# Circular Motion

```Chapter 10. Uniform Circular Motion
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
&copy;
2007
Centripetal
forces keep
these children
moving in a
circular path.
Objectives: After completing this
module, you should be able to:
• Apply your knowledge of centripetal
acceleration and centripetal force to the
solution of problems in circular motion.
• Define and apply concepts of frequency and
period, and relate them to linear speed.
• Solve problems involving banking angles, the
conical pendulum, and the vertical circle.
Uniform Circular Motion
Uniform circular motion is motion along a
circular path in which there is no change in
speed, only a change in direction.
Fc
v
Constant velocity
tangent to path.
Constant force
toward center.
Question: Is there an outward force on the ball?
Uniform Circular Motion (Cont.)
The question of an outward force can be
resolved by asking what happens when the
string breaks!
Ball moves tangent to
v
path, NOT outward as
might be expected.
When central force is removed,
ball continues in straight line.
Centripetal force is needed to change direction.
Examples of Centripetal Force
You
You are
are sitting
sitting on
on the
the seat
seat next
next to
to
the
the outside
outside door.
door. What
What isis the
the
direction
direction of
of the
the resultant
resultant force
force on
on
you
you as
as you
you turn?
turn? Is
Is itit away
away from
from
center
center or
or toward
toward center
center of
of the
the turn?
turn?
• Car going around a
curve.
Fc
Force ON you is toward the center.
Car Example Continued
Reaction
Fc
F’
The centripetal
force is exerted
BY the door ON
you. (Centrally)
There
There isis an
an outward
outward force,
force, but
but itit does
does not
not act
act
ON
ON you.
you. It
It isis the
the reaction
reaction force
force exerted
exerted BY
BY
you
you ON
ON the
the door.
door. It
It affects
affects only
only the
the door.
door.
Another Example
Disappearing
platform at fair.
R
Fc
What exerts the centripetal force in this
example and on what does it act?
The
The centripetal
centripetal force
force isis exerted
exerted BY
BY the
the wall
wall
ON
ON the
the man.
man. AA reaction
reaction force
force isis exerted
exerted by
by
the
the man
man on
on the
the wall,
wall, but
but that
that does
does not
not
determine
determine the
the motion
motion of
of the
the man.
man.
Spin Cycle on a Washer
How is the water removed
from clothes during the
spin cycle of a washer?
Think carefully before answering . . . Does the
centripetal force throw water off the clothes?
NO.
NO. Actually,
Actually, itit isis the
the LACK
LACK of
of aa force
force that
that
allows
allows the
the water
water to
to leave
leave the
the clothes
clothes
through
through holes
holes in
in the
the circular
circular wall
wall of
of the
the
rotating
rotating washer.
washer.
Centripetal Acceleration
Consider ball moving at constant speed v in a
horizontal circle of radius R at end of string tied
to peg on center of table. (Assume zero friction.)
Fc
W
v
R
n
Force Fc and
acceleration ac
toward center.
W=n
Deriving Central Acceleration
Consider initial velocity at A and final velocity at B:
vf
vf B
R
vo
A
-vo v
R
s v
o
Deriving Acceleration (Cont.)
v
Definition: ac =
Similar
Triangles
ac =
v
t
=
Centripetal
Centripetal
acceleration:
acceleration:
v
v
vs
Rt
vf
-vo v
t
=
=
R
s
R
s v
o
mass m
vv
R
2
v
ac  ;
R
mv
Fc  mac 
R
2
Example 1: A 3-kg rock swings in a circle
of radius 5 m. If its constant speed is 8
m/s, what is the centripetal acceleration?
2
v
v
m
m = 3 kg
ac 
R
R
R = 5 m; v = 8 m/s
2
(8 m/s)
2
ac 
 12.8 m/s
5m
mv
Fc  mac 
R
2
F = (3 kg)(12.8 m/s2)
FFcc == 38.4
38.4 NN
Example 2: A skater moves with 15 m/s in a
circle of radius 30 m. The ice exerts a
central force of 450 N. What is the mass of
the skater?
