Download Chapter 5 - Physics@Brock

Chapter 5: Dynamics of Uniform Circular
Motion
Tuesday, September 17, 2013
10:00 PM
Rotational Kinematics
We'll discuss the basics of rotational kinematics in this chapter; the kinematics
equations for constant angular acceleration are discussed in Chapter 8. The
basic quantities of rotational kinematics are angular position, angular
displacement, angular velocity, and angular acceleration; period and frequency.
Let's first discuss the period and frequency for circular motion:
• period T: amount of time needed to complete one revolution (unit: s,
min, h, etc.)
• frequency f: number of revolutions per unit time (unit: s -1 (i.e.,
"revolutions per second", which is the same as "cycles per second"), rpm,
etc.)
• basic formula relating T and f:
Remember that the basic formula relating period and frequency is only valid
when T and f have compatible units. For example, if you would like T to be
measured in minutes (which really means minutes per revolution), then f
must be measured in revolutions per minute (rpm).
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Example: Your car's tachometer reads 2500 rpm. How long does it take
for the engine to make one cycle?
Solution:
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Thus, the engine completes one cycle every 24 thousands of a
second.
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Example: The earth rotates once on its axis every day.
Determine the earth's frequency in rpm.
Solution:
Degrees and radians
Angular position is measured in degrees or radians; here is a review of angle
measure in degrees and radians; remember that the radian is a "unitless" unit:
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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,
Using these two conversion factors allows one to convert from
degrees to radians, or vice versa. For example,
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Angular displacement and angular velocity
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Connecting linear and angular kinematic quantities
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Centripetal acceleration
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Centripetal acceleration
A confusing example: driving around a circular curve in a car.
When an object moves in a circle at a constant speed, is the net
force acting on the object zero? If not, what is the direction of the
net force acting on the object?
https://www.youtube.com/watch?v=dxmedyNZ_8s
We previously discussed the fact that we viscerally feel that we are
thrown outwards when we go around a curve in a car; however, the
acceleration is actually inward. It's understandable why this is
confusing; our bodies are telling us that something is pushing us
outwards, yet the following arguments indicate that the
acceleration is actually inwards. It's difficult for the brain to
understand something when the body feels the opposite. (You can
understand why it took a few millennia for the greatest minds in
humanity to figure out what is going on here.)
The vector diagrams below argue that the acceleration in uniform
(that is, constant speed) circular motion is really towards the centre
of the circle. (As we argued in a previous chapter, the acceleration is
not exactly towards the centre if the speed is increasing or
decreasing in circular motion.) Because the acceleration is exactly
towards the centre in uniform circular motion, it is called
centripetal acceleration ("centripetal" just means "towards the
centre").
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The following figure is relevant (and the conclusions are
valid) for circular motion with CONSTANT speed only:
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• formula for centripetal acceleration
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Let's derive a neat formula for the magnitude of the centripetal
acceleration. In Part (a) of the figure, let s represent the length of
the circular arc (dashed arc), which is nearly the same as the
length of the vector labelled d. Then,
From the first figure in Part (c),
We can't be sure of this formula for centripetal acceleration because
of the approximations used, and because some of the quantities
were average values. Nevertheless, the formula is correct, although it
will take calculus to prove that it is exact (calculus lovers are
welcome to read the proof that is further below in the notes).
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Getting an intuitive feel for the formula for centripetal acceleration:
Why does the formula for centripetal acceleration that we just wrote
down have the square of the speed in the numerator, and not just
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down have the square of the speed in the numerator, and not just
the speed? One way to see this is using what is called dimensional
analysis. In essence, you check the units on both sides of the
equation and the units match if the numerator of the right side has
the square of the speed, but not just the speed. Such unit checks
cannot prove that the formula is correct (the formal proof is below,
in blue), but they can help you quickly spot incorrect formulas. Thus,
if the units of each term of an equation match, the formula is not
necessarily correct (some unitless factors might be missing), but if
the units of each term are not the same then you can be sure the
formula is incorrect.
