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Uniform Circular Motion
Chapter 5
Expectations
After Chapter 5, students will:
 understand that an object in circular motion
continually accelerates, even though its speed may
not change.
 understand very firmly that an object moving along a
curved path is not in equilibrium.
 never again use an obscenity like “centrifugal force.”
 perform calculations involving the relationships
among centripetal force, centripetal acceleration,
speed, and path radius.
Expectations
After Chapter 5, students will:
 use angular velocity, period, and radian angle
measures to solve problems.
 analyze situations in which centripetal forces are
frictional or gravitational in nature.
 solve problems involving circular motion in a vertical
plane.
Uniform Circular Motion
If an object travels on a circular path, and its speed is
constant, it is performing uniform circular motion.
v
In one complete journey
around its circular path,
its angular displacement
is 2p radians.
2pr
T
v

r
Radian Angle Measurement
The radian measure of an angle is the length of the arc it
subtends, divided by the radius:
S  r
or
S

r
Thus, the radian (rad) is the
unitless ratio of two lengths.
A circle contains 2p of them.
r
S

Angular Velocity
Average angular velocity is the angular displacement
divided by the time interval in which it occurred.
x  vt
x  r (small  )
vt  r

v
 
t
r

r
x
v
Angular Velocity
The units of angular velocity are:


t
 rad
s
The radian, however, is not a “real” unit, in the sense
that it is dimensionless. So angular velocity has the
dimensions of reciprocal time.
Angular velocity can also be expressed in terms of other
units: degrees/s, revolutions/min, etc.
Centripetal Acceleration
The velocity of an object in uniform circular motion is
always changing.
The magnitude of the velocity (speed) is constant ... but
the direction of the velocity changes continually.
If the direction of the velocity were constant, the object
would move in a straight line, not in a circular path.
Centripetal Acceleration
Centripetal: “center-seeking”
small - angle : v  vt 
v vt
aC 

 v
t
t
v
but  
so
r
2
v v
aC  v  
r r
SI units: m/s2
v1
t
v0
v
t
v0
v1
Centripetal Force
To produce a centripetal acceleration requires a
centripetal force, according to Newton’s second law:
 v 2  mv2
FC  maC  m  
r
 r 
Please keep in mind that the object moving in a circular
path is not in equilibrium. A net force acts on it: the
centripetal force. There is no offsetting “centrifugal”
force. There is no “centrifugal” force at all. None.
Never has been; never will be.
Centripetal Force
Where do centripetal forces come from?
 Gravitational (a moon orbiting a planet)
 Frictional (a race car going around a flat curve)
 Tension (a stone whirled on a string)
 Normal (clothing in a washing machine during its
spin cycle)
 Electrical (electrons orbiting an atomic nucleus)
 Combinations (a race car going around a banked turn
– frictional and normal)
Gravitational Centripetal Force
2
For objects orbiting the Earth:
mv
mM E
FC 
G 2
r
r
This equation can be solved for various things:
v
GM E
r
r
2pr
noting that v 
,
T
4p 2 r 3
T
GM E
and
GM E
v2
orbital period
2
T
GM E
3
r
4p 2
Frictional Centripetal Force
A car travels around a turn of radius r.
The centripetal force required for the turn
is provided by the static frictional
force:
mv2
FC 
 S N
r
N  mg (vertical equilibriu m), so
mv2
  S mg
r
N
FC
INDIANA
314159
mg
Frictionless Banked Turn
If the turn is banked by an angle  and there is no
friction:
N
N cos  mg (vertical equilibriu m)
mg
N
cos
2
mv
mg sin 
FC 
 N sin  
r
cos
v2
be solved for a
 g tan  (can

desired quantity)
r

N cos 
N sin 
A
IAN
IND 159
314
mg
Banked Turn With Friction
If there is friction, the situation becomes more complicated.
The car does not accelerate in the
N
N cos 

Y direction, so:
 Fy  0  N cos  mg  FS sin 
N sin 
but FS   S N , so
N cos   S N sin   mg
mg
N
cos   S sin 
FS cos 

A
IAN
IND 159
4
31
FS sin 
FS

mg
Banked Turn With Friction
In the X direction, the car does accelerate.
mv2
 Fx  r  FS cos  N sin 
again, FS   S N and
N

N cos 
N sin 
mg
N
. Substituti ng :
cos    S sin 
g  S cos   sin   v 2

cos    S sin 
r
which can be solved for any
of several quantities .
FS cos 

A
IAN
IND 159
4
31
FS sin 
FS

mg
Circular Motion in a Vertical Plane
Depending on location, the weight force provides none,
some, or all of the centripetal force.
2
N3
mv3
N 3  mg 
r
mg
N2
r
mg
N1
mv1
N1  mg 
r
2
mg
mv2
N2 
r
2