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Gravity ...................................................................................................................................................... 155 Center of Gravity ................................................................................................................................... 157 Centripetal Acceleration .......................................................................................................................... 162 Circular Motion ..................................................................................................................................... 162 Linear Speed and Circular Motion ........................................................................................................ 163 Force and Circular Motion .................................................................................................................... 164 Centripetal Force and Gravity ............................................................................................................... 169 Satellite Stuff ............................................................................................................................................ 174 Orbital Equations .................................................................................................................................. 176 Period of Satellites ................................................................................................................................ 177 Gravity in Orbit ...................................................................................................................................... 180 Gravity Wrap‐up ....................................................................................................................................... 184 AP Physics – Gravity
Sir Isaac Newton is credited with the discovery of gravity. Now, of course we know that he didn’t
really discover the thing – let’s face it, people knew about gravity for as long as there have been
people. Gravity didn’t have to be discovered for crying out loud! Your basic tiny little toddler
figures it out in the first few months of life. So what is the big deal with Newton and gravity?
Well, what Newton did was to describe gravity, extend its effects from the surface of the earth
(which everyone understood) out into space to explain the behavior of the planets (which nobody
suspected), and to formulate a mathematical law that accurately described the force of gravity
between objects with mass.
Perhaps the greatest success of his theory of gravity was to successfully explain the motion of the
heavens – planets, moons, etc.
Newton’s theory is very simple. Gravity is a force of attraction between any two objects that have
mass. Two objects sitting on a desktop attract each other with a force that we call gravity. They
don’t go flying together because gravity is a very weak force and is only significant when one or the
other of the masses is enormous – planet size. This is why we aren’t attracted to objects that we
pass on our daily wanderings.
The size of the force of gravity is given by this equation:
mm
F  G 12 2
r
Newton’s law of gravity
G is the universal gravitational constant.
G  6.67 x 1011
N  m2
kg 2
m1 is the mass of one of the bodies and m2 is the mass of the other body.
r is the distance between the two bodies.
The value of G is the same everywhere throughout the universe.
On the old AP Physics Test the equation is written as:
FG  
G m1m2
r2
155
Newton’s law of gravity is an inverse-square Law. This means that the force of gravity gets
smaller or larger by the square of the distance. The force is directly proportional to the masses, so
if the mass of one of the objects doubles, the force of gravity would double. But if the distance
doubled, the force of gravity would decrease by a factor of four. That’s because it decreases by the
square of the distance. Inverse-square laws are very common in physics. We’ll see more of them in
our explorations.
To give the particulars of the theory:
Universal Law of Gravitation
1.
2.
3.
4.
5.
6.
Gravitational force is a field force between two particles -- in all mediums.
Force varies as the inverse square of the distance
Force is proportional to mass of objects.
The gravity force acts from the center of the two objects.
The gravitational force is always attractive.
The gravitational force cannot be shielded or canceled.
Solving gravity problems is quite simple. Let’s do one.

A girl, Brandy (42.5 kg), sits 1.50 m from a boy (63.0 kg), George. What is the force of gravity
between them? (This will tell us how attracted they are to each other.)
F G
2  42.5 kg 63.0 kg




11 N  m
F   6.67 x 10


kg 2 
1.50 m 2

m1m2
r2
F  7940 x 1011 N

7.94 x 108 N
You can see that this is a tiny force, one so small that Brandy will never,
the Physics Kahuna is sure, notice its presence. George must generate
some other attractive force if there is to be a relationship between the two
of them.
One useful application of Newton’s law of gravity was to weigh Earth – this allowed physicists to
make an accurate determination of the Earth’s mass. Let’s do that.

Find the mass of the earth. re = 6.38 x 106 m. We will use a 10.0 kg mass in our solution (not
that it matters), the other mass will be that of the earth.
F G
m1mE
r2
This is the force of gravity using Newton’s law. But we know that the
force of gravity must also equal F = mg, so we set them equal to one
another:
156
m1g  G
m1mEarth
r2
The mass of the object cancels out:
mE 
m
s
2

6.38 x 106 m

2

m 
11 kg  m

 6.67 x 10
 10.0 kg 
s2
kg 2 

mE  59.8 x 1023 kg
2

mE
r2
gr 2
mE 
G
Solve for the mass of the earth
9.8
g G
 59.8 x 1023 kg

5.98 x 1024 kg
5.98 x 1024 kg
Center of Gravity: A stone resting on the ground is acted upon by gravity.
In fact, every
atom within the rock is attracted to the earth. The sum of all these forces is the thing that we call
weight. Wouldn’t it be a real pain to have to add up every single vector for every single atom in an
object in order to find out what it weighed? Well fortunately, we don’t have to do that. Nature sort
of does it for us. Basically the vectors all add together to give us one big vector, the weight vector,
which is directed down and which acts at a point where all the weight appears to be concentrated.
This point is called the center of gravity.
For an object like a uniform sphere, the center of gravity, or CG, is located at its geometric center.
Irregular objects, like, say, your average baseball bat would have their CG located closer to the fat
end of the bat rather than the center.
The dot represents the CG of the objects in the drawings below:
The weight of these objects, e.g., the vector we call weight, is applied at the CG.
When a ball is thrown in the air, it follows a parabolic path. Actually, what follows this parabolic
path is the CG of the ball.
This is also true for irregular
objects that are thrown. If you
throw a crescent wrench so that it
spins when you release it, it
seems to follow a very erratic
path. Its CG, however, traces out
a nice smooth parabola.
157
All irregular objects rotate about their CG when thrown.
Let’s do some more gravity problems.

You weigh 567 N on earth. How much would you weigh on Mars?
First we must calculate the mass by using the weight on earth:
W  mg
m
W
g

kg  m  1
m  567

s 2  9.8 m

s2



  57.86 kg



Now we need only find the force of gravity between our person and Mars. For the distance we use
the radius of Mars, since the person is on the surface of Mars and the force of gravity acts at the
center of the planet.
mm
F  G 12 2
r



2  6.42 x 10 23 kg  57.86 kg 

11 N  m
  6.67 x 10
2 
2


kg


3.37 x 106 m



218 N
What is the acceleration of gravity on Venus?
We know:
F  ma
mm
m2 a  G 1 2 2
r
mm
F  G 12 2
r
and



2  4.88 x 10 24 kg

11 kg  m  m
a   6.67 x 10
2
2 
2


s
kg


 6.06 x 106 m

 0.886 x 101
m
s2

8.86
m
s2
158
The acceleration on Venus would be slightly smaller than the acceleration on earth.
AP Question:
We’re revisiting one we already looked over, but this time we can analyze it on
Mars.

The Sojourner rover vehicle was used to explore the surface of
Mars as part of the Pathfinder mission in 1997. Use the data in the
tables below to answer the questions that follow.
Sojourner Data
Mass of Sojourner vehicle:
Wheel diameter:
Stored energy available:
Power required for driving under average conditions:
Land speed:
11.5kg
0.13 m
5.4 x l05 J
10 W
6.7 x 10-3 m/s
Mars Data
Radius:
Mass:
0.53 x Earth's radius
0.11 x Earth's mass
(a) Determine the acceleration due to gravity at the surface of Mars in terms of g, the acceleration
due to gravity at the surface of Earth.
Force and mass are directly proportional to each other. Force is inversely proportional to the radius
squared.
We set up the proportions for g, the acceleration of gravity on earth.
g Mars  g
mMars
 rMars 2
m 0.11
g Mars  9.8 2
s  0.532

m
3.8 2
s
159
(b) Calculate Sojourner's weight on the surface of Mars.
w  mg mars
m

 11.5 kg  3.84 2  
s 

44 N
(c) Assume that when leaving the Pathfinder spacecraft Sojourner rolls down a ramp inclined at
20 to the horizontal. The ramp must be lightweight but strong enough to support Sojourner.
Calculate the minimum normal force that must be supplied by the ramp.
1. Draw a FBD:
n  mg Mars cos  0
n  mg Mars cos
m

n  11.5 kg  3.8 2  cos 20.0o 
41 N
s 

n
mg
sin 
Mars
(d) What is the net force on Sojourner as it travels
across the Martian surface at constant velocity?
Justify your answer.


mg Mars
Zero. If it is traveling at constant velocity in the x
direction then there is no acceleration and no net
force. . The force of gravity is cancelled by the normal force resulting in no motion in the y
direction.
(e) Determine the maximum distance that Sojourner can travel on a horizontal Martian surface
using its stored energy.



