7.4 Angular Acceleration
There is in fact a second type of acceleration (in addition to
centripetal acceleration) that can be associated with circular
motion. Angular acceleration occurs when the circular motion is
not uniform; that is, if the tangential speed is increasing or
decreasing. In the last lecture, we often referred to a merry-goround or a turntable. These undergo UCM most of the time, but
at some point they have to come to a stop and then start up
again. During these periods of time, they have a non-zero
angular acceleration.
Average angular acceleration is, not surprisingly,
∆𝜔
𝛼=
∆𝑡
(that’s a Greek lower-case “alpha”).
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Slide 1
7.4 Angular Acceleration
In most cases, we are only concerned with constant angular
acceleration. Then if 𝜔 = 𝜔0 at t = 0, we can write
𝛼=
𝜔−𝜔0
𝑡
or
𝜔 = 𝜔0 + 𝛼𝑡
(note the resemblance to the translational case from chapter 2,
∆𝑣
where 𝑎 = and 𝑣 = 𝑣0 + 𝑎𝑡).
∆𝑡
As with arc length and angle (𝑠 = 𝑟𝜃), and tangential and angular
speeds (𝑣 = 𝑟𝜔), we can relate the tangential acceleration to
the angular acceleration: 𝑎𝑡 = 𝑟𝛼.
In the text (and the notes from last class), we dropped the
subscript on the tangential speed 𝑣𝑡 , since this is the only
possible type of speed involved with circular motion. We can not
do the same thing here, since we must distinguish 𝑎𝑡 from 𝑎𝑐 .
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Slide 2
7.4 Angular Acceleration
Note carefully that centripetal acceleration is
necessary for circular motion, but tangential
acceleration is not. The two acceleration
vectors are perpendicular to each other at any
moment, and the total acceleration vector is
a = a𝑡 t + a𝑟 r, where t and r are unit vectors
in the tangential and radial directions.
The table shown
here illustrates
similarities
between
translational and
angular motion.
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Slide 3
Problem #1: Fan Blades
WBL EX 7.49
The blades of a fan running at low speed turn at 250 rpm. When the fan is
switched to high speed, the rotation rate increases uniformly to 350 rpm in 5.75 s.
a) What is the magnitude of the angular acceleration of the blades?
b) How many revolutions do the blades go through while the fan is accelerating?
Solution: In class
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Slide 4
Problem #2: Dizzy Hamster
The hamster shown in this video has decided to undergo some nonuniform circular motion
(FOR SCIENCE!) Let’s analyze the hamster’s kinematics during the time it’s spinning. The
hamster’s name is Steve. Steve starts spinning at the 31-second mark of the video and falls
out at the 36-second mark. During this time, Steve’s angular velocity decreases in
magnitude from 25.0 rad/s to 12.5 rad/s. We will assume that the angular acceleration is
constant during this time and that Steve’s 0.15-kg body exists entirely at a radius of 5 cm.
a) What are the signs of ω and α?
b) What is the constant angular acceleration,
α?
c) How many revolutions does Steve make?
d) What is the total distance that Steve travels
while he is spinning?
e) What is Steve’s maximum tangential speed?
f) What is the largest magnitude of centripetal
acceleration that Steve experiences?
g) What is the largest magnitude of centripetal
force that Steve experiences?
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Slide 5
7.5 Newton’s Law of Gravitation
At this point in the course, our knowledge of gravity is simply
that “it produces an acceleration of magnitude 9.80 m/s,
directed downward”, and that “this magnitude depends weakly
on one’s location on the surface of the Earth, and that it doesn’t
change much up to altitudes of 100 km or so”.
That second statement is a bit annoying. Why should classical
mechanics produce results that are only valid in particular
regions of the universe? Well…it doesn’t.
Newton’s Universal Law of Gravitation tells us what sort of
gravitational forces are present among any masses at any
distances from each other. It’s valid for planets, moons, stars,
and even galaxies.
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Slide 6
7.5 Newton’s Law of Gravitation
The law provides a relationship for the gravitational interaction
between two point masses 𝑚1 and 𝑚2 , separated by a distance r.