Draw and label sketch
Fc R
mv 2
; m 2
Fc 
v = 15 m/s
R
v
Fc R
450 N
30 m
m=?
Speed skater
(450 N)(30 m)
m
2
(15 m/s)
m
m == 60.0
60.0 kg
kg
Example 3. The wall exerts a 600 N force on
an 80-kg person moving at 4 m/s on a
circular platform. What is the radius of the
circular path?
Draw and label sketch
Newton’s 2nd law
for circular motion:
m = 80 kg;
v = 4 m/s2
Fc = 600 N
2
mv
mv
F
; r
r
F
r=?
(80 kg)(4 m/s)
r
600 N
2
rr =
= 2.13
2.13 m
m
2
Car Negotiating a Flat Turn
v
Fc
R
What is the direction of the
force ON the car?
Ans. Toward Center
This central force is exerted
BY the road ON the car.
Car Negotiating a Flat Turn
v
Fc
R
Is there also an outward force
acting ON the car?
Ans. No, but the car does exert a
outward reaction force ON the road.
Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
m
Fc
R
n
fs
Fc = fs
R
v
mg
The
The central
central force
force FFCC and
and the
the friction
friction force
force ffss
are
are not
not two
two different
different forces
forces that
that are
are equal.
equal.
There
There isis just
just one
one force
force on
on the
the car.
car. The
The nature
nature
of
of this
this central
central force
force isis static
static friction.
friction.
Finding
Finding the
the maximum
maximum speed
speed for
for
negotiating
negotiating aa turn
turn without
without slipping.
slipping.
n
fs
Fc = fs
m
v
R
Fc
R
mg
The car is on the verge of slipping when FC is
equal to the maximum force of static friction fs.
Fc = fs
Fc =
mv2
R
fs = smg
Maximum
Maximum speed
speed without
without slipping
slipping (Cont.)
(Cont.)
n
Fc = fs
fs
R
mv2
R
mg
v=
m
v
Fc
R
= smg
sgR
Velocity
Velocity vv isis maximum
maximum
speed
speed for
for no
no slipping.
slipping.
Example 4: A car negotiates a turn of
radius 70 m when the coefficient of static
friction is 0.7. What is the maximum
speed to avoid slipping?
m
v
Fc
Fc =
R
s = 0.7
mv2
R
fs = smg
From which: v =
sgR
g = 9.8 m/s2; R = 70 m
v   s gR  (0.7)(9.8)(70 m) vv == 21.9
21.9 m/s
m/s
Optimum Banking Angle
Fc
m
R
v
fs
w
By banking a curve at the
optimum angle, the normal
force n can provide the
necessary centripetal force
without the need for a
friction force.
n

slow speed
n
w
fs

fast speed
fs = 0
w
n

optimum
Free-body Diagram
Acceleration a is toward the
center. Set x axis along
the direction of ac , i. e.,
horizontal (left to right).
n
x
mg

n

n cos 

n
n sin 
mg

mg
+ ac
Optimum Banking Angle (Cont.)
n cos 
n
n
mg

n sin 

Apply
Newton’s 2nd
Law to x and
y axes.
mg
mv2
Fx = mac
n sin  
Fy = 0
n cos = mg
R
Optimum Banking Angle (Cont.)
n cos 
n
n

n
sin 
tan  
n cos 
n
sin

mg 
mg
n sin  
mv2
R
n cos  = mg
2
mv
2
v
R
tan  

mg
gR
1
Optimum Banking Angle (Cont.)
n
mg

Optimum Banking
Angle 
n cos 
n

n sin 
mg
2
v
tan  
gR
Example 5: A car negotiates a turn of
radius 80 m. What is the optimum
banking angle for this curve if the speed
is to be equal to 12 m/s?
n
tan  =
mg
n cos 


n
n sin 
mg
v2
gR
=
(12 m/s)2
(9.8 m/s2)(80 m)
tan =
 0.184
 = 10.40
How might you 2find the
mv
centripetal
FC force on the
car, knowing R
its mass?