1 
W
W
P
t
 5.4 x 105 J 
 5.4 x 104 s

J
t
P
 10 
s 

x
m
v
x  vt  6.7 x 103
5.4 x 104 s 
360 m
t
s


(f). Suppose that 0.010% of the power for driving is expended against atmospheric drag as
Sojourner travels on the Martian surface. Calculate the magnitude of the drag force.
P   0.00010 10W   0.001W
P  Fv
F
 0.001W  
P

v
6.7  103 m s


0.149 N
160
Dear Doctor Science,
If, theoretically, you were at the exact center of the earth and you dropped
a ball, what would it do?
-- Rick McGuire from Birmingham, MI
Dr. Science responds:
It would fly into your face at great velocity. Most probably, it would damage
one eye and, if your reflexes were slow enough, could permanently blind you.
That's why I can't urge you strongly enough to leave experiments like this to
trained scientists! We've been there before. We know what to expect in
situations like this. I wouldn't think of going to the center of the earth without
protective eyewear, nor would I tamper with terrestrial forces without first
having done my homework! The Center of the Earth, man! Didn't you see the
movie starring James Mason and Pat Boone? Sure, it had a happy ending -- all
movies did back then. But the reality of Pat Boone taking a shower surrounded
by glowing Technicolor rocks...the mind can only stand so much.
Dear Cecil:
Why are there high tides twice a day when the earth rotates beneath the moon
only once a day? In diagrams it appears the moon's gravity causes the earth's
oceans to bulge (creating a high tide) not only on the side toward the moon,
but also on the side away from the moon. I've heard some unconvincing
explanations for this, including: "the water on the far side is flung away from
the earth" (why?); "the moon attracts the earth, and the water on the far side
is left behind" (why isn't the water on the far side attracted too?); and "the
earth and the moon both revolve around a common point" (I know that, but
what does that have to do with the question?). Please help.
--Kathleen Hunt, Brookline, Massachusetts
Cecil replies:
Not to discourage you, Kathleen, but this makes 22 questions from you in three
months. Think quality, not quantity. This isn't a scrap drive.
The following homely metaphor is sometimes used to explain why there are two
tides: the earth and moon, which are really dual planets, are like two figure skaters
spinning around one another while holding hands. Centrifugal force naturally tends
to pull them apart, but their clasped hands (i.e., their mutual gravitational
attraction) keep them together. Similarly, centrifugal force tends to fling the ocean
outward on the side of the earth away from the moon. On the near side, the water is
tugged moonward by lunar gravity.
There's just one problem with this explanation. It's wrong. Cecil has consulted with
the physics division of the Straight Dope Science Advisory Board and is satisfied
that centrifugal force (OK, inertia, if you want to get technical) has nothing to do
with why there are two tides.
The real reason is this. The pull of gravity drops off rapidly with distance. Lunar
gravity tugs on the side of the earth facing it a lot, on the earth itself a medium
amount, and on the opposite side of the earth relatively little. In short, the far-side
tide is a result of the moon attracting the earth, leaving the ocean behind. Which,
looking back at your letter, I guess you already knew and didn't find convincing. If
so, Kathleen, come on. Would I lie to you?
161
AP Physics – Centripetal Acceleration
All of our motion studies thus far have dealt with straight-line stuff. We haven’t dealt with things
changing direction during their travel. This type of motion is called angular motion. A special
case of angular motion is circular motion (when things travel around in circles).
We know that velocity and acceleration are both vectors. Recall that vectors have both magnitude
and direction. Change either one and you have changed the motion. We know from the first law
that objects in motion stay in motion unless an outside force acts to change that motion. We’ve
studied all about forces that change the magnitude of the velocity. Now we’ll deal with forces that
change the direction of a motion.
Circular Motion:
Objects that have circular motion are moving in a circular path around a
central point which is called the axis or center of spin. If the axis is within the body, we say it is
rotating. If the axis is outside the body, then the object is revolving. The earth rotates around its
axis, which causes day and night. The earth revolves around the sun, which causes the four seasons
during the year (the time it takes the earth to revolve around the sun one time).
An object undergoing circular motion has an angular velocity and undergoes an angular
displacement.
Angular displacement can be measured in several units: revolutions (the number of times the thing
makes a circle), degrees, or radians. These are the most common units. We will deal only with
revolutions.
The angular velocity is the rate of change in angular displacement.
angular velocity 
angular displacement
time
The Greek lower case letter omega is used as the symbol for angular velocity. Here is the equation:


t
The units for angular velocity we will use is:



rpm
t

rev
min
.
A record is rotating with an angular velocity of 45 rpm. If it rotates for 35 seconds, how many
rotations does it make?


t
  t 45
 1 min 
rev
 35 s  
 
min
 60 s 
26 rev
162
Linear Speed and Circular Motion:
When an object is rotating all points on the object have the same angular velocity. But at the same
time each point is also tracing out a circle. If you divide the time it takes to make one revolution
into the circumference of the circle that is traced out you get the speed of the point. We call this the
linear or tangential speed. The linear velocity is not constant as its direction changes every instant.
Calvin’s father is correct about the two points on a record. But, unlike Calvin, this should make
perfect sense to you, the sophisticated advanced student of physics. The further a point is from the
center of spin, the larger will be its linear speed.
This is true for the earth as well. You have a much larger linear speed if you are at the equator than
you would have if you were at one of the poles.
We can figure out the linear speed of a point using the angular velocity.
Speed is distance divided by time, the time it takes to make one revolution is:

t

t
t



1



Let the angular displacement be 1 revolution:
This is the period of the rotation, the time it takes to make
one revolution.
So we now have the time. Here is the equation for speed:
v
x
t
The distance is the circumference of the circular path, 2 r . We plug this in and also the value we
found for the time.
163
v
d
t

2 r
1
 2 r

v  2 r
So the linear speed is given by:
Dear Doctor Science,
Why do you have to rotate the tires on your car? Don't they rotate when you
drive on them?
-- Kenneth Kuller from Minneapolis MN
Dr. Science responds:
No, Ken, they don't. What appears to be a spinning tire is
actually an optical illusion. Car tires stay put and slide on a
special greasy fluid they excrete as the rub along the road.
Even wagon wheels turn backwards, as anyone who has
seen reruns of "Bonanza" can attest. Again, appearances
are deceiving, because in order to make the wagons appear
to move forward, they had to run the film backwards
h
h h
All h
k
i i
ill
Force and Circular Motion:
A ball is attached to a string. The ball is whirled around and
around. You can picture this right? In the drawing, we are looking down on the path of the ball
from above.

If the string breaks at the point shown, what will be the path of the ball? The Physics Kahuna will
give you three choices:
A
B
C
164
In order for an object to undergo circular motion, a force must act. Picture an object that has some
velocity. What will happen to it if no forces act on it? Well, according to the first law, it will
continue to move with a constant velocity. It will follow a straight-line path.
To make it change direction a force must act on it. In order to make it changes direction constantly,
a force must act on it constantly.
What is the direction of the force needed to do this? Well, when you spin something in a circle,
what do you have to do? You just pull it towards the center as you go around and around.
The object gets accelerated towards the center. We call this the centripetal acceleration. The
equation for the centripetal acceleration is:
v2
ac 
r
ac is the centripetal acceleration, v is the linear or tangential speed, and r is the radius
of the circular path.
This equation will be provided to you for the AP Physics Test.

