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Chapter 5
Dynamics of Uniform Circular
Motion
Circular Motion
The effect of an object’s acceleration on its velocity depends on
their relative directions.
If the acceleration is in the same direction as the velocity i.e.,
parallel to the velocity, the magnitude of the velocity increases i.e.,
the object moves faster.
If the acceleration is in the opposite direction as the velocity i.e.,
antiparallel to the velocity, the magnitude of the velocity decreases
i.e., the object moves slower.
If the acceleration is non-parallel to the velocity, The direction
(and possibly the magnitude) of the velocity changes.
In the special case where the acceleration is perpendicular ( ^ ) to
the velocity the object will follow a uniform circular path. The
direction of the velocity changes while the magnitude of the
velocity remains constant.
The acceleration is called the centripetal acceleration which means
“center-seeking” and always points toward the center of the
circular path.
The magnitude of the centripetal acceleration, ac, depends on the
magnitude of the object’s velocity and the radius of the circular
path.
2
v
ac 
r
If an object travels in a uniform circle there must be a net force
acting on it to produce the centripetal acceleration. This net force
is called the centripetal force.
From Newton’s Second Law:
Fnet  ma
Fc  mac
2
v
Fc  m r
NOTE :The centripetal force is not a new type of force but rather the
net force that results in uniform circular motion.
The source of the centripetal force varies from case to case.
Examples of Sources of Centripetal Force:
1. Car travelling around a curve: Friction between tires and road
2. Planet travelling around the sun: Gravitational force
NOTE: Planetary orbits are not truly circular but elliptical.
3. Electron travelling around the nucleus of an atom: Electrostatic
force
4. Ball on a string being swung in a circle: Tensional force
5. Plane circling an airport: Component of lift on the wing
Consider a car following a circular curve in the broad.
F
c

st
at
r
ic
fr
ic
t
v
io
n
Static friction between the tires and the road must supply the
necessary centripetal force.
2
v
Fc   s mg  m r
2
v
s g  r
The maximum speed at which the car can safely travel around the
curve is given by:
vmax   s gr
The dependence on friction can be reduced by “banking” the curve.
FN
FN cos
FN
FN cos


FN sin 
fs cos
fs

mg
FNsin can supply some
of the centripetal force.


mg
fscos can also supply some of the
centripetal force.
Fc  FN sin   fs cos 

F
cos

mg
N


mg

FN  cos 
FN cos

mg

mg
FN sin 
sin
cos


FN sin  mgtan
fs   s FN

f   mg
s cos 
 s

f cos    mg cos 
s cos 
 s

f cos    mg
s
s
Fc  mgtan    s mg
mv2  mg tan   
s
r
v 2  g tan    
s
r
v max  tan    s gr
w/o “banking
vmax   s gr
v max,banking
vmax,w / o banking 
v max,banking
v max,w / o banking

 tan  s gr
 sgr
 tan  s 
tan

 
 1
s
 s

The track at the Daytona National Speedway is banked at 31°. The
typical coefficient of friction between rubber and asphalt is 0.5. How
much faster can cars travel around the track due to its banking?
v max,banking
vmax,w / o banking

tan 31  1  1.48
0.5
About 50% faster
Satellites
For a satellite in a circular orbit the gravitational force provides
the required centripetal force.
FG  Fc
msmo msv2
G 2 
rs,o
rs,o
Where :
m s  mass of satellite
m o = mass of object it is orbiting
rs,o  distance from satellite to the center
of the orbited body
The velocity of a satellite in a stable circular is given by:
v
Gm o
rs,o
What is the velocity of a satellite orbiting above the earth,
me=5.98x1024kg and re=6.38x106m, at an altitude of
(a) 5mi. (8.05x103m),
(b) 1000mi. (1.61x106m)?
(a) rs,o  re  h  6.38  10 6 m  8.05  10 3 m
rs,o  6.39  10 m
6
11
2
Nm  5.98  10 24 kg
2
v
kg
6
6.39  10 m
v  7.9  103 m  17, 700 mi
s
hr
Gm o
6.67

10

rs,o
(b) rs,o  re  h  6.38  10 6 m  1.61  10 6 m
rs,o  7.99  10 m
6
v
Gm o
rs,o
2
24
Nm
2  5.98  10 kg
 6.67  10
kg
7.99  10 6 m
11
v  7.06  10 m  15, 800 mi
s
hr
3
Frequently the time required for a satellite to complete one orbit is
of interest. This time is called the period, T, of the satellite’s orbit.
Since the satellite is travelling in a circle one orbit equals a distance
of: 2pr.
Since in a circular orbit the velocity of the satellite is constant:
2pr
T v
Substituting v 
gives :
2 p r3 2
T
Gm o
Gm o
rs,o
What is the orbital period of the moon whose average distance
from the earth is 3.84x108m.
2pr
T
Gmo
32

2 p(3.84  10 8 m)3 2
2
11 Nm
24
6.67  10

5.98

10
kg
2
kg
T  2.367  106 s  27.3 days
This period of time is called the Sidereal Month. It is not the period
of time on which our calendar months are based.
Upon what period of time are our calendar months based?
A geosynchronous satellite remains directly above a given point on
the earth’s surface. Therefore, its period must be the same as the
earth’s period of rotation (24 hours = 8.64x104s)
At what altitude must a geosynchronous satellite be placed?
Historical Models of the Solar System
Aristotelean-Geocentric Model:
The earth is stationary and at the center of the Solar System. All
celestial bodies revolve around the earth in perfectly circular orbits.
Copernican-Heliocentric Model:
The sun is at the center of the Solar System. The earth is a planet
that revolves around the sun in a circular orbit. The earth spins on
its axis.
Planetary
Orbits
Terrestrial Planets
Orbits
Earth-Mars
Orbits
Kepler’s Laws of Planetary Motion
In the late 15th and early17th centuries, Johannes Kepler a believer
in the Copernican sun-centered universe derived three laws of
planetary motion.
Kepler used the extremely precise measurements of the position of
Mars made by the Danish astronomer Tycho Brahe.
Kepler’s First Law
I. The orbits of the planets are ellipses, with the Sun at one focus of
the ellipse.
Kepler’s Second Law
II. The line joining the planet to the Sun sweeps out equal areas in
equal times as the planet travels around the ellipse.
A planet moves faster when it is nearer the Sun. Thus, a planet
executes elliptical motion with constantly changing angular speed
as it moves about its orbit. The point of nearest approach of the
planet to the Sun is termed perihelion; the point of greatest
separation is termed aphelion. Hence, by Kepler's second law, the
planet moves fastest when it is near perihelion and slowest when it
is near aphelion.
Kepler’s Third Law
II. The ratio of the squares of the revolutionary periods for two
planets is equal to the ratio of the cubes of their semimajor axes:
T12
T22
3 
R1 R 32
2pr
T
Gm o
32
T2 
r3
2p
 constant
Gm o
Kepler's Third Law implies that the period for a planet to orbit the
Sun increases rapidly with the radius of its orbit. Thus, we find that
Mercury, the innermost planet, takes only 88 days to orbit the Sun
but the outermost planet (Pluto) requires 248 years to do the same.