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Chapter Five Lecture
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Circle, Round & Round, & Gravity
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 Formulas:
 Constants:
G = 6.67´10-11 N-m2/kg2.
Demonstrations:
- Twirl ball on a string
- Twirl water in a coffee can
Main Ideas:
- Uniform circular motion
- Non-uniform circular motion
- Universal Gravitation
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Uniform Circular Motion
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 We have talked about motion
where the acceleration is either zero or is a constant magnitude in a constant direction. We will now talk about cases where the acceleration is a constant magnitude and always
directed perpendicular to the direction of motion. This leads to uniform circular motion - as we shall see.
Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path. Recall that we said that
velocity is a vector, so even if an object has a constant speed, but is changing direction it is accelerating. Something following a uniform
circular path is always changing direction, so it is always accelerating. Let's see what the magnitude and direction of its acceleration is.
v1 = v2 and r = r and angles are the same, so the triangles are the same and
(Book describes one way of getting the direction. Also, the direction can be seen by looking at when Dv
changes by 90°.) The direction is always toward the center of the circle.
This is called the
centripetal acceleration (center seeking). It is the acceleration whose direction is toward the center of the circle, and
whose magnitude is equal to v2/r. What would happen if the object did not have this acceleration? (It would go straight by Newton's first law).
So everything moving at a uniform speed in a circle is acceleration by this amount. Note that the acceleration is not constant. The magnitude of
the acceleration is, but its direction is changing and acceleration is a vector.
What is required, according to Newton's second law, for there to be
acceleration? There must be a net force in the direction of the acceleration. So,
It takes a force of magnitude mv2/r directed towards the center of the
circle in order for an object to remain in a circular motion. This is called centripetal force. It is not a magic new force. It is simply a force
applied by a number of different methods: A string twirling a ball, Friction of tires on a road, Friction of a seat in a car. There is no such
thing as centrifugal force (center-fleeing). That is a pseudo-force that we say we feel because we are not going in a straight line. If there were
such a force, a ball on a string would fly outward after being let go. Instead it goes in a straight line (according to Newton's first law).
Demonstration: Ball on a string.
Let's look at the example of circular motion for a car turning on a level, flat road.
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Problem: A car turns a corner on a road of radius 50.0 m. If the
coefficient of static friction ms = 0.90, what is the maximum speed the car can negotiate the turn without sliding?
Draw a free-diagram (right).
Vertical Forces: Horizontal Forces:
Remembering that
 , we can plug this
into the friction equation.
and we also have the normal for Fn, which can be used.
What if the road is icy and friction ms = 0.1?
Then v = 7.0 m/s = 15.7 miles/hour.
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NOTE: We did not need the mass of the car. A designer of the road
does not need to worry about the mass of the vehicle (as far as curves are concerned, not as far as wear and tear on the road), but does need to worry about how good the tires on cars are (ms). The designer does need to worry about where the center of gravity of a vehicle is or it might tip
over, which we will talk about in chapter 9.
NOTE: The centripetal acceleration always equals the sum of the forces
for uniform circular motion. The important forces are those pointing inward (outward). So while doing problems with circular motion, the two
important axes are tangent to the circle and perpendicular to the circle.
Remember our example of a car turning a corner. There is a way to make
a sharper turn with a lower coefficient of friction. We can do that by banking the corner. Why does this work? - Because now, part of the normal force is used to provide the centripetal force.
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Problem: Let's first ask: What is the angle we need to bank the curve so
that a car doesn't need any friction to stay on the curve.
Note, here we use the normal x and y axes because the acceleration is along the horizontal.
 Along y direction:
Along x direction:
so since we have two equations that solve Fn we can set the two
equations equal to each other.
So to design a road with radius 50 m for a speed of 15 m/s the road must be banked at an angle of 25°.
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Problem: Suppose, I go back to the first case of a road with ms=0.9, but now I add a banked turn of 15°. How fast can I
safely go around the curve now? (r=50m)
 After drawing a free body diagram, we
can apply Newton's Laws again. We must also remember that
In y direction:
In x direction:
 As before, we can set both Fn's equal.
Here we go!
(It was 21 m/s with no banking).
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Demonstration - Twirl a coffee can around on a string with a ball or water in it.
If the string is 1 m long, what is the minimum velocity to spin without the water falling out?
At top:
Since one revolutions is a distance of pr
=(3.14)(1m)=3.14m, then the minimum speed not to have water fall with a one meter string is one revolution in one second, or mathematically
What is the tension force at the bottom for such a case?
This is more than just mg.
Normal is less at top or may be even zero. Think about a roller coaster.
The normal (what you "feel") is less at top. The normal is strongest at the bottom. You feel the seat pressing up into you on a roller coaster.
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Problem: I rotate a ball at an angle of q=30o. What is the centripetal acceleration? If the string is 1 meter long, how fast is it
rotating?
 In the vertical y-direction:
In the horizontal x-direction:
Set the two equations of T equal to each other
To determine the velocity, we simple remember the definition of centripetal acceleration.
(circumference is 2pr = 3.14 m), so it should take 3.14 m/1.68 m/s =
1.86 seconds to go one revolution - or 5 revolutions in 9.35 seconds. Note that r is not 1 meter, but (1m) sinq.