A rotating object has a linear speed of 1.5 m/s. It undergoes a centripetal acceleration of 3.6
m/s2. What is the radius of the mass's circular motion?
ac 
2
v
r
r
2
v
ac

 m  1
r  1.5  
s   3.6 m


s2

2


 



0.62 m
The force that brings about this acceleration is called the centripetal force. Its direction is also
towards the center of the circular path. Centripetal means "center seeking". The centripetal force
changes the direction of the object’s velocity vector. Without it, there would be no circular path.
The centripetal force is merely a convenient name for the net force that is towards the center. It is
always caused by something – it could be caused by the force of gravity, the reaction force between
the control surfaces of an airplane with the air, etc.
When you rotate a ball around your head in a circle, the centripetal force is supplied by the tension
in the string.
What is the source of the centripetal force that causes a racecar to travel in a circular path on the
racetrack?
The force is brought about by the tires pushing on the racetrack. The friction between the
road and the tire is very important, so race tires are designed to maximize friction.
What is the source of the centripetal force required to make the earth revolve around the sun?
165
This is where the apple falling on Newton story fits in. Before Newton no one could explain the
orbits of the planets and moons. Newton, the story goes, was relaxing under an apple tree
pondering the problem of the moon’s orbit. He knew that there had to be a force acting on the
moon to accelerating it towards the earth, but had no idea what was the source of the force. Then he
saw an apple fall and the simple solution struck him like the old thunderbolt. Just as the earth’s
gravity reached out and made the apple fall, so it reached out and made the moon fall. Thus, the
force that keeps the planets and moon following their orbital paths is gravity.
The AP Test equation sheet will not give you the equation for centripetal force. It does give you the
equation for centripetal acceleration. It also gives you the equation for the second law. Using these
two equations you can easily derive the formula for centripetal force. Here’s how to do it:
F  ma
v
r
so plug in the value of the centripetal acceleration: aC  2
 v2 
F  ma  m  
 r 
mv 2
FC 
r
That’s all there is to it.
Something Completely Different:








At the 1939 World's Fair, Dupont introduced nylon stockings to the world. Dupont
chemists discovered nylon while trying to produce artificial silk. On May 15,
1940, they were available in stores. Nineteen years later, Glen Raven Mills of
North Carolina introduced panty hose, eventually developing a seamless model
just in time for the advent of the miniskirt in 1965.
In 1970, Hanes creatively packaged panty hose in plastic eggs in super markets,
drugstores, and convenience stores - places where they had never been available
before. Sheer Energy was introduced in 1973 as the first L'eggs panty hose made
with sheer spandex yarn.
The brand-name "L'eggs" is a combination of the words legs and eggs, with an
apostrophe added to make the wordplay idiot-proof.
In 1991, L'eggs replaced the plastic egg with a cardboard package – no doubt it
was cheaper. While the plastic eggs were recyclable and used for arts-and-crafts
projects, the new box, using 38 percent less material and made from recycled
paper, allows 33 percent more containers to fit into the store-display rack and is
still rounded at the top like an egg.
In the 1970s, when Peter Lynch, the most successful money manager in America,
noticed his wife Carolyn bringing L'eggs panty hose home from the supermarket,
his Fidelity Magellan fund bought stock in Hanes. The value of its shares rose
nearly 600 percent.
L'eggs supported the women of the 1994 and 1996 United States Olympic Teams.
L'eggs is the best-selling panty hose in America.
97 percent of all supermarkets, drugstores, and convenience stores in the United
States carry L'eggs panty hose.
166

1.2 kg stone is attached to a 1.3 m line and swung in a circle. If it has a linear speed of 13 m/s,
what is the centripetal force?
mv 2
FC 
r

 m
1.2 kg 13 
 s

1.3 m
2

160 N
A car is traveling at a constant speed and makes a turn with a radius of 50.0 m. Its speed is15.0
m/s. Find the minimum coefficient of friction needed to
keep the car traveling along the path.
n
Let’s look at the FBD:
The frictional force must equal the centripetal force.
FC 
The centripetal force is given by:
mv
r
2
Fc
We know that this must equal the frictional force. We also
know that the frictional force is:
f
mg
f s  s N
Assume the road is flat, so
n = mg
Set the two equal to each other and solve for the coefficient of friction:
 s mg 
mv 2
r
s 
v2
gr
2
m
1

 s  15.0 
s  
m 

9.8

 50.0 m
s2 



0.459
A child twirls a yo yo. If angle of the cord with the vertical is 30.0, find ac.
0
Look at the forces in the y direction:
 Fy = 0
T cos  mg  0 T 
mg
cos
167
The horizontal component of T is the centripetal force.
FC  T sin 
FC 
so
T
Plug into equation for T:
mg
sin 
cos
FC  mg tan 
maC  mg tan 
m
aC  9.8 2 tan 30.0o
s
We know that:
FC  maC
0
aC  g tan 
aC 
m
5.66 2
s
mg
Centrifugal force: You may have heard of the centripetal force
before you studied physics. It is possible. Most people don’t use the term however. Instead they
talk about the centrifugal force. Just what the heck is that?
Okay, you’ve seen the word “centrifugal force”, now forget it! It’s terribly bad form for a physicist
such as yourself to use such a naïve term. Here’s the deal. The centrifugal force is the thing that
people blame for the feeling that things seem to be pushed away from the center of spin during
rotation. You place a coin on a turntable and then spin it really fast. What happens to the coin? In
your mind you picture the coin flying straight away form the center of the record. The individual
who had not studied physics would say this was because of the centrifugal force.
Here is an important concept:
The centrifugal force is a fictional force. There is no actual force that is
pushing away from the center of a rotating system.
You feel this “centripetal force” when you are a passenger in a car that makes a turn. When the car
enters the turn, you feel as if you have been pushed into the door, away from the center of the
circular path the car is making. So you think, “Hey, I’m being pushed into the door so there must
be a force pushing me away from the center of the turn.”
That’s certainly how you feel at any rate.
Remember the problem at the beginning of this distinguished paper? About the path a ball would
follow if the string were to break? You had you three of your basic choices:
168
A
B
C
The correct choice, you’ve hopefully (actually, the Physics Kahuna should say “it is to be hoped”,
but that sounds very pompous, so we won’t say that) figured out that the correct path is B. Why?
Well at the point in the circle where the string breaks, the ball has a velocity that is tangent to the
circular path. The string is providing the centripetal force – pulling the ball towards the center. The
ball wants to follow the tangential path because of the first law, but the string won’t let it. The
string, via the tension it exerts, pulls the ball towards the center, changing the direction of its motion
and making it follow the circular path. When the string breaks there is no longer any force to
change its direction, so the ball travels in a straight-line path that is tangent to the circle as in the B
drawing in accordance with the first law of motion.
This is sort of what is going on in the car with the passenger.
The passenger wants to travel in a straight-line path at a constant speed in accordance with the law,
the first law to be exact. The car however has different ideas. It decides to go in a circular path.
It’s the tires pushing it toward the center exerting a force to make it all happen. So the car changes
direction, but you, the passenger, do not. No force is acting on you. So you go forward in your
original direction until you push into the door, which then pushes you toward the center and you
then go in a circle as well. The third law rears it head – you push into the door and the door pushes
into you. You feel like you are being pushed into the door, even though there is no real force doing
this. It’s just the reaction force to you pushing into the door. This is the so called centrifugal force,
this sensation of being pushed away from the center.
If the centrifugal force was real, i.e., there was a force pushing you away from the center of the
circular path, then if the door were to suddenly pop open you would fly straight away from the
center of the circular path. But of course that does not happen. You would travel in a tangential
path. The centripetal force is real, the centrifugal force is not.
Centripetal Force and Gravity: The Physics Kahuna did a silly demonstration involving
a bucket of water that was spinning in a vertical circle. The water stayed in the bucket and did not
fall out.
So what was the deal? Does spinning something
in a vertical circle somehow cancel out gravity?
Well, no, gravity is a force that cannot be stopped
or canceled. It is always there, anytime you have
the appropriate masses.
The water does fall, it falls but the bucket falls
169
with it and catches it.
This only works if the bucket is moving fast enough to catch the water. If the bucket is too slow,
then the water will fall out of it.
The minimum linear speed for this is called the critical velocity.
Critical velocity  minimum velocity for an object to travel in vertical circle
and maintain its circular path against the force of gravity.
The same thing is needed for satellites in orbit around the earth or planets in orbit around the sun.
They too must travel at the critical velocity.
The critical velocity formula is not provided on the AP Test, but it is very simple to figure out.
You just set the centripetal force equal to the weight of the object that is in circular motion. If the
two forces are equal, then the object won’t be able to “fall out” of the bucket.
mv 2
FC 
and
F  mg
Set them equal to each other:
r
mg 
mv 2
r
v 2  gr
So here is the critical velocity

