1. CIRCULAR MOTION
SPECIFIC LEARNING OUTCOMES
1. Recognize that even though a body in circular motion may have a constant speed,
it velocity is changing and hence it is accelerating.
2. Demonstrate that the acceleration is towards the centre, i.e. centripetal.
3. Deduce that all things in circular motion must have a centripetal force.
4. Show graphically using velocity vectors that the acceleration is towards the centre.
5. Remember the equations for uniform circular motion questions, and use them to
solve problems:
Read
Chapter 12 (p139 to 145)

2. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration
(accelerometer on turntable)
The speed of an object in circular motion:
• is constant in size but .....
• changing in direction
Drawing activity

3. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration
(accelerometer on turntable)
The speed of an object in circular motion:
• is constant in size but .....
• changing in direction THEREFORE ..... the object’s velocity is
changing continuously
Drawing activity

4. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration
(accelerometer on turntable)
The speed of an object in circular motion:
• is constant in size but .....
• changing in direction THEREFORE ..... the object’s velocity is
changing continuously
THEREFORE An object in circular motion is always accelerating
Drawing activity

5. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration
(accelerometer on turntable)
The speed of an object in circular motion:
• is constant in size but .....
• changing in direction THEREFORE ..... the object’s velocity is
changing continuously
THEREFORE An object in circular motion is always accelerating
The size of the acceleration is given by a = ∆v/∆t where ∆v is the size of the change
in velocity vector. Our prior observations indicated that ∆v is constant in size
~
therefore we can expect a to also be constant in size.
~
The direction of the acceleration will be the same as the direction of the change in
velocity (a = ∆v/∆t). Since, from observation, the direction of ∆v is always towards
~ ~ ~
the centre we can expect the direction of a to also be towards the centre.
~
Drawing activity

6. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration
(accelerometer on turntable)
The speed of an object in circular motion:
• is constant in size but .....
• changing in direction THEREFORE ..... the object’s velocity is
changing continuously
THEREFORE An object in circular motion is always accelerating
The size of the acceleration is given by a = ∆v/∆t where ∆v is the size of the change
in velocity vector. Our prior observations indicated that ∆v is constant in size
~
therefore we can expect a to also be constant in size.
~
The direction of the acceleration will be the same as the direction of the change in
velocity (a = ∆v/∆t). Since, from observation, the direction of ∆v is always towards
~ ~ ~
the centre we can expect the direction of a to also be towards the centre.
~
Features of this acceleration:
Size is constant
Direction is towards the centre of the circular path
Drawing activity

7. “Feel the force” CENTRIPETAL FORCE
Since an accelerating object will always do so in response to an unbalanced force,
every object that travels in uniform circular motion must have an unbalanced force
acting on it which is in the same direction as the acceleration. The unbalanced force
acing on an object in uniform circular motion is called Centripetal force, Fcent
Features of this force:
Size is constant
Direction is towards the centre of the circular path
If the speed of the circular motion is increased then the force required to change the
direction of the velocity will need to increase :
Example - the carousel
__________
Label the forces
force
__________
__________ force
force
Time for a complete rotation = 3s Time for a complete rotation = 2s

8. Flash cards EQUATIONS FOR UNIFORM CIRCULAR
Consider a mass in uniform circular motion with speed v and
a radius of circular path, r:
v
F
Acceleration and force
. m At any point:
velocity is at a
Experiment shows that: acent = v2 tangent to the
r circular path
Fcent = mv2
r since F = ma
Period and frequency
• The Period, T is the time the object takes to move through one complete revolution.
(Unit: second, s)
• The frequency, f is the number of revolutions performed per second.
(Unit: Hertz, Hz or s-1))
T=1 f=1
and
f T

9. Speed (in terms of period)
In one complete revolution the distance travelled = the circumference of the circular
path, C = 2πr
The time taken for a complete revolution is the period, T
=> v = 2πr
T
Acceleration (in terms of period)
Substituting v = 2πr into acent = v2 gives acent = (2πr/T)2
T r r
acent = 4π2r
=>
T2
Force (in terms of period)
Since F = ma => Fcent = macent
Fcent = m4π2r
=>
T2

10. Examples
1. The Apollo 11 space capsule was placed in a parking orbit around the Earth before
moving onwards to the Moon. The radius of the orbit was 6.56 x 104 m and the
mass of the capsule was 4.4 x 104 kg.
to the moon
moon
parking orbit Earth
(a) If the centripetal force on the capsule was 407 kN while it was in the parking
orbit, what was its acceleration?
(b) What was the speed of the capsule in the parking orbit?
(c) How long did it take the capsule to complete one orbit?

11. 2. A game of swing-ball is played with a 100 g ball. The effective radius of the circular
path of the ball is 1.4 m. Find the tension in the string (centripetal force) when the
ball has a velocity of:
(a) 7.5 ms-1
(b) 15 ms-1
3. A string has a breaking strain of 320 N. Find the maximum speed that a mass can
be whirled around in uniform circular motion with a radius of 0.45 m if the mass is
0.2 kg.
4. An object is in uniform circular motion, tracing an angle of 30o every 0.010 s. Find:
(a) the period of this motion.
(b) the frequency of this motion
If the radius of the object’s path is doubled but the period remains the same, what
happens to:
(c) its speed?
(d) its acceleration?