It states that every particle in the universe has an attractive
gravitational interaction with every other particle in the
universe. These forces of mutual interaction are equal and
opposite Newton’s 3rd-law pairs (F12 = −F21 ).
Furthermore, the law states that the magnitude of
the gravitational force depends inversely on the
square of the distance between the masses (inversesquare forces tend to be rather prominent in
universes with three spatial dimensions). Finally, the
force depends on the product of the two masses
themselves.
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Slide 7
7.5 Newton’s Law of Gravitation
All together, we can write Newton’s law of gravitation
mathematically as
𝐹𝑔 =
𝐺𝑚1 𝑚2
,
2
𝑟
𝐺 = 6.67 ×
N∙m2
−11
10
.
kg 2
(𝐺 is called the universal gravitational constant).
From the 1/r2 dependence, we see that not only is
gravity a non-contact force, it acts over an infinitely
long range. Masses that are separated by hundreds
of billions of km still exhibit a gravitational attraction, albeit a
weak one.
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Slide 8
7.5 Newton’s Law of Gravitation
Of course, most interesting masses are not point masses; they have a
spatial extent. For homogeneous spheres, the preceding equation is
valid if we consider all of the mass to exist at the center of the sphere
(and if we only wish to calculate 𝐹𝑔 outside of the spheres’ surfaces).
Proving this requires calculus, but the figure should at least convince
you that the direction of F𝑔 makes sense.
It should be noted that Newton didn’t have a good way of determining
the value of the proportionality constant 𝐺. It was first measured by
Henry Cavendish about 100 years after Newton first published the
inverse-square law. Even today, 𝐺
remains one of the “least-precise”
physical constants. Some physicists
would argue that even the three
significant digits shown on the previous
slide are one too many.
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Slide 9
7.5 Newton’s Law of Gravitation
Well, now we have two equations for the gravitational force:
𝐹𝑔 =
𝐺𝑚1 𝑚2
𝑟2
and
𝐹𝑔 = 𝑚𝑔
How are they related?
Let’s assume that the two masses are a (relatively small mass) 𝑚
and (a planet’s mass) 𝑀. Newton’s 2nd law can then be used to
analyze the acceleration of 𝑚 due to the gravitational force:
𝐺𝑚𝑀
𝐹𝑛𝑒𝑡 = 𝑚𝑎 → 2 = 𝑚𝑎𝑔
𝑟
Thus, the acceleration due to gravity at any distance 𝑟 from the
planet’s center (assuming that it’s outside of the planet’s surface)
is
𝑎𝑔 =
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𝐺𝑀
𝑟2
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Slide 10
7.5 Newton’s Law of Gravitation
What if this acceleration is being measured by someone at the
Earth’s surface? In this case, 𝑀 = 𝑀𝐸 = 5.98 × 1024 kg (the
mass of the Earth), and 𝑟 = 𝑅𝐸 = 6.371 × 106 m (the radius of
the Earth). Plugging in the numbers, we get
𝑎𝑔 =
𝐺𝑀𝐸
2
𝑅𝐸
=
(6.67×10−11 N∙m2 /kg 2 )(5 .98×1024 kg)
6.371×106 m 2
= 9.82 m/s 2
This is the value of g that we have been using since chapter 2.
Well, almost. We use 9.80. There’s various reasons – primarily,
the fact that the Earth isn’t perfectly spherical, so the value of 𝑅𝐸
shown above is just an approximation. Note that our equation
for 𝑎𝑔 does not involve m, the mass of the object. This agrees
with our experimental observation that acceleration due to freefall is independent of the mass of the falling object.
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Slide 11
7.5 Newton’s Law of Gravitation
We are now in a position to calculate the magnitude of
gravitational acceleration at higher altitudes. If h is the altitude,
then the distance from the Earth’s center is 𝑅𝐸 + ℎ, and
𝐺𝑀𝐸
𝑎𝑔 =
𝑅𝐸 + ℎ 2
Since 𝑅𝐸 ≈ 6370 km, an altitude of a few hundred km is required
in order to achieve a noticeable change in 𝑎𝑔 . Later, we will
learn why astronauts appear “weightless”, despite the data
below.
