Transcript Physics Chapter 7

Physics Chapter 7
Circular Motion and Gravitation
Section 7.1
Day1
Objectives:
• Solve problems involving centripetal
acceleration.
• Solve problems involving centripetal force.
• Explain how the apparent existence of an
outward force in circular motion can be
explained as inertia resisting the
centripetal force.
Section 7.1
Circular Motion
• Circular motion = any object revolving
about a single axis
• Axis of rotation = a line
• Centripetal vs. centrifugal
• Centripetal = center seeking
• Centrifugal = center fleeing
• Tangential speed = speed of an object that
is tangent to the object’s circular path
• When tangential speed is constant, the
object is moving with uniform circular
motion.
• Tangential speed depends on the distance
from the object to the center.
• Example: 2 carousel horses (outer horse
has greater tangential speed)
Centripetal acceleration is due to a
change in direction.
• Remember that acceleration can change due to
a change in magnitude of velocity or due to a
change in direction or both.
• Centripetal acceleration = acceleration directed
toward the center of a circular path due to a
change in direction
• ac = vt2
r
• Centripetal acceleration = (tangential speed)2
radius of circular path
A test car moves at a constant speed around a
circular track. If the car is 48.2 m from the track’s
center and has a centripetal acceleration of 8.05
m/s2, what is the car’s tangential speed?
r= 48.2 m
vt = ?
ac = 8.05 m/s2
Practice A p. 236
1. A rope attaches a tire to an overhanging tree
limb. A girl swinging on the tire has a centripetal
acceleration of 3.0 m/s2. If the length of the rope
is 2.1 m, what is the girl’s tangential speed?
2.
As a young boy swings a yo-yo parallel to the ground
and above his head, the yo-yo has a centripetal
acceleration of 250 m/s2. if the yo-yo’s string is 0.50 m
long, what is the yo-yo’s tangential speed?
3. A dog sits 1.5 m from the center of a merry-goround. The merry-go-round is set in motion, and
the dog’s tangential speed is 1.5 m/s. What is
the dog’s centripetal acceleration?
4. A race car moving along a circular track has a
centripetal acceleration of 15.4 m/s2. if the car
has a tangential speed of 30.0 m/s, what is the
distance between the car and the center of the
track?
• Tangential acceleration = an acceleration due to
a change in speed of a circular object
• Centripetal force = net force directed toward the
center of an object’s circular path
• Fc = mac
• Fc = mvt2
r
• Centripetal force = (tangential speed)2
radius of circular path
• Centripetal force is necessary for circular
motion.
A pilot is flying a small plane at 56.6 m/s in a
circular path with a radius of 188.5 m. The
centripetal force needed to maintain the plane’s
circular motion is 1.89 x 10 4 N. What is the
plane’s mass?
vt = 56.6 m/s
Fc = mvt2
r
r = 188.5 m
Fc = 1.89 x 104 N
Practice B p. 238
1. A 2.10 m rope attaches a tire to an
overhanging tree limb. A girl swinging on the
tire has a tangential speed of 2.50 m/s. If the
magnitude of the centripetal force is 88.0 N,
what is the girl’s mass?
2. A bicyclist is riding at a tangential speed of 13.2
m/s around a circular track. The magnitude of
the centripetal force is 377 N, and the combined
mass of the bicycle and the rider is 86.5 kg.
What is the track’s radius?
3. A dog sits 1.50 m from the center of a merrygo-round and revolves at a tangential speed of
1.80 m/s. If the dog’s mass is 18.5 kg, what is
the magnitude of the centripetal force on the
dog?
4. A 905 kg car travels around a circular track with
a circumference of 3.25 km. If the magnitude of
the centripetal force is 2140 N, what is the car’s
tangential speed?
Section 7.2
Objectives:
• Explain how Newton’s law of universal
gravitation accounts for various
phenomena, including satellite and
planetary orbits, falling objects, and the
tides.
• Apply Newton’s law of universal gravitation
to solve problems.
Newton’s Law of Universal
Gravitation
• Newton realized that gravitational force
was the centripetal force holding the
planets in orbit.
• Orbiting objects are in free fall.
• Free fall = the motion of a body when only
the force of gravity is acting on the body
• Free-fall acceleration on Earth’s surface =
9.81 m/s2
Orbiting Objects
• Newton realized that if an object were
projected at just the right speed, the object
would fall down toward Earth in just the
same way that Earth curved out from
under it.
• Gravitational force depends on mass and
distance.
