Chapter 8: Newton’s Laws Applied to
Circular Motion
Centrifugal Force is Fictitious?
“Center FLEEing”
Factual = Centripetal Force
Ffictitious = Centrifugal Force
Centrifugal Force is Fictitious?
“Center FLEEing”
Suppose there is a lady bug in the can. There is a centripetal force
acting on the bug, transmitted to her feet by the can. Her feet push
back on the can producing a centrifugal “fictitious” force, that acts
like artificial gravity.
Centripetal & Centrifugal Force
Depends on Your Reference Frame
Outside Observer
(non-rotating frame) sees
Centripetal Force pulling
can in a circle.
Center-seeking
Center-fleeing
Inside Observer
(rotating reference frame)
feels Centrifugal Force
pushing them against the
can.
Centrifugal Force is Fictitious?
The centrifugal force is a real
effect. Objects in a rotating frame
feel a centrifugal force acting on
them, trying to push them out. This
is due to your inertia – the fact that
your mass does not want to go in a
circle. The centrifugal force is
called ‘fictitious’ because it isn’t
due to any real force – it is only due
to the fact that you are rotating.
The centripetal force is ‘real’
because it is due to something
acting on you like a string or a car.
The Earth rotates once per day around its axis as shown.
Assuming the Earth is a sphere, is the rotational speed at Santa
Rosa greater or less than the speed at the equator?
366 m/s
464 m/s
What is the total acceleration
acting on a person in Santa Rosa?
The vector sum.
ac
g
Is your apparent weight as
measured on a spring scale more
at the Equator or at Santa Rosa?
ac
g
Since you are standing on the Earth
(and not in the can) the centrifugal force tends
to throw you off the Earth. You weigh less
where the centripetal force is greatest because
that is also where the centrifugal force is
greatest – the force that tends to throw you
out of a rotating reference frame.
Artificial Gravity
How fast would the space station segments A and B have to rotate in
order to produce an artificial gravity of 1 g?
vA  56m / s ~ 115mph
vB  104m / s ~ 210mph
Can the two segments be connected?
“Coriolis Force”
• This is an apparent
force caused by
changing the radial
position of an object in
a rotating coordinate
system
• The result of the
rotation is the curved
path of the ball
“Coriolis Force”
• This is an apparent
force caused by
changing the radial
position of an object in
a rotating coordinate
system
• The result of the
rotation is the curved
path of the ball
Coriolis Effect
Translational and Rotational
Kinematics For CONSTANT
Accelerations ONLY
Total Acceleration & Force
v2
ar  aC  
r
d v
at 
dt
a  a a
2
r
2
t
F  F  F
2
r
2
Horizontal Circle: Constant Speed & Acceleration
Vertical Circle: Changing Speed & Acceleration
Important: Inside vs Outside the
Rotating Frame
Motion in a Horizontal Circle
Looking down:
• The speed at which the
object moves depends on
the mass of the object and
the tension in the cord. It
is constant!
• The centripetal force is
supplied by the tension.
mv
 Fc  T  mac  r
2
Tr
v
m
Motion in a Horizontal Circle
Horizontal (Flat) Curve
• The force of static friction
supplies the centripetal
force
mv
 Fc  f  r
2
• The maximum speed at
which the car can
negotiate the curve is
v   gr
• Note, this does not depend
on the mass of the car
F
y
 N  mg  0
f   mg
Horizontal (Flat) Curve
3. A highway curve has a radius of 0.14 km and is
unbanked. A car weighing 12 kN goes around the
curve at a speed of 24 m/s without slipping. What
is the magnitude of the horizontal force of the
road on the car? What is μ? Draw FBD.
a.
12 kN
b.
17 kN
c.
13 kN
d.
5.0 kN
e.
49 kN
Banked Curve
These are designed with friction
equaling zero - there is a
component of the normal force
that supplies the centripetal force
that keeps the car moving in a
circle.
2
mv
 Fr  n sin   r
 Fy  n cos  mg  0
Dividing:
2
v
tan  
rg
Banked Curve
A race car travels 40 m/s around a banked (45 with
the horizontal) circular (radius = 0.20 km) track.
What is the magnitude of the resultant force on the
80-kg driver of this car?
a.
0.68 kN
b.
0.64 kN
c.
0.72 kN
d.
0.76 kN
e.
0.52 kN
Hints for HW Problem
Determine the range of speeds a car can have
without slipping up or down the road when it
is banked AND has friciton.
If the car is about to slip down the incline, f
is directed up the incline. This would
happen at a minimum speed.
When the car is about to slip up the incline, f
is directed down the incline. This would
happen at a maximum speed.
Vertical Circle with Non-Uniform Speed
Where is the speed Max? Min?
Where is the Tension Max? Min?
Vertical Circle with Non-Uniform Speed
• The tension at the bottom is a maximum
• The tension at the top is a minimum
• Look at radial and tangential:
 F  mg sin  ma
t
t
at  g sin 
mv 2
 Fr  T  mg cos  R
 v2

T  m   g cos 
R

Vertical Circle: Mass on a String
A 0.40-kg mass attached to the end of a string swings
in a vertical circle having a radius of 1.8 m. At an
instant when the string makes an angle of 40 degrees
below the horizontal, the speed of the mass is 5.0 m/s.
What is the magnitude of the tension in the string at
this instant? Draw the FBD.
a.
9.5 N
b.
3.0 N
c.
8.1 N
d.
5.6 N
e.
4.7 N
Vertical Circle: Mass on a String
A 0.30-kg mass attached to the end of
a string swings in a vertical circle
(R = 1.6 m), as shown. At an instant
when  = 50, the tension in the string
is 8.0 N. What is the magnitude of the
total force on the mass at this instant?
a.
5.6 N
b.
6.0 N
c.
6.5 N
d.
5.1 N
e.
2.2 N
Hint: F  Fr2  F2
Minimum Speed for Vertical
Circular Motion
What is the minimum speed so that
the ball can go in the circle?
That is, when T = 0 at the top?
At the top:
  180
 v2