Altitude (km)
Altitude Example
𝒂𝒈 (m/s2)
0
8.8
36.6
400
35,700
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9.82
9.80
9.71
8.70
0.225
Mean Earth surface
Mt. Everest
Highest crewed balloon
Space shuttle orbit
Communications satellite
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Slide 12
7.5 Newton’s Law of Gravitation
In chapter 5, we learned that an object at a height h above a
(zero-potential-energy) reference level has a potential energy
𝑈 = 𝑚𝑔ℎ. However, this was valid only if the gravitational
force was constant over the distance h. As we now know, if h is
very large, Fg is not constant; it varies with position.
Using fairly simple calculus (not repeated here), it can be shown
that the gravitational potential energy of two point masses
separated by a distance r is
𝐺𝑚1 𝑚2
𝑈=−
𝑟
(this definition requires that we set U = 0 when the objects are
infinitely far apart; i.e. 𝑟 → ∞). If we consider a mass m at an
altitude of h above the Earth’s surface, then
𝐺𝑚𝑀𝐸
𝑈=−
𝑅𝐸 + ℎ 2
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Slide 13
7.5 Newton’s Law of Gravitation
We say that the Earth (in fact,
any mass) exists in a negative
gravitational potential energy
well. As h increases, U
becomes less negative.
Thus, when gravity does negative work (an object
moves higher in the well) or positive work (an
object falls lower in the well), there is a change in
potential energy. This energy change will be
beneficial in solving some upcoming problems.
Our equation for U can also be extended to the case of three or
more masses. For example,
𝑈 = 𝑈12 + 𝑈23 + 𝑈31 =
PC141 Intersession 2013
𝐺𝑚1 𝑚2
−
𝑟12
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𝐺𝑚2 𝑚3
−
𝑟23
𝐺𝑚3 𝑚1
−
𝑟31
Slide 14
7.5 Newton’s Law of Gravitation
For example, when considering the total mechanical energy of a
mass m1 in orbit about a mass m2, we have
𝐸 =𝐾+𝑈 =
1
𝑚1 𝑣 2
2
−
𝐺𝑚1 𝑚2
.
𝑟
Rearranging, we have
𝑣=
2
𝐺𝑚1 𝑚2
𝐸+
𝑚1
𝑟
Although we like to think that the Earth’s orbit about the sun is
circular, it is actually slightly elliptical (more on that later). At
perihelion (the point of closest approach), the speed of the Earth
in its orbit is slightly faster than at aphelion (the farthest point).
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Slide 15
Problem #3: High-Altitude Gravitational Acceleration
WBL LP 7.17
Compared with its value on the Earth’s surface, the value of the
acceleration due to gravity at an altitude of one Earth radius is
A
the same
B
two times as great
C
one-half as great
D
one-fourth as great
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Slide 16
Problem #4: Orbit of the Moon
WBL EX 7.59/7.63
The moon orbits the Earth with an approximately circular orbit of radius 3.85 x 105
km, and an orbital period of 27.5 days.
a)
b)
c)
d)
e)
Use this data to calculate the mass of the Earth
What is the moon’s tangential speed?
What is the moon’s kinetic energy?
What is the Earth-moon system’s potential energy?
What is the system’s total energy?
Solution: In class
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Slide 17
Problem #5: Gravitational Potential Energy
WBL EX 7.64
a) What is the gravitational potential energy of the
configuration shown in the figure, if all of the masses are
1.0 kg?
b) What is the gravitational force per unit mass at the center
of the configuration?
Solution: In class
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Slide 18
7.6 Kepler’s Laws and Earth Satellites
In the very early 1600s (pre-dating Newton), Johannes Kepler
produced three empirical laws that governed planetary motion.
With the benefit of Newton’s law of gravitation, Kepler’s laws can
now be derived using just a few lines of algebra. We present the
laws here. Note that they do not only apply to planetary orbits.
Any system consisting of a body revolving about a much more
massive body due to an inverse-square force conforms to Kepler’s
laws.