Newton’s Law of Universal
Gravitation
Fg = (G) m1 m2
r2
Gravitational force = constant x mass 1 x mass 2
(distance between masses)2
G = constant of universal gravitation
= 6.673 x 10-11 N•m2
kg2
Mass of Object 1 (kg)
Mass of Object 2
(kg)
Separation Distance
(m)
Football Player 100 kg
Earth
5.98 x1024 kg
6.38 x 106 m
(on surface)
Earth
5.98 x1024 kg
6.38 x 106 m
(on surface)
Earth
5.98 x1024 kg
6.60 x 106 m
(low-height orbit)
d. Physics Student 70 kg
Physics Student
70 kg
1m
e. Physics Student 70 kg
Physics Student
70 kg
0.2 m
f. Physics Student 70 kg
Physics Book
1 kg
1m
g. Physics Student 70 kg
Moon 7.34 x 1022 kg
1.71 x 106 m (on surface)
h. Physics Student 70 kg
Jupiter 1.901 x 1027 kg
6.98 x 107 m (on surface)
a.
b. Ballerina 40 kg
c.
Physics Student
70 kg
www.glenbrook.k12.il.us/.../circles/u6l3c.html
Force of Gravity
(N)
Find the distance between a 0.300 kg billiard
ball and a 0.400 kg billiard ball if the
magnitude of th egravitational force between
them is 8.92 x 10-11 N.
Practice C p. 242
1. What must be the distance between two 0.800
kg balls if the magnitude of the gravitational
force between them is equal to that in the
sample problem?
2. Mars has a mass of about 6.4 x 1023 kg, and
its moon Phobos has a mass of 9.6 x 1015 kg.
If the magnitude of the gravitational force
between the two bodies is 4.6 x 1015 N, how
far apart are Mars and Phobos?
3. Find the magnitude of the gravitational force a
66.5 kg person would experience while standing
on the surface of each of the following planets:
Planet
Mass
Radius
a. Earth
5.97 x 1024 kg
6.38 x 106 m
b. Mars
6.42 x 1023 kg
3.40 x 106 m
c. Pluto
1.25 x 1022 kg
1.20 x 106 m
Newton’s law of gravitation
accounts for ocean tides.
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On the side of Earth that is nearest to the moon, the moon’s
gravitational force is greater than it is at Earth’s center.
This is because gravitational force decreases with distance.
The water is pulled toward the moon, creating an outward bulge.
On the opposite side of Earth, the gravitational force is less than it is
at the center.
On this side, all mass is still pulled toward the moon, but the water is
pulled least.
This creates another outward bulge.
Two high tides take place each day because when Earth rotates one
full time, any given point on Earth will pass through both bulges.
When the sun and moon are in line, the combined effect produces a
greater-than- usual high tide called a spring tide.
When the sun and moon are at right angles, the result is a lower-thannormal high tide called a neap tide.
Each revolution of the moon around Earth corresponds to two spring
tides and two neap tides.
www.this-magic-sea.com/TIDE.HTM
• Other things influencing tides: depth of
ocean, Earth’s tilt, rotation, friction
between ocean floor and water, and sun.
Why the moon not the sun?
• Tidal forces arise from the differences
between the gravitational forces at Earth’s
near surface, center, and far surface.
• As the distance (r) between two bodies
increases, the tidal force decreases as
1/r3. For this reason, the sun, which has a
greater mass than the moon does, has
less effect on the Earth’s ocean tides than
the moon does.
• Tidal forces are exerted on all substances.
• The amount of distortion depends on the
elasticity of the body under gravitational
influence.
• If an orbiting body moves too close to a
more massive body, the tidal force on the
orbiting body may be large enough to
break the body apart.
• The distance at which tidal forces can
become destructive is called Roche’s limit.
• Gravity is a field force.
• Gravitational force is an interaction between a
mass and the gravitational field created by other
masses.
• Gravitational field strength equals free-fall
acceleration.
• Weight changes with location.
• Gravitational mass equals inertial mass.
• Gravitational mass = weight/ gravitational field
strength at the location of measurement
• Fg = (G) m1 m2
r2
• F = ma
inertial mass
Section 7.3
Day2
Objectives:
• Describe Kepler’s laws of planetary
motion.
• Relate Newton’s mathematical analysis of
gravitational force to the elliptical planetary
orbits proposed by Kepler.
• Solve problems involving orbital speed
and period.
Motion in Space
• In 1543 – Polish Astronomer – Nicolaus
Copernicus proposed that the earth and
other planets orbit the sun in perfect
circles
• Kepler (1571- 1630) 3 laws of planetary
motion
Kepler’s 3 Laws of Planetary
Motion
1. Each planet travels in an elliptical orbit around
the sun, and the sun is at one of the focal
points.
