Transcript centripetal force

Circular Motion
Reviewing….
Rotating
Turning about an
internal axis
Revolving
Turning about an
external axis
Linear speed, v
How far you go in a certain amount of time
Miles per hour, meters per second
Rotational (angular) speed,
How many times you go around in a certain
amount of time
Revolutions per minute, rotations per hour,
radians per second
Which horse has a
larger linear speed
on a merry go round,
one on the outside or
one on the inside?
Outside.
Which horse has a
greater rotational
speed?
Neither, all the horses
complete the circle in
the same amount of
time.
The number of revolutions per second is
called the frequency, f.
Frequency is measured in Hertz, Hz.
The time it takes to go all the way around
once is called the period, T.
Frequency is related to period by
f=1/T
How do you find the velocity if it is not
directly provided?
Velocity = distance / time
In circular motion, the distance traveled is all
around the circle… the circumference.
The circumference = 2pr
So…
v = 2pr / T
Uniform Circular Motion, UCM: moving in a circle
with a constant speed.
Question: Is there a constant velocity when an
object moves in a circle with a constant speed?
No, the direction changes, therefore the velocity
changes.
If the velocity changed, the object is actually
ACCELERATING even while moving at the
same speed.
Now on to some new things…..
Suppose an object was moving in a straight line with some
velocity, v.
According to Newton’s 1st Law of Motion, “An object in motion
continues that motion unless a net external force acts on
it”.
If you want the object to move in a circle, some force must
push or pull it towards the center of the circle.
A force that pushes or
pulls an object towards
the center of a circle is
called a centripetal force
Centripetal means “center
seeking”
According to Newton’s 2nd Law, SF = ma, If
there is a centripetal force, there must be
a centripetal acceleration.
ac = v2 / r
Where r is the radius of the circle and v is
the velocity of the object.
Centripetal force
Since SF= ma, the net centripetal force is
given by
2
v
SF  m
r
Lots of forces can help in pushing or pulling
an object towards (or away from) the
center of a circle. Sometimes it takes
more than one force to get an object to
move in uniform circular motion.
Centripetal force is NOT a new kind of force.
If an object moves in a circle (or an arc),
there must be at least one force that is
acting toward the center of the circle.
When can these forces be
centripetal forces?
Gravity?
Moon revolving around the
Earth
Tension?
Twirling a pail at the end of
a string
Friction?
Cars rounding a curve.
Air Resistance (“Lift”)?
Airplane or birds flying in a
circle.
Normal?
Riders in a carnival ride
What happens if the string breaks?
Which way will the ball move?
The ball will continue to move in a
straight line path that is “tangent”
to the circle.
Tension in a string as a
centripetal force
A student twirls a rock around
and around in a horizontal
circle at the end of the string.
The only force that contributes
to a NET centripetal force is
the tension in the string.
Example
A boy twirls a ½ kg rock in a horizontal circle on the end
of a 1.6 meter long string. If the velocity of the rock was
4 m/s, what is the Tension in the string?
m = ½ kg
r = 1.6 m
2
v = 4 m/s
The only centripetal force is Tension.
T = m v2 / r
T = ½ 42 / 1.6
T=5N
v
SF  m
r
Example
How fast was the ½ kg rock moving if the Tension
was 10 N and the string was 1.6 m long?
m = ½ kg
r = 1.6 m
T = 10 N
T = mv2 / r
Tr/m = v2
10 x 1.6 / .5 = v2
v = 5.7 m/s
Friction along a surface as a centripetal force
A 1500 kg race car goes
around a curve at 45 m/s. If
the radius of the curve is
100 m, how much friction is
require to keep the car on
the track? What is m, the
coefficient of friction?
m = 1500 kg
v = 45 m/s
r = 100 m
The centripetal force is friction.
v2
SF  m
r
2
f = mv /r
f = 1500 x 452 / 100
f = 30375 N
f = mN
m= f / N but what is N?
N = mg = 15000 N
m = 30375 N / 15000 N
m = 2.02
The Normal force
In some cases the normal force can contribute to the net centripetal
force.
For example, on the carnival ride where the riders stand against the
walls of the circular room and the floor drops out! And yet, the rider
does not slide down!
Draw the free-body diagram!
What keeps them from sliding down?
The wall pushes against the rider toward the center of the circle.
2
SFcentripetal
v
m
r
N = m v2 / r
Also, if he is not sliding down, we know
f = mg and therefore…..mN = mg
We can combine those two equations
(divide one by the other!)
f
N
mg
Vertical loops
Twirling a rock at the end of a string in a
vertical loop.