v  gr
v  gr
A carnival ride travels in a vertical circle. If the ride has a radius of 4.5 m, what is the critical
velocity?
v  rg
m

 4.5 m  9.8 2 
s 


m2
44.1 2
s

6.6
m
s
AP Test Question Time:
 A heavy ball swings at the end of a string as shown here, with
negligible air resistance. Point P is the lowest point reached by
the ball in its motion, and point Q is one of the two highest
points.
(a) On the following diagrams draw and label vectors that
could represent the velocity and acceleration of the ball at points P and Q. If a vector is zero,
explicitly state this fact. The dashed lines indicate horizontal and vertical directions.
i. Point P
170
ii. Point Q
(b) After several swings, the string breaks. The mass of the string and air resistance are
negligible. On the following diagrams, sketch the path of the ball if the
break occurs when the ball is at point P or point Q. In each case, briefly
describe the motion of the ball after the break.
at P
i. Point P
ii. Point Q
at Q
Okay, here we go. Let’s answer the questions.
(a) On the following diagrams draw and label vectors that could represent the velocity and
acceleration of the ball at points P and Q. If a vector is zero, explicitly state this fact. The
dashed lines indicate horizontal and vertical directions.
i. Point P
ac
v
Many foolish AP Physics students forgot about the centripetal acceleration when they
labeled the ball at point P. You wouln’d forget something like that would you?
ii. Point Q
at Q
At point Q the ball has come to rest (for an instant) and so its velocity is
zero. The only acceleration acting on it is the acceleration of gravity.
(d) After several swings, the string breaks. The mass of the string and air resistance are
negligible. On the following diagrams, sketch the path of the ball if the break occurs when
the ball is at point P or point Q. In each case, briefly describe the motion of the ball after the
break.
171
i. Point P
at P
At point P the ball is at its lowest point in the path. When the centripetal force disipears (the
broken string, right?) the ball continues to move with the velocity it had at this point, which is
horizontal. So its path looks like the projectiles we studied that started out with a horizontal
velocity.
ii. Point Q
at Q
At point Q the ball is at rest so when the string breaks it simply falls straight down.
Dear Doctor Science,
If the earth spins on its axis while it rotates around the sun, why don't Iget dizzy
more often?
-- Star Wholethumb from Eugene OR
Dr. Science responds:
You can only get so dizzy and then you break through the
other side, and progress towards undizziness. This process
usually reaches its peak in the first year of life, which is why
babies fall down so often. If the earth were to stop spinning
and circling the sun, even newborn babies would be able to
walk just fine, but the rest of us who have adapted to this
amusement park ride would suffer horribly from motion
sickness. So it's a trade-off, us or them. Until newborns get
172
Dear Straight Dope:
What is down? If the Earth were a perfect sphere, then a line perpendicular to the tangent
of any point on the Earth's surface would go through the core and would certainly be
down. But since the Earth is not a perfect sphere, is down the perpendicular to the tangent
or the most direct line to the core? While you're at it, what's up?
--Andrew Mattison, Waltham, Massachusetts
SDSTAFF Dex replies:
For this one we turned to our physicist/penguinist friend Karen. Her reply:
Let's examine the two possibilities. If I drop an object, that object goes down. Therefore, I
define down as the direction an object goes when dropped.
From gravitational physics, we know that two objects will gravitationally attract each other
according to their centers of mass, therefore, down is directly towards the center of mass of the
earth, which for our purposes is the geometrical center of the earth. (The "core" of the earth is
quite large.)
Now what if you define down as the direction perpendicular to the tangent to the earth? (In
geometry this is called the "normal.") Won't work. First, as you rightly point out, typically
there's an angle between the direction of gravity and the normal. Therefore, when you drop an
object, it will not fall "down" but at some angle to "down." That's too weird.
Second, I can think of no instance where anyone would need to know the direction of the
normal. If you do care about the surface of the globe and its curvature, you are lots more
sensitive to the local terrain--for example, a contractor building a house on the side of a hill.
Third, that contractor is mostly concerned with the effects of gravity tending to pull his house
down. So if all he knew were the normal, he would have to get some paper and a calculator and
figure out his latitude and calculate the direction of "gravitational down," similar to the
rigmarole people go through to convert magnetic north to true north. And why bother? It is very
easy to measure "gravitational down" (even on a hillside) with a plumb bob, and it's the most
relevant concept anyway. (As opposed to magnetic/true north where you only measure magnetic
north and you only care about true north.)
OK, so that's the theory. The real Earth is another story. When people hear the earth is an oblate
spheroid, they tend to think "hamburger bun" when they should be thinking "billiard ball." The
earth shrunk to the size of a billiard ball would be out-of-round by +/-0.004", whereas the
tolerance for billiard balls according to the Billiard Congress of America is +/-0.005"
(http://www.bca-pool.com/rules/equip.htm ). I think you'll agree that the press has really blown
the flattening of the earth's poles all out of proportion. The normal misses the earth's geometrical
center by only 21.5 km (earth's inner core has a diameter of 1200 km.) If you had two poles as
tall as the Empire State building, and you took them to latitude 45 degrees (where the difference
between ""gravitational down"" and the normal is greatest) and the bottoms of the poles both hit
the earth at the same spot, but one pointed along ""gravitational down"" and one pointed along
the normal, the tops of the poles would be separated by only 5 feet.
Fine, Karen. Blame the press for everything.
--SDSTAFF Dex
Straight Dope Science Advisory Board
173
AP Physics – Satellite Stuff
One of Sir Isaac Newton’s greatest claims to fame is his explanation of how the planets orbit the
sun. That and the ability to compute the orbits, orbital speeds, orbital periods, etc. Before Newton,
the motion of the heavens was a mystery. After Newton, the motion of the heavens was an
explainable physical phenomenon.
Let’s do us a “mind experiment”. This is an experiment where you think instead of do. Anyway,
picture a cannon that is set to fire horizontally. What does the path of the projectile look like?
Path of short range projectile
The projectile will follow a curved path. This is because it is being accelerated downwards by the
force of gravity. The greater the velocity of the projectile, the farther it will go before it strikes the
Earth.
The Earth, however, is not flat, although over short distances we can pretend that it is. So what
actually is happening is that the projectile moves over and falls to the ground on a curved surface.
So we have a curved path and a curved surface. We have to take the curvature of the Earth into
account when firing long range projectiles. Possible paths would look like this:
Path of long range projectile
Again, the greater the velocity of the projectile, the greater the range. Newton showed that if the
velocity was great enough, the curving path of the falling projectile would match the curved surface
of the earth and the falling projectile would never actually hit the Earth. Here is a drawing of this.
Path of projectile with same curvature
as surface of the earth
Newton imagined a mountain on the earth that was so high that its summit was outside of the
earth’s atmosphere (this eliminates friction with air). On top of the mountain is a powerful cannon.
The cannon fires a projectile horizontally. The projectile follows a curved path and eventually hits
the earth. Now we add more gunpowder to the charge and fire another cannonball. This cannonball
will travel a greater distance before it too hit the surface of the earth. We keep firing the gun with a
bigger and bigger charge. The cannonball goes further and further before it strikes the earth.
174
Eventually the velocity is great enough so that the curved path of the projectile matches the curved
surface of the earth and the cannonball never gets closer to the planet’s surface. It keeps falling
forever.
Earth
Path of projectile fired
with larger and larger
charges
Earth
With just the right velocity,
the projectile never reaches
the earth’s surface
That basically is your orbit.
The earth is not flat. It is a sphere and its surface has a fairly constant curvature. The surface drops
4.9 m in 8 000 m of horizontal travel.
If we launch a cannonball with a velocity of 8 000 m/s, it will fall a distance of 4.9 m and travel a
horizontal distance of 8 000 m in one second. This means that it will stay at the same height above
the earth’s surface throughout its path. Of course, if we did this near the surface, we’d have the air
slowing the projectile down. We’d also have to worry about the cannonball running into houses and
mountains and trees and so forth. Above the atmosphere, however, all these impediments are
eliminated.
175
The orbit of the everyday celestial object is described by a combination of the law of gravity,
Newton’s three laws, and the stuff we just learned about, circular motion.
Sherlock Holmes is the most portrayed character on film,
having been played by 72 actors in 204 films. The historical
character most represented in films is Napoleon Bonaparte,
with 194 film portrayals. Abraham Lincoln is the U.S.
President to be portrayed most on film, with 136 films
featuring actors playing the role.
Orbital Equations:
Let us assume that the orbit of a satellite about the earth (or any other
massive body) is a circle. Most orbits, the Physics Kahuna must point out, are not circles but are
instead ellipses. This was discovered by Johanes Keppler in the 1600’s. But let’s keep it simple
and look at a circular orbit.
In order to have a circular path, a centripetal force is required. This is supplied by the force of
gravity between the two bodies.
So we can set the centripetal force equal to Newton’s law of gravity:
F G
m1m2
r2
gravity
Set them equal to one another:
m2v 2
FC 
r
G
m1 m2
r2
centripetal force
m2v 2

