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Physics Beyond 2000
Chapter 3
Circular Motion
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/ucm.html
Uniform Circular Motion
• The path is the circumference of a circle
with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.
r
v
Uniform Circular Motion
• The path is the circumference of a circle
with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.
r
Uniform Circular Motion
• The path is the circumference of a circle
with constant radius r.
• The speed is a constant.
• The direction of motion changes with time.
v
v
r
v
v
Uniform Circular Motion
• Period T is the time needed to complete one
cycle.
• Frequency f is the number of cycles
completed in one second.
1
f 
T
r
Uniform Circular Motion
• Example 1
Uniform Circular Motion
• Angular displacement 
– It is the angle, in radian, that the
object turns.

r
Angular displacement
• Angular displacement 
– It is the angle, in radian, that the
object turns.
s
Length of arc
s = r.

r
Angular displacement
• After one complete cycle,
angular displacement
θ= 2π
Angular displacement
• Example 2
Angular speed
• Definition of average angular speed, ωav

 av 
t
– Δθis the angular displacement
– Δt is the time taken
Angular speed

 av 
t
If we consider one complete cycle,
Δθ= 2π and Δt = T
then ωav = 2πf
Angular speed
• Example 3
Instantaneous angular speed
 d
  lim

t 0 t
dt
Instantaneous angular speed
Example 4
Angular speed and linear speed
Δs
B
v
A
Δθ
O
r
X
• When the object moves from A to B at
linear speed v,
• Δs = r. Δθ  v = r. ω
Angular speed and linear speed
• Example 5
Centripetal acceleration
• In a uniform circular motion, the velocity v
changes in direction but not in magnitude.
• It requires an acceleration a to change the
direction of the velocity but not the
v
magnitude.
• The acceleration must be always
a
perpendicular to the velocity.
Centripetal acceleration
In time Δt, the object moves from
A to B.
  
vB
B
 v vB  v A
a

t
t
vA
A
Δθ
O
r
|vB| = |vA| = v
X



a.t  vB  (vA )
Centripetal acceleration
In time Δt, the object moves from
A to B.
vB
B



a.t  vB  (vA )
vA
A
Δθ
O
r
|vB| = |vA| = v

vB
X

 vA

a.t
Centripetal acceleration
In time Δt, the object moves from
A to B.
vB
B
Note that the triangle is an
isosceles triangle.
vA
A
Δθ
O
r
|vB| = |vA| = v
X

 vA
Δθ

vB

a.t
Centripetal acceleration
In time Δt, the object moves from
A to B.
v. Δθ= a. Δt
v
B
v


A
 a = v.
t
B
A
Δθ
O
r
|vB| = |vA| = v
X

 vA
Δθ

vB

a.t
Centripetal acceleration

a  v.
 v.
t
 a  r. 2
vB
B
v2
or a 
r
vA
A
Δθ
O
r
|vB| = |vA| = v
X

 vA
Δθ

vB

a.t
Centripetal acceleration
The acceleration is pointing
to the centre of the circle.
vB
B
vA
A
Δθ
O
r
|vB| = |vA| = v
X

 vA
Δθ

vB

a.t
Centripetal acceleration
v
a
O
r
The acceleration is pointing
to the centre of the circle.
The magnitude of the 2
v
acceleration is
r
or r. ω2
X
Centripetal acceleration
In this motion, though the magnitude
of the acceleration does not change,
its direction changes with time.
So the motion is of variable acceleration.
v
v
a
O
r
X
a
O
r
X
Centripetal acceleration
• Example 6
Centripetal force
• Force produces acceleration.


