1. Circular Motion http://www.scicomics.com/uploads/centripetal_acceleration.jpg

2. Rotation vs Revolution Rotation occurs when an object rotates around an internal axis –ex: the earth rotates around its axis (internal) every day Revolution occurs when an object rotates around an external axis –ex: the earth revolves around the sun (external axis) every year

3. Rotational Speed also called angular speed equals the number of rotations per time unit –All parts of the earth rotate about its axis in the same amount of time…thus all parts of the earth have the same rotation …or the same number of rotations per time (revolutions per minute or RPM) When a twirling ice skater extends her arms outward, her rotational speed decreases.

4. Tangential Velocity While the speed of the object is constant, its velocity is changing. Velocity vector and position vector for an object executing uniform circular motion

5. Tangential Speed Tangential speed is directly proportional to rotational speed and the distance from the axis Tangential ~ Radial x Rotational Speed distance Speed -at the center of the earth…there is no tangential speed, (zero distance) but there is rotational speed

6. twice as fast fast What type of speed is changing – rotational or tangential?

7. Mathematical Relationship between tangential and rotational speed v = r ω v = tangential speed r = radial distance ω = rotational speed

8. Accelerometer The deflection of the flame will be in the direction of the acceleration. This is because the hot gases of the flame are less massive and thus have less inertia than the cooler gases which surround

9. Centripetal Force the force required to change the direction of a moving object v = tangential speed http://webpages.uah.edu/~wilderd/resources.html http://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm F c = mv 2 / r

10. Centripetal Force center–seeking or toward the center any force that causes an object to follow a circular path –think about the force you feel when riding a fast circular ride at an amusement park…without that force you would continue to move in a straight line…being “launched” form the ride

11. Centripetal Force As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion. As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion. As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

12. Centrifugal Force…False Force center-fleeing or away from the center An apparent outward force that acts on a rotating object…it is due to inertia…an object’s desire to continue moving in a straight path. The sensation of being thrown outward is attributable to the idea of inertia, rather than the idea of force.

13. Centrifugal force holds people on walls in Round Up ride

14. The Fictitious Force When a ball swings on a string, many people would say that "CENTRIFUGAL FORCE" keeps the string taut. –There is actually no force pulling the ball out. –There is nothing out there pulling the ball! –All the sideways and back and forth forces you feel while riding in a car are FICTITIOUS, they are the results of your inertia resisting acceleration http://www.scigolf.com/scigolf/myths/myth2.htm

15. The Spin Cycle in a Washing Machine The water goes through the holes at a tangent to the tub, while the clothes are moved inward in a circle. http://www.scigolf.com/scigolf/myths/myth2.htm

16. The Fictitious “Force”

17. http://www.geo.umn.edu/courses/1006/Fall01_night/Coriolis.html

18. The centripetal force of an aircraft is the resultant force of the two other forces acting on the aircraft: the weight of the aircraft and the lift of the aircraft.

19. Tether Ball Twirl a ball, at a constant speed, at the end of a rope. change of direction = acceleration http://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm http://www.hoopsplus.com/nss- folder/outdoorequipment/tether.jpg

20. Direction of Accelerated Motion Use vector addition to determine the resultant velocity: –It is acting inward; the same as the force! (Newton’s 2 nd ) http://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm

21. a c = v 2 / r a c = centripetal acceleration v = tangential velocity r = radius The greater an object’s velocity, the greater the centripetal acceleration. The smaller the curve of the path being traveled, the greater the centripetal acceleration. http://library.thinkquest.org/27948/carousel.html http://www.college- cram.com/study/physics/presentations/472http://www.college- cram.com/study/physics/presentations/472 http://www2.ignatius.edu/faculty/decarlo/Centripetal%20force.htm

22. Trial # 1 A 4.0 kg ball is attached to 0.70 meter string and spun at 2.0 meters/sec. Calculate a) centripetal acceleration a c = v 2 /r a c = (2.0 m/s) 2 /(0.7 m) a c = 5.7 m/s 2 b) centripetal force F c = ma c F c = (4.0 kg)(5.71 m/s2) F c = 23 N

23. Double the speed from the previous problem and perform the same calculations A 4.0 kg ball is attached to 0.70 meter string and spun at 4.0 meters/sec. Calculate a) centripetal acceleration (23 m/s 2 ) b) centripetal force (92 N) c) what happened to the force when the speed was doubled?

24. http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/circmot/rht.html

25. http://www.coolschool.ca/lor/PH12/unit4/U04L01.htm

26. FBD Consider an example of a person riding a roller coaster through a circular section of the track, a "loop-the-loop." At the TOP: net force to the center = N + mg N + mg = m(v2/r) N = m(v2/r) - mg At the BOTTOM: net force to the center = N - mg N - mg = m(v2/r) N = m(v2/r) + mg

27. Roller Coaster Physics –Many high-speed roller coasters use banked turns rather than the flat ones that are safe for slower speeds. Banking the turns in a roller coaster gives you the feeling of being pushed into your seat rather than being thrown to the side of the car. –acceleration stress: http://library.thinkquest.org/TQ0313004/Accel ertion_stress.html be sure to mouse over roller coaster photo http://library.thinkquest.org/TQ0313004/Accel ertion_stress.html

28. Physics of Roller Coasters and look at the forces on a victim -- er, I mean "guest" -- as the roller coaster goes through the top of a hill, the bottom of a valley, or at the top of a loop.

29. FBD Roller Coaster FBD

30. Opportunity: While driving to work you pass over a "crest" in the road that has a radius of 30 meters. How fast would you need to be traveling to experience apparent "weightlessness" while passing over the crest?