OK, that's good, but it would be better to have an intuitive feel for
why the square of the speed appears in the numerator. Think of it in
this way: For a car trip around a circular track at a constant speed,
the "job" of the acceleration vector is to change the direction of the
velocity vector. If you double the speed, then the change in velocity
over a certain time interval is twice as great, because both the initial
and final velocity vectors are twice as long as before. However, the
same distance is traversed in half the time. Because the acceleration
is the change in velocity divided by the time interval, by doubling the
speed going around the circle you pick up a factor of 2 because the
change in velocity has doubled, and also a factor of 2 because the
time interval is half as much. Thus, doubling the speed around the
circle requires a centripetal acceleration that is 4 times as great.
You can readily understand why the radius of the circle appears in
the denominator by thinking about going around a circular track that
is very large. Because the curve of such a large circle is so gentle, the
rate at which the velocity vector turns is very small, and so the
centripetal acceleration is also very small. The larger the circle, the
smaller the centripetal acceleration for a given speed.
The discussion in the last few paragraphs is based on the assumption
that the motion around a circle is at a constant speed; if the speed is
not constant, then the situation is more complicated. We'll get into
some of the complications in a later chapter, and we'll save other
complications for second-year physics.
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Example: A flywheel with radius 30 cm rotates at a constant rate of 200
rpm.
a. Determine the period of rotation.
b. Determine the speed of a point on the rim of the flywheel.
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b. Determine the speed of a point on the rim of the flywheel.
c. Determine the acceleration of a point on the rim of the flywheel.
We can also simply say that the period is T = 0.3 s, because the "per
revolution" is understood as part of the definition of the period.
(b) The distance d covered by a point on the rim of the wheel in one
revolution is
Because the speed is constant, the speed of a point on the rim of the
wheel is
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For calculus lovers only, here is a precise derivation of the formula for
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For calculus lovers only, here is a precise derivation of the formula for
centripetal acceleration. For an object moving in a circle of radius r at a
constant speed, the position function can be written as
Now differentiate twice to obtain an expression for the acceleration,
using the chain rule. Remember that theta is a function of time and the
radius of the circle is constant. You'll also note that we've used a fact
from vector calculus, that to differentiate a vector function you just
differentiate each component separately. You might like to think about
why this is true. (Further discussion to come in second-year calculus.)
Note that the acceleration is in the direction opposite to the position vector.
The position vector points from the centre of the circle to the location of the
moving object; thus, the acceleration vector points towards the centre of
the circle, which is why it's called a "centripetal" (centre-seeking)
acceleration. The magnitude of the centripetal acceleration is
Note that
Thus
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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.
Also note that static friction between the tires and the road is the origin of the force
pushing the vehicle towards the centre of the circle (why is it not kinetic friction?).
Because the static friction force points towards the centre of the circle, most
textbooks call it a centripetal force. Note, however, that this is not a new kind of
force, but is simply a reminder that the force acts towards the centre of the circle. In
this example, it's a static friction force, nothing new.
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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 (at the specified speed).
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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:
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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:
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.
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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, using a string
of reasonable length. How fast do you have to swing a bucket around so
that the water doesn't fall out?
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• 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.)
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Newton's law of gravity
Example: gravitational force between Earth and Moon
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As I mentioned in class, one of Newton's greatest achievements was putting
the finishing touches on a complete transformation in the way we humans
look at the universe, building on the work of Copernicus, Kepler (who built
on the work of Brahe), and Galileo, among many others. Previously the
prevailing viewpoint was that there were two realms, the Earthly (corrupt,
rotten, imperfect in every way) and the Heavenly (perfect in every way, the
realm of the Gods). Newton used his laws of motion together with his law
of gravity to show that the same force that pulled apples to Earth when
they fell from apple trees also pulled the Moon to the Earth. He did this by
calculating the acceleration of the Moon in two ways. The first way was to
use the formula for centripetal acceleration (valid because the Moon's orbit
is nearly circular and the Moon travels at nearly a constant speed), together
with the known values of the radius of the Moon's orbit and the Moon's
period.
