3. 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.
4. 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?
5. 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.
6. Examples of Centripetal
Force
You are sitting on the seat next to
You are sitting on the seat next to
the outside door. What is the
the outside door. What is the
direction of the resultant force on
direction of the resultant force on
you as you turn? Is it away from
you as you turn? Is it away from
center or toward center of the turn?
center or toward center of the turn?
• Car going around a
curve.
Fc
Force ON you is toward the center.
7. Car Example Continued
Reaction
Fc
F’
The centripetal
force is exerted
BY the door ON
you. (Centrally)
There is an outward force, but it does not act
There is an outward force, but it does not act
ON you. It is the reaction force exerted BY you
ON you. It is the reaction force exerted BY you
ON the door. It affects only the door.
ON the door. It affects only the door.
8. Another Example
Disappearing
platform at fair.
R
Fc
What exerts the centripetal force in this
example and on what does it act?
The centripetal force is exerted BY the wall
The centripetal force is exerted BY the wall
ON the man. A reaction force is exerted by
ON the man. A reaction force is exerted by
the man on the wall, but that does not
the man on the wall, but that does not
determine the motion of the man.
determine the motion of the man.
9. 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. Actually, it is the LACK of a force that
NO. Actually, it is the LACK of a force that
allows the water to leave the clothes through
allows the water to leave the clothes through
holes in the circular wall of the rotating
holes in the circular wall of the rotating
washer.
washer.
10. 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.)
n
Fc
R
v
W
Force Fc and
acceleration ac
toward center. W
=n
12. Deriving Acceleration (Cont.)
Definition: ac =
Similar
Triangles
ac =
∆v
t
=
Centripetal
Centripetal
acceleration:
acceleration:
∆v
v
=
vs
=
Rt
vf
∆v
-vo ∆v
t
R
s
R
s v
o
mass m
vv
R
2
v
ac = ;
R
mv
Fc = mac =
R
2
13. 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)
Fcc = 38.4 N
F = 38.4 N
14. 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
Fc =
; m= 2
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 = 60.0 kg
m = 60.0 kg
15. 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 m
= 2.13 m
2
16. 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.
17. 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.
18. Car Negotiating a Flat Turn
m
Fc
R
The centripetal force Fc is
that of static friction fs:
n
Fc = fs
fs
R
v
mg
The central force FC and the friction force ffss are
The central force FC and the friction force are
not two different forces that are equal. There is
not two different forces that are equal. There is
just one force on the car. The nature of this
just one force on the car. The nature of this
central force is static friction.
central force is static friction.
19. Finding the maximum speed for
Finding the maximum speed for
negotiating a turn without slipping.
negotiating a turn without 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 = µsmg
20. Maximum speed without slipping (Cont.)
Maximum speed without slipping (Cont.)
n
Fc = fs
fs
R
R
mg
m
v
Fc
mv2
v=
R
= µsmg
µsgR
Velocity v is maximum
Velocity v is maximum
speed for no slipping.
speed for no slipping.
21. 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
µs = 0.7
mv2
R
From which:
fs = µsmg
v=
µsgR
g = 9.8 m/s2; R = 70 m
v = µ s gR = (0.7)(9.8)(70 m)
v = 21.9 m/s
v = 21.9 m/s
22. Optimum Banking Angle
Fc
m
v
fs
w
R
n
θ
slow speed
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.
fs = 0
n
n
w
fs
θ
fast speed
w
θ
optimum
23. 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
θ
n
θ
n cos θ
θ
n
n sin θ
mg
θ
mg
+ ac
24. Optimum Banking Angle (Cont.)
n cos θ
n
n
mg
θ
n sin θ
θ
Apply
Newton’s 2nd
Law to x and y
axes.
mg
ΣFx = mac
ΣFy = 0
n sin θ =
mv2
R
n cos θ = mg
25. Optimum Banking Angle (Cont.)
n cos θ
n
n
θ
n sin θ
tan θ =
n cos θ
n sin θ
mg θ
mg
n sin θ =
mv
2
R
n cos θ = mg
2
mv
2
R = v
tan θ =
mg
gR
1
26. Optimum Banking Angle (Cont.)
n
mg
θ
Optimum Banking
Angle θ
n cos θ
n
θ
n sin θ
mg
2
v
tan θ =
gR
27. 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
mg
n cos θ
tan θ =
θ
θ
n
n sin θ
mg
v2
gR
=
tan θ = 0.184
(12 m/s)2
(9.8 m/s2)(80 m)
θ = 10.40
How might you 2 the
find
mv on the
centripetal force
FC =
car, knowing its mass?
R
28. 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 θ
T
L θ
R
T
θ
h
mg
T sin θ
Note: The inward component of tension
T sin θ gives the needed central force.
29. Angle θ and velocity v:
T cos θ
T
L θ
θ
h
mg
R
Solve two
equations
to find
angle θ
T
T sin θ =
T sin θ
mv2
R
T cos θ = mg
tan θ =
v2
gR
30. 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?
θ = 300
T
L θ
R
h
1. Draw & label sketch.
2. Recall formula for pendulum.
2
v
tan θ =
gR
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R=5m
31. Example 6(Cont.): Find v for θ = 300
4. Use given info to find the
velocity at 300.
