1. Circular Motion Accelerated requires Fnet. Acceleration called ac.
Fnet called centripetal force.

2. The term “Centripetal Force” decribes the direction of Fnet & acceleration. It is NOT itself a force.
Any force applied at 90o to displacement or forces can cause curved or circular motion.
Fc is the amount of force required to keep an object of mass m, moving at speed v, in a circle of radius r.

3. Circular or Curved Motion Equations
T = time to complete a cycle (sec)
f = number of cycles in unit time (hz)
T = 1/f
vc = d/t
= 2pr/T
ac = v2/r
Fc = mac = mv2/r.

4. Centripetal force and acceleration may be caused by:
gravity -
friction –
Tension in a rope or cord-
Felc –
F mag
R (Fn) –
Etc.

5. 1. Write an equation for centripetal acceleration in terms of r and T.
Substituting 2pr for v in T
accl eq. a = v2 r
ac = 4p2r
T2.

6. 2. The distance from moon’s center to earth’s center = 3. 8 x 108 m
2. The distance from moon’s center to earth’s center = 3.8 x 108 m. Moon’s ac = 2.8 x 10-3 m/s2. What is moon’s period?
2.3 x 106 seconds
~ 27 days.

7. The car is turning due to the inward force, you feel as though you are being forced leftward or outward. The car is beginning its turning motion (to the right) while you continue in a straight line path.

8. Direction Velocity Tangent to circular path
Direction Velocity Tangent to circular path. Accl & force toward center of circular path.

9. Mud sticks to tire.

11. Circular or Curved Motion Equations
T = time to complete a cycle (sec)
f = number of cycles in unit time (hz)
T = 1/f
vc = d/t
= 2pr/T
ac = v2/r
Fc = mac = mv2/r.

12. Sometimes a measured in g’s. Multiples of Earth’s a of gravity.
1g = 9.81 m/s2.
2g = 19.6 m/s2.
3g = 29.4 m/s2. .
etc.

13. An 80-kg astronaut experience a force of 2890-N when orbiting Earth
An 80-kg astronaut experience a force of 2890-N when orbiting Earth. How many g’s does he feel?
ac = F/m
2890 N / 80-kg = 36 m/s2.
36 m/s2 / 9.81 m/s2
= 3.7 g.

14. Vertical Circles

15. Swing keys in vertical circle. Gravity Fg acts vertically
Swing keys in vertical circle. Gravity Fg acts vertically. Sketch Free body diagrams top and bottom. What about a cart on a track?

16. At the top, Fc is combo of weight & Fn. Write the Fnet equation
At the top, Fc is combo of weight & Fn. Write the Fnet equation. Take center as +. Fn W
Fnet = Fc = mv2/r = mg + Fn.

17. Minimum v occurs just as Fn = 0, when you lose contact with your seat, you begin to fall. Fc = mv2/r = mg + Fn When Fn = 0, mv2/r = mg. So mass is irrelevant. Rearrange, solve for vmin, v = (gr)1/2. v depends on radius only.

18. R (FN) may also be called "apparent weight" this is the force of the seat on the rider and also describes what the rider "feels" .
If R is less than the rider's weight, rider will fall out. R might become negative, a safety restraint system -- seat belts, lap bars, shoulder restraints, are needed.

19. Tension in Ropes Swinging Keys
To find the required tension at the top of the arc: Write a free body Fnet expression:
Fc = T + mg.
Fc – mg = T
mv2/r – mg = T.
The tension is equivalent to R in the ride.

20. At the bottom the vel is max & Fc = T – W. Fc = T – mg. mv2/r = T – mg
At the bottom the vel is max & Fc = T – W Fc = T – mg. mv2/r = T – mg. The necessary tension to swing in a circle is: mv2/r + mg = T. W. T.

21. Ex 1: A 10 kg package is swinging in a vertical circle of radius 5
Ex 1: A 10 kg package is swinging in a vertical circle of radius 5.0 m at the end of a rope. If the ac is 5.0 m/s2, a. Sketch the free body diagram at the bottom of the loop. b. what is the tension in the rope when the package is at the bottom of the loop?

22. m (v2/r + g) = T. m(ac + g) = T. 10kg (5.0 m/s2 + 10 m/s2.) 150 N.
mg. T.
Fc = T – mg.
mv2/r + mg = T.
m (v2/r + g) = T.
m(ac + g) = T.
10kg (5.0 m/s m/s2.)
150 N.

23. Ex 2. A motorcyclist rides in a vertical circle of 20. 0 m radius
Ex 2. A motorcyclist rides in a vertical circle of 20.0 m radius. What is the minimum speed he must have at the top to complete the loop?

24. 14 m/s.
Hwk Packet.

25. Banked Curves

26. Wall of Death

27. Banked Curves.
Flat road only friction acts supplies Fc to pull car into curves.
Banked curves add R component toward the center.
If the curve is banked, there’s a critical speed vc, Ff =0, the car can turn without slipping out. 

28. Free Body Diagram Banked Curve

29. Normal Force Components sinq = Fc n tan q = Fc mg
This is NOT the same vector sketch as for Fll.

30. Fc = mac points horizontally toward the center of the curve
Fc = mac points horizontally toward the center of the curve. This component of the normal is supplying the Fc to keep the car moving through the banked curve.

31. If friction is present both the horizontal component of N and the Friction force contribute to Fc.
FN tan θ + Ff = Fc.

32. Read Hamper (purple book) pg 40- 42. Do handout problems Ex 2
Read Hamper (purple book) pg Do handout problems Ex 2.4 prb 1 – 5 and pg 143 #3.

33. IB Problem Packet Banked Curve