Transcript v - Purdue Physics

This Week
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Circular motion
Going round the bend
Riding in a ferris wheel, the vomit comet
Gravitation
Our solar system, satellites (Direct TV)
• The tides, Dark matter, Space Elevator
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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 an object that is not moving in a straight line
is moving along the arc of a 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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Circular motion
a = v2/r
F = ma = mv2/r
If the velocity of an object changes direction then the object
experiences an acceleration and a force is required.
This is centripetal acceleration and force and is directed toward
the center of the circle.
This is the effect you feel rounding a corner in a car
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Balance of forces
We need to understand the forces
that are acting horizontally and vertically.
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.
Tv = W = mg
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Th = mv2/r
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Cars
Ff
Ff
Rear
Above
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 a force.
For a banked track there is a velocity for
which no friction is required.
Ff
W = mg
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Vertical circles
N
v
g
W = mg
mg – N = mv2/r
If v = 0 then N = mg
+ is always
toward the
center of
the circle
At the bottom
N - mg = mv2/r
As v increases N becomes
smaller
When v2/r = g the car becomes
weightless. Same as the “vomit
comet”
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Ferris wheel
Physics 214 Fall 2011
At the top
Mg – N = mv2/r
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Free fall versus Weightless
Near the earths surface there is a force F = mg which is called the
weight of an object. Since the force of gravity is always present
then this force always is present. There are situations where if we
were standing on a scale the apparent weight, the reading on the
scale, could be smaller or larger than the real weight.
• In an accelerating elevator
In the case of free fall our apparent weight is zero
• In an elevator and the cable snaps.
The case of being weightless corresponds to the case where one
is moving in a circle and the force of gravity is exactly equal to the
centripetal force required. Since the force of gravity is vertical the
type of cases where weightlessness happens are
•At the top of a ferris wheel if mg = mv2/r
•At the top of a hump back bridge if mg = mv2/r
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Planes and satellites
Vomit Comet
This uses a plane moving in a circle so that mg = mv2/r
Although this only occurs exactly at one point in the arc the
The trajectory of the plane is designed so that the effect of gravity
is very small for a much longer time
http://www.youtube.com/watch?v=xrjosOjvlOc
http://www.avweb.com/news/airman/184318-1.html
Space station
As a satellite circles the earth in a stable orbit we
have all the time that the gravitational force exactly
equals the centripetal force since the center of the
earth is the center of the circle and the direction of
gravity. So everything and everybody is weightless
all the time independent of mass. That is
toothbrushes and people are both weightless.
GmM/r2 = mv2/r or GM/r2 = v2/r
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Gravitation and the planets
Astronomy began as soon as man was able to observe the sky
and records exist going back several thousand years. In
particular 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 signalling the flooding of the Nile.
Copernicus was the first person to advocate a sun centered solar
system. Followed by Galileo who used the first telescopes
Tycho Brahe was the most famous naked eye astronomer.
Kepler, his assistant used the data to draw quantitative
conclusions.
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Keplers Laws
1) Orbits are ellipses
2) The radius vector sweeps out
areas in equal times equal
3) T2 proportional to r3
T is the period which for the earth
is one year and r is the average radius
For circular motion with constant velocity v
The circumference of a circle is 2πR and the
Period T = 2πr/v
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Newton and Gravitation
Newton developed the Law of
Gravitation
force between two objects is
F = GM1m2/r2 .
The constant of
proportionality was measured
by Cavendish after more than 100 years
G = 6.67 x 10-11 N.m2/kg2.
Since at the earths surface
mg = GmMe/r2 the experiment
measured the mass of the earth
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http://www.physics.purdue.
edu/class/applets/Newtons
Cannon/newtmtn.html
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Planetary orbits
For a simple circular orbit
GmM/r2 = mv2/r
T2/r3 = 4π2/GMs
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 a satellite.