F  m.a
•Centripetal force produces centripetal
acceleration.
2
v
Fc  m.
r
Centripetal force
2
v
Fc  m.
v
r
or
Fc
O
r
X
Fc  mr
2
•The force Fc is
pointing to the centre
of the circle.
•The force Fc is
perpendicular to the
direction of the
velocity.
Centripetal force
2
v
Fc  m.
v
r
or
Fc
O
r
X
Fc  mr
2
To keep the object moving in
a circle of radius r and speed
v, it is necessary to have a
net force, the centripetal
force, acting on the object.
Centripetal force
Example 7
Centripetal Force: Example 7
Fc
The man is in circular motion.
The net force on the man
= Fc.
Centripetal Force: Example 7
There are two forces
on the man.
N = normal contact
force
W = weight
Fc
W
N
W – N = Fc
mg – N = Fc
N = mg - Fc
Centripetal force
• If the provided force =
2
v
Fc  m.
r
then the object is kept in a uniform circular
motion.
r
Centripetal force
• If the provided force >
2
v
Fc  m.
r
then the object is circulating towards the
centre.
r
Centripetal force
• If the provided force <
2
v
Fc  m.
r
then the object is circulating away from the
centre.
r
Whirling a ball with a string
in a horizontal circle
Top View
Whirling a ball with a string
in a horizontal circle
There is force Fc acting on the ball along
the string.
There is force F acting at the centre along
the string.
F
Top View
Fc
v
Whirling a ball with a string
in a horizontal circle
The two forces Fc and F are action and
reaction pair.
v
Fc
Top View
v
F
F
Fc
Whirling a ball with a string
in a horizontal circle
What happens if Fc suddenly disappears?
(e.g. the string breaks.)
v
Fc
Top View
v
F
F
Fc
Whirling a ball with a string
in a horizontal circle
What happens if Fc suddenly disappears?
(e.g. the string breaks.)
v
Whirling a ball with a string
in a horizontal circle
What happens if Fc suddenly disappears?
(e.g. the string breaks.)
It is moving away tangent to the circle.
Example 8
What is the source of the centripetal force?
m=0.4kg
Fc
r = 0.5 m
In circular motion
M=0.5kg
http://www.dipmat.unict.it/vpl/ntnujava/circularMotion/circular3D_e.html
In equilibrium.
Centripetal force
2
v
Fc  m.
r
or
Fc  mr
2
Centripetal force
• Fc  m
• Fc  v2
1
Fc 
r
Uniform motion
in a horizontal circle

R
v
Uniform motion
in a horizontal circle
Uniform motion
in a horizontal circle
Uniform motion
in a horizontal circle
Uniform motion
in a horizontal circle
Uniform motion
in a horizontal circle
The object is
under two
forces, the tension T
and the weight mg.

T
R
mg
Uniform motion
in a horizontal circle
The net force on
the object is
the centripetal
force because
the object is
moving in
a circle.

 
T  m.g  Fc

T
R
Fc
mg
Uniform motion
in a horizontal circle
T.cos = mg
and
mv2
T . sin  

R

 
T  m.g  Fc
T
R

Fc
mg
Uniform motion
in a horizontal circle
2
v
tan  
Rg

T

 
T  m.g  Fc
R

Fc
mg
Uniform motion
in a horizontal circle
2
v
tan  
Rg

T
R

Fc
For a faster speed v, the angle θtends to increase.
mg
Experiment
• To verify the equation for centripetal force
2
v
Fc  m.
r
or
Fc  mr
2
To verify the equation for
centripetal force
• Whirl the bob in a horizontal circle with string.
• The other end of the string is tied to some hanging
weight.
• Maintain the hanging weight in equilibrium.
hollow plastic
tube
bob
hanging
weight
To verify the equation for
centripetal force
L = length of the string in motion
m = mass of the bob
M = mass of the hanging weight
ω= angular velocity of the bob
θ= angle that the string makes with horizontal
θ
M
hanging
weight
L
bob
m
To verify the equation for
centripetal force
T = tension on the string (tensions on both ends are equal if there
is not any friction between the string and the tube.)
Mg = weight of the hanging weight
mg = weight of the bob
θ
T
bob
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
The hanging weight is in equilibrium.
T = Mg ----------- (1)
θ
T
bob
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
The bob is in circular motion with angular
velocity ω.
Fc = m.r. ω2 ----------- (2)
θ
L
T
bob
r
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
The net force on the bob is equal to the centripetal force.
 

Fc  T  mg
(3)
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
Resolve the forces on the bob into vertical and
horizontal components.
Fc = T.cosθ ------------- (4)
and mg = T.sinθ------------ (5)
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
Also
cosθ=
r
L
(6)
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
From equations (1), (2), (4), (5) and (6),
find ωin terms of L, m, M and g.
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
To verify the equation for
centripetal force
Mg
 
mL
2
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
Measure M and m before the
experiment.
Mg
 
mL
2
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
Measure ω during the
experiment.
ω = Number of revolution × 2π÷ time
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
Verify the following equation.
Mg
 
mL
2
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
Problem: How to measure L?
Refer to the textbook for the skill.
Mg
 