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Non-Uniform Circular Motion
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 Motion in a circle, which is not at a constant speed, is called non-uniform
circular motion. For this to occur there must be a component of acceleration perpendicular to the direction of motion as well as parallel to the direction of motion. There is the
centripetal acceleration, ac, and the tangential acceleration aT. Since acceleration is a vector, the total acceleration is the vector sum of the two.
a = ac + aT so the magnitude of a is
If the speed is increasing, then it is possible to have a constant magnitude of aT, but not of ac.
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Problem: Suppose I go from 0 to 22 m/s in around a quarter circle turn with a radius of
87 m. Assuming constant tangential acceleration, what is aT, and ac when my velocity is 15 m/s?
The distance is (0.25)(2pr) = 137 m.
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Note that the tangential acceleration changes as the tangential velocity changes.
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Newton's Law of Universal Acceleration
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 Newton realized that gravity worked over a distance (the apple?), and
he proposed that gravity worked over large distances, even from the earth to the moon, or between heavenly bodies.
He proposed his famous law of universal gravitation.
The direction is toward the two objects. What does this mean?
The force never dies out. Every body in the universe feels gravity from every other body in the universe.
 The larger the mass, the stronger the
force. The force between two objects is the same (equal and opposite). The accelerations are different due to their
different masses. If I double the mass of an object, the force doubles. If I double the distance of an object, the force is cut to one quarter of its value.
The value of G is
 . We can measure the distance from
the center of an object when the object is spherical. (This is why Newton developed calculus).
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 Problem: What is the gravitational force on the moon
from the earth?
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This force is a vector, as are all forces, and must be added as vectors, by components.
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Problem: Suppose I place three billiard balls, all of mass .300 kg on a
table at the corners of a right triangle (.3 m, .4 m, .5 m). What is the net gravitational force on the ball at the corner from the other two
balls? (Not from the Earth). Even billiard balls exert a gravitational force on each other.
The total force is
With an angle of
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Example 5-10 in the book is just like this, but instead of 3 billiard balls,
the three objects are the earth, the moon and the sun. The book shows FEM = 1.99´1020N and FSM = 4.34´1020N. Why is the force between the moon and the sun greater than the force between the moon and the
Earth? If so, the moon is rotating around the sun. It is!
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Gravity Near the Earth
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What is force of gravity near the surface of the Earth? If I am
standing on the Earth, we have said that the force is "mg". That must come from Newton's Law of Universal Gravitation. Let's see, how far am I from the center of the Earth?
How far above the earth must I go for this force to be half of this value?
which is 9.02´106 m - 6.38´106 m = 2.64´106 m above the earth.
 Large objects near us influence what we perceive as the acceleration of gravity. Near
a mountain, there is mass pulling us toward the mountain.
Most of the mass is still underneath us, so the difference is very small, but it is
there. Also if you go to higher elevations, you are farther from the center of the earth so g goes down very slightly, as well.
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Satellites & "Weightlessness"
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 Suppose I am bouncing on a trampoline or bungee jumping off a bridge. What forces are acting on me? -
Gravity and air resistance. Let's disregard air resistance for a moment. The only force acting on me is gravity. We say that I am in free fall. Now consider an astronaut in
orbit, or the moon orbiting the earth. What force is acting on it? Only gravity. These are in free fall as well. They are falling toward the Earth. The moon and astronaut are in
free fall around the Earth. The only force acting on them is the force of gravity. Why don't they hit the Earth? Because they are moving tangent to the Earth and according to Newton's First Law
, they continue in motion. There is no tangential force, so they are in free fall. We call this weightlessness, but it is not really. There is a weight (force=mg) pulling the object down.
The object is still experiencing a force and it IS accelerating. The acceleration is always perpendicular to the velocity
so the velocity (vector) is changing, but the speed (scalar) is not. We call it weightless, because if we were to stand on a scale in free fall, our
weight would read 0. Why? - Because the scale would be falling as well. The scale would exert no normal force on us and thus read 0.0N.
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 Problem
: What is the speed of a satellite in orbit around the Earth at a distance of 12,200km above the surface of the Earth?
(Note: the universal law of gravity uses distance from the center of the Earth. The radius of the Earth is R=6.38´106m.
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All satellites at the same radius have the same speed, regardless of their
mass. Often we write the speed as a function of the period, where the period is the time it takes to make one orbit.
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Problem: How long does it take the above satellite to circle the Earth?
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Kepler's Laws
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 Long before Newton proposed his three laws of motion
and his universal law of gravitation, Johannes Kepler had developed three laws that describe the motion of the planets. Newton showed that his law of universal gravitation
predicted Kepler's Three Laws. It was an amazing prediction and helped convince people that Newton's law was correct. They are
- All planets move in elliptical orbits with the Sun at on focal point.
- A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.
- The square of the orbital period ("T") of any planet is proportional to the cube of the average distance ("R") from the planet to the Sun.
or if we take the ratio of two planets
For this to work we must measure the period and the average radius
of two orbiting objects around the same object, like earth and mars around the Sun, but not earth around the sun and moon around the earth.
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Problem: Neptune is an average distance of 4.5´109 km from the Sun. What is the length of Neptune's year? The earth is 1.5´108 km from the Sun.
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Problem: Determine the mass of the Earth from the known distance and
period of the moon. Distance from the Earth to the moon is 3.84´108m
and the period of the moon to orbit the Earth is 27.4 days = 2.37´106s.
Look at units remembering that 1 Newton = 1kg-m/s2
 So the units work out correctly. |
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