r
Notice how the mass of the object canceled out.
176
v2 
Gm1
r
v
Gm1
r
This gives us an equation for the orbital velocity:
Gm
r
v
The mass, m, in the equation is the mass of the body being orbited. If we are talking about a planet
orbiting the sun, then the mass we would use would be that of the sun. The mass of the satellite
cancels out, so its mass is not involved in the orbital velocity equation at all.
The equation for the orbital velocity will not be given you on the AP Physics Test. So be prepared
to derive it if you need it.
Period of satellite: This is another simple derivation job.
The period of a satellite is T, the
time to make one orbit. What would be the period of the earth around the sun?
Let’s develop the equation for the period of a satellite. We’ll use the equation for distance and
solve it for the time:
v
x
t
t
x
v
d, the distance traveled is the circumference of the orbit. We know that it would be:
x  2 r
So we can plug that in to the equation we solved for time:
t
2 r
v
but v is also given by the equation we just derived for the orbital velocity:
v
Gm
r
If we plug the orbital velocity into our working equation, i.e., put them together, we get:
t
2 r
Gm
r
177
Square both sides:
t2 
4 2r 2
Gm
r
Clean up everything up nice and neatlike using our potent algebra skills:
t
4 2r 3
t 
Gm
2 2 r
t  4 r
2
2
Gm
4 2r 3
Gm
r3
Gm
t  2
And we end up with an equation for the period of a satellite. Again the mass in the thing is the mass
of the body being orbited:
r3
Gm
t  2
Now let’s do some exciting problems.

What is orbital velocity of the earth around the sun? The sun has a mass of 1.99 x 1030kg, the
mean distance from the earth to the sun is 1.50 x 1011 m.
v

Gm
r

v
m2
8.85 x 10 2
s
6.67 x 10
11
8
kg m  m 2

s 2kg 2
1.99 x 10
2.97 x 104
30
kg
 1.50 x110
11
m
m
s
A satellite is in a low earth orbit, some 250 km above the earth's surface. rearth is 6.37 x 106 m
and mearth = 5.98 x 1024 kg. Find the period of the satellite in minutes.
r is the radius of the earth plus y, the height of the satellite.
 103 m 
y  250 km 

 1 km 
r  rearth  y
 0.25 x 106 m
r  6.37 x 106 m + 0.25 x 106m = 6.62 x 106 m
178
T  2
3
r
Gm

 2
6.67 x 1011
T  2
290.12 x 1018 2
s
39.89 x 1013
T  2
72.73 x 104 s 2


3

kg m m 2
5.98 x 1024 kg
2
2
s kg

7.273 x 105 s 2
 53.57 x 102 s
 1 min 
T  5357 s 

 60 s 

 2
6.62 x 106 m
 5357 s
89.3 min
The moon has a period of 28 days. If the earth's mass is 5.98 x 1024 kg, how far is the moon
from the earth?
 24 h  3600 s 
T  28 day 


 1 day  1 h 
T  2
r3
Gm
square both sides:
T  2.42 x 106 s
2
r3 
T  4 

 Gm 
2
Solve for r3:
 1 
 1 
r 3  T 2Gm  2  r  3 T 2Gm  2 
 4 
 4 

r  3 2.42106 s

2
6.67 x 1011


kg m m 2
 1 
24
5.98
x
10
kg
 2
s 2 kg 2
 4 
r  3 5.923 x 1025 m3
convert 5.923 x 1025 to something times 1024 (we chose 1024 because 24 is divisible by 3)
r  3 59.23 x 1024 m3

