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• Francis Marlon Cabredo
• Mary Minette Geñorga
• Mary Judith Verdejo
VCircular motion:
Motion of a body along a circular
path is known as circular motion.
Uniform circular motion:
Motion of a body long a circular path
with the constant speed is called
uniform circular motion.
▪Uniform Circular
Motion is the
motion of an object
in a circle with a
constant or
uniform speed
SPEED:
▪ In UCM, distance= the circumference of
the circle
▪ T(period) is the time (or number of
seconds) to make one revolution
PROBLEM 1: Tire Balancing Machine
• The wheel of a car has a radius of
0.29m and is being rotated at
0.072 sec/rev 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 .
SOLUTION
sm
s
m
T
r
v /25
072.0
)29.0(22


ANSWER
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.
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.
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.
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
CENTRIPETAL ACCELERATION
FORMULA:
r
v
ac
2

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..
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 
14
CENTRIPETAL FORCE
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 .
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
F ma
F
mv
r


2
•APPLYING NEWTON’S
SECOND
LAW OF MOTION
CENTRIPETAL FORCE
Centripetal Force
ac=v2/r
F=ma
mv2
___
r
Newton
N=kg m/s2
UNIT
USED:
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?
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
• 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.
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
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 ?
ANSWER:
• the object would move along
the straight line between
points O and A, not on the
circular arc between points O
and P.
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.
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.
27
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
Assignment: • Solve the problems in the
Practice Math in yourbook,
page 19. (1/2 crosswise pad
paper)
• Study for the Long Test on,
July 18.
Topics: Types of Forces, Friction,
Laws of Motion, Uniform
Circular Motion
AP PHYSICS Circular Motion Mrs. Coyle
Practical ad Explorational Physics Padua et al
You and The Natural world Physics Navaza, Valdes
INTERNET
YOUTUBE
GOOGLE……..
SOURCES:
• Francis Marlon Cabredo
• Mary Minette Geñorga
• Mary Judith Verdejo
APPLICATIONS
PROBLEMS

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Uniform circular motion

  • 1. • Francis Marlon Cabredo • Mary Minette Geñorga • Mary Judith Verdejo
  • 2. VCircular motion: Motion of a body along a circular path is known as circular motion. Uniform circular motion: Motion of a body long a circular path with the constant speed is called uniform circular motion.
  • 3. ▪Uniform Circular Motion is the motion of an object in a circle with a constant or uniform speed
  • 4. SPEED: ▪ In UCM, distance= the circumference of the circle ▪ T(period) is the time (or number of seconds) to make one revolution
  • 5. PROBLEM 1: Tire Balancing Machine • The wheel of a car has a radius of 0.29m and is being rotated at 0.072 sec/rev 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 .
  • 7. 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.
  • 8. 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.
  • 9. 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.
  • 10. 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
  • 12. 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..
  • 15. 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 .
  • 16. 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
  • 17. F ma F mv r   2 •APPLYING NEWTON’S SECOND LAW OF MOTION CENTRIPETAL FORCE Centripetal Force ac=v2/r F=ma mv2 ___ r
  • 19. 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?
  • 20. 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
  • 21. • 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.
  • 22. 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
  • 23. 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 ?
  • 24. ANSWER: • the object would move along the straight line between points O and A, not on the circular arc between points O and P.
  • 25. 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.
  • 26. 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.
  • 27. 27 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
  • 28. Assignment: • Solve the problems in the Practice Math in yourbook, page 19. (1/2 crosswise pad paper) • Study for the Long Test on, July 18. Topics: Types of Forces, Friction, Laws of Motion, Uniform Circular Motion
  • 29. AP PHYSICS Circular Motion Mrs. Coyle Practical ad Explorational Physics Padua et al You and The Natural world Physics Navaza, Valdes INTERNET YOUTUBE GOOGLE…….. SOURCES: • Francis Marlon Cabredo • Mary Minette Geñorga • Mary Judith Verdejo APPLICATIONS PROBLEMS