T  m   g cos   0
R

v  gR
Minimal Speed to JUST get around the circle only depends on R!
ROOT GRRRRRRRR
QuickCheck 8.11
Loop d’ Loops: Inside the Vertical Loop
A roller coaster car does a loopthe-loop. Which of the free-body
diagrams shows the forces on
the car at the top of the loop?
Rolling friction can be neglected.
Slide 8-82
QuickCheck 8.11
Loop d’ Loops: Inside the Vertical Loop
A roller coaster car does a loopthe-loop. Which of the free-body
diagrams shows the forces on
the car at the top of the loop?
Rolling friction can be neglected.
The track is above the car, so
the normal force of the track
pushes down.
Slide 8-83
Loop d’ Loops: Inside the Vertical Loop
A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom
of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What
is the magnitude of the force of the track on the car at the bottom of the dip?
a.
3.2 kN
b.
8.1 kN
c.
4.9 kN
d.
1.7 kN
e.
5.3 kN
Loop d’ Loops: Inside the Vertical Loop
Minimum Speed to get to the Top.
What is the minimum speed so that the
car barely make it around the loop the
riders are upside down and feel
weightless ? R = 10.0m
QuickCheck 8.10
Humps in the Road: Outside the Vertical Loop
A car that’s out of gas coasts
over the top of a hill at a steady
20 m/s. Assume air resistance
is negligible. Which free-body
diagram describes the car at
this instant?
Slide 8-80
QuickCheck 8.10
Humps in the Road: Outside the Vertical Loop
A car that’s out of gas coasts
over the top of a hill at a steady
20 m/s. Assume air resistance
is negligible. Which free-body
diagram describes the car at
this instant?
Now the centripetal
acceleration points down.
Slide 8-81
Humps in the Road
Outside the Vertical Loop
A roller-coaster car has a mass of 500 kg when fully
loaded with passengers. The car passes over a hill of
radius 15 m, as shown. At the top of the hill, the car has
a speed of 8.0 m/s. What is the force of the track on the
car at the top of the hill?
a.
7.0 kN up
b.
7.0 kN down
c.
2.8 kN down
d.
2.8 kN up
e.
5.6 kN down
Maximum Speed for Vertical
Circular Motion
Humps in the Road
What is the maximum speed the car can have as it
passes this highest point without losing contact with
the road?
:
Max speed without losing contact MEANS:
n
Take : n  0
Therefore:
2
mv
mg
r
v  gr
mg
Maximum Speed to not loose contact with road only depends on R!
ROOT GRRRRRRRR
What is the maximum speed
the vehicle can have at B and
still remain on the track?
Hump in the Road
Suppose that a 1 800-kg car
passes over a bump in a
roadway that follows the arc
of a circle of radius 20.4 m.
(a) What force does the road
exert on the car as the car
passes the highest point of
the bump if the car travels at
30.0 km/h? (b) What If?
What is the maximum speed
the car can have as it passes
this highest point without
losing contact with the road?
Suppose that a 1 800-kg car passes over a
bump in a roadway that follows the arc of a
circle of radius 20.4 m.
Chapter 6 Problem#51
a) What force does the road exert on the car as the car passes
the highest point of the bump if the car travels at 30.0 km/h?
 1 h   1000 m 
v   30 km h 
 8.33 m s



 3 600 s  1 km 
:
n
 Fy  m ay
2
mv
n  m g  
r
Minus because
ay is pointing
down.
2

8.
33
m
s


v
2 


n  m  g    1800 kg 9.8 m s 
r
20.4 m 




2
 1.15  104 N up
mg
+
Suppose that a 1 800-kg car passes over a
bump in a roadway that follows the arc of a
circle of radius 20.4 m.
Chapter 6 Problem#51
(b) What If? What is the maximum speed the car can
have as it passes this highest point without losing
contact with the road?
:
Take : n  0
 Fy  m ay
n
2
mv
n  m g  
r
m v2
mg
r
v  gr 


mg
9.8 m s2  20.4 m   14.1 m s  50.9 km h
Vertical Motion: Constant Speed.
Loop-the-Loop: UCM
• This is an example of a
vertical circle with
constant motion. The
constant motion is
maintained by an engine.
Where is the force on the pilot the greatest, at
the top or the bottom of circle?
Is the force greater or less than her weight?
Vertical Motion: Constant Speed.
Loop-the-Loop: UCM
• This is an example of a
vertical circle with constant
motion. The constant motion
is maintained by an engine.
• At the bottom of the loop,
the upward force
experienced by the pilot is
greater than at the top and is
greater that her weight:
F
y
 nbot
2
v
 mg  m
r
nbot

v2 
 mg  1  
 rg 
Vertical Motion: Constant Speed.
Loop-the-Loop: UCM
• At the top of the circle,
the force exerted on the
pilot is less than her
weight:
F
y
 nbot
2
v
 mg  m
r
v

 mg   1
 rg 
2
ntop
Vertical Motion: Constant Speed.
Loop-the-Loop: UCM
An airplane moves 140 m/s as it travels around a
vertical circular loop which has a 1.0-km radius. What
is the magnitude of the resultant force on the 70-kg
pilot of this plane at the bottom of this loop?
a.
2.1 kN
b.
1.4 kN
c.
0.69 kN
d.
1.5 kN
e.
1.3 kN