Chapter 7: Uniform Circular Motion
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Uniform circular motion
• motion of an
object in a
circle with a
constant or
uniform speed
• constant
change in
direction
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Uniform Circular Motion: Period
Object repeatedly
finds itself back where
it started.
The time it takes to
travel one “cycle” is
the “period”.
distance = rate  tim e
tim e =
T=
distance
rate

2r
v
2r
v
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Quantifying Acceleration: Magnitude
v
v
v t

r
v t
2
v 
a 
r
v
t

v
2
r
Centripetal Acceleration
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Applying Newton’s 2nd Law:
F  ma
F 
mv
2
Centripetal Force
r
Always points toward center of circle.
(Always changing direction!)
Centripetal force is the magnitude of the force
required to maintain uniform circular motion.
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Direction of Centripetal Force,
Acceleration and Velocity
Without a centripetal
force, an object in
motion continues along
a straight-line path.
Without a centripetal
force, an object in
motion continues along
a straight-line path.
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Direction of Centripetal Force,
Acceleration and Velocity
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What if velocity decreases?
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What if mass decreases?
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What if radius decreases?
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What provides the centripetal force?
• Tension
• Gravity
• Friction
• Normal Force
Centripetal force is NOT a new “force”. It is simply a
way of quantifying the magnitude of the force
required to maintain a certain speed around a circular
path of a certain radius.
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Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles of identical diameter but one
object goes around the circle twice as fast as the
other. The force required to keep the faster object
on the circular path is
The answer is E. As the
A. the same as
velocity increases the
B. one fourth of
centripetal force required to
maintain the circle increases
C. half of
as the square of the speed.
D. twice
E. four times
the force required to keep the slower object on the path.12
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles with the same speed. The
diameter of one circle is half of the diameter of
the other. The force required to keep the object
on the smaller circular path is
A. the same as
The answer is D. The centripetal force needed
B. one fourth of to maintain the circular motion of an object is
inversely proportional to the radius of the circle.
C. half of
Everybody knows that it is harder to navigate a
D. twice
sharp turn than a wide turn.
E. four times
the force required to keep the object on the larger
path.
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Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in horizontal circles of
identical diameter and speed but one object has twice the
mass of the other. The force required to keep the more
massive object on the circular path is
A. the same as
B. one fourth of
Answer: D.The mass is directly
C. half of
proportional to centripetal force.
D. twice
E. four times
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Tension Can Yield a Centripetal Acceleration:
If the person doubles the
speed of the airplane,
what happens to the
tension in the cable?
F = ma 
mv
2
r
Doubling the speed, quadruples the force (i.e.
tension) required to keep the plane in uniform circular
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motion.
Friction Can Yield a Centripetal Acceleration:
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Car Traveling Around a Circular Track
Friction provides the centripetal acceleration
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Friction Can Yield a Centripetal Acceleration
FN
fs
Force
X
Y
W
0
-mg
FN
0
FN
fs
-sFN
0
Sum
ma
0
W
What is the maximum
speed that a car can use
around a curve of radius
“r”?
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Maximum Velocity
F
 0   m g  FN
y
FN  m g
F
mv

 ma   
x
 
v  s
v 
mg
X
Y
W
0
-mg
FN
0
FN
FC
-sFN
0
Sum
ma
0
2
r
2
m ax
s
Force
m ax
s
m ax
m ax
s
mg
gr
gr
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Centripetal Force: Question
A car travels at a constant
speed around two curves.
Where is the car most likely to
skid? Why?
F = ma 
mv
2
r
Smaller radius: larger force
required to keep it in uniform
circular motion.
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Gravity Can Yield a Centripetal Acceleration:
Hubble Space Telescope
orbits at an altitude of 598 km
(height above Earth’s surface).
What is its orbital speed?
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Gravity and Centripetal Acceleration:
Centripetal acceleration provided by gravitational force
G m ME
R
2

mv
2
R
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Gravity and Centripetal Acceleration:
G m ME
R
2

mv
2
R
Solve for the velocity….
G m ME  R
2
v 
2
mR
G  ME
2
v 
R
v 
G  ME
R
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Hubble Space Telescope:
v 
v 
GM E
R E  598 km
( 6 .67  10
 11
m kg s )  (5.974  10
3
-1 -2
24
kg)
6,976,000 m
v  7 ,600 m / s
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Banked Curves
Q: Why exit ramps in highways are banked?
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Banked Curves
Q: Why exit ramps in highways are banked?
A: To increase the centripetal force for the higher exit speed.
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The Normal Force Can Yield a Centripetal Acceleration:
Engineers have learned to “bank” curves so that
cars can safely travel around the curve without
relying on friction at all to supply the centripetal
acceleration.
How many forces are
acting on the car (assuming
no friction)?
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Banked Curves
Why exit ramps in highways are banked?
FN cosq = mg
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Banked Curves
Why exit ramps in highways are banked?
FN cosq = mg
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The Normal Force as a Centripetal Force:
Two: Gravity and Normal
Force
X
Y
W
0
-mg
FN
Sum
FNsinq FNcosq
ma
0
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The Normal Force as a Centripetal Force:
F
y
  m g  F N co s q  0
FN 
F
x
mg
co s q
mg
co s q
 F N sin q  m a 
 sin q 
tan q 
v
mv
mv
2
r
2
r
2
gr
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The Normal Force and Centripetal Acceleration:
How to bank a curve…
tan q 
v
2
gr
…so that you don’t rely on friction at all!!
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Artifical Gravity
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Vertical Circular Motion
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Vertical Circular Motion
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The End!
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