12. 5. In a circular motion experiment, a mass is whirled around a horizontal circle of
radius 0.5 m. A student times four revolutions to take 1.5 s.
sinker
tube
(a) Calculate the speed of the mass around the circle
(b) What is the direction of the velocity of the mass?
(c) Calculate the centripetal acceleration of the mass.
(d) What is the direction of this acceleration?
(e) How does the value of the centripetal acceleration compare to the acceleration of
gravity?
Ex.12A All Q’s

13. CIRCULAR MOTION SUMMARY
An object in circular motion is traveling along a circular path with constant
speed. The object’s motion can be described in more detail using vectors:
Velocity, v
~
• constant in size but constantly changing in its direction.
• At any point P, along the path of motion, the direction is at a tangent to the
circular path.
Force, F
~
• A force at right angles to the velocity causes the direction of the velocity to
change. It cannot cause the size of the velocity to change. This force is constant in
size.
• The direction of the force is towards the centre of the circular path.
Acceleration, ~
a
• Acceleration is towards the centre of the circular path because the object is
responding to an unbalanced force that acts towards the centre.
• Acceleration is constant in size because the object is responding to a force that is
constant in size.

14. v = velocity (ms-1)
v ~
~
m
F = Force (N)
F ~
~
P a = acceleration (ms-2)
a ~
~
r r = radius of the circular path (m)
(measured from the centre of the path to
the centre of mass of the object.
m = mass of the object (kg)
Period and frequency
Motion that takes place in regular cycles is called
periodic motion.
• The period, T is the time it takes for the object to
complete a full cycle. (units: s) v=2πr
=>
• The circumference, C is the distance travelled in T
one cycle. C = 2πr (units, m)
• The frequency, f is the number
of cycles completed in one s. T=1 f=1
and
(units: s-1 or Hz) f T

15. Fundamental equations
acent = v2
r
Use these equations when the speed is given
Fcent = mv2
since F = ma
r
Derived equations
acent = 4π2r
T2 Use these equations when the period is given
Fcent = m4π2r since F = ma
T2

16. Term Definition
Velocity speed in a given direction
a constant acceleration directed towards the centre of the circular path
Centripetal acceleration
for an object which is in uniform circular motion
a single force would have the same effect as all the actual forces that
Unbalanced force act on an object. The unbalanced force is responsible for the object’s
acceleration
a constant force directed towards the centre of the circular path for an
Centripetal force
object which is in uniform circular motion.
a line at right angles to the radius of a circle that touches the circle at
Tangent one point only.
Period the time taken for an object to complete a single revolution.
Frequency the number of rotations per second.
Circumference the total length of the circular path
Revolution the single completion of a rotation
Constant unchanging
Fundamental unable to expressed in a simpler form
Derived obtained from another concept

17. 12 PHYSICS CIRCULAR MOTION ASSIGNMENT Name
1. A car is travelling around a bend in the road and for a few seconds is in uniform
circular motion.
(a) The centripetal force is being provided by the road. Name this force.
________________________________________________________________
The car passes over a patch of oil while it is rounding the bend.
(b) Describe the path the car will take after it hits the oil patch and explain why this
happens in terms of the forces acting.
________________________________________________________________
________________________________________________________________
________________________________________________________________
2. An object in uniform circular motion completes 10 revolutions in 0.4 seconds
(a) Find the frequency of this motion.
________________________________________________________________
________________________________________________________________
(b) Find the period of this motion.
________________________________________________________________
_______________________________________________________________

18. 3. A big wheel at a fair spins in a circular path of radius 20 m. Once the wheel has
reached a steady speed, a student times each revolution at 13 seconds.
(a) Calculate the circumference of the big wheel.
______________________________________
(b) Hence calculate the speed of the big wheel.
______________________________________
______________________________________
(c) Calculate the centripetal acceleration of each passenger.
________________________________________________________________
________________________________________________________________
4. In a circular motion experiment, a mass is whirled around a horizontal circle which
has a 0.50 m radius. A student time 4 revolutions to take 2.0 s.
(a) Calculate the speed of the mass around the circle.
________________________________________
________________________________________
(b) What is the direction of the velocity of the mass?
________________________________________
________________________________________
(c) Calculate the centripetal acceleration of the mass.
________________________________________
________________________________________

19. (d) What is the direction of this acceleration?
_______________________________________________________________
(e) How does the value of the centripetal acceleration compare to the acceleration
of gravity? ______________________________________________________
5. Two Aquinas students go to a fun park for a day where they pay to drive carts
around a circular track. The track has a radius of 31.8 m and once the carts are at a
maximum speed they complete a lap in 16 s.
(a) What is the frequency of the cart’s motion when
travelling at maximum speed?
__________________________________________
(b) When travelling at maximum speed, calculate the
speed of the cart.
__________________________________________
__________________________________________
(c) Calculate the acceleration of the cart when travelling at maximum speed.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________

20. The cart has a mass of 150 kg and one of the students, Chris has a mass of 75 kg.
(d) Calculate the size of the force acting on Chris and his cart at maximum speed.
________________________________________________________________
________________________________________________________________
________________________________________________________________
(e) Chris drives over a patch of oil and loses control of his cart whilst travelling at
this maximum speed. On the diagram, draw his path after driving through the
oil.
6. Jon and Ana are two ice-skaters. In a practiced skating move, Jon spins Ana around
in a horizontal circle.
Ana moves in a circle
as shown: Jon
Ana
(a) Draw an arrow on the diagram to show the direction of the tension force that
Jon’s arm exerts on Ana at the instant shown.
(b) If the radius of the circle is 0.95 m and the tension force in Jon’s arm is
5.00 x 102 N, calculate the speed with which Ana (55 kg) is travelling around the
circle. Give your answer to the correct number of significant figures.
________________________________________________________________
________________________________________________________________
________________________________________________________________

21. (d) While Ana is still moving in a circle on the ice, Jon lets her go.
(i) Describe her velocity (speed and direction) after he releases her.
______________________________________________________________
(ii) Explain why Ana travels with this velocity.
______________________________________________________________
______________________________________________________________
______________________________________________________________
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