The Conical Pendulum
A conical pendulum consists of a mass m
revolving in a horizontal circle of radius R
at the end of a cord of length L.
T cos 
L 
T
R
T

h
mg
T sin 
Note: The inward component of tension
T sin  gives the needed central force.
Angle  and velocity v:
T cos 
L 
T

h
mg
R
Solve two
equations
to find
angle 
T
T sin  
T sin 
mv2
R
T cos = mg
tan  =
v2
gR
Example 6: A 2-kg mass swings in a
horizontal circle at the end of a cord of
length 10 m. What is the constant
speed of the mass if the rope makes an
angle of 300 with the vertical?

T
L 
R
h
1. Draw &amp; label sketch.
2. Recall formula for pendulum.
2
v
tan  
gR
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R=5m
Example 6(Cont.): Find v for  = 300
4. Use given info to find the
velocity at 300.
R=5m
Solve for
v=?

g = 9.8 m/s2
T
v  gR tan 
v  gR tan 
v  (9.8 m/s )(5 m) tan 30
2
h
R
v2
tan  
gR
2
L 
R=5m
0
vv == 5.32
5.32 m/s
m/s
Example 7: Now find the tension T in the
cord if m = 2 kg,  = 300, and L = 10 m.
T cos 
L 
2 kg
T
T=

h
mg
R
Fy = 0:
mg
cos 
T cos  - mg = 0;
=
T
(2 kg)(9.8 m/s2)
cos 300
T sin 
T cos  = mg
TT == 22.6
22.6 NN
Example 8: Find the centripetal force Fc
for the previous example.
 = 300
2 kg
T cos 
L 
h
T Fc

mg
R
T
T sin 
m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N
Fc =
mv2
R
or Fc = T sin 300
FFcc == 11.3
11.3 NN
Swinging Seats at the Fair
This problem is identical
to the other examples
except for finding R.
b
L 
T
h
d
R=d+b
R
tan  =
R = L sin  + b
v2
gR
and
v=
gR tan 
Example 9. If b = 5 m and L = 10 m, what
will be the speed if the angle  = 260?
v2
tan  =
R=d+b
gR
L  b
d = (10 m) sin 260 = 4.38 m T
d
R = 4.38 m + 5 m = 9.38 m
R
v  gR tan 
2
v  gR tan 
v  (9.8 m/s 2 )(9.38 m) tan 260
vv == 6.70
6.70 m/s
m/s
Motion in a Vertical Circle
v
Consider the forces on a
ball attached to a string as
it moves in a vertical loop.
+
v
Note also that the positive
direction is always along
acceleration, i.e., toward
the center of the circle.
v
+
v
Bottom
Tmg
+
T
T
T
mg
+
+mg
mg
T
v
Top
ofSide
Path
Left
mg
Top Right
Right
Top
Tension is
Weight
hasasno
Maximum
minimum
Weight
causes
Weight
has
no
effect
on
T
tension
T, W
Bottom
weight
helps
small
decrease
effect on T
opposes
F
force
Ftension
in
Tc
c
Note changes as you click
the mouse to show new
positions.
+
v
10 N
T
As an exercise, assume
that a central force of
Fc = 40 N is required to
maintain circular motion
of a ball and W = 10 N.
+
R
T
+
10 N
v
The
The tension
tension TT must
must
so that
that central
central
resultant
resultant isis 40
40 NN..
At top: 10 N + T = 40 N
TT =
?_N
= _30
Bottom: T – 10 N = 40 N
TT==__?___
50 N
Motion in a Vertical Circle
v
mg
T
mv2
Resultant force
Fc =
toward center
R
R
v
mg + T =
AT TOP:
+
mg
T
Consider TOP of circle:
T=
mv2
R
mv2
R
- mg
Vertical Circle; Mass at bottom
v
T
Resultant force
toward center
R
v
mg
T - mg =
mg
+
R
Consider bottom of circle:
AT Bottom:
T
Fc =
mv2
T=
mv2
R
mv2
R
+ mg
Visual Aid: Assume that the centripetal force
required to maintain circular motion is 20 N.