mL
2
θ
L
r
T
bob
Fc
mg
T
Mg
hanging
weight
Note that the bob must have a net force, the
centripetal force, on it in order to keep it in a
circular motion.
As a matter of fact, the bob is not in equilibrium.
It is in a motion with variable acceleration.
θ
r
L
T
bob
Fc
mg
Leaning on a vertical cylinder
r
Place an object on the inner
wall of the cylinder.
The cylinder starts to rotate
about its axis.
Leaning on a vertical cylinder
ω
r
As the cylinder rotates, the
object performs a circular
motion. At a certain angular
velocity ω, the static friction
may be sufficient to support
the object on the wall without
touching the ground.
Leaning on a vertical cylinder
There are 3 forces on the
object.
ω
f
r
N
W
N = normal reaction
W = weight of the object
= mg
f = static friction
= μs.N
where μs is the coefficient
of static friction.
Leaning on a vertical cylinder
ω
f
r
N
As the object is in a circular
motion, the net force must
be the centripetal force.
N = mrω2
and
μsN≧mg
W
Note that the static friction (f) cancels the weight (W).
But the left hand side on the second equation is the
limiting static friction which is the maximum friction.
Leaning on a vertical cylinder
Solve the two equations.
We have
ω
f
r
N
g

s r
and
W
 min
g

s r
Rounding a Bend
http://oldsci.eiu.edu/physics/DDavis/1150/05UCMGrav/Curve.html
Rounding a Bend
r
A car turns round
a corner.
It is a circular motion
with a path of radius
of curvature r.
Rounding a Bend
v
r
A car turns round
a corner.
It is a circular motion
with a path of radius
of curvature r.
Rounding a Bend
http://plabpc.csustan.edu/general/tutorials/CircularMotion/CentripetalAcceleration.htm
Rounding a Bend
v
r
Fc
It requires a
centripetal force
for the circular
motion.
Rounding a Bend
v
r
Fc
How comes
the centripetal
force?
It may come
from the friction
or the normal
reaction.
Level Road without Banking
v
r
Fc
Fc comes from the
static friction fs between
tyres and the road.
The speed v of the
car must not exceed
.
 s gr
where μs is the coefficient of static friction.
Level Road without Banking
v
r
Fc
vmax =
 s gr
Note:
vmax is independent
of the mass of the car.
Force on the passenger
• The passenger needs a centripetal force for
turning round the corner with the car.
• The normal contact force from the car is
the centripetal force.
Example 9
• To find the coefficient of static friction.
v
r
Fc
Don’t rely on friction!
• When the road condition changes (e.g.
on a rainy day), μs becomes very small.
Even a slow speed may exceed the safety
limit. vmax =  gr
s
Banked Road
• Design a banked road
which is inclined to the
centre.
Banked Road
• Design a banked road which is inclined to
the centre.
R
r
θ
W
The car is moving
forward (into the
plane) with
velocity v and is
turning left.
The radius of
curvature is r.
Banked Road
• The centripetal force comes from
the normal contact force R.
R
Fc
r
θ
W
Banked Road
• The centripetal force comes from
the normal contact force R.
Note that Fc is the horizontal component of R.
R
Fc
r
θ
W
Banked Road
• In ideal case, friction is not necessary.
2
v
The ideal banking angle is tan  
rg
R
r
Fc
θ
W
Example 10
• Find the ideal banking angle of a road.
2
v
tan  
rg
• The ideal speed is
rg. tan 
Banked Road
• In non-ideal case, friction f is needed.
• Speed is too slow, less than the ideal speed.
R
f
Fc
r
θ
W
Banked Road
• In non-ideal case, friction f is needed.
• Speed is too fast, more than the ideal speed.
R
f
Fc
r
θ
W
Railway
• When there is a bend, the railway is banked.
• This avoids having lateral force on the rail.
Aircraft
• When an airplane moves in a horizontal
circular path in air, it must tilt about its own
axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc. Back view of
FL
r
W
θ
the car.
The aircarft is moving
forward (into the
plane) with
velocity v.
Aircraft
• When an airplane moves in a horizontal
circular path in air, it must tilt about its own
axis an angle θ.
Aircraft
• When an airplane moves in a horizontal
circular path in air, it must tilt about its own
axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc.
FL
r
Fc
θ
W
Note that
Fc is horizontal.
Aircraft
• When an airplane moves in a horizontal
circular path in air, it must tilt about its own
axis an angle θ.
• The horizontal component of the lift force
FL is the centripetal force Fc.
FL
r
Fc
θ
W
2
v
tan  
gr
Example 11
• Find the speed of the aircraft.
2
v
tan  
gr
Bicycle on a Level Road
• When turning round a corner, it needs
centripetal force.
• Like a car bending round a corner, the
centripetal force comes from the static
friction between the tyres and the road.
Bicycle on a Level Road
Unlike a car, the bike inclines towards the
centre to avoid toppling.
vertical

