”
the Moon’s orbit to move the orbit further
away and slow down the Earth’s rotation.
This ongoing effect is measurable (3.8
cm/yr). Also geologically measurable
(Devonian seashells with growth layers
showing ~400 days per year!!)
Someday far in the future, the Earth’s spin
may get locked to the Moon’s orbit!
Week 4
anim0012.mov
X
E+M Centerof Mass is 80
times closer to Earth since Earth
is 80 times heavier than Moon
Earth & Moon both orbit around
their mutual CofM
http://www.sfgate.com/getoutside/1996/jun/tides.html
whytides.gif
Physics 214
Spring 2009
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Dark Matter
For the orbit of a body of mass m about a much more massive
M is the
body of mass M: GmM/r2 = mv2/r and GM/r = v2.
mass inside the orbit. If we look at stars in motion in galaxies
we find that the orbital speeds do NOT fall off as quickly as you’d
expect from the VISIBLE (shining) stars. There’s extra invisible
mass boosting the effective M inside each orbit. We now know
that 25% of our Universe is Dark Matter, only abaout 5% is
“baryonic” matter (made of nuclei and atoms) and only about 1%
– 2% of the Universe is made of shining stars!!
v~1/sqrt(r)
Week 4
Physics 214
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Space Elevator (courtesy of Arthur C. Clarke)
station
v
counterweight
100,000km
Week 4
•The Space Elevator is a thin ribbon, with a cross-section area roughly half that of a
pencil, extending from a ship-borne anchor to a counterweight well beyond geosynchronous orbit. – and over ¼ of the way from Earth to Moon
•The ribbon is kept taut due to the rotation of the earth (and that of the counterweight
around the earth). At its bottom, it pulls up on the anchor with a force of about 20
tons.
•Electric vehicles, called climbers, ascend the ribbon using electricity generated by
solar panels (i.e. photovoltaic) facing a ground based booster light beam (laser.)
•In addition to lifting payloads from earth to orbit, the elevator can also release them
directly into lunar-injection or earth-escape trajectories.
•The baseline system weighs about 1500 tons (including counterweight) and can carry
up to 15 ton payloads, easily one per day.
•The ribbon is 62,000 miles long, about 3 feet wide, and is thinner than a sheet of
paper. It is made out of a carbon nanotube composite material.
•The climbers travel at a steady 200 kilometers per hour (120 MPH), do not undergo
accelerations and vibrations, can carry large and fragile payloads, and have no
propellant stored onboard.
•The climbers are driven by earth based lasers.
•Orbital debris are avoided by moving the floating anchor ship on Earth, and the
ribbon itself is made resilient to local space debris damage.
•The elevator can increase its own payload capacity by adding ribbon layers to itself.
There is no limit on how large a Space Elevator can be!
Physics 214
Spring 2009
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Space Elevator (courtesy of Arthur C. Clarke)
station
v
counterweight
100,000km
•The Space Station can be beyond geosynchronous
height (as shown) and can launch objects with greater
than escape velocity. This is because at its location, it is
LESS gravitationally bound to the Earth AND it has MORE
than the Velocity for a circular orbit AT THAT DISTANCE.
•Basically, the “free orbit” from the blue Space Station
is not actually an orbit, it won’t even close into an ellipse,
the path is too fast and straight for that.
If the ribbon breaks, the station and the counterweight
will go flying off away from the earth and not return.
Week 4
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Space Elevator (courtesy of Arthur C. Clarke)
station
v
counterweight
100,000km
•Engineering note: The strongest steel cable is much too
heavy, for its strength, to do this job. A 10 mile long steel
cable suspended (magically) near the Earth’s surface,
would break under its own weight. We need something
WAY lighter (per unit of strength) to make the space
elevator.
•Carbon nanotubes are incredibly strong. They are like
rolled-up tubes of single-layer graphite (a hexagonal
network of Carbon atoms, like chicken wire.) The
Carbon-Carbon bond achieves this (somewhat the way
diamond, which is a crystal of pure carbon, is so hard.)
