1. SACE Stage 2 Physics
Circular Motion

2. Speed/Velocity in a Circle
Consider an object moving in a circle around a specific origin. The DISTANCE the object covers in ONE REVOLUTION is called the CIRCUMFERENCE. The TIME that it takes to cover this distance is called the PERIOD.
Speed is the MAGNITUDE of the velocity. And while the speed may be constant, the VELOCITY is NOT. Since velocity is a vector with BOTH magnitude AND direction, we see that the direction o the velocity is ALWAYS changing.
We call this velocity, TANGENTIAL velocity as its direction is draw TANGENT to the circle.

3. Tangential Speed/Velocity
Note :The linear speed of the ball at any given instant is always directed tangent to the circular path.

4. Force causing the circular motion
In order for an object to move along a circular path, there must be a force acting on the object to change its direction of motion.
It is the force that is acting towards the centre.
We will come back to it later.

5. Circular Motion
Uniform Circular Motion

6. Circular Motion Uniform Circular Motion
Uniform circular motion occurs when an object moves in a circle with constant speed.
Consider a point mass orbiting in a circle at constant speed and attached to a string. r vi vf 
-vI- v

7. Circular Motion Uniform Circular Motionvi vf 
-vI- v
The change in velocity can be found by vector subtraction
v = vf vi
= vf + ( - vi)
View Circular motion and Centripetal Acceleration as a limit

8. Drawing the Directions correctly
So for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and the acceleration would be drawn TOWARDS the CENTER.
To find the MAGNITUDES of each we have:

9. Centripetal Acceleration
Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circle

10. Uniform Circular Motion
Period of Revolution
Proof
The period (T) is the time taken (s) for one complete revolution.
1 revolution is equivalent to a displacement equal to the circumference of the circle ie. Ds = 2r.
Hence but since Ds = 2r and Dt=T, then and hence,

11. Centripetal Acceleration
The velocity and acceleration vectors are always perpendicular to each other.
Although the speed of an object undergoing uniform circular motion remains constant, the body accelerates.
Since a body undergoing uniform circular motion maintains a constant speed, we must find the acceleration of this body using

12. Circular Motion Uniform Circular Motion Note.
Acceleration is always at right angles to the velocity. Has no affect on speed.
Consider an object moving in a circle. The acceleration of the object is constant and dependant upon its speed and the radius at which it is travelling:
Acts radially towards the centre of the circle.

13. Circular Motion Uniform Circular Motion
eg. a mass on a string travelling through half a revolution, with radius 10 cm.
(i) at speed 10 m s-1 (ii) at speed 20 m s-1
vT = - 10 m s-1
vT = 10 m s-1
vT = - 20 m s-1
vT = 20 m s-1

14. Circular Motion Uniform Circular Motion
(1) For mass traveling at 10ms-1,
(2) For mass traveling at 20ms-1,
Note, If we double the linear velocity, the centripetal acceleration is quadrupled.

15. Circular Motion Example 1
A car traveling at 30kph is traveling around a bend of radius 100m.
What is its centripetal acceleration?
What will be its centripetal acceleration if,
(i) it travels, at the same speed, around a bend of radius 70m?
(ii) it travels at 60kph around a bend of radius 25m?

16. Circular Motion
Example 1
What is its centripetal acceleration?

17. Circular Motion Example 1
(2) What will be its centripetal acceleration if,
(i) it travels, at the same speed, around a bend of radius 70m?
(ii) it travels at 60kph around a bend of radius 25m?
(i)
(ii)

18. Circular Motion Example 2
The moon orbits the Earth in an approximately circular path with a mean radius of 384,000km. The moon completes one revolution in 27.3 days. Find,
The orbital speed of the moon.
The centripetal acceleration of the moon.

19. Circular Motion Example 2 The orbital speed of the moon.
(2) The centripetal acceleration of the moon.

20. Uniform Circular Motion
Force Causing the Centripetal Acceleration
To cause an acceleration, there must be a net external force acting on an object. To make an object move in a circle, you need to pull (exert a force) on the object.
From Newton's Second Law Fresultant = ma =
ac  acts towards the centre of the circle and directed along a radius.
This force is applied on the object towards the centre of the circle to keep the object moving in a circle.
Check question: What happens if the inward directed force stops acting?

21. Centripetal Force
Comparing with N.S.L
Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.

22. Uniform Circular Motion
Examples of Circular Motion FG
Sun
Planet
F (=FG)
(a) Planets orbiting the sun: Planets move in a circular path due to the gravitational force acting on the planet.

23. Uniform Circular Motion
Examples of Circular Motion
proton (+) FE
electron (-)
F (=FE)
(b) The electrostatic attraction between the proton and the electron provides the necessary force to enable rotation in a circular orbit.

24. Uniform Circular Motion
Example 1
A car is traveling on a flat road around a bend of radius 80m at a speed of 120kph. If the mass of the car is 1.4t, what is the average force between each tyre and the road?

25. Circular Motion Application – Banking of Road Curves Case (a)
A car stationary on a horizontal road.
The normal force directed perpendicular to the road surface and upwards.
This normal force is equal and opposite to the gravitational force acting on the car. (Assuming the car is at rest and not sinking into the ground)

26. Circular Motion Application – Banking of Road Curves Case (b)
A car travelling on a horizontal road around a circular bend.
Friction provides the inwards directed force enabling the car to travel around the bend. Since the inward force is given by
the faster the car travels, the greater the magnitude of the required force.
The frictional force will have a fixed value, if the speed is high enough friction will not be sufficient to enable the car to travel around the bend, and it will slide off the road.
View movie: Forces on a car on level track various views

27. Circular Motion Application – Banking of Road Curves Case (c)
A car travelling on a horizontal road around a banked circular bend.
The normal force is greater than the gravitational force acting on the car. (In turning around a bend, the car “presses” into the surface of the road, increasing the normal force)
The faster the car travels, then the greater the force directed towards the centre of the turning circle.
The component of the normal force directed in towards the centre of the circle supplies the force required for rotation in a circle. (Provided the speed of the car does not become so large that the magnitude of this component is exceeded.) View Movie: Forces on a car on banked track various views

28. Circular Motion Derivation of The normal force, Horizontal Component
Horizontal component of normal force, magnitude given by
The normal force,
Horizontal Component
This force is created when the car travels around the bend and presses into the road.
Vertical component
Opposite to that of the weight of the car.
Vertical component of the normal force equal & opposite to weight (mg)
Normal force on vehicle by road (90o to road) 
Weight = force of gravity 

29. Circular Motion Derivation of
Hence from the geometry of the diagram, where  is the banking angle

30. APPLICATION: THE BANKING OF ROAD CURVES
Normal force
Drag force (friction)
Applied force (engine)
Gravitational force (weight)

32. Lateral friction force
Drag force (friction)
Lateral friction force
Applied force (engine)
Normal force
Lateral friction force
Gravitational force (weight)

33. APPLICATION: THE BANKING OF ROAD CURVES
FV normal = mg
Fnormal