3.90 x 108 m
179
Useless Trivia Facts:
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15 million blood cells are destroyed in the human body every second.
3,000 teens start smoking every day in the United States.
The average person spends two weeks of their life kissing.
Scientists in Australia's Parkes Observatory thought they had positive proof of alien life,
when they began picking up radio-waves from space. However, after investigation, the
radio emissions were traced to a microwave in the building.
40 people are sent to the hospital for dog bites every minute.
In Japan, 20% of all publications sold are comic books.
The average day is actually 23 hours, 56 minutes and 4.09 seconds. We have a leap year
every four years to make up for this shortfall.
In the next seven days, 800 Americans will be injured by their jewelry.
Disneyland opened in 1955.
It is estimated that at any one time, 0.7% of the world's population are drunk.
If you divide the Great Pyramid's perimeter by two times it's height, you get pi to the
fifteenth digit.
The pupil of the eye expands as much as 45 percent when a person looks at something
pleasing.
50,000 of the cells in your body will die and be replaced with new cells all while you
have been reading this sentence.
67.5% of men wear briefs instead of boxers.
75% of people wash from top to bottom in the shower.
The average four year-old child asks over four hundred questions a day.
80% of all body heat escapes through the head.
Lightning strikes the Empire States Building more than 50 times a year.
85% of the population can curl their tongue into a tube.
A fetus acquires fingerprints at the age of three months.
The Future's Museum in Sweden features a scale model of the solar system. The sun is
105 meters in diameter and the planets range from 5mm to 6 km from the ‘sun’. The
exhibit also features the nearest star, Proxima Centauri. The distance to the star is also to
scale. It is located in the Museum of Victoria. . . . in Australia.
Gravity in Orbit:
We all know that the astronauts in space orbiting the earth are "weightless".
Does this mean that there is no gravity in space? Well, no. Most of our spacecraft are in pretty low
orbits. The distance between the astronauts and the earth is not that much greater than when they
are on the earth. Their weight is only about ten percent less than it is on earth. So why are they
weightless?
The space shuttle and everything in it are falling towards the earth. It is in a state of freefall. We
know that everything falls at the same rate, so everything in the space shuttle is falling at the same
speed. Because of this there is no relative motion between the space shuttle and everything in it.
There is no sense of up or down and the astronauts no longer feel the force of gravity. It’s like
being inside an elevator that is falling down the elevator shaft (the cable broke or something). In a
normal elevator, one that isn’t falling, gravity exerts a downward force on everything. If you stand
180
on a bathroom weight scale, you push down on it and it reads out your weight. But now the cable
breaks. You are still standing on the scale, but the elevator, the scale, and you are all falling down
accelerating at 9.8 meters per second squared. You no longer exert a force on the scale – it is
falling at the same speed that you are. It now reads zero.
Any objects in the elevator would appear to be weightless. If you held a ball outward and then
released it, it would not appear to fall down (since it is already falling). It would appear to float in
space in front of you. You would think, “Hey, cool, there’s no gravity in the elevator. Neat!”
It would be pretty neat too. Until the elevator hits the
bottom of the elevator shaft.
Okay, let’s transfer this idea to the space shuttle. It is,
in effect, a falling elevator, one with the advantage of
not crashing into the floor – it never hits the earth!
That’s why the astronauts are “weightless”.
Free Falling in an
elevator
181
Dear Cecil:
Tycho Brahe, the astronomer, is buried here in Prague in a church. That's not so
unusual. What is rather strange is the persistent rumor that he had a silver nose,
something to do with a duel. Did it tie on around his face with silk strings? Did it have
little hooks that went around his ears, like eyeglasses? Was it surgically attached? I'm
truly curious. I also wonder if it tarnished, and whether he had to polish it. Did it make
a note like a whistle when he blew his nose? --Raymond Johnston, chief copy editor,
Prague Post
Cecil replies:
Glad to be of help, Ray. I know how tough it is for journalists to come by this kind of
information. What you heard was no rumor. Tycho Brahe (1546-1601), the father (or at least
the godfather) of modern astronomy, really did wear an artificial nose, owing to the fact that
the real one had been sliced off in a duel. You may think: this does not sound like the
scientists I know. Tycho Brahe, however, was no ordinary stargazer.
It happened in 1566 while the 20-year-old Tycho was studying at the University of
Rostock in Germany. Attending a dance at a professor's house, he got into a quarrel
with one Manderup Parsbjerg, like himself a member of the Danish gentry. Over a
woman? Nah--tradition has it that the two were fighting over some fine point of
mathematics. (My guess: Fermat's Next-to-Last Theorem, which posits that 2 + 2 = 5
for very large values of 2.) Friends separated them, but they got into it again at a
Christmas party a couple weeks later and decided to take it outside in the form of a
duel. Unfortunately for Tycho the duel was conducted in pitch darkness with swords.
Parsbjerg, a little quicker off the dime, succeeded in slicing off the bridge
(apparently) of Tycho's nose. Reconstructive surgery then being in a primitive state,
Tycho concealed the damage as best he could with an artificial bridge made of
precious metals. He carried some nose goop with him always, either to polish the
nose or to glue it more firmly in place. But no hooks or string, and probably no
whistling either.
High-handed and irascible, Tycho Brahe was the kind of guy who got into duels. Luckily he
was also a genius. Fascinated by the stars since his youth, he discovered that existing
astronomical tables were grossly inaccurate and set about making his own meticulous
observations of the heavens, a project that occupied him for most of his life. To keep him
from going abroad, the king of Denmark and Norway gave Tycho a prodigious quantity of
cash ($5 billion in today's money, by one estimate) and his own island. There Tycho
constructed an observatory where for 20 years he compiled the impressive body of
astronomical data that his assistant Johannes Kepler subsequently used to deduce the laws of
planetary motion. All this, mind you, with the naked eye; the telescope hadn't yet been
invented.
To give you a further indication of the type of guy we're dealing with there, Tycho
didn't marry the mother of his eight children, employed a dwarf as a jester, kept a pet
elk (which died after breaking a leg while going downstairs drunk), dabbled in
alchemy, and tyrannized the local peasantry. After his royal patron died of excessive
drink he managed to tick off everyone in Denmark, had his subsidies revoked, and
eventually found it wise to leave the country. Having relocated to Prague, he died
after drinking heavily at dinner, obviously a pretty common fate in those days.
182
Tycho's tomb was reopened in 1901 and his remains were examined by medical experts.
The nasal opening of the skull was rimmed with green, a sign of exposure to copper.
Presumably this came from the artificial nose, which supposedly had been made of silver
or gold. The experts put the best face on this, as it were, saying that Tycho was an expert
in metallurgy and probably wanted an alloy that was durable and skin colored. Sure, guys.
I say Mr. Astronomy got nicked in the nose department twice.
--CECIL ADAMS
Dear Doctor Science,
IF YOU’RE SO SMART TELL ME HOW THE INERTIAL DAMPENERS WORK
ON THE USS ENTERPRISE-NCC-1701-D? BET I STUMPED YOU ON THAT
ONE.
-- JON STIGLER from EADS, TN
Dr. Science responds:
They didn't work, and that was the big problem with the series. The ship slipped into
hyperdrive whenever it felt like it, sending the crew zipping across the galaxy in a
nanosecond. This made it especially hard on the navigators, who were constantly coming
up with novel ways to get their bearings, including monitoring talk radio from nearby
planets in the hope of learning their approximate location. Eventually, captain and crew
developed a pervasive sense of chronic apathy, and instead of piloting their craft, spent
their time watching reruns of the first Star Trek, where Jim and Spock traded profundities
on the planet of scantily clad women.
183
AP Physics – Gravity Wrap-up
You get three equations to work with, and only three. Everything else you have to develop from
these three equations (also have to use some other ones). Here they are:
FG  
Gm1m2
r2
Newton’s law of gravity.
v2
ac 
r
Centripetal acceleration equation.
 F  FNet  ma
Second law.
Here’s what you have to be able to do:
1.
You should understand the uniform circular motion of a particle so you can:
a. Relate the radius of the circle and the speed or rate of revolution of the particle to the
magnitude of the centripetal acceleration.
v2
ac 
r
You have this equation to work with:
In this equation, v is the linear speed, so it’s fairly easy to relate it to the radius (r) or to
the centripetal acceleration (ac). You just use the old equation as it is given. If r gets
bigger, then the centripetal acceleration gets smaller, etc.
The rate of revolution is the rate that the thing rotates. This would be the number of
rotations divided by the time it took to do them. This is called the angular velocity,  .
Unfortunately, you don’t got no equation for this. Just remember that any rate is simply a
quantity divided by time, so the angular velocity is simply:


