Chapter 6: Circular Motion, Orbits, and
Gravity
Tuesday, September 17, 2013
10:00 PM
Circular Motion
Rotational kinematics
We'll discuss the basics of rotational kinematics in this chapter; the kinematics
equations for constant angular acceleration are discussed in Chapter 7. The
basic quantities of rotational kinematics are angular position, angular
displacement, angular velocity, and angular acceleration.
angular position … measured in degrees or radians
review of angle measure in degrees and radians; remember that the radian is a
"unitless" unit
Remember that for a complete circle, the circumference is
and therefore the angle of a complete circle is
But we know that for a complete circle,  = 360°. Thus,
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Using these two conversion factors allows one to convert from
degrees to radians, or vice versa. For example,
Angular displacement and angular velocity
Connecting linear and angular kinematic quantities
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Example: maximum speed for a car turning around a curve on a level
road with friction
A car of mass 1200 kg moves around a curve on level ground that has a
a radius of 20 m. Determine the maximum speed for which the car can
safely move around the curve. The coefficient of friction is 0.5.
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Thus, the maximum acceleration that friction between the tires and the road can
produce is 4.9 m/s2.
Notice that the maximum safe speed through the curve is independent of the mass
of the vehicle; thus, a speed limit sign can be used that is appropriate for all
vehicles, whether they are light motorcycles or heavy transport trucks.
Example: banking angle for a highway curve
Determine the ideal banking angle for a highway curve that has a (horizontal)
radius of 20 m. Suppose that the typical driving speed around the curve is
about equal to the speed determined in the previous example.
Solution: There is no friction on the road; presumably it's very icy.
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Notice that the ideal banking angle is independent of the mass of the
vehicle; this is nice. It means that one can design a banked highway that will
be appropriate for all vehicles, no matter their mass, so it will be just as safe
for light motorcycles and heavy transport trucks.
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Example: apparent weight for motion in a vertical circle
Consider a car going around a vertical "loop-the-loop" of radius 3 m. Determine
the minimum speed the car needs to make it through the loop.
Alternatively: Consider a bucket of water spinning in a vertical circle and
determine the minimum speed (or angular speed) so that the water does not fall
out of the bucket.
Solution: Draw a free-body diagram for the car when it is at the top of the loop:
This means that if the car is to complete the circle, the force must be
provided by the normal force from the loop and gravity. As the speed
increases, the normal force has to increase to provide the necessary force.
On the other hand, if the speed of the car decreases, then the normal force
will also decrease, until at a critical speed, the weight of the car will be
sufficient to provide the centripetal force. If the speed were to decrease
below this critical minimum value, the car will leave the loop and crash down.
Thus, the minimum speed for the car to make it through the loop
corresponds to n = 0. Setting n = 0 and solving for the speed of the car, we
obtain:
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This may not seem like a very high speed, but remember that the loop is
not very big. I once saw a "cirque" stunt where motorcycles were flying
around the inside of a spherical metal structure, and the radius might
have been this big and the speeds seemed about this fast or a bit faster.
For a much bigger loop, a larger speed is required.
Now solve the problem of the water in the bucket yourself. How fast do
you have to swing a bucket around so that the water doesn't fall out?
• centrifuges
Read about centrifuges in the text book; they provide a nice practical example
of circular motion. (Also, you'll think about physics the next time you use a
lettuce spinner, which is a sort of centrifuge.)
Newton's law of gravity
Example: gravitational force between Earth and Moon
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Newton's law of gravity is an inverse-square force law, and has the same structure
as Coulomb's law for the force between two charged particles at rest. The diagram
above is intended to illustrate that the force decreases by a factor of 4 when the
distance between the objects doubles.
Example: gravitational force between Earth and a small object of mass m at the
Earth's surface
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This provides insight into our assumption earlier in the course that the
acceleration due to gravity g is constant; we can see by the previous equation
that this is not exactly true. Close to the Earth's surface it is approximately true,
but as you move away from the Earth's surface the value of the acceleration
due to gravity decreases.
The equation above also gives us a way to "weigh" the Earth. The acceleration
due to gravity can be measured in a laboratory (in fact you did so in the
pendulum experiment in this course), and so can the gravitational constant G
(look up the famous Cavendish experiment for details). The radius of the Earth
can be determined using an ingenious geometrical method first devised by
Eratosthenes (you can also look this up); then the previous equation can be
solved for the mass of the Earth.