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Slide 19
7.6 Kepler’s Laws and Earth Satellites
Kepler’s First Law (the law of orbits)
Planets move in elliptical orbits, with the sun at one of the focal points.
This fact was alluded to a few slides ago. Absolutely nothing
about Newton’s Law of Gravitation suggests that the orbits need
to be circular; the orbits are in fact ellipses (at least in the case of
closed orbits. For open orbits, in which a body approaches the
sun and then leaves, never to return, the path is a hyperbola or
parabola).
As you may recall, a circle is a special case
of ellipse for which the two focal points
coincide. Most planets have orbits that
are nearly circular; the Earth’s perihelion
and aphelion differ by only about 3% of
the average Earth-sun separation.
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Slide 20
7.6 Kepler’s Laws and Earth Satellites
Kepler’s Second Law (the law of areas)
A line from the sun to a planet sweeps out equal areas in equal times.
This is illustrated in the figure below. Since the speed of the
planet is greatest when it is closer to the sun (slide 15), the arc
lengths 𝑠1 and 𝑠2 swept out during the same time duration are
not equal in length. However, the two areas 𝐴1 and 𝐴2 are
equal. This can be proven fairly easily using techniques from
chapter 8, but we won’t attempt it here.
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Slide 21
7.6 Kepler’s Laws and Earth Satellites
Kepler’s Third Law (the law of periods)
The square of the orbital period of a planet is directly proportional to the
cube of the average distance of the planet from the sun. That is, 𝑻𝟐 ∝ 𝒓𝟑 .
This law can be proven fairly easily in the special case of a
circular orbit. Since the centripetal force is supplied by the force
of gravity, these forces are equal. Using Ms and mp as the masses
of the sun and the planet, respectively,
𝑚𝑝 𝑣 2 𝐺𝑚𝑝 𝑀𝑠
=
𝑟
𝑟2
and thus the speed of the orbiting planet (constant, in a circle) is
𝑣=
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𝐺𝑀𝑠
𝑟
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Slide 22
7.6 Kepler’s Laws and Earth Satellites
Kepler’s Third Law (the law of periods) cont’
However, we also know from last lecture that the speed is
related to the period as 𝑣 = 2𝜋𝑟/𝑇. Therefore,
2𝜋𝑟
=
𝑇
𝐺𝑀𝑠
𝑟
Squaring both sides and solving for T gives
𝑇2 =
4𝜋 2 3
𝑟
𝐺𝑀𝑠
Which is a mathematical statement of the law. It’s important to
note that the proportionality constant (the term in the brackets)
depends only on the sun’s mass. In this way, we can easily
calculate the masses of various planets by observing the orbital
radii and periods of their moons.
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Slide 23
7.6 Kepler’s Laws and Earth Satellites
Kepler’s Third Law (the law of periods) cont’
This law can be used to determine the required orbital altitude of
a geosynchronous satellite. This is a satellite whose orbital
period is one day, such that it’s always overhead of the same
point on the Earth’s surface†. These orbits are useful for
communications satellites, since the relay stations on the Earth’s
surface don’t have to continuously re-aim. A good example can
be found on p. 242 of the text.
†
I’ve skipped a whole lot of physics here. A geosynchronous satellite can only remain overhead of the
same point on the Earth’s surface if that point lies on the equator – this is called a geostationary orbit. A
geosynchronous orbit that isn’t geostationary does move relative to this point during each day, but not by
a large degree.
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Slide 24
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites
We conclude this chapter with a few words about man-made
satellites and the physics that governs their flight and orbits.
First, consider this “thought experiment”. From chapter 2, we
know that when an object is projected upward, it slowly loses
speed (since acceleration is downward). Eventually, it will come
to an instantaneous stop, after which it will fall back downward
(accelerating as it goes). In this case, we assumed that the
acceleration was constant.
What if acceleration became weaker as the object climbed to
higher altitudes? Is it possible for the object to have a
sufficiently high initial speed for it to “outrun” the weakening
acceleration, such that it never comes to a stop? The answer is
yes!