t
Where  is the angular displacement, which will most likely be the number of revolutions
that have been made?
What happens to the centripetal acceleration if the velocity is increased and you want to
keep the same radius? If the radius increases, but the velocity stays the same, what
happens to the centripetal acceleration?
184
You should be able to derive an equation that relates linear speed to angular velocity. We
did that in your gravity handout. The Physics Kahuna showed you how to develop the
equation:
v  2 r
Basically, the velocity is proportional to the angular velocity. So whatever happens when the
linear speed increases also happens if the angular velocity increases.
You might be able to get away without developing the equation at all. Using this concept,
it is a simple matter to relate the rate of revolution of the particle to the magnitude of the
centripetal acceleration.
b. Describe the direction of the particle’s velocity and acceleration at any instant during
the motion.
The acceleration is always towards the center.
The velocity is always tangent to the object’s path.
Path of Orbit
ac
Planet
Moon
v
For any type of circular motion, such as the moon orbiting a planet, the direction of the
centripetal acceleration is towards the center and the velocity direction is a tangent to the
path.
c. Determine the components of the velocity and acceleration vectors at any instant and
sketch or identify graphs of these quantities.
This is a simple matter – use trigonometry to resolve the acceleration and or velocity
vector into its components. Again, graphing these quantities is quite simple.
185
1
v
v
5 vx
2
ac
vy
1
4
3
5
2
t
4
3
Above is a drawing of an object that is following a circular path. The velocity and centripetal
acceleration vectors are shown at five different positions. To the right of that is a graph of
velocity Vs time for the same motion. Please now figure out how the graph works.
The velocity is either only up and down or only right or left at the four cardinal positions, 1,
3, 4, and 5. So the speed is either all vx or all vy. For all other positions on the path, there
will be both a horizontal and vertical component for the velocity. Can you see how the speed
v is the sum of the squares of the components? Think about till you can.
Here is a graph of the centripetal acceleration Vs time for the same set of points.
ac
4
2
ay
3
2
5
t
ax
1
2. Students should be able to analyze situations in which a body moves with specified
acceleration under the influence of one or more forces so they can determine the
magnitude and direction of the net force, or of one of the forces that makes up the net
force, in situations such as the following:
186
a. Motion in a horizontal circle (e.g., mass on a rotating merry-go-round, or car
rounding a banked curve).
The main thing here is using the second law and the centripetal acceleration to find the
centripetal force.
F  ma
v2
ac 
r
 v2 
Fc  m  
 r 
mv 2
Fc 
r
The centripetal force can then be used to calculate the frictional force acting on a car
traveling in a circle or some such thing.
In a previous unit we looked at several problems along this line.
The centripetal force will be used in several other sections. One that really requires it is when
we study a charged particle moving through a uniform magnetic field. You’ll see how this
works later on.
b. Motion in a vertical circle (e.g., mass swinging on the end of a string, cart rolling
down a curved track, rider on a Ferris wheel).
Draw a free body diagram. Analyze the forces. The centripetal force will always end up
being the net force acting on the body and will always be directed to the center of the
circular path. There are two forces working on the vertical circle – the weight of the
object and the centripetal force. At the very top for the instant of time that the body is up
there, the sum of the vertical forces must be zero since the body is moving sideways and
not up and down. So you can solve for the tension which will turn out to be The idea here
is that the centripetal force must equal the weight of the object. From this you can
calculate an equation for the minimum velocity a body needs to travel along a vertical
circular path. The equation is:
v
gr
3. You should know Newton’s Law of Universal Gravitation so you can:
a.
Determine the force that one spherically symmetrical mass exerts on another.
This is pie. You just use Newton’s law of gravity.
b.
Determine the strength of the gravitational field at a specified point outside a
spherically symmetrical mass.
This is another slice of really good pumpkin pie with whipped cream on top (yummy).
Again, merely apply good old Newton’s law of gravity.
4. You should understand the motion of a body in orbit under the influence of gravitational
forces so you can, for a circular orbit:
187
a. Recognize that the motion does not depend on the body’s mass, describe
qualitatively how the velocity, period of revolution, and centripetal acceleration
depend upon the radius of the orbit, and derive expressions for the velocity and
period of revolution in such an orbit.
This is a derivation special. The Physics Kahuna demonstrated how to do all of this in a
previous unit.
You have to derive the equation for centripetal force and set it equal to the force of gravity
from Newton’s law of gravity. Then you solve for the velocity. This gives you the orbital
velocity.
v
Gm1
r
To find the period, you use the equation for velocity,
v
x
and the
t
equation for the orbital velocity we just found. The distance x is the circumference of the
circle. Which is 2 r . Put them together and you end up with:
t  2
r3
Gm
We did a number of problems that required developing and using these equations.
188
AP Test Examples:

From 1997:
To study circular motion, two students use the hand-held
device shown, which consists of a rod on which a spring scale
is attached. A polished glass tube attached at the top serves
as a guide for a light cord attached the spring scale. A ball of
mass 0.200 kg is attached to the other end of the cord. One
student swings the ball around at constant speed in a
horizontal circle with a radius of 0.500 m. Assume friction
and air resistance are negligible.
a.
Explain how the students, by using a timer and the
information given above, can determine the speed of the ball as it is revolving.
Measure radius and calculate circumference. Use stop watch to measure the period
(time of one revolution). Use v 
x
to find the speed of the ball.
t
b.
How much work is done by the cord in one revolution? Explain how you arrived at your
answer.
No work. The motion is perpendicular to the force.
c.
The speed of the ball is determined to be 3.7 m/s. Assuming that the cord is horizontal as
it swings, calculate the expected tension in the cord.
FT  FC
FC
d.
2
0.200kg  3.7 m s 



0.500 m
mv 2
FC 
r
5.5 N
The actual tension in the cord as measured by the spring scale is 5.8 N. What is the
percent difference between this measured value of the tension and the value calculated in
part (c)?
% diff 
e.
F  ma
v2
ac 
r
5.8 N  5.5 N
100%   5.5%
5.5 N
The students find that, despite their best efforts, they cannot swing the ball so that the cord
remains exactly horizontal. On the picture of the ball below, draw vectors to represent the
forces acting on the ball and identify the force that each vector represents.
Explain why it is not possible for the ball to swing so that the cord remains exactly
horizontal.
The tension in the string, t, provides the centripetal force. mg is
the weight of the ball.
There is always a downward force acting on the ball, its weight,
mg. So there must be a component of the tension in the opposite
t
mg
189
direction since the ball is not moving in a vertical direction. The sum of the vertical
forces must be zero.
f. Calculate the angle that the cord makes with the horizontal.

2
  0.200kg  9.8 m s
1  mg 
1
  sin 
  sin 
kg  m
T



5.8
s2








19.8o
From 2001:

Z
A ball of mass M is attached to a string of length R and negligible mass.
The ball moves clockwise in a vertical circle, as shown to the right.
When the ball is at point P, the string is horizontal. Point Q is at the
bottom of the circle and point Z is at the top of the circle. Air resistance is P
negligible. Express all algebraic answers in terms of the given quantities
and fundamental constants.
(a) On the figures below, draw and label all the forces exerted on the ball
when it is at points P and Q, respectively.
t
P
M
Q
Side View
t
mg
Q
mg
(b) Derive an expression for vmin, the minimum speed the ball can have at point Z without leaving
the circular path.
FC  t  mg but the tension is zero
F  ma
Fc  mac
mv 2
 mg
r
v2
g
r
Fc 
FC  mg
mv 2
Derive the equation for centripetal force.
r
v  rg
v  Rg
(c) The maximum tension the string can have without breaking is Tmax. Derive an expression for
vmax, the maximum speed the ball can have at point Q without breaking the string.
The Tension in the string is the upward force and the weight is the downward force, the sum of
these two forces is ma so:
ma  TMax  mg
But the acceleration acting on the system is the centripetal acceleration.
190
v2
m  TMax  mg
r
v
v2 
r
TM ax  mg 
m
r
TMax  mg 
m