The same formula can be used to determine the acceleration due to gravity on
other planets, moons, asteroids, etc. Just replace the mass and radius of Earth
by the mass and radius of the other planet. Also note that some books call the
acceleration due to gravity at the surface (i.e., "g") by the term "surface
gravity."
orbital motion of a satellite around Earth … direction of gravitational forces at
various points of the orbit … gravitational acceleration is approximately constant
near surface, but the direction is clearly not constant over larger scales, nor is
the magnitude constant over larger scales
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Weightlessness in space
satellites in orbit are in free fall … hence occupants are weightless … check
the textbook for details
Kepler's third law of planetary motion
Using Newton's law of gravity and Newton's second law of motion, we can
derive Kepler's third law of planetary motion.
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If the orbit of the planet is elliptical instead of circular, a more complex
analysis shows that Kepler's third law is still valid provided that we use the
"semi-major axis" of the orbit in place of r. The semi-major axis of the
elliptical orbit is the distance from the centre of the ellipse to the most
distant point on the orbit.
Example: Use Kepler's third law of planetary motion to determine the distance
between the Earth and Sun, given that the mass of the Sun is about 2 × 1030 kg.
Solution: Make sure to convert the period of the Earth into seconds:
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Dark matter: One of today's unsolved puzzles about the universe
As we discussed in class, if you are deep below the surface of the
Earth, let's say in a very deep mine shaft, your weight is less than at
the surface of the Earth. Only the mass "interior" to you (i.e., at radii
smaller than yours) is effective in exerting a force on you; the force
exerted on you by the mass of the Earth that is at larger radii cancels.
This means that if you were anywhere inside a hollow spherical shell
of mass, provided the shell has constant density, the gravitational
force on you is zero. (If you wish to learn more about this, look up
"Gauss's law" for gravity (there is a version of Gauss's law for
electrostatic forces as well); to understand the mathematical
argument, you'll need to have some integral calculus under your belt.)
The same principle can be applied to the motions of stars in our
galaxy. If you analyze the motion of stars at various positions in our
galaxy, you can deduce the amount of mass in the galaxy that lies
closer to the galactic centre than the given star (using Newton's law
of gravity and Newton's laws of motion). Repeating this kind of
analysis for many stars gives us a good idea for the distribution of
mass in the galaxy.
And this leads to a puzzle: The amount of mass that we detect by
usual means (regular light telescopes, radio telescopes, etc.) is
nowhere near enough to account for the mass that we know must be
there by analyzing motions in the galaxy. That is, the "visual matter"
does not account for all the matter that must be present; there must
be some "dark matter."
What on Earth can this dark matter be? Nobody knows. It is highly
unlikely that it could be simply ordinary matter that can't be detected
(such as "cold" dust particles or gas, abandoned TV sets, etc.), so
scientists have turned to more speculative possibilities. Maybe dark
matter is some exotic new form of matter. If so, such forms of matter
have not been detected in laboratories, which leaves us no closer to
resolving the puzzle.
This is an example of the type of unresolved puzzle that is found at
the frontier of every branch of science. There are always unsolved
puzzles, which means there is always room for new ideas, and
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puzzles, which means there is always room for new ideas, and
creative researchers have plenty of opportunities for making
interesting new discoveries. Maybe one of you will devote the time
and work necessary to reach one of the research frontiers; it will take
a lot of time and work to reach the frontier, but for the right kind of
person (i.e., one who is persistent and willing to put up with a certain
amount of failure and frustration) the journey will be a lot of fun and
very satisfying.
Geostationary satellite orbits
It's convenient to have communications satellites that orbit Earth above
its equator with a period equal to Earth's rotational period; in this way,
they "hover" over the same geographical point on Earth. Using Kepler's
third law we can calculate the radius of the orbit of such "geostationary"
satellites.
This is the distance from the centre of the Earth, so the distance of such a
satellite from the surface of the Earth is 6400 km less, which is 35,850 km
above the Earth's surface.
The International Space Station orbits Earth at an altitude of about 400
km, which is considered "low Earth orbit;" geosynchronous satellites are
in "high Earth orbit."
Additional exercises:
Page 196, Exercise 7: A turntable rotates counterclockwise at 78 rpm. A
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speck of dust on the turntable is at  = 0.45 rad at t = 0 s. Determine the
angle of the speck at t = 8.0 s. (The result should be between 0 and 2.)
Solution:
Page 197, Exercise 27: A satellite orbiting the Moon very near the
surface has a period of 110 min. Use this information, together with
the radius of the Moon (which is 1.74 × 106 m), and the mass of the
Moon (which is 7.36 × 1022 kg), to calculate the free-fall acceleration
on the Moon's surface.
Solution:
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This means that the surface gravity (which is another word for the free-fall
acceleration at the surface) on the Moon is about 1/6 as much as the
surface gravity on the Earth. How would this change life for you if you
lived on the Moon for a while?
Page 197, Exercise 45: A 500 g ball swings in a vertical circle at the end
of a 1.5-m-long string. When the ball is at the bottom of the circle, the
tension in the string is 15 N. Determine the speed of the ball at this
point.
Solution:
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