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Slide 25
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites cont’
We can show this using conservation of energy. Suppose that a
spacecraft is initially on the surface of the Earth, and that it is
given a speed which is sufficient for it to escape the Earth’s
potential well – that is, it only slows to a stop when it is infinitely
far away. Conservation of energy tells us that
𝐾𝑖 + 𝑈𝑖 = 𝐾𝑓 + 𝑈𝑓
1
2 (the terminology will become clear in a
In this case, 𝐾𝑖 = 𝑚𝑣𝑒𝑠𝑐
2
𝐺𝑚𝑀𝐸
moment), and 𝑈𝑖 = −
. Furthermore, 𝐾𝑓 = 𝑈𝑓 = 0 (why?)
𝑅𝐸
Rearranging, we find that
𝑣𝑒𝑠𝑐 =
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2𝐺𝑀𝐸
=
𝑅𝐸
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2𝑔𝑅𝐸
Slide 26
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites cont’
𝑣𝑒𝑠𝑐 is called the escape speed (roughly 11.2 km/s) - the initial
speed needed to escape from the surface of the Earth (without
falling back toward it). It depends only on the Earth’s
parameters, not those of the spacecraft. Of course, we can
calculate escape speeds from any other planet / moon / asteroid
/ etc., as long as we know its mass and radius.
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Slide 27
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites - Tangential Speed and “Zero-Gravity”
Earlier in this lecture, we saw that 𝑎𝑔 ≈ 8.70 m/s2 at the altitude
of the space shuttle (≈ 400 km). This may seem strange, because
video of astronauts aboard the shuttle seems to show that they
are weightless. What’s going on here?
Remember that any orbiting body must have a tangential speed
– if we simply placed the space shuttle at an altitude of 400 km
and released it, it would come plummeting back to Earth. The
tangential speed depends on the desired orbital altitude.
Recycling our Kepler’s 3rd law math,
𝑚𝑣 2
𝑟
=
𝐺𝑚𝑀𝐸
𝑟2
→𝑣=
𝐺𝑀𝐸
𝑟
(as usual, r is the distance from the center of the Earth).
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Slide 28
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites - Tangential Speed and “Zero-Gravity” cont’
This tangential speed sees to it that the satellite continuously
“falls” around the Earth. If an insufficient tangential speed is
reached, the satellite will spiral inward toward the Earth,
eventually burning up in the
atmosphere (due to the rather
extreme drag force…kinetic energy
is converted to thermal energy!). If
the tangential speed is too great,
the satellite simply achieves an
elliptical orbit. If 𝑣 > 𝑣𝑒𝑠𝑐 , the
satellite leaves its orbit and moves
off into space.
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Slide 29
7.6 Kepler’s Laws and Earth Satellites
Earth’s Satellites - Tangential Speed and “Zero-Gravity” cont’
The reason that astronauts aboard the space shuttle appear to
be weightless is simply that the shuttle and its occupants are
accelerating toward the Earth at the same rate. It’s the same
effect that would occur if a person was inside an elevator when
the cable snapped. The effect can be safely replicated near the
Earth’s surface in an airplane that flies in
a parabolic arc (or unsafely replicated in
an airplane that has stalled!)
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Slide 30
Problem #6: Kepler Orbits
WBL LP 7.19
As a planet moves in its elliptical orbit…
A
…its speed is constant
B
…its distance from the sun is constant
C
…it moves faster when it is closer to the sun
D
…it moves slower when it is closer to the sun
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Slide 31
Problem #7: Satellite About Venus
WBL EX 7.71
Venus has a rotational period of 243 days and a mass of 4.87 x 1024 kg. What
would be the altitude of a “venosynchronous” satellite?
Solution: In class
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Slide 32
Problem #8: Asteroid Belt
WBL EX 7.69
The asteroid belt that lies between Mars and Jupiter may be the debris of a planet
that broke apart, or that was not able to form as a result of Jupiter’s strong
gravitation. An average asteroid in the belt has an orbital period (about the sun)
of about 5.0 years. Approximately how far from the sun would this “fifth” planet
have been?
Solution: In class
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Slide 33