2. An imaginary line drawn from the sun to any
planet sweeps out equal areas in equal areas
in equal time intervals.
3. The square of a plantet’s orbital period (T2) is
proportional to the cube of the average
distance (r3) between the planet and the sun,
or T2 α r3.
• First Law – orbits are elipses not circles
• Second Law – planets travel faster when they
are closer to the sun
• Third Law – orbital period = time a planet takes
to finish one full revolution (T) and distance (r) =
mean distance between the planet and the sun
(this also applies to satellites orbiting Earth,
including the moon) T12 = r13
T22 r23
• Kepler’s laws are consistent with Newton’s law
of gravitation.
• Kepler’s third law describes orbital period
• Another way of stating Kepler’s third law:
T2 = 4π2 • r3
Gm
• Assume a circular path. This is a close
approximation for planets in our solar system,
because most are nearly circular (except
Mercury and Pluto)
Period and Speed of an Object in
Circular Orbit
• T = 2π2
• v=G
•
•
•
•
•
r3
Gm
m
r
T = orbital period
v = orbital speed
r = mean radius
m = mass of central object
G = gravitational constant
Practice D p. 251
Use chart p. 250
1. Find the orbital speed and period that the
Magellan satellite from the sample
problem would have at the same mean
altitude above Earth, Jupiter, and Earth’s
moon.
2. At what distance above Earth would a
satellite have a period of 125 min?
Section Review p. 253
5. Find the orbital speed and period of
Earth’s moon. The average distance
between the centers of Earth and the
moon is 3.84 x 108 m.
Weight and Weightlessness
• What is weight?
• If a friend pushes down on you while you are on
the scale, what will happen?
• The scale reading is a reading of the normal
force acting on you. Why?
• In an elevator, the normal force will be smaller
as the elevator accelerates downward.
• If the elevator’s acceleration were equal to freefall, you would experience apparent
weightlessness.
• Astronauts in orbit experience apparent
weightlessness.
• The force due to gravity keeps the
astronauts and shuttle in orbit, but the
astronauts feel weightless because no
normal force is acting on them.
• Our bodies rely on this. Prolonged
weightlessness produces weakened
muscles and brittle bones.
• Astronauts in orbit experience apparent
weightlessness.
• Actual weightlessness occurs only in deep
space, far from stars and planets.
• Gravity is never entirely absent, but if an object
is far enough from any masses, gravity can
become negligible.
• If this were the case, the object would drift
through space in a straight line at constant
speed.
Section 7.4
Objectives:
• Distinguish between torque and force.
• Calculate the magnitude of a torque on an
object.
• Identify the six types of simple machines.
• Calculate the mechanical advantage of a
simple machine.
Torque and Simple Machines
• Torque = a quantity that measures the
ability of a force to rotate an object around
some axis
• Torque depends on the force and the lever
arm.
• Lever arm = the perpendicular distance
from the axis of rotation to a line drawn
along the direction of the force.
The Inclined Plane: Often referred to as a 'ramp' the
inclined plane allows you to multiply your force over a
longer distance. In other words, you exert less force
but for a longer distance. You do the same amount of
work, it just seems easier because you spread it over
time.
The Wedge: A wedge works in a similar way to the
inclinded plane, only it is forced into an object to
prevent it from moving or to split it into pieces. A knife
is a common use of the wedge.
The Screw: The screw is really just an inclined plane
wrapped around a rod. It too can be used to move a
load (like a corkscrew) or to 'split' and object (like a
carpenter's screw).
www.coolschool.ca/lor/SC9/unit16/U16L04.htm
The Lever: The lever is simply a bar
supported at a single point called the
fulcrum. The positioning of the fulcrum
changes the mechanical advantage of the
lever. Look at how you can manipulate the
position of the fulcrum relative to the
heavier weight to lift the 200g mass with
only 100g of force...
The Wheel and Axle: Any large disk (the
wheel) attached to a small diameter shaft
or rod (the axle) can give you mechanical
advantage. Turning a screw with a
screwdriver is a simple example of a wheel
and axle. Can you think of others we use
everyday?
The Pulley: A pulley is any rope or cable
looped around a support. A very simple
pulley system would be a rope thrown over
a branch to hoist something into the air.
Often, pulleys incorporate a wheel and axle
system to reduce the friction on the rope
and the support.
www.coolschool.ca/lor/SC9/unit16/U16L04.htm