Week 4
Physics 214
Spring 2009
24
Ch 5 QUIZ Question
On a rotating platform, spinning at a certain constant
rate, you move three times farther from the center. How
many times more (or less) friction does it take to keep
you from slipping off the turntable?
A. 1/9
B. 1/3
C. same
D. 3
E. 9
Note that both v and r change when you move. Think about how the distance
around the circle changes when r increases, and remember that you go
around once in the same amount of time wherever you are on the turntable.
Week 4
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Ch 5 QUIZ Question
On a rotating platform, spinning at a certain constant
rate, you move three times farther from the center. How
many times more (or less) friction does it take to keep
you from slipping off the turntable?
A. 1/9
B. 1/3
C. same
D. 3
E. 9
Note that both v and r change when you move. Think about how the distance
around the circle changes when r increases, and remember that you go
around once in the same amount of time wherever you are on the turntable.
New v (call it v’) is three times faster. New r (call it r’) is three times bigger
New mv2/r = m(3v)2/3r = (9/3)mv2/r : Three times as much Fc is needed.
Week 4
Physics 214
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1C-05 Velocity of Rifle Bullet
MEASURING SPEEDS OF OBJECTS MOVING VERY FAST MAY NOT BE
DIFFICULT.
THIS TECHNIQUE IS THE SAME ONE USED TO MEASURE SPEEDS OF
MOLECULES.
How can we
measure the
speed of a bullet ?
We know that the distance between two disks is L. If the second disk rotates
an angle Δθ before the bullet arrives, the time taken by this rotation is
t = Δθ / 2πn , where 2πn is the angular frequency of the shaft, in radians/s.
Therefore we come up with v = L / t = 2πn L / Δθ .
Note: θ is in radians, and n by itself is the revolution rate, in revols./s
or Hz.
Week 4
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1D-02 Conical Pendulum
T sin(θ) = mv2/R
T cos(θ) = mg
v = sqrt( gR tan(θ) )
‘cuz tan() = v2/gR
Could
you find
the NET
force?
Note that T’s cancel,
and m’s cancel.
Which of the m’s is inertial?
Which of the m’s is grav.?
Period of the pendulum
τ= 2πR/v,
where R = L / sin(θ)
τ= 2πsqrt( Lcos(θ)/g )
NET FORCE IS TOWARD THE CENTER OF THE CIRCULAR
PATH
Week 4
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1D-03 Demonstrations of Central
Force
THE SHAPES/SURFACES OF SEMI-RIGID OBJECTS BECOME MORE
CURVED TO PROVIDE GREATER CENTRAL FORCES DURING ROTATION.
T
What will happen
when it is subjected
to forces during
rotation ?
θ
T
θ
Fc= mv2/R
The force comes
from the strips
wanting to un-flex
Week 4
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1D-04 Radial Acceleration &
Tangential Velocity
Once the string is
cut, where is the
ball going?
AT ANY INSTANT, THE VELOCITY VECTOR OF THE BALL IS
DIRECTED ALONG THE TANGENT.
AT THE INSTANT WHEN THE BLADE CUTS THE STRING, THE
BALL’S VELOCITY IS HORIZONTAL SO IT ACTS LIKE A
HORIZONTALLY LAUNCHED PROJECTILE AND LANDS IN THE
CATCH BOX.
Week 4
Physics 214
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30
1D-05 Twirling Wine Glass
m
v
WHAT IS THE PHYSICS
THAT KEEPS THE
WINE FROM SPILLING ?
g
Same as
N + mg = mv2/R
string
N>0
THE GLASS WANTS TO MOVE ALONG THE TANGENT TO
THE CIRCLE AND THE REACTION FORCE OF THE PLATE
AND GRAVITY PROVIDE THE CENTRIPETAL FORCE TO
KEEP IT IN THE CIRCLE (and the wine in the glass).
