1. Uniform Circular Motion
Physics 12

2. Uniform Circular Motion
object is moving at a constant speed but changing directions
acceleration occurs due to direction changes

3. Circular Motion is Periodic
The motion occurs at a regular time interval called the period (T). The period is the time for one complete cycle.
(sinusoidal curve)

4. Calculating Speed Constant speed: v = d/t
In a circle, d = circumference and t is the period

5. Example
A mass on a string is swung in a circle of radius 1.0m at a speed of 5.0 m/s. What is the period of the revolution?
1.3 s

6. Centripetal Acceleration
In order for an object to follow a circular path, a force needs to be applied in order to accelerate the object
Although the magnitude of the velocity may remain constant, the direction of the velocity will be constantly changing
As a result, this force will provide a centripetal acceleration towards the centre of the circular path

8. Centripetal Acceleration
caused by a centre-seeking (centripetal) force that acts toward the centre of the circular path

9. Diagrams… r1 and r2 represent POSITION at two different time
v1 and v2 represent VELOCITY at two difference positions, each tangent to the path.
Now use vector addition**
Note that the resultant vector is direction TOWARDS the CENTRE of the circle.

10. How can we calculate centripetal acceleration (Derivation)?
Acceleration on a v-t graph = slope

11. Centripetal Acceleration (derivation on page 552)
Where v is the magnitude of the velocity (or speed) in m/s
r is the radius of the circle (in m)
and ac is the centripetal acceleration (m/s2)

12. Example
A mass on a string is swung in a circle of radius 0.75m at 7.0 m/s. What is its rate of acceleration?
65 m/s2 (towards centre of circle)

13. Centripetal Force
Circular motion occurs due to the presence of a centripetal (centre-seeking) force that pulls toward the centre of the path.
This force is UNBALANCED, therefore causing acceleration (Newton’s 2nd Law) directed toward the centre of the circle.

14. Centripetal Force
The centripetal force works to oppose the object’s inertia.
NOTE: At one time inertia was mistakenly referred to as centrifugal force. There is no such thing as centrifugal force – it is an illusion caused by inertia.

15. Notes: Centripetal Force is not really a specific force
Any force that causes circular motion is a centripetal force – it isn’t one type of force.
Examples: Mass on a string is really Tension.
Planetary Orbit is really Fg.
Car rounding a bend is really Ff.

16. Centripetal Force
Like the centripetal acceleration, the centripetal force is always directed towards the centre of the circle
The centripetal force can be calculated using Newton’s Second Law of Motion

18. NOTE: T is the inverse of f (remember from waves) so you can also manipulate the equation that way as well!
T = 1/f

19. Problem – horizontal circle
A student attempts to spin a rubber stopper (m=.050kg) in a horizontal circle with a radius of .75m. If the stopper completes 2.5 revolutions every second, determine the following:
The centripetal acceleration
The centripetal force

20. Remember
You only need to look at the forces that are affecting the acceleration.
Gravity does not affect a horizontal circular motion as it acts 90’ to the plane of motion.

21. The stopper will cover a distance that is 2
The stopper will cover a distance that is 2.5 times the circumference of the circle every second
Determine the circumference
Multiply by 2.5
Use the distance and time (one second) to calculate the speed of the stopper

22. Use the speed and radius to determine the centripetal acceleration
Then use the centripetal acceleration and mass to determine the centripetal force

23. Try This
A certain string can withstand a maximum tension of 55N before breaking. What is the maximum speed at which we can safely swing a 1.0kg mass on a piece of this string 0.50 m long in a horizontal circle?
5.2m/s

24. Road Design
You are responsible to determine the speed limit for a turn on the highway. The radius of the turn is 55m and the coefficient of static friction between the tires and the road is 0.90.
Find the maximum speed at which a vehicle can safely navigate the turn
If the road is wet and the coefficient drops to 0.50, how does this change the maximum speed

25. Diagrams

26. The maximum speed at which a vehicle can safely navigate the turn

27. Coefficient drops to 0.50, how does this change the maximum speed

28. Try This
A car of mass 1250kg goes around a corner of radius 15.0m. Find the minimum coefficient of friction between the tires and the road required to safely navigate the turn at that speed.
u = 0.979

29. Page 559
15, 16 ,18

30. Problem – vertical circle
A student is on a carnival ride that spins in a vertical circle.
Determine the minimum speed that the ride must travel in order to keep the student safe if the radius of the ride is 3.5m.
Determine the maximum force the student experiences during the ride (in terms of number of times the gravitational force)

31. Vertical Circle
While travelling in a vertical circle, gravity must be considered in the solution
While at the top of the circle, gravity acts towards the centre of the circle and provides some of the centripetal force
While at the bottom of the circle, gravity acts away from the centre of the circle and the force applied to the object must overcome both gravity and provide the centripetal force

32. Vertical Circle
To determine the minimum velocity required, use the centripetal force equal to the gravitational force (as any slower than this and the student would fall to the ground). This means consider the centripetal force to ONLY be Fg (no Ftens)
To determine the maximum force the student experiences, consider the bottom of the ride when gravity must be overcome

33. At the top of the circle, set the gravitational force (weight) equal to the centripetal force
Solve for velocity

34. At the bottom of the circle, the net force is equal to the sum of the gravitational force and the centripetal force
Solve for number of times the acceleration due to gravity

35. Practice Problems Page 559 15-19