For a geosynchronous orbit
period is 24 hours and height above the
earths surface is 22,000miles
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Calculations
GmMe/r2 = mv2/r where Me is the mass of the earth
and m the mass of a satellite.
At the earths surface mg = GmMe/Re2
v
r
v2 = gRe2 / r for a stable orbit
So at the earths surface
V = ~ 17500mph
For a synchronous satellite at a height of 22,000
miles
V = ~6500mph
V decreases with distance.
The moon is losing energy because of the tides so
it is moving away from us.
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Summary of Chapter 5
Circular motion and centripetal acceleration and
force.
Fc = mv2/r
Ferris wheel, car around a corner or over a hill.
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
T2/r3 = 4π2/GMs
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Examples of circular motion
Vertical motion
Looking down
N
N
v
v
W = mg
mg – N =
N = mv2/r
mv2/r
Side
N
mg
N - mg = mv2/r
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v
T
mg
Ff
mg
mg = Ff
Physics 214 Fall 2011
mg + T = mv2/r top
T - mg = mv2/r bottom
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1D-02 Conical Pendulum
Could
you find
the NET
force?
T sin(θ) = mv2/R
T cos(θ) = mg
v = sqrt( gR tan(θ) )
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
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1D-03 Demonstrations of Central
Force
What will happen
T
T
THE SHAPES/SURFACES
OF
SEMI-RIGID
OBJECTS
BECOME MORE
when it is
CURVED
TO PROVIDE
subjected
to forcesGREATER CENTRAL FORCES DURING ROTATION.
during rotation ?
θ
θ
2T cos (θ)= mv2/R
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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.
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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
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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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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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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 after the
string breaks (as seen from
above)? Explain.
4
•
3
2
A
1
Path number 3
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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
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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.
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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
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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
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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
gmoo
Wmoon = m gmoon
n
gmoon = 1/6 gearth
Wmoon = m 1/6 gearth = 1/6 m gearth = 1/6 (180 lb)
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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
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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 2r 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 = 2r/t = 2(12m)/8s = 9.42 m/s
Fcent
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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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 = 2r, 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 = 2(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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Ch 5 CP 6
The period of the moons orbit about the earth is 27.3 days,
but the average time between full moons is 29.3 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
Moon orbital period = 27.3 days = 0M
0M/0E = 27.3/364  0.075
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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 inline w/ sun and earth, moon must move thru same fraction of
orbit (see diagram (iii)).
0.075 (27.3 days)  2 days.
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Moon and tides
anim0012.mov
Tides are dominantly due to the
gravitational force exerted by the
moon. Since the earth and moon are
rotating this effect also plays a role.
The moon is locked to the earth so
that we always see the same face.
Because of the friction generated by
tides the moon is losing energy and
moving away from the earth.
4/4/2017
http://www.sfgate.com/getoutside/1996/jun/tides.html
whytides.gif
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Dark Matter
For the orbit of a body of mass m about a much more massive body of
mass M GmM/r2 = mv2/r and
GM/r = v2. In fact M is the mass inside the orbit, that is the sun could be
nearly as big as the orbit of the earth and it would not change anything. If
we look at stars in motion in galaxies we find there is not enough normal
matter to provide the necessary gravitational force. We believe this is
caused by a new form of matter, which we call dark matter, and it
comprises 25% of the energy in our Universe.(normal matter = 4.4%)
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Space Elevator
http://www.youtube.com/watch?v=F2UZDHHDhog
http://www.pbs.org/wgbh/nova/sciencenow/3401/02.html
station
v
counterweight
100,000km
4/4/2017
•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.
•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 and a ground based booster light beam.
•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 anchor ship, 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!
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Trajectories to other planets
To launch a space craft from earth to say
Mars or Jupiter is quite complicated
since all the planets are moving including
the earth. One needs to be able to
calculate a trajectory that minimizes the
amount of fuel required. That is why in
nearly all launches there are specific time
windows which are optimum.
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