r
horizontal
The bike is moving
into the plane at
speed v and is turning
left.
The radius of curvature
is r.
Bicycle on a Level Road
What is the angle of tilt  ?
vertical

r
horizontal
The bike is moving
into the plane at
speed v and is turning
left.
The radius of curvature
is r.
Bicycle on a Level Road
Forces on the bike: weight W, normal contact
force R and static friction fs.
vertical
G  R
h
fs
horizontal
W
Note that W
acts at the centre
of mass G of the
bike.
h is the height of
G from the ground.
Bicycle on a Level Road
R = mg ------- (1)
vertical
G  R
h
fs
horizontal
W
mv2
fs 
----------- (2)
r
R balances W.
fs is the centripetal
force.
Bicycle on a Level Road
mv2
fs 
----------- (2)
r
R = mg ------- (1)
vertical
G  R
h
fs
horizontal
W
In order not to
topple, the moment
about G must be zero.
About G,
clockwise moment
= anticlockwise moment
Bicycle on a Level Road
R = mg ------- (1)
h
fs
mv2
fs 
----------- (2)
r
About G,
vertical
clockwise moment
G  R = anticlockwise moment
fs.h = R.a
2
v
 tan  
rg
horizontal
W
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
r
The car is moving
into the plane at
speed v and is turning
left.
The radius of curvature
is r.
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
R1
R2
G
r
f1
W
f2
W acts at
the centre of mass
G of the car.
Tilt of a Car in Circular Motion
Forces on the car: frictions f1 and f2, normal
contact forces R1 and R2, weight W.
R1
R2
G
r
f1
W
f2
We are going
to compare
R1 and R2.
Tilt of a Car in Circular Motion
R1
R2
G
r
f1
f2
W
L L
Let 2L be the
separation
between
the left and right
tyres.
Tilt of a Car in Circular Motion
R1
r
R2
G
h
f1
f2
W
L L
Let h be the
height of the
centre of mass
G from the
ground.
Tilt of a Car in Circular Motion
R1
r
R2
G
h
f1
f2
W
L L
Without toppling,
the moment about
G must be zero.
About G,
clockwise moments
=
anticlockwise moments
Tilt of a Car in Circular Motion
h
R2  R1  .( f1  f 2 )
L
R1
R2
So R2 > R1
r
G
h
f1
f2
W
L L
Tilt of a Car in Circular Motion
As R2 > R1, the springs on the
right are compressed more.
R1
R2
The car tilts right
when it turns
left.
G
r
f1
W
f2
http://www.sciencejoywagon.com/physicszone/lesson/03circ/centrif/centrif.htm
Uniform Motion
in a Vertical Circle
• The path is the circumference of a vertical
circle with constant radius r.
• The speed is v, a constant.
• The mass of the object is m.
Uniform Motion
in a Vertical Circle
• The path is the circumference of a vertical
circle with constant radius r.
• The speed is v, a constant.
• The mass of the object is m.
Uniform Motion
in a Vertical Circle
mv
The centripetal force is Fc 
r
2
How comes the centripetal force?
r
O F
c
v
Uniform Motion
in a Vertical Circle
The centripetal force comes
from the tension T and the weight
of the mass W or mg.
r
O F
c
v
At the highest position
v
Fc
O
r
mg
T1
T1 + mg = Fc
and
2
mv
Fc 
r

mv
T1  mg 
r
2
At the highest position
v
Fc
O
r
mv
T1  mg 
r

2
mg
T1
2
mv
T1 
 mg
r
At the highest position
v
2
Fc
O
r
mv
T1 
 mg
r
What would happen
mg
T1
m
if v2 =
?
r
At the lowest position
r
O
Fc
v
T2
mg
T2 - mg = Fc
and
2
mv
Fc 
r

