Transcript Dynamics of Uniform Circular Motion

CHAPTER 5
Uniform Circular Motion
 Uniform circular motion is the motion
of an object traveling at a constant
speed on a circular path.
 If T (period) is the time it takes for an
object to travel once around a circle, of
radius, r, then the velocity of the object
is given by:
2r
vt 
T
 Where vt is the tangential velocity of
the object
Acceleration in Uniform Circular Motion
 An object that moves at constant speed in a
circular path is accelerating.
 This acceleration is known as Centripetal
Acceleration (ac) and it is directed towards
the center of the circular path.
 This object traveling in a circular path also
experiences a force that is also directed
towards the center of the circular path
 This force is known as Centripetal Force
(Fc)
 Therefore change in velocity of this object
is also directed toward the center of the
circular path
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Centripetal Acceleration
 The magnitude of the centripetal
2
t
v
aC 
r
 Where :
 vt= tangential speed
 r= radius of the circle.
 The direction of the force is toward
the center of the circle.
© 2012 OpenStax College
acceleration is given by the formula
Centripetal Force
 The force causing constant circular
motion is given by:
mvt
Fc  maC 
r
2
 This force is not a new force. It just
tells us the amount of force that
must be provided, by a tension,
gravity, etc., in order for an object to
move in a circle.
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Banked Curves:

© 2012 OpenStax College
Banked Curves
2
v
Fc  FN sin   m
r
FN cos  mg
Putting the equations together:
2
v
FN sin   m
2
v
r
FN sin  m r
FN cos  mg
FN cos 

mg
v2
tan  
rg
or
v  rg tan 
***This equation
can be used to
calculate the speed
of an object
traveling on a
banked curve
Example problem #1

 Trebuchet project:
 2-part project
 Both test grades
 Part 1 (research, design, build) due 12/07/16
 Part 2 (launch and review) due 12/09 or 12/12
 3 to 4 people in a team
 Opportunities for extra credit points 
Example problem #2
 The Moon (mass = 7.36 x 1022 kg) orbits the Earth at a range of 3.84 x 105
km, with a period of 28 days. What is the magnitude of the force that
maintains the circular motion of the Moon?
 𝑮: 𝑚 = 7.6𝑥1022𝑘𝑔, 𝑟 = 3.84𝑥108𝑚, 𝑇 = 2.42𝑥106𝑠
 𝑼: 𝐹𝑐 =?
 𝑬:
𝐹𝑐 =
𝑚𝑣𝑡2
𝑟
 S: 𝐹𝑐 =
𝑚(2𝜋𝑟)2
𝑟𝑇2

7.36×1022 (2𝜋×3.84×108 )2
3.84×108 2.42×106 2
𝐹𝑐 =
 S: 𝟏. 𝟗 𝒙 𝟏𝟎𝟐𝟎 𝑵.
Newton’s Law of Universal Gravitation
 Gravitational force is the mutual force of attraction
between particles of matter
 There is a gravitational force between any 2 objects
 The larger the object…the more pull it has
 For example, there is a gravitational force between 2
pencils
 If the objects are larger, they will have a larger
gravitational field
 Our gravitational field is extremely small (due to
our small mass) relative to the earth’s gravitational
field (which has a much larger mass)
 As a result, we don’t feel the gravitational force between
us and “small” objects
Gravitational force is the force that keeps
planets in orbit and keeps them from
coasting off in a straight line
Gravitational force is an attractive force
It depends on the distance between two objects
and
The magnitude of the masses
Increase the distance-decrease the gravity
 Gravitational force is directly proportional to the product of the two
masses involved
m1m2
Fg  G 2
r
 Gravitational force is inversely proportional to the square of the
distance of separation
 𝐹𝑔 =
𝑚1 𝑚2
𝐺 2
𝑟
 Fg = the gravitational force
 m1 and m2 = the masses being observed
 r = the distance between the two masses ( in meters)
 G = the universal gravity constant
 The
constant is: G = 6.67 x
2
𝑁∙𝑚
10-11 2
𝑘𝑔
Inverse square of distance – complete the
chart
Original New
distance distance
10
20
20
10
50
5
8
64
25
5
𝑵𝒆𝒘
reduce
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍
𝒄𝒉𝒂𝒏𝒈𝒆
𝒊𝒏 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
20
2
10
1
10
1
20
2
inverse
1
2
2
1
Square
Change
in Fg
¼x
4x
Example problem #1
 Calculate the gravitational force between Mrs. Akibola and
her computer monitor when they are 0.45m apart. Mrs.
Akibola’s weight is 736 N and the weight of the computer
is 98.1 N
 G: r = 0.45 m,
m1 =
 U: Fg = ?
 E: 𝐹𝑔 =
 𝑆:
𝐹𝑔 =
𝑚1 ×𝑚2
𝐺
𝑟−2
6.67 x 10
11
× 75 x 10
0.452
 S: 2.4 x 10-7 N
736
=
9.81
75 kg ,
m2 =
98.1
9.81
= 10 kg
Satellites in Circular Orbits
There is only one speed that a satellite can have if the
satellite is to remain in an orbit with a fixed radius.
Finding the speed of an Earth Satellite
write these equations at the bottom of your notes!!
 Centripetal acceleration:
 Centripetal force:
 The force is the force of Gravity
 The mass of the satellite divides
out, so it doesn’t matter.
v2
ac 
r
mSatellite v 2
F  mSatellite a 
r
GM EarthmSatellite mSatellitev 2

2
r
r
GM Earth v 2

2
r
r
Finding the speed of an Earth
Satellite (Continued)
 Solving for v:
GM Earth
GM Earth
v 
or v 
r
r
2
Newton’s Cannon
 Newton compared the
motion of a falling object
to the motion of the
Moon.
 An object in orbit is
actually falling toward
the Earth, at just the
same rate that the Earth
curves away from it.
Source: Brian Brondel, Newton Cannon.svg,
Wikimedia Commons,
http://en.wikipedia.org/wiki/File:Newton_Cannon
.svg
Gravitational Field Strength of a Point Mass
(or Spherically Symmetric Mass)
write these equations at the bottom of your notes!!
Mm
FG  G 2
r
Fg
Mm
g
G 2
m
r m
M
g G 2
r
 The gravity field around a point
mass, or a spherically symmetric
mass, depends only on the mass
and the distance away from the
center.
Satellites in Circular Orbits
Example #2: Orbital Speed of the Hubble Space Telescope
Determine the speed of the Hubble Space Telescope orbiting
at a height of 598 km above the earth’s surface.
GM E
v
r
v
6.67 10 5.98 10 
11
6.38 10  598 10
6
24
3
 7.56 10 m s
3
16900 mi h 
Vertical Circular Motion
2
1
v
FN 1  mg  m
r
2
v2
FN 2  m
r
FN 4
2
4
v
m
r
2
3
v
FN 3  mg  m
r
Kepler’s contributions to astronomy
 Used Tycho’s data to formulate three laws of
planetary motion.
 The planets move in elliptical orbits with the Sun
at one center of the ellipse.
 The planets trace out equal areas in equal times.
 The square of a planet’s period is proportional to
the cube of its distance from the sun. T2 ~ r3
Kepler’s First and Second Law
 Here is an animated visualization of Kepler’s first two laws.
http://www.surendranath.org/Applets/Dynamics/Kepler/Kepler1Applet.
html