Here's the first calculation of the Moon's acceleration:
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For the second calculation of the Moon's acceleration, Newton combined his
law of gravity with his second law of motion, roughly as follows:
When Newton found that the two calculations for the acceleration of the
Moon "agreed, pretty nearly" he knew that he was on to something: His
new law of gravity was good! Together with his laws of motion, a new
system of mechanics (which we now call Newtonian mechanics) was now
available and could be used to understand not just motions on Earth but in
the Heavens as well. There are not two realms, as the ancient and medieval
thinkers believed, but just one realm; the laws of physics are universal, and
work just as well in all parts of the universe.
(I have oversimplified Newton's second calculation somewhat, but you get
the idea. I believe Newton did not simply use the mass of the Earth, but
rather he made use of the fact (that we'll get to very shortly in these notes)
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rather he made use of the fact (that we'll get to very shortly in these notes)
that the value of g is related to the mass and radius of the Earth to form a
ratio. What he really compared was the value of the acceleration of the
Moon to the acceleration due to gravity on the surface of the Earth; when
he found that the acceleration diminishes as the inverse square of the
distance he knew he had it right!)
This is more in line with Newton's second calculation, but as you can see
it's very similar to the calculation we displayed above. We avoided this
calculation because we hadn't yet got to the connection between g and
G, but we'll get to it shortly.
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The following diagram is meant to help us understand the "inverse-square"
nature of the gravitational force law. The same inverse-square nature is present
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nature of the gravitational force law. The same inverse-square nature is present
in the decrease in the intensity of light or sound from a point source, which is
the inspiration for the diagram. Imagine "spraying" a substance, such as paint
from a paint-sprayer. If the spray is uniform, the area covered by the spray
increases by a factor of 4 if the distance from the sprayer doubles; similarly, the
area covered by the spray increases by a factor of 9 if the distance from the
sprayer triples. In general, the area covered increases by a factor of r2 when the
distance increases by a factor of r.
But, if the area covered doubles, then the "intensity" of the spray decreases by a
factor of 2, because the amount of spray doesn't change, it's just spread over a
larger area.
The gravitational force behaves similarly; as you get farther away from the
source by a factor of r, the gravitational force decreases by a factor of r2,
because it is spread over an area that has increased by a factor of r2.
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, as we'll learn
in second semester. The diagram above is intended to illustrate that the force
decreases by a factor of 4 when the distance between the objects increases by a
factor of 2.
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Surface gravity of a planet
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 formula can be
used to determine the mass of the Earth by inserting the experimentally
determined values for the other quantities.
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."
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Example: Determine the Moon's surface gravity.
Solution: Look up the following data in the textbook:
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Solution: Look up the following data in the textbook:
Then the surface gravity of the Moon is
Thus, the surface gravity of the Moon is about one-sixth the surface gravity
of the Earth. What are the consequences of this? What would it be like to
walk about on the Moon?
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Example: Determine the acceleration due to gravity (a) 1 km above the Earth's
surface and (b) 100 km above the Earth's surface. Compare each value to the
value of the acceleration due to gravity at the Earth's surface.
Solution: Look up the following data in the textbook:
Using these data, the acceleration due to gravity at the Earth's surface is
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(a) The acceleration due to gravity 1 km above the Earth's surface is
(b) The acceleration due to gravity 100 km above the Earth's surface is
At a height of 1 km above the Earth's surface, the value of the
acceleration due to gravity has decreased by 0.031%, and so it is quite
reasonable for ordinary calculations to assume that the acceleration due
to gravity is constant over this range of distances, as we did in Chapters 2
and 3. At a height of 100 km above the Earth's surface, the value of the
acceleration due to gravity has decreased by 3.1%, and so it may not be a
reasonable assumption to treat it as constant over this range of
distances.
These calculations also emphasize the point that gravity does not
suddenly disappear when you are out in space; spacecraft in orbit
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suddenly disappear when you are out in space; spacecraft in orbit
around the Earth are in free fall, and so the apparent weight of objects
and people in orbit is zero. However, their real weight is not zero, as the
Earth's gravity is still quite strong at altitudes at which satellites orbit.
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Satellite motion
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;" that is
their apparent weight is zero, even though their real weight is not zero …
check the textbook for details (It should really be called apparent
weightlessness; the objects still have weight, because there is still a
gravitational force acting on them, but because they are in free fall they
feel weightless.)
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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.