R=5m
Solve for
v=?
θ = 300
g = 9.8 m/s2
T
v2
tan θ =
gR
v = gR tan θ
2
h
R
v = gR tan θ
v = (9.8 m/s )(5 m) tan 30
2
L θ
R=5m
0
v = 5.32 m/s
v = 5.32 m/s
32. Example 7: Now find the tension T in the
cord if m = 2 kg, θ = 300, and L = 10 m.
T cos θ
2 kg
T
L θ
T=
θ
h
mg
R
ΣFy = 0:
mg
=
cos θ
T
T cos θ - mg = 0;
(2 kg)(9.8 m/s2)
cos 300
T sin θ
T cos θ = mg
T = 22.6 N
T = 22.6 N
33. Example 8: Find the centripetal force Fc
for the previous example.
θ = 300
2 kg
T cos θ
L θ
h
T Fc
R
T
θ
mg
T sin θ
m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N
Fc =
mv2
R
or Fc = T sin 300
Fcc = 11.3 N
F = 11.3 N
34. Swinging Seats at the Fair
This problem is identical
to the other examples
except for finding R.
b
T
L θ
d
h
R=d+b
R
tan θ =
R = L sin θ + b
v2
gR
and
v=
gR tan θ
35. Example 9. If b = 5 m and L = 10 m, what
will be the speed if the angle θ = 260?
v2
tan θ =
R=d+b
gR
L θ b
d = (10 m) sin 260 = 4.38 m
T
d
R = 4.38 m + 5 m = 9.38 m
R
v = gR tan θ
v = gR tan θ
2
v = (9.8 m/s )(9.38 m) tan 26
2
0
v = 6.70 m/s
v = 6.70 m/s
36. 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
mg
+T T
T
T
mg
+
mg +mg
T
v
Top of Path
Left Side
mgTop
Top Right
Tension is
Right no
Weight causes
Maximum
minimum as
Weight has no
Weight hasT
effect T,
tensionon W
Bottom
weight helps
small decrease
effect on T
opposes T
inFc forceFc
tension
Note changes as you click
the mouse to show new
positions.
+
37. 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 tension T must
The tension T must
adjust so that central
adjust so that central
resultant is 40 N..
resultant is 40 N
At top: 10 N + T = 40 N
T = _?_N
T = 30
Bottom: T – 10 N = 40 N
T = __?___
T = 50 N
38. Motion in a Vertical Circle
v
mg
T
Resultant force F = mv2
c
toward center
R
R
v
mg + T =
AT TOP:
+
mg
T
Consider TOP of circle:
T=
mv2
R
mv2
R
- mg
39. Vertical Circle; Mass at bottom
v
T
Resultant force
toward center
R
v
mg
AT Bottom:
T
mg
+
Fc =
mv2
R
Consider bottom of circle:
T - mg =
T=
mv2
R
mv2
R
+ mg
40. Visual Aid: Assume that the centripetal force
required to maintain circular motion is 20 N.
Further assume that the weight is 5 N.
v
R
v
FC = 20 N at top
AND at bottom.
mv 2
FC =
= 20 N
R
Resultant central force FC
at every point in path!
FC = 20 N
Weight vector W is
downward at every point.
W = 5 N, down
41. Visual Aid: The resultant force (20 N) is the
vector sum of T and W at ANY point in path.
W
T
T
Top: T + W = FC
+
+
v
W
T + 5 N = 20 N
R
v
FC = 20 N at top
AND at bottom.
T = 20 N - 5 N = 15 N
Bottom:
T - W = FC
T - 5 N = 20 N
T = 20 N + 5 N = 25 N
42. For Motion in Circle
v
AT TOP:
R
+ T=
mg
mv2
R
T
v
AT BOTTOM:
T
mg
+
T=
mv2
R
+ mg
- mg
43. 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
v
At Top:
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
T = 5.40 N
T = 5.40 N
44. 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
T = 44.6 N
T = 44.6 N
45. Example 12: What is the critical speed vc at
the top, if the 2-kg mass is to continue in a
circle of radius 8 m?
0
mv2
v
At Top:
mg + T =
mg
R
vc occurs when T = 0
R
T
mv2
vc = gR
mg =
v
R
v=
gR =
(9.8 m/s2)(8 m)
vcc= 8.85 m/s
v = 8.85 m/s
46. The Loop-the-Loop
Same as cord, n replaces T
v
AT TOP:
mg
R
v
AT BOTTOM:
n
mg
+
+
n=
mv2
R
n
mv2
n=
+ mg
R
- mg
47. The Ferris Wheel
v
AT TOP:
n
R
v
+
mg
AT BOTTOM:
n
mg
+
n
mg
- n=
n = mg -
mv2
=
+ mg
R
mv2
R
mv2
R
48. Example 13: What is the
apparent weight of a 60-kg
person as she moves
through the highest point
when R = 45 m and the
speed at that point is 6 m/s?
n
mg
Apparent weight will be the
normal force at the top:
mv2
mg - n =
n = mg
R
(60 kg)(6 m/s) 2
2
n = 60 kg(9.8 m/s ) −
45 m
+
v
R
v
-
mv2
R
n = 540 N
n = 540 N
