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Goals for Chapter 5
Chapter 5
Dynamics of Uniform Circular Motion
• To understand the dynamics of circular
motion.
• To study the motion of satellites in orbit as a
special case of circular motion where the
centripetal force is due to gravitational pull of
the earth.
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5.1 Uniform Circular Motion
Uniform circular motion is the motion of an object
traveling at a constant speed on a circular path.
Example: A tire-balancing machine
The wheel of a car has a radius of 0.29m and it being
rotated at 830 revolutions per minute (rpm) on a tirebalancing machine. Determine the speed at which the
outer edge of the wheel is moving.
1
 1.2 10 3 min revolution
830 revolutions min
T  1.2 10 3 min  0.072 s
speed:
Let T be the time it takes for the object
to travel once around the circle.
v
2r
T
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v
2  r 2  0.29 m 

 25 m s
T
0 .072 s
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5.2 Centripetal Acceleration
5.2 Centripetal Acceleration
In uniform circular motion, the speed is constant, but
the direction of the velocity vector is not constant.
v vt

r
v
v v 2

t
r
    90 and     90
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 
The direction of the centripetal acceleration
is towards the center of the circle; in the
same direction as the change in velocity.
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ac 
v2
r
Acceleration in uniform circular motion
Is Orlando a good inertial reference frame ?
• Uniform circular motion results in acceleration:
– magnitude:
a = v2 / R
– direction:
- R (towards center of circle)
The velocity vector
changes direction,
not magnitude
travels the circumference of the circle
in the time T for one revolution
• Is Orlando accelerating ?
• YES!
Orlando is on earth.
–
The earth is rotating.
• What is the centripetal acceleration of Orlando ?
T = 1 day = 8.64 x 104 sec,
R ~ RE = 6.4 x 106 Meter .
v
2R
T
v 2 1  2 R   2 
 
   R
R R T   T 
2
a
2
speed:
2R
v
T
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• Plug this in: a = .034 m/s2 ( ~ 1/300 g)
• Close enough to 0 that we will ignore it.
• Orlando is a pretty good inertial reference frame.
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Example: Effect of radius of curvature
The bobsled track contains turns
with radii of 33 m and 24 m. Find
the centripetal acceleration at each
turn for a speed of 34 m/s. Express
answers as multiples of g  9.8 m s2 .
a c  v2 r
ac 
34 m s 2
33 m
34 m s 

 35 m s 2  3.6g
5.3 Centripetal Force
Recall Newton’s Second Law
When a net external force acts on an object of
mass m, the acceleration that results is directly
proportional to the net force and has a
magnitude that is inversely proportional to the
mass. The direction of the acceleration is the
same as the direction of the net force.

a
2
ac
24 m
 48 m s  4.9g
2
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m


 F  ma
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5.3 Centripetal Force
Thus, in uniform circular motion there must be a net
force to produce the centripetal acceleration.
The centripetal force is the name given to the net force
required to keep an object moving on a circular path.
Example: The effect of speed on centripetal force
The model airplane has a mass of 0.90 kg and moves at
constant speed on a circle that is parallel to the ground.
The path of the airplane and the guideline lie in the
same horizontal plane because the weight of the plane is
balanced by the lift generated by its wings. Find the
tension in the 17 m guideline for a speed of 19 m/s.
The direction of the centripetal force always points
toward the center of the circle and continually changes
direction as the object moves.
Fc  T  m
v2
r
T  0.90 kg 
v2
Fc  ma c  m
r
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
F

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19 m s 2
17 m
 19 N
Example: A high-speed carnival ride
Two cars on a curving road
The passengers in a carnival ride travel in a circle with radius 5.0 m.
They complete one circle in a time T = 4.0 s. What is their acceleration?
What is the centripetal force if the mass of a passenger is 90 kg ?
v
2 R
T
a
v2
R
2 ( 5 .0 m )
 7 .9 m / s
4 .0 s
2
2
v
( 7 .9 )
a 

 12 m / s 2
R
5 .0
F rad  m  a  90  12  1080 N
v 
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Flat curves
• A small car with mass m and a large car with
mass 2m drive around a highway curve of radius
R with the same speed v. As they travel around
the curve, what is their acceleration?
• (a) equal.
• (b) along the direction of motion.
• (c) in the ratio of 2 to 1.
• (d) zero.
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Example: Rounding a flat curve
On an unbanked curve, the static frictional force
provides the centripetal force.
n: normal force
n  (mg )  0
A car rounds a flat, unbanked curve
with radius R. a) If the coefficient of
static friction between tires and road
is s, derive an expression for the
maximum speed vmax at which the
driver can take the curve without
sliding. b) What is vmax for R = 250 m
and s = 0.90 ?
v max
R
2
v2
fs  m
R
 s mg  m
fs  s n
v max   s gR
v max  0.90  9.8 m s 2  250m  47 m s
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5.4 Banked curves
On a frictionless banked curve, the centripetal force is the
horizontal component of the normal force. The vertical
component of the normal force balances the car’s weight.
FN sin   m
v2
r
FN cos   mg
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tan  
v2
rg
Rounding a banked curve - Daytona speedway
5.5 Satellite in circular orbits
The turns at the Daytona
International Speedway have a
maximum radius of 316 m and are
steely banked at 31 degrees.
Suppose these turns were
frictionless. As what speed would
the cars have to travel around them?
v2
Rg
tan  
v
There is only one speed that a satellite can have if the
satellite is to remain in an orbit with a fixed radius.
v  R g tan 
316 m  9.8 m

s 2 tan 31
 43 m s 96 mph 
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Circular orbits - projectile at the ‘right’ velocity
5.5 Satellite in circular orbits