At the top of the loop, both the Tension
and the weight point towards the
center of the circle!
SF = T + mg = mv2/r
At the bottom of the loop, the Tension
points toward the center, the weight
away from the center:
SF = T – mg = mv2/r
What about an object on a vertical
track?
At the top of the track, both the
Normal force (the track pushing
against the ball) and the weight
point down toward the center of
the circle, therefore, they are
both positive:
SF = N + mg = mv2/r
At the bottom of the track, the
Normal force points toward the
center and the weight points
away from the center:
SF = N – mg = mv2/r
Loop
the
Loop
What is the minimum speed that a rider must be moving at in order to
complete a loop the loop of radius 12 m?
At the top of the loop, both the Normal force and weight point towards
the center of the circle, so
SFcentripetal = N + mg = mv2 / r
However, at the minimum required speed, contact is lost for a moment at
the top of the loop, so that…
The Normal force goes all the way to ZERO.
The weight is the only centripetal force when the rider is moving at the
minimum required speed.
v2
SF  m
r
2
mg = mv /r
g = v2/r
v2 = rg
v2 = 12 x 10
v = 10.95 m/s
“Artificial Gravity”
Occupants of a space station feel weightless because they lack a
support (Normal) force pushing up against their feet. By spinning
the station as just the right speed, they will experience a “simulated
gravity” when the Normal force of the floor pushing up on their feet
becomes a centripetal force. The closer their centripetal
acceleration, v2/r is to g, the acceleration due to Earth’s gravity, the
more they feel the sensation of normal weight.
Artificial Gravity Example…
A circular rotating spacestation has a radius
of 40 m. What linear velocity, v, must be
maintained along the outer edge, to maintain
a sense of “normal” gravity?
We want the centripetal acceleration, v2/r
(due to the rotation) to be the same as the
acceleration due to gravity on Earth- 9.8 m/s2
v2/r = 9.8 m.s2
v2 = 9.8 x 40
v = 19.8 m/s
Newton’s
Universal
Law of
Gravitation
The Gravitational Force
Newton’s Universal Law of Gravitation
states that every particle in the universe
exerts an attractive force on every other
particle.
FG
m1m 2
d
2
Where “G” is the “universal gravitational constant”
G = 6.67 x 10-11
FG
m1m 2
d
2
What happens to the Force if one of the masses is doubled?
It is also doubled.
What happens to the Force if both of the masses were
doubled?
It will be four times larger.
What happens to the Force if one of the masses is doubled
and then other one is halved?
The Force will remain the same.
This is an “inverse square” law, since the
Force is proportional to the inverse of the
distance squared.
m1m 2
1
FG

d2
d2
Example:
At twice the distance, the gravitational
force between two objects would be
less. How much less?
1
1
F 2  2 ?
d
2
Two objects are separated by some
distance, d. How would the
gravitational force differ if the distance
was tripled?
1/9 the original force
What if the distance was 4d?
1/16 the original force
5d?
10d?
½ d?
4 times the original force
Example: Two masses of 8 kg and 12 kg
are separated by 1.5 m. What is the
gravitational force they exert on each
other?
m1m 2
FG
2
G = 6.67 x 10-11
d
How do you enter all those numbers in
your calculator? Use your exponent
button for “G”!!
6.67E-11*8*12÷1.52 =
F = 2.85 x 10-9 N
What is the gravitational force between a
600 kg mass and a 850 kg mass if they are
0.4 meters apart?
FG
m1m 2
d
2
F = 2.126 x 10-4 N
G = 6.67 x 10-11
Example: Two masses of 3 x 103 kg and
1.8 x 1015 kg are separated by
d = 1.4 x 1021m. What is the gravitational
force they exert on each other?
m1m 2
FG
2
G = 6.67 x 10-11
d
How do you enter all those numbers in
your calculator? Use your exponent
button!!
6.67E-11*3E3*1.8E15÷1.4E21 2 =
F = 1.84 x 10-34 N
If the gravitational force between a 75 kg mass
and a 120 kg mass is 4.2 x 10-4N, how far apart
are they?
FG
m1m 2
d
2
G = 6.67 x 10-11
What’s the shortcut to get d2 out of the
denominator?
Trade places with F!!
d = 0.0378 m
NET Gravitational Force
Two masses pull on the central mass.
How would you get the NET gravitational
force?
Subtract the two forces.
NET Gravitational Force
Two masses pull on the left mass.
How would you get the NET gravitational
force?
Add the two forces. (Be careful about your
distances!)
NET Gravitational Force
Two masses pull on
the mass at the
origin.