2
mv
T2 
 mg
r
Note that T2 is always positive.
At any other positions
O
r
There are three
forces on the mass.
T3 F v
T3 is the tension from
θ
the rod,
F is force from the rod
mg
and
mg is the weight of
the mass
At any other positions
O
r
θ
Fc
v
The net force
is the centripetal
force Fc.
mv
Fc 
r
2
At any other positions
O
r
θ
Along the radial direction,
T3
v
θ
mg.cosθ
Fc = T3 – mg.cosθ
At any other positions
So
O
r
θ
2
T3
v
θ
mg.cosθ
mv
T3  mg. cos  
r
2
mv
T3 
 mg. cos 
r
Uniform Motion
in a Vertical Circle
2
mv
 mg
• At the highest position, T1 
r
2
mv
• At the lowest position, T2 
 mg
r
2
mv
• At any other
T3 
 mg. cos 
positions,
r
Non-uniform Motion
in a Vertical Plane
• The motion of an object coasting
along a vertical “ looping-the-loop”.
• Its speed would change at different
positions.
• The principle is also applied to whirling
mass with a string in a vertical plane.
Looping the loop
• Mass of the marble is m.
• Radius of the loop is r.
• The marble starts at the lowest position
with speed vo.
r
vo
O
Looping the loop
• Note that there is change in kinetic energy
and gravitational potential energy.
• Assume that energy is conserved.
r
vo
O
vo
http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/energy/ce.html
Looping the loop
• The speed v of the marble changes on the
loop.
• The centripetal force changes on the loop.
r
v
O
v
v
At the lowest position
• The speed v1 of the marble is vo.
• The centripetal force Fc comes from the
normal contact force N1 and the weight of
the marble mg.
r
O
N1
mg
vo
At the lowest position
2
o
mv
N1  mg 
r
2
o
mv
 mg
 N1 
r
r
O
N1
mg
vo
Below the centre
2
2
2
2
mv
mv
N 2  mg. cos 
 mg. cos
 N2 
r
r
r
O
N2
v2
θ
mg
Below the centre
From conservation of energy,
v  v  2 gr (1  cos )
2
2
2
o
r
vo
O
N2
v2
θ
mg
h = r(1-cosθ)
Above the centre
2
3
2
3
mv
mv
N 3  mg. cos  
 mg. cos 
 N3 
r
r
v3
r
N3
ψ
O
mg
Above the centre
From conservation of energy,
v  v  2 gr (1  cos )
2
3
2
o
v3
r
vo
ψ
O
N3
mg
h=r(1+cosψ)
At the highest position
2
4
mv
N 4  mg 
r
2
4
mv
 mg
 N4 
r
v4
N4 mg
O
r
At the highest position
From conservation of energy,
v  v  4 gr
2
4
2
o
v4
N4 mg
vo
O
r
h = 2r
Completing the Circle
• For the marble to reach the highest position,
2
4
mv
N4 
 mg  0
r
and

v  v  4 gr  gr 
2
4
2
o
v4  gr
vo  5gr
Completing the Circle
vo  2 gr
• The marble cannot move up the loop and oscillates
like a pendulum.
vo
r
vo2
h
r
2g
Completing the Circle
vo  2 gr
• The marble cannot move up the loop and oscillates
like a pendulum.
vo
r
vo2
h
r
2g
Completing the Circle
2 gr  vo  5gr
• The marble rises up to height more than r and is
projected away.
2r  h  r
vo
r
Completing the Circle
vo  5gr
r
vo
O
Whirling freely with a rod
• The ball moves and passes its loop with its
own initial energy.
vo
Whirling freely with a rod
• A light rod would not be loosen.
• The light rod can provide tension or
compression depending on the case.
vo
Whirling freely with a rod
The minimum vo is for the marble to just
reach the top.
From conservation of energy, minimum vo= 2
v=0
2r
h
min vo
gr
Whirling freely with a rod
mv
F  mg. cos 
r
2
v  2 gr (1  cos )
2
v
θ
2r
θ
F
mg
min
vo
h=r(1+cosθ)
Changing from tension to
compression
When F = 0, the force changes from
tension (F>0) to compression (F<0).
v
θo
2r
θo
min
mg
vo
h=r(1+cosθo)
Changing from tension to
compression
Prove that θo = 48.2o when F = 0.
v
θo
2r
θo
min
mg
vo
h=r(1+cosθo)
Example 12
• Whirling a bucket of water in a vertical
circle.
Example 12
• Water does not flow out when the bucket is
at the top position.
v
mg
r
Centrifuge
• It is a device to
separate solid or
liquid particles of
different densities
by rotating
them in a tube in a
horizontal circle.
Centrifuge
axis of
rotation
ω
• ω is the angular velocity.
• r is the distance of the small
portion of liquid from the axis
of rotation.
r
Centrifuge
axis of
rotation
ω
The small portion is in uniform
circular motion.
The centripetal force comes
from the pressure difference ΔP.
r
ΔP
Centrifuge
axis of
rotation
ω
The small portion is replaced by
another portion of smaller density.
The centripetal force is not enough
to support its uniform circular
motion.
ΔP
r
Centrifuge
axis of
rotation
As a result, this portion of smaller
density moves towards the central
axis.
ω
r
ΔP
Centrifuge
axis of
rotation
Portion of larger density moves
away from the central axis.
ω
r
ΔP
Centrifuge
• Study p.51 and 52 for the mathematical
deduction.