Uniform Circular Motion credits to Internet sources for information and data

1. • Francis Marlon Cabredo
• Mary Minette Geñorga
• Mary Judith Verdejo

2. ▪Uniform Circular
Motion is the
motion of an object
in a circle with a
constant or
uniform speed

3. SPEED:
▪ In UCM, distance= the circumference of
the circle
▪ T(period) is the time (or number of
seconds) to make one revolution

4. PERIOD and
FREQUENCY

5. • Period (T) -> the time it takes for an object to make one
complete revolution {#sec/rev}
PERIOD and FREQUENCY
• Frequency (f)  the number of revolutions completed by
an object in a given time {#rev/sec}
Units: Hertz (Hz) or 1 rev/sec
rpm (#rev/min); rps (#rev/sec)

6. PERIOD and FREQUENCY

7. SPEED expressed in
PERIOD and FREQUENCY
FORMULA(s):

8. PROBLEM 1: Tire Balancing Machine
• The wheel of a car has a radius of
0.29m and is being rotated at 830
revolutions per minute on a tire-
balancing machine. Determine the
speed (in m/s) at which the outer edge
of the wheel is moving. Use 3.14 as the
value for .

9. SOLUTION#1:
f=830 revolutions in one minute
revolution
srevolution
min/102.1
min/830
1 3

T= 0.072s
Given:
r=0.29 m
Find:
T
V
1.2x10-3 min___________
revolution x 60 sec___________
1 min

10. SOLUTION
sm
s
m
T
r
v /25
072.0
)29.0(22


ANSWER

11. SOLUTION #2:
830 revolutions in one minute
v= 25 m/s
Given:
r=0.29 m
Find:
V
2(3.14)(0.29 m)(830 revolutions)
x 1 min___________
60 sec
_______________________
1 minute

12. VELOCITY
▪ The velocity of an object in UCM has a constant
magnitude and a constant change in direction
▪ The object does not have constant velocity since
its direction changes at every point along the
circle.
▪ Also, The velocity is always tangent to the path of
the object.

13. In UCM, the velocity is always tangent to the path of the
object. The tangent specifies the direction of the motion.
VA
VD
VB
VC
At a time t0, the car is located
at point A with a velocity of
VA, this is tangent to the circle
at point A.
This means that…
The direction of the velocity of
the Car at point A is due west.

14. CENTRIPETAL ACCELERATION
▪ in UCM, an object has no constant velocity.
Therefore, If there is a change in velocity, then that must mean that an
object has acceleration.
Centripetal Acceleration. This acceleration has
constant magnitude but changing direction and is
directed radially inwards.

15. CENTRIPETAL ACCELERATION
• Symbol: ac
• Vector
• It is the rate of change of
Tangential velocity
• Always perpendicular to the
path of the motion.
• Points toward the center
of the circle.
VA
VD
VB
VC
A
B
C
D

16. CENTRIPETAL ACCELERATION
FORMULA(S):
22
2
22
4
4
rf
T
r
r
v
ac 



17. PROBLEM 2: Centripetal Acceleration
The bobsled track at the 1994
Olympics in Norway, contained turns
with radii of 33 m and 24 m, as the
figure illustrates. Find the centripetal
acceleration at each turn for a speed
of 34 m/s, a speed that was achieved
in the two-man event..

18. From ac=v2/r
Radius=33m
Radius=24m
2
2
/35
33
)/34(
sm
m
sm
ac 
2
2
/48
24
)/34(
sm
m
sm
ac 

20. (a) The velocity is due south, and
the acceleration is due west.
(b) The velocity is due west, and
the acceleration is due north.
(c) ac=5.48 m/s2

21. 21
CENTRIPETAL FORCE

22. NEWTON’S SECOND LAW OF MOTION STATES
THAT ALL ACCELERATIONS ARE CAUSED BY A
NET FORCE ACTING ON AN OBJECT. IN THE
CASE OF UCM, THE NET FORCE IS A SPECIAL
FORCE CALLED THE CENTRIPETAL FORCE .

23. CENTRIPETAL IS LATIN FOR "CENTER SEEKING".
IT IS THE INWARD NET FORCE WHICH
KEEPS AN OBJECT MOVING WITH A
UNIFORM VELOCITY ALONG A CIRCULAR
PATH. THIS FORCE IS DIRECTED ALONG
THE RADIUS TOWARDS THE CENTER.
CENTRIPETAL FORCE

24. F ma
F
mv
r


2
•APPLYING NEWTON’S
SECOND
LAW OF MOTION
CENTRIPETAL FORCE
Centripetal Force
ac=v2/r
F=ma
mv2
___
r

25. Newton
N=kg m/s2
UNIT USED:

26. PROBLEM 3A: Centripetal Force
• A 400-g rock attached to a 1.0-m string
is whirled in a horizontal circle at a
constant speed of 10.0m/s. Neglecting
the effects of gravity, what is the
centripetal force acting on the rock?