Further assume that the weight is 5 N.
v
mv 2
FC 
 20 N
R
Resultant central force FC
at every point in path!
R
v
FC = 20 N at top
AND at bottom.
FC = 20 N
Weight vector W is
downward at every point.
W = 5 N, down
Visual Aid: The resultant force (20 N) is the
vector sum of T and W at ANY point in path.
W
T
+
T
W
+
v
Top: T + W = FC
T + 5 N = 20 N
R
T = 20 N - 5 N = 15 N
v
FC = 20 N at top
AND at bottom.
Bottom:
T - W = FC
T - 5 N = 20 N
T = 20 N + 5 N = 25 N
For Motion in Circle
v
AT TOP:
R
+ T=
mg
mv2
R
- mg
T
v
AT BOTTOM:
T
mg
+
T=
mv2
R
+ mg
Example 10: A 2-kg rock swings in a vertical
circle of radius 8 m. The speed of the rock as it
passes its highest point is 10 m/s. What is
tension T in rope?
2
At Top:
v
mg
T
T=
R
v
mg + T =
mv2
mv
- mg
R
R
2
(2 kg)(10 m/s)
2
T
 2 kg(9.8 m/s )
8m
T = 25 N - 19.6 N
TT == 5.40
5.40 NN
Example 11: A 2-kg rock swings in a vertical
circle of radius 8 m. The speed of the rock as it
passes its lowest point is 10 m/s. What is
tension T in rope?
2
At Bottom:
v
R
T
mg
T=
v
T - mg =
mv2
mv
+ mg
R
R
2
(2 kg)(10 m/s)
2
T
 2 kg(9.8 m/s )
8m
T = 25 N + 19.6 N
TT == 44.6
44.6 NN
Example 12: What is the critical speed vc at
the top, if the 2-kg mass is to continue in a
0
2
mv
v
At Top:
mg + T =
mg
R
vc occurs when T = 0
R
T
mv2
mg =
vc = gR
v
R
v=
gR =
(9.8 m/s2)(8 m)
vvcc== 8.85
8.85 m/s
m/s
The Loop-the-Loop
Same as cord, n replaces T
v
AT TOP:
mg
R
v
AT BOTTOM:
n
mg
+
+
n=
mv2
R
n
n=
mv2
R
+ mg
- mg
The Ferris Wheel
v
AT TOP:
n
R
v
+
mg
AT BOTTOM:
n
mg
+
n=
mg
- n=
n = mg mv2
R
+ mg
mv2
R
mv2
R
Example 13: What is the
n
apparent weight of a 60-kg
person as she moves
v
through the highest point
mg
when R = 45 m and the
speed at that point is 6 m/s?
Apparent weight will be the
normal force at the top:
mv2
mg - n =
n = mg
R
2
(60
kg)(6
m/s)
2
n  60 kg(9.8 m/s ) 
45 m
-
+
R
v
mv2
R
nn == 540
540 NN
Summary
Centripetal
Centripetal
acceleration:
acceleration:
v=
sgR
Conical
pendulum:
2
v
ac  ;
R
mv
Fc  mac 
R
v2
tan  = gR
v=
gR tan 
2
Summary: Motion in Circle
v
AT TOP:
+
mg
R
v
2
mv
T=
- mg
R
T
AT BOTTOM:
T
mg
+
2
mv
T=
+ mg
R
Summary: Ferris Wheel
v
AT TOP:
n
R
v
+
mg
AT BOTTOM:
n
mg
+
n=
mg
- n=
n = mg mv2
R
+ mg
mv2
R
mv2
R
CONCLUSION: Chapter 10
Uniform Circular Motion
```