31. Answer: As seen in the following diagram, the NET FORCE to the center equals mg - N. If we let the normal approach 0 to represent apparent weightlessness, then

32. Roller Coaster G Forces http://www.glenbrook.k12.il.us/gbssci/phys/mm edia/circmot/rcd.html

33. Roller Coaster Problem Anna Litical is riding on The Demon at Great America. Anna experiences a downwards acceleration of 15.6 m/s 2 at the top of the loop and an upwards acceleration of 26.3 m/s 2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 864 kg roller coaster car. Given Info: m = 864 kg a top = 15.6 m/s2, down a bottom = 26.3 m/s2, up Find: F n at top and bottom http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2b.html

34. Solution to Roller Coaster Problem Bottom of Loop F net =ma F net = (864 kg)(26.3 m/s2, up) F net = 22 723 N, up Top of Loop F net = ma F net = (864 kg)(15.6 m/s2, down) F net = 13478 N, down Fn must be greater than the Fg by 22723 N in order to supply a net upwards force of 22723 N. Thus, F n = F g + F net F n = 31190 N Fn and Fg together must combine together (i.e., add up) to supply the required inwards net force of 13478 N. Thus, F n = F net - F g F n = 5011 N

35. Various Directions of Acceleration http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2b.html

36. Roller Coaster Problem # 2 Anna Litical is riding on The American Eagle at Great America. Anna is moving at 18.9 m/s over the top of a hill which has a radius of curvature of 12.7 m. Use Newton's second law to determine the magnitude of the applied force of the track pulling down upon Anna's 621 kg roller coaster car. Given Info: m = 621 kg v = 18.9 m/s r = 12.7 m Find: F A at top of hill http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2b.html

37. Solution: Accerleration a = v 2 / r a = (18.9 m/s) 2 / (12.7 m) a = 28.1 m/s 2 http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l1b.html

38. Solution: Force F net = ma F net = (621 kg) (28.1 m/s 2, down) F net = 17467 N, down F A = F net - F g F A = 11381 N F A and F g must combine together (i.e., add up) to supply the required downwards net force of 17467 N.

39. Practice An air puck of mass 0.025 kg is tied to a string and allowed to revolve in a circle of radius 1.0 m on a frictionless horizontal surface. The other end of the string passes through a hole in the center of the surface, and a mass of 1.0 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves on the surface. –A. what is the magnitude of the force that maintains circular motion acting on the puck? –B. What is the linear speed of the puck? http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html

40. Answer m p = 0.025 kg r = 1.0 m m = 1.0 kg g = 9.81 m/s 2 F c = F g = mg = (1.0 kg)(9.91 m/s 2 ) = 9.8 N F c = m p v 2 r v = 2.0 x 1.0 1 m/s

41. Practice An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between the person and the wall is s, and the radius of the cylinder is r. Draw the force diagram! http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html

42. Force Diagram http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html F n = m v 2 / r F f = mg F f = F n

43. http://visual.physics.tamu.edu/vp218/LabManual/Figure10_11.jpg

44. Gravity: force of attraction between two objects. Newton noted that the moon was falling toward the Earth. The moon falls along a curved path. It acts as a projectile. The moon has a component of velocity parallel to the Earth’s surface known as tangential velocity.

45. The force of gravitational attraction decreases as you move further from the center of the Earth. The moon orbits the Earth, just as the Earth & other planets orbit the Sun. The Sun is the center of the solar system.

46. Universal Gravitation Law of Universal Gravitation: Every object attracts every other object with a force that for any two objects is directly proportional to the mass of each object.

47. The rest is just for Fun and Interest: –Ocean Tides are due to the differences in the gravitational pull of the moon on opposite sides of the Earth. Variations in sun, moon and earth alignment causes low and high tides. http://oceanlink.island.net/oinfo/tides/tides.html http://home.hiwaay.net/~krcool/Astro/moon/moontides/

48. Tides SPRING TIDES: the sun and moon are aligned, there are exceptionally strong gravitational forces, causing very high and very low tides. NEAP TIDES: the sun and moon are not aligned, the gravitational forces cancel each other out, and the tides are not as dramatically high and low.

49. Center of Gravity (CG) Located at the object’s average position of weight (or balance point). For a symmetrical object such as a ball. The CG is located in the center. For an irregular shaped object (a bat), the CG is located towards the more massive end.

50. Toppling (falling over) If the CG is located above the area of support, the object will not fall. If it is outside the support area, the object will fall over. Neutral, Stable and Unstable Equilibrium.

51. Torque The turning force used to turn a door knob or tighten a bolt produces torque. Torque is defined as force times length of lever arm. For greatest torque, force should be applied perpendicular to the lever arm.

52. Notice that the unit of measurement for torque contains a distance (meter) and a force (Newton). To calculate the torque needed to turn something, multiply the force ┴ and the distance from the center of the object you are trying to turn. Torque = Force ┴ x distance

53. Challenge: Our moon moves in a circle so its direction changes continually. Therefore the moon accelerates due to directional changes; this is the centripetal acceleration. –Calculate the speed and the centripetal acceleration of the moon. a c = v 2 / r –What do you know? What do you need to know? How can you find the unknown, needed information? http://www.iop.org/activity/education/Teachin g_Resources/Teaching%20Advanced%20Ph ysics/Fields/Images%20400/img_tb_4641.gif

54. The Solution Use available sources to find the radius of moon and period of moon Determine the moon’s speed: d = 2 π r therefore: v = d / T (T = period or time for one revolution) v = 2 π r / T v = [2(3.14) ( 3.84 x 108 m ) ]/ (27.32 da)][da/24h][h/3600s] v = 1 022 m / s Calculate the acceleration of the moon: a c = v 2 / r a c = (1 022 m / s) 2 / 3.84 x 108 m a c = 2.72 x 10-3 m / s 2