27. SOLUTION:
Given:
V=10.0 m/s
R=1.0 m
M=400g= 0.4 kg
Find: Fc
(0.4 kg)(10 m/s)2___________
1.0 m
mv2___
r
40 N
40 kg m/s2

28. PROBLEM 3B: Centripetal Force
• How long does it take a 50 kg runner to
run a circular track starting and ending
at the same point, if the radius of the
track is 30m and a force of 68 N keeps
him running at constant speed in the
circular path?

29. SOLUTION:
Given:
M=50 kg
R=30 m
Fc=68 N
Find: T
mv2___
r
29. 50 s
2r__
T
m
( )
2
___________
r
___________
T2
___________T2
________T

30. • FORCES SUCH AS THE GRAVITATIONAL FORCE (w=mg), TENSION
FORCE (tied to a string; pushing/ pulling), FRICTIONAL FORCE (Ex. a car
turning) and Normal force (on a surface) can be the centripetal forces
• GRAVITATIONAL FORCE. For satellites in orbit around a planet, the
centripetal force is supplied by gravity.
• TENSILE FORCE. For an object swinging around on the end of a rope in a
horizontal plane, the centripetal force on the object is supplied by the
tension of the rope.

31. DRAWING A FREE BODY
DIAGRAM CAN HELP YOU
UNDERSTAND THE PROBLEM
BETTER.

32. • A free body diagram, is a pictorial device, used to analyze the forces
and moments acting on a body.
What is included:
1. The body
2. The external forces: These are indicated by labelled arrows.
The forces acting on the object include friction, gravity, normal
force, drag, tension, or a human force due to pushing or
pulling.
FREE BODY DIAGRAM
FORCE DIAGRAM/

36. FG

38. FN
FG

40. FN
FG
FT

42. FN
FG
FTFf

44. FREE BODY DIAGRAM
FN
FG
FTFf

45. PROBLEM 4: Friction as Centripetal Force
A car is going around in a circular road.
Given R, Ff b/w the tires and the road
and m ; Find v.

46. FREE BODY DIAGRAM
FN
(on a
surface)
Fg (weight)
Ff (friction)
Ff is the force going to the
direction of the
acceleration, therefore:
It is the one causing the
centripetal acceleration

47. ANSWER
V=10 m/s

48. THE CENTRIFUGAL FORCE ACTS AWAY FROM THE CENTER.
THE WORD ITSELF MEANS “FLEEING FROM THE CENTER” .
THIS FORCE IS A FICTITIOUS FORCE. IT DOESN’T ACT
ON A BODY IN MOTION, BUT ONLY ON NON-INERTIAL
COORDINATE SYSTEMS SUCH AS A ROTATING ONE.
CENTRIFUGAL FORCE

49. PROBLEM 5: Which way will the object go?
• An object on a guideline is in uniform circular
motion. The object is symbolized by a dot, and at
point O it is release suddenly from its circular path.
• If the guideline is cut suddenly, will the object move
along OA or OP ?

50. ANSWER:
• the object would move along the
straight line between points O
and A, not on the circular arc
between points O and P.

51. SOLUTION:
• NEWTON’S LAW OF MOTION:
“An object continues in a state of rest/ motion at a constant
speed unless compelled to changes to its net force.”
 When the object was suddenly released from its path, there was no longer a net
force (i.e. centripetal force) being applied to the object.
 In the absence of a net force, the object will continue to move at
a constant speed but, along a straight line in the direction it had
at the time of release.

52. CENTRIFUGAL FORCE vs. INERTIA
If you let go of the rope (or
the rope breaks) the object
will no longer be kept in
that circular path and it will
be free to fly off on a
tangent.

53. 53
Uniform circular motion
emphasizes that
1.The speed, or the magnitude of the velocity vector, is constant.
2. Direction of the vector is not constant.
3. Change in direction, means acceleration
4. “Centripetal acceleration” , points toward the center of the
circle.
5. “Centripetal Force” is the net force that causes centripetal
acceleration

55. (a)
T= 1__
f

56. (b)
Hertz (Hz) = 1 rev/sec

57. (c)

58. r
v
ac
2

(d)

59. 5g=0.005 kg
(e)
mv2___
r

60. AP PHYSICS Circular Motion Mrs. Coyle
Practical ad Explorational Physics Padua et al
You and The Natural world Physics Navaza, Valdes
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SOURCES:
• Francis Marlon Cabredo
• Mary Minette Geñorga
• Mary Judith Verdejo
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