R
TM ax  Mg 
M
(d) Suppose that the string breaks at the instant the ball is at point P. Describe the motion of the ball
immediately after the string breaks.
The ball would go straight up, slowing down as it is accelerated downward by gravity, it will stop,
then fall down, accelerating as it goes. Its motion is tangential to the path, which is straight up.
From 1999:
5.
A coin C of mass 0.0050 kg is placed on a horizontal disk at a
distance of 0.14 m from the center, as shown below. The disk
rotates at a constant rate in a counterclockwise direction as
seen from above. The coin does not slip, and the time it takes
for the coin to make a complete revolution is 1.5 s.
a.
b.
The figure to the right shows the disk and coin as viewed from
above. Draw and label vectors on the figure below to show
the instantaneous acceleration and linear velocity vectors for
the coin when it is at the position shown.
Determine the linear speed of the coin.
v
a
The coin makes one revolution is a time of 1.5 s. So the distance
it travels in that time is the circumference of the circle it makes,
which is 2 r .
v
c.
x
t

2 r
t

1.5 s 

0.586
m
s
The rate of rotation of the disk is gradually increased. The coefficient of static friction
between the coin and the disk is 0.50. Determine the linear speed of the coin when it just
begins to slip.
Fc  f
v  r g
d.
2  0.14 m 
m

v2
  mg
r
 0.14 m  0.5  9.8

m

s2 

0.828
m
s
If the experiment in part (c) were repeated with a second, identical coin glued to the top of
the first coin, how would this affect the answer to part (c)? Explain your reasoning.
There would be no change. The mass cancels out.
191
From 1995:

Part of the track of an amusement park roller coaster is shaped as shown below. A safety bar is
oriented length-wise along the top of each car. In one roller coaster car, a small 0.10 kilogram
ball is suspended from this bar by a short length of light, inextensible string.
a.
Initially, the car is at rest at point A.
i.
On the diagram to the right, draw and label all the forces acting on the
0.10-kilogram ball.
ii. Calculate the tension in the string.
t  mg  0 t  mg
m

 0.10 kg  9.8 2  
s 

t
mg
0.98 N
The car is then accelerated horizontally, goes up a 30° incline, goes down a 30° incline, and
then goes around a vertical circular loop of radius 25 meters. For each of the four situations
described in parts (B) to (E), do all three of the following. In each situation, assume that the
ball has stopped swinging back and forth. 1) Determine the horizontal component Th of the
tension in the string in newtons and record your answer in the space provided. 2) Determine
the vertical component Tv of the tension in the string in newtons and record your answer in the
space provided. 3) Show on the adjacent diagram the approximate direction of the string with
respect to the vertical. The dashed line shows the vertical in each situation.
b. The car is at point B moving horizontally to the right with an acceleration of 5.0 m/s .
m

Th  ma  0.10 kg  5.0 2  
s 

Tv  mg
c.
m

 0.10 kg  9.8 2  
s 

0.49
0.98
m
s2
m
s2
The car is at point C and is being pulled up the 30° incline
with a constant speed of 30 m/s.
Th  0
Tv  mg
d.
m

 0.10 kg  9.8 2  
s 

0.98
m
s2
The car is at point D moving down the incline with an
acceleration of 5.0 m/s2.
m

Th  ma cos  0.10 kg  5.0 2  cos 30o 
s 

0.43 N
192
The sum of the forces in the y direction must equal maV . Therefore:
mg  Tv  mav
Tv  mg  mav
TV
mg
The acceleration in the vertical direction is ma sin  .
av

a
m
m


TV  mg  ma sin   0.10 kg  9.8 2   0.10 kg  5.0 2  sin 30o 
s 
s 


e.
T
0.73 N
The car is at point E moving upside down with an instantaneous speed of 25 m/s and no
tangential acceleration at the top of the vertical loop of radius 25 m.
TH  0
In the vertical direction, we have a centripetal FC , the
tension, and the weight of the ball. The centripetal force
must equal the sum of the other two forces.
FC  TV  mg
TV  FC  mg
mv 2

 mg
r
2
 m
0.10 kg  25 
 s   0.10 kg  9.8 m 
TV 



s2 
 25 m 

TV 
1.5 N
193























More Foolish Trivia:
A jogger's heel strikes the ground 1,500 times per mile.
A lifetime supply of all the vitamins you need has a mass of only about
225 g (eight ounces).
3 out of 4 optometrists wear eyeglasses.
Computer users, on average, blink seven times per minute.
Dark circles under the eye are an inherited trait.
The iris membrane controls the amount of light that enters your eye.
In 1992, when EuroDisney first opened in France, park visitors beat up
some of the costumed characters (oh please not Snow White!) because
at the time most people had been against the park being built.
Until the 1960's men with long hair were not allowed to enter
Disneyland.
Eiffel Tower was the tallest structure in the world before the
construction of the Empire State Building in 1930.
A B-25 bomber airplane crashed into the 79th floor of the Empire State
Building on July 28, 1945.
Sunbeams that shine down through clouds are called crepuscular rays.
The Apollo 11 Lunar Module had only 20 seconds of fuel left when it
landed on the moon in 1969.
Kite flying is a professional sport in Thailand.
A normal raindrop falls at about 7 miles per hour.
A penny whistle has six finger holes.
The human eye sees everything upside-down, but the brain turns it
right side up.
A shell constitutes 12 percent of an egg's weight.
A silicon chip a quarter-inch square has the data processing capacity of
the original 1949 ENIAC computer, which occupied a city block.
All the dirt from the foundation to build the former World Trade
Center in NYC was dumped into the Hudson River to form the
community now known as Battery City Park.
Rice paper does not contain rice or any rice products.
The height of the Eiffel Tower varies as much as six inches depending
on the temperature.
At April 2000, Hong Kong had 392,000 faxlines - one of the highest
rates of business fax use in the world.
The only part of the human body that has no blood supply is the
cornea. It takes its oxygen directly from the air.
194
(CNN) -- On Saturday the Earth reaches its farthest point from the sun during its orbit, but
the average global temperature actually goes up this time of year. What gives?
Like the other planets in the solar system, the Earth moves in a lopsided orbit around the sun. In
July, our planet reaches the most distant point of its annual revolution, called aphelion. In January,
it dips to its nearest position, or perihelion.
Compared to other planets, the Earth's orbit is only slightly askew. It has an eccentricity of 1.7
percent, while Mars' is more than 9 percent and Mercury's more than 20 percent.
Looking at a page-sized map, an observer might mistake the Earth's path around the sun for an exact
circle. Scientists know better.
"When we're closest the sun, the distance is 147.8 million kilometers [91.8 million miles]. This
weekend we will be 152.6 million kilometers [94.8 million miles] away, a 5 million kilometer (3.1
million mile) difference," University of Florida astronomer George Lebo said.
The greater the distance, the less intense the rays of the sun.
"Averaged over the globe, sunlight falling on Earth at aphelion is about 7 percent less intense than it
is at perihelion," NASA climate researcher Roy Spencer said.
Then why is it so warm now in the Northern Hemisphere? The reason is that the Earth tilts 23.5
degrees on its axis, and because of where it is in its orbit, the northern half is the portion most
exposed to the sun during the summer months.
The conditions are reversed in the Southern Hemisphere, which basks in the solar rays half a year
later, when the South Pole tips more toward the sun.
Overall, the entire globe averages higher temperatures when the Earth is farther from the sun. The
planet is about 4 degrees F (2.3 degrees C) warmer during aphelion than perihelion, NASA
scientists said.
The reason? Continents and oceans are distributed unevenly. The north has more land and the south
more water. When sunlight strikes the former, it heats up more quickly than the latter, which can
absorb more heat before rising in temperature.
In other words, Earth is slightly warmer in July because the sun is shining on all the northern
continents, which heat up rather easily. In January, the sun focuses its rays on the southern oceans,
which have higher heat capacity.
Another quirk of physics contributes to even more planetary heat this time of year. During aphelion,
planets move in their orbits more slowly than during perihelion.
As a consequence, the summer in the north is several days longer than in the south, affording the
sun more time to cook the landmasses in the Northern Hemisphere.
195