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Example: Given the distance from the Earth to the Sun (1.5 × 10 11 m), use
Newton's second law of motion and Newton's law of gravity to determine
the mass of the Sun.
Solution:
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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
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frontier of every branch of science. There are always unsolved 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.
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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."
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Additional exercises:
Example: A turntable rotates counterclockwise at 78 rpm. A 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:
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Example:
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:
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?
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Example: A 500 g ball swings in a vertical circle at the end of a 1.5-mlong 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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Here's an interesting question: How does the Earth's gravity vary as you go
inside the Earth? You might imagine going down a very deep mine shaft
and wonder about the magnitude of the acceleration due to gravity at the
bottom of the mine shaft. Measurements have been made, and they are
in accord with the following reasoning, which requires calculus to
understand in full mathematical detail.
The first point to understand is that the gravitational force inside a hollow
spherical shell with constant density is zero. Calculus lovers can go here
for a proof:
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html
or here:
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm
Non-calculus people can carefully examine the following diagram, taken
from the previous link. Consider an imaginary double-cone drawn from
the point P and intersecting the shell in the two areas indicated. By
symmetry, the force on a small mass at P due to the mass at each red area
is directed along the red arrows. The mass at A 2 is closer to P than the
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is directed along the red arrows. The mass at A 2 is closer to P than the
mass at A1, but the amount of mass at A2 is less than the amount of mass
at A1. Both influences cancel exactly, and so the force on a small mass at P
from each of the masses at the red areas is equal and opposite, and so the
net force on the small mass at P due to the masses at the red areas is zero.
(In a little more detail, the area is proportional to the square of the
distance from P, but the gravitational force at P due to the mass at the
area is inversely proportional to the square of the distance to P.) The same
argument applies no matter which double-cone we draw, and so the net
force at P due to the entire mass of the spherical shell is zero.
Conclusion: The gravitational force anywhere inside a spherical shell of
constant density is zero.
If you are at the bottom of a deep mineshaft, then, the gravitational force
on you due to the mass of the Earth that lies at radii greater than your
radius is zero. This is because the part of the Earth farther from the centre
than you can be considered to be a collection of thin spherical shells, and
you lie inside all of these shells. We've concluded that at any position inside
a spherical shell of constant density the gravitational force is zero, so the
force on you due to the mass of the Earth that lies at greater radii than you
is zero. The net gravitational force on you can be calculated by considering
only the mass that is interior to you (shaded in red in the following figure).
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Let's suppose that you are at a distance r from the centre of the Earth. How
much of the Earth's total mass lies closer to the centre than you (shaded in
red)? Assuming that the Earth's density is constant, the mass M that is closer
to the Earth's centre than you is
If your mass is m, your weight at the bottom of the mine shaft is
mg, which is equal to the gravitational force exerted on you by the
mass M of the part of the Earth interior to your position:
Thus, the acceleration due to gravity at the bottom of the mine
shaft is
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Inserting the expression for M derived above, we obtain a
convenient expression for the acceleration due to gravity at the
bottom of the mine shaft:
This is interesting: The acceleration due to gravity in a mine shaft
inside the Earth varies linearly with distance from the centre of the
Earth. Using the expression for the acceleration due to gravity at the
surface of the Earth,
we can write the acceleration due to gravity in the mine shaft as
Thus, if you're half way down to the centre of the Earth, the
acceleration due to gravity would be half of the surface value.
Similarly, if you've only gone down 1 percent of the Earth's radius,
the acceleration due to gravity will be 1 percent less than the
surface value. Of course, for any realistic mine shaft, the actual
change in g is very small indeed. For example, if the mine shaft is 2
km deep, then the value of g at the bottom of the mine shaft will
be 6398/6400 times the surface value, which amounts to a little
more than 3/100 of a percent difference.
We learned earlier in the chapter that the acceleration due to
gravity decreases as 1/r2 as you go away from the surface of the
Earth, so a graph of the variation of g both inside and outside the
Earth looks something like this:
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I haven't drawn the graph very well, but you can see the linear
portion inside the Earth, and then the portion outside the Earth is
supposed to decrease asymptotically towards zero, never reaching
zero.
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