Week 4
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Q8 For a ball twirled in a horizontal circle
at the end of a string, does the vertical
component of the force exerted by the
string produce the centripetal acceleration
of the ball? Explain.
Vertical component balances the weight
Horizontal component provides the acceleration
N
Q9 A car travels around a flat (unbanked)
curve with constant speed.
Rear
Ff
A. Show all of the forces acting on the car.
B. What is the direction of the net force act.
mg
The force acts toward the center of the turn circle
Week 4
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Q10 Is there a maximum speed at which the car in question 9 will
be able to negotiate the curve? If so, what factors determine this
maximum speed? Explain.
Yes. The friction between the tires and the road
Q11 If a curve is banked, is it possible for a car to negotiate the
curve even when the frictional force is zero due to very slick
ice? Explain.
Yes there is just one speed. If the car moves too slowly it
will slide down. If it moves to fast it will slide up.
Week 4
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Q12 If a ball is whirled in a vertical circle with constant speed, at
what point in the circle, if any, is the tension in the string the
greatest? Explain. (Hint: Compare this situation to the Ferris wheel
described in section 5.2).
The tension is the greatest at the bottom because the
string has to support the weight and provide the force
for the centripetal acceleration.
Q19 Does a planet moving in an elliptical orbit about the sun
move fastest when it is farthest from the sun or when it is nearest
to the sun? Explain by referring to one of Kepler’s laws.
When it is nearest
Week 4
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Q20 Does the sun exert a larger force on the earth than that exerted
on the sun by the earth? Explain.
The magnitude of the forces is the same
they are a reaction/action pair
Q23 Two masses are separated by a distance r. If this distance is
doubled, is the force of interaction between the two masses
doubled, halved, or changed by some other amount? Explain.
The force reduces by a factor of 4
Week 4
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35
Ch 5 E 14
The acceleration of gravity at the surface of the moon is
about 1/6 that at the surface of the Earth (9.8 m/s2). What
is the weight of an astronaut standing on the moon
whose weight on earth is 180 lb?
Wearth = m gearth = 180 lb
gmoon
Wmoon = m gmoon
gmoon = 1/6 gearth
Wmoon = m 1/6 gearth = 1/6 m gearth = 1/6 (180 lb) = 30 lb
Week 4
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Ch 5 E 16
Time between high tides = 12 hrs 25 minutes.
High tide occurs at 3:30 PM one afternoon.
a) When is high tide the next afternoon
b) When are low tides the next day?
T= 12hrs 25min
a) 3:30 PM + 2 (12 hrs 25 min)
high tide
= 3:30 PM + 24 hrs + 50 min
= 4:20 PM
t
low tide
T= 12hrs 25min
b) Low tide the next day = 4:20 PM - 6 hr 12 min 30 s = 10:07:30 AM
2nd Low tide = 10:07:30 AM + 12 hrs 25 min = 10:32:30 PM
Week 4
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37
1D-07 Paper Saw
THE RADIAL FORCES HOLDING THE PAPER TOGETHER MAKE
THE PAPER RIGID.
Is paper
more rigid
than wood ?
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38
1D-08 Ball in Ring
Is the ball leaving
in a straight line or
continuing this
circular path?
THE FORCE WHICH KEEPS THE BALL MOVING CIRCULAR IS
PROVIDED BY THE RING. ONCE THE FORCE IS REMOVED,
THE BALL CONTINUES IN A STRAIGHT LINE, ACCORDING
TO NEWTON’S FIRST LAW.