How would you get
the NET
gravitational force?
Pythagorize the two
forces.
2nd tan for angle.
Cavendish and “G”, the gravitational constant
Henry Cavendish, a British scientist,
first devised an experiment to
determine “G” in 1797.
He suspended two small known
masses from a “torsion wire” of
which he knew the strength.
These two small masses were
gravitationally attracted to two
large known masses, which
caused the wire to twist until the
torsion force was balanced by the
gravitational force. Because he
knew the strength of the torsion
force, he also knew the strength
of the gravitational force.
With known masses, known Force,
and known distance, the only
“unknown” left was G!
FG
m1m 2
d
2
* You need to know who
first determined “G”,
Finding “g”
Weight is the gravitational force a planet
exerts.
Weight = Gravitational Force
mg = G
m1planet
m2
2
d
“g”, the acceleration due to gravity can be found by
canceling an “m”.
The distance, d, is measured from the center of the
planet to the location of interest. (often, the radius)
The acceleration due to gravity, “g”, is also called the
“gravitational field strength”.
How large is “g” on the planet
Venus, which has a mass of
4.87 x 1024 kg and has a radius
of 6,050,000 meters?
m planet
gG
2
d
6.67E -11x 4.87 E24 ÷ 6,050,0002
=
g = 8.87 m/s2
Example: An asteroid
of radius 500 m has a
mass of 6.5 x 1013 kg.
What is the
gravitational field
strength at its
surface?
gG
m planet
d
2
6.67E -11 x 6.5 E13 ÷
5002 =
g = 0.0173 m/s2
Aristotle
Geocentric
universe
384 BC
“geocentric” – Earth centered universe…… WRONG!
Ptolemy, 83 AD
Ptolemy (also geocentric universe)
presented his astronomical models in
convenient tables, which could be
used to compute the future or past
position of the planets, the Sun, and
Moon, the rising and setting of the
stars, and eclipses of the Sun and
Moon. His model showed the planets
turning in small circles as they
orbited the Earth!
The tables actually produced fairly
good predictions, but his model and
his geocentric universe was…..
WRONG!
Ptolemy was also the first to use
latitude and longitude lines.
Copernicus 1473
heliocentric universe
“sun-centered” universe
Although others
before him had
proposed that the
planets orbit the sun
rather than the Earth,
Copernicus was the
first to publish
mathematical
evidence
• Tycho Brahe
• 1546
• Built “The
Castle of the
Stars”
• Had an accident
in a duel
• Died an unusual
death…
Johannes Kepler
1571
A mathematician
hired as Brahe’s
assistant
• Wrote Three
Laws of
Planetary Motion
1. The Law of Orbits:
All planets have
elliptical orbits with
the sun at one
focus.
Planets’ orbits are only
slightly elliptical
Comets have highly elliptical orbits
2. The Law of Areas:
A line that connects
a planet to the sun
sweeps out equal
areas in equal times.
(Planets move faster
when they are
closer to the sun.)
Perigeeclosest
distance
Apogeemost
distance
location
3. The Law of Periods: The
square of the period of a
planet is proportional to the
cube of its average orbital
radius. (this is easy IF you
measure using some different
units than we usually use)
(earth years) T2 = r3 (AU)
An AU is an “astronomical
unit” and is the distance from
the Sun to the Earth.
Kepler’s Third Law in fundamental units of
seconds and meters (it’s a LOT more
complicated…
We know that the centripetal force
for satellites in circular orbit is
the gravitational force.
Solve this for v of a satellite in
CIRCULAR ORBIT.
We also know the velocity for any
circular motion.
Set those two values for v equal v circular orbit
and solve for the period T.
Cancelling and rearranging give
us Kepler’s Third Law with the
standard units of meters and
seconds.
2
This law will work for circular
orbits around ANY body as
long as you know the mass of
that body.
2
Mm
v
G 2 m
d
r

GM
r
2pr
v
T
4p
3
T (
)r
GM
2
But… let’s stick to the easier units for now…
Venus is located 0.72
AU’s from the Sun.
How many years
does it take Venus to
orbit the sun?
T2 = a3
X cubed
T  a  0.72 
3
0.61 years
Math button or
use housetop
3
If it takes an asteroid 5
Earth years to orbit
the sun, how far is the
asteroid from the sun?
Math button
T2 = a3
3
3 2
a T  5 
2
2.92 AU
cube root
“Newton’s Cannon”
If you fire a rocket horizontally from the top
of a very high mountain, gravity will pull it
towards the center of the Earth and it will
go a certain distance before it hits the
ground.