 F  ma
• Increase projectile velocity: when Fg balances  orbit
• When v is large enough, you achieve escape velocity
Gm s m E m s v 2

r2
r
v
Gm E
r
T
Larger period T
corresponds to larger orbit
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T
2r 3 / 2
Gm E
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Example: Orbital speed of the Hubble space telescope
Example: Orbital radius for synchronous satellites
What is the height H above
the earth’s surface at which
all synchronous satellites
(regardless of mass) must
be placed in orbit ?
Determine the speed of the Hubble Space Telescope
orbiting at a height of 598 km above the earth’s surface.
Gm E
r
v
v
6.67 10
11
T  24 hours

N  m 2 kg 2 5.98 10 24 kg
6.38 106 m  598 103 m
 7.56 103 m s
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2r
r
 2r
v
G mE

16,900 mi h 
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T
2r 3 2
G mE
r3 2  T
G mE
2
Example: Global Positioning System(24 satellites)
r3 2  T
 T 
r  
 2 
G mE
2
T  24 hours
 8.64 104 s 

r  
2


2
3
2
3
G m E  13
These satellites are all synchronous satellites, having
a period of exactly 24 hours.
32
T  24 hours
 86400 s


N  m2
 6.67 10 11
5.98 10 24 kg 
2
kg


r  4.23 107 m
1
2r
G mE
T
3
since earth’s radius is
R E  6.38  106 m
H  r  R E  4.23 107 m  6.38 106 m
 3.59  107 m  35900 km
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Acceleration in uniform circular motion
• Uniform circular motion results in acceleration:
– magnitude:
ac = v2 / R
– direction:
- R (towards center of circle)
The velocity vector
changes direction,
not magnitude
Circular Motion Dynamics
Fnet  m
G
2
v
r
m  mE
v2

m
r2
r
travels the circumference of the circle
in the time T for one revolution
Satellite in circular
(earth) orbit
v
speed:
Gm E
r
ac 
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v
r
v
2R
T
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5.6 Apparent weightlessness and artificial gravity
Masses of Astronomical Objects
Fg  G
Where
Mm
R2
During free-fall, the elevator, the person, and the scale all accelerate
downward with the same acceleration due to gravity. The apparent
weight of the person is zero. When the person is in a orbiting space
station, he also experiences apparent weightlessness.
G = 6.67 x 10 -11 m3 kg-1 s-2
• But |Fg| = m a  m
2
M ( 2 )
G 2 
R
R
T2
2 r 3 2
T
GM
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2
v2
R
M: central mass
m: mass of satellite
For Fg all objects accelerate
with same acceleration,
regardless of their mass!
If the orbit radius r and
the period T of a center
mass can be measured,
we can calculate the
mass of the object !
M
( 2 ) 2
T2 G
R3
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Example: Artificial Gravity
At what speed must the surface of the space station
rotate so that the astronaut experiences a push on his feet
equal to his weight on earth? The radius is 1700 m.
5.7 Vertical circular motion
Motor cycle stunt driver going around a vertical circular track
normal force !
FN 3  mg  m
v2
Fc  m
 mg
r
FN 4
v  rg

1700 m  9.80 m
s2 
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as the cycle goes
around the normal
force changes
FN 2  m
FN1  mg  m
v12
r
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Example:
Skiing overMotion
a bump
Example:
in a Circle
• A skier of mass m goes over a mogul having a radius
of curvature R. How fast can she go without leaving
the ground?
Example:
Skiing overMotion
a bump
Example:
v
• For N = 0:
mg N
v
R
R
mg N
(a)
v  mRg
(b) v 
Rg
m
(c) v 
RR
Rg
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Example 16, Centripetal force
Ball A is attached to one end of a rigid massless rod, while an identical ball B is
attached to the center of the rod, as shown. Each ball has a mass of 0.5 kg, and
the length of each half of the rod is L = 0.4 m. Each ball is in uniform circular
motion. Ball A travel at a constant speed of 5.0 m/s. Find the tension in each
half of the rod.
2
TA 
mv A
15.6 N
2L
2
mvB
 7.8
L
TB  23.4 N
TB  TA 
in a Circle
• mv2 / R = mg - N
v
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v
r
v2
m 4
r
centripetal force is
the sum of all the
force components
oriented along the
radial direction
 129m / s
2
3
Rg
v 22
r