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Ch 5 CP 2
A Ferris wheel with radius 12 m makes one complete
rotation every 8 seconds.
a) Rider travels distance 2πr every rotation. What speed do riders
move at?
b) What is the magnitude of their centripetal acceleration?
c) For a 40 kg rider, what is magnitude of centripetal force to keep him
moving in a circle? Is his weight large enough to provide this
centripetal force at the top of the cycle?
d) What is the magnitude of the normal force exerted by the seat on the
rider at the top?
e) What would happen if the Ferris wheel is going so fast the weight of
the rider is not sufficient to provide the centripetal force at the top?
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Ch 5 CP 2 (con’t)
a) S = d/t = 2πr/t = 2π(12m)/8s = 9.42 m/s
Fcent
r=
12m
b) acent = v2/r = s2/r = (9.42m/s)2/12m = 7.40 m/s2
c) Fcent = m v2/r = m acent = (40 kg)(7.40 m/s2) = 296 N
W = mg = (40 kg)(9.8 m/s2) = 392 N
Yes, his weight is larger than the centripetal force required.
d) W – Nf = 296
N = 96 newtons
N
e) rider is ejected
W
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41
Ch 5 CP 4
A passenger in a rollover accident turns through a radius
of 3.0m in the seat of the vehicle making a complete turn in
1 sec.
a) Circumference = 2πr, what is speed of passenger?
b) What is centripetal acceleration? Compare it to gravity (9.8
m/s2)
c) Passenger has mass = 60 kg, what is centripetal force
required to produce the acceleration? Compare it to
passengers weight.
a) s = d/t = 2π(3.0m)/1 = 19m/s
b) a = v2/r = s2/r = (19 m/s)2/3m = 118 m/s2 = 12g
3m
c) F = ma = (60 kg)(118 m/s2) = 7080 N
F = ma = m (12 g) = 12 mg = 12 weight
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42
Ch 5 CP 6
The period of the moon’s orbit about the earth is 27.3 days,
but the average time between full moons is 29.5 days. The
difference is due to the Earth’s rotation about the Sun.
a) Through what fraction of its total orbital period does the
Earth move in one period of the moons orbit?
b) Sketch the sun, earth & moon at full moon condition.
Sketch again 27.3 days later. Is this a full moon?
c) How much farther does the moon have to move to be in
full moon condition? Show that it is approx. 2 days.
a) Earth orbital period = 365 days = 0E (I’d have called it TE)
Moon orbital period = 27.3 days = 0M ( “ TM)
0M/0E = 27.3/365.25 ≈ 0.0747
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Ch 5 CP 6 (con’t)
b)
(i)
(ii)
S
E
M
E
M
S
E
(iii)
M
Day 0
Full Moon
27.3 Days Later
This is not a full moon.
This is the next
full moon.
S
c) For moon to achieve full moon condition, it must sit along the line
connecting sun & earth. In part (a) we found that the earth has
moved thru 0.075 of its full orbit in 27.3 days (see diagram (ii)). To
be in-line w/ sun and earth, moon must move thru same fraction of
orbit (see diagram (iii)). This answer is only approx. (but not too
bad) since in 2 days the earth-sun line has advanced even further!
0.0747 (27.3 days) ≈ 2.04 days.
Apparent lunar month ≈ 29.34
Week
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44
days
Spring 2009
Ch 5 CP 6 (con’t)
b)
(i)
(ii)
S
E
M
E
M
S
M
E
(iii)
S
Day 0
Full Moon
27.3 Days Later
This is not a full moon.
This is the next
full moon.
Fraction of 2 pi
radians angle
Fraction of
a revolution
-------------
c) Exact solution: θE = 2π(t/365.25) = θM = 2π(t/27.3 - 1) with t in days
Here, earth makes a fraction of a revolution, and the moon makes one full revolution
plus the SAME fraction of the next revolution, to line up with the Sun-Earth
direction. We have to subtract off the full revolution to compare angles.
Solve: 1 + t/365.25 = t/27.3; 1 = -0.002738 t + 0.036630 t = 0.033892 t
t = 29.505 days
compare with approx. 29.34 days previous slide
Week 4
Physics 214
Spring 2009
45