“Newton’s Cannon”
The faster the rocket is launched, the further
it will go as it falls until it hits the ground.
“Newton’s Cannon”
If it is launched fast enough, the pathway of the
rocket as it falls will exactly match the curvature
of the Earth. The satellite will continue to fall
and fall and fall, but it will never fall to the
ground. It goes into orbit about the Earth!
“Newton’s Cannon”
The horizontal speed of the orbiting rocket must be
very high and it must maintain that speed or it will
fall into the Earth. Satellites in orbit are always
FALLING… and FALLING… and FALLING…
“Newton’s Cannon”
This is why the astronauts appear to be
weightless. There is in fact plenty of gravity
at the elevation of the space shuttle and
space station, but since they are always
falling, they appear to be in zero gravityweightless.
Satellite Motion
The gravitational force provides the
centripetal force for an orbiting satellite in
circular orbit.
G
m1m 2
d
2
2
v
m
r
v sat  G
mplanet
d
For any object moving in a circle the velocity is
given by
v = circumference / time = 2pr / T
Where T is the period of the motion – the time to go
around once.
Therefore, for a satellite in circular orbit:
v sat  G
mplanet
d
= 2pr / T
If a rocket is launched VERTICALLY from the surface
of a planet, is it true that what goes up must come
down?
If we throw a ball into the air, it reaches some highest
point and then gravity pulls it back down.
If we throw it faster, it will go higher before it comes
back down.
However, if we throw it fast enough, it can escape
the gravitational pull of the Earth and keep moving
upward.
The speed at which that will occur is called the
ESCAPE SPEED. The escape speed for a planet
is given by
mplanet
ESCAPE SPEED = v escape  2G
d
Geosynchronous Satellites
A geosynchronous satellite is one whose orbital period is
the same as Earth’s rotational period.
So, as Earth rotates once every 24 hours, the satellite
orbits the Earth once every 24 hours.
This means that when the orbit lies entirely over the
equator, the satellite remains stationary relative to the
Earth's surface and the antennae’s do not have to “track”
it continually. These satellites are used for
communications, and intelligence!
How many satellites are in Earth’s orbit?
One natural satellite
Over 8000 artificial satellites!
There are two main processes constantly
going on in the super massive stars:
nuclear fusion (which tends to blow the
star's hydrogen outward from the star's
center) and
gravitation (which tends to pull all
hydrogen back in the direction it had come).
These two processes balance one another
until all the star's hydrogen is exhausted,
allowing gravitation to take over.
Once gravitation dominates, the star
becomes unstable and starts to collapse.
Once a super massive star starts to
collapse, it does not stop, and the star (and
ultimately its atoms) will cave inward upon
itself, resulting in the formation of a black
hole (Hewitt 186).
When a star has exhausted its
fuel supply, gravitational forces
crush the star to one of three
possible outcomes:
1) The star shrinks and
stabilizes into a white dwarf.
2) The star crunches into a
neutron star.
3) The star collapses to a black
hole.
A star less than 1.4 times the
mass of the sun will become a
white dwarf. A star between 1.4
and 3 times the mass of the
sun will become a neutron star.
It's only those stars greater
than 3 times the mass of the
sun that become black holes
upon collapse.
How do you “see” a black hole
when it can’t be seen??
When a star collapses and changes
into a black hole, the strength of its
gravitational field still remains the
same as it had been before the
collapse. Therefore the planets in orbit
would not be affected. The planets
would continue in their orbits as usual
and would not be drawn into the black
hole. Because black holes do not give
off any light, the planets would appear
to be orbiting around nothing. There is
reason to believe that the planets
could just be orbiting about a star that
is too faint to be seen, but there is an
equal chance that a black hole could
be present (Hewitt 187).
Because the gravity of a black hole is so intense,
dust particles from nearby stars and dust clouds are
pulled into the black hole. As the dust particles
speed and heat up, they emit x-rays. Objects that
emit x-rays can be detected by x-ray telescopes
outside of the Earth's atmosphere (Miller).
Black holes can also be detected through a technique
called gravity lensing. Gravity lensing occurs when a
massive object, in this case a black hole, passes between a
star and the Earth. The black hole acts as a lens when its
gravity bends the star's light rays and focuses them on the
Earth. From an observer's point of view on the Earth, the
star would appears to brighten or to be distorted or to be in
a different part of the sky.
The event horizon is the boundary around a black
hole where gravity has become so strong that
nothing- not even light- can escape. The
escape velocity at the event horizon = c. The
escape velocity inside the event horizon > c,
therefore, escape is impossible. The event
horizon is the point of no return.