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# Uniform Circular Motion

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Uniform Circular Motion
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### Uniform Circular Motion

1. 1. • Francis Marlon Cabredo• Mary Minette Geñorga• Mary Judith Verdejo
2. 2. ▪Uniform CircularMotion is themotion of an objectin a circle with aconstant oruniform speed
3. 3. SPEED:▪ In UCM, distance= the circumference ofthe circle▪ T(period) is the time (or number ofseconds) to make one revolution
4. 4. PERIOD andFREQUENCY
5. 5. • Period (T) -> the time it takes for an object to make onecomplete revolution {#sec/rev}PERIOD and FREQUENCY• Frequency (f)  the number of revolutions completed byan object in a given time {#rev/sec}Units: Hertz (Hz) or 1 rev/secrpm (#rev/min); rps (#rev/sec)
6. 6. PERIOD and FREQUENCY
7. 7. SPEED expressed inPERIOD and FREQUENCYFORMULA(s):
8. 8. PROBLEM 1: Tire Balancing Machine• The wheel of a car has a radius of0.29m and is being rotated at 830revolutions per minute on a tire-balancing machine. Determine thespeed (in m/s) at which the outer edgeof the wheel is moving. Use 3.14 as thevalue for .
9. 9. SOLUTION#1:f=830 revolutions in one minuterevolutionsrevolutionmin/102.1min/8301 3T= 0.072sGiven:r=0.29 mFind:TV1.2x10-3 min___________revolution x 60 sec___________1 min
10. 10. SOLUTIONsmsmTrv /25072.0)29.0(22ANSWER
11. 11. SOLUTION #2:830 revolutions in one minutev= 25 m/sGiven:r=0.29 mFind:V2(3.14)(0.29 m)(830 revolutions)x 1 min___________60 sec_______________________1 minute
12. 12. VELOCITY▪ The velocity of an object in UCM has a constantmagnitude and a constant change in direction▪ The object does not have constant velocity sinceits direction changes at every point along thecircle.▪ Also, The velocity is always tangent to the path ofthe object.
13. 13. In UCM, the velocity is always tangent to the path of theobject. The tangent specifies the direction of the motion.VAVDVBVCAt a time t0, the car is locatedat point A with a velocity ofVA, this is tangent to the circleat point A.This means that…The direction of the velocity ofthe Car at point A is due west.
14. 14. CENTRIPETAL ACCELERATION▪ in UCM, an object has no constant velocity.Therefore, If there is a change in velocity, then that must mean that anobject has acceleration.Centripetal Acceleration. This acceleration hasconstant magnitude but changing direction and isdirected radially inwards.
15. 15. CENTRIPETAL ACCELERATION• Symbol: ac• Vector• It is the rate of change ofTangential velocity• Always perpendicular to thepath of the motion.• Points toward the centerof the circle.VAVDVBVCABCD
16. 16. CENTRIPETAL ACCELERATIONFORMULA(S):2222244rfTrrvac 
17. 17. PROBLEM 2: Centripetal AccelerationThe bobsled track at the 1994Olympics in Norway, contained turnswith radii of 33 m and 24 m, as thefigure illustrates. Find the centripetalacceleration at each turn for a speedof 34 m/s, a speed that was achievedin the two-man event..
18. 18. From ac=v2/rRadius=33mRadius=24m22/3533)/34(smmsmac 22/4824)/34(smmsmac 
19. 19. (a) The velocity is due south, andthe acceleration is due west.(b) The velocity is due west, andthe acceleration is due north.(c) ac=5.48 m/s2
20. 20. 21CENTRIPETAL FORCE
21. 21. NEWTON’S SECOND LAW OF MOTION STATESTHAT ALL ACCELERATIONS ARE CAUSED BY ANET FORCE ACTING ON AN OBJECT. IN THECASE OF UCM, THE NET FORCE IS A SPECIALFORCE CALLED THE CENTRIPETAL FORCE .
22. 22. CENTRIPETAL IS LATIN FOR "CENTER SEEKING".IT IS THE INWARD NET FORCE WHICHKEEPS AN OBJECT MOVING WITH AUNIFORM VELOCITY ALONG A CIRCULARPATH. THIS FORCE IS DIRECTED ALONGTHE RADIUS TOWARDS THE CENTER.CENTRIPETAL FORCE
23. 23. F maFmvr2•APPLYING NEWTON’SSECONDLAW OF MOTIONCENTRIPETAL FORCECentripetal Forceac=v2/rF=mamv2___r
24. 24. NewtonN=kg m/s2UNIT USED:
25. 25. PROBLEM 3A: Centripetal Force• A 400-g rock attached to a 1.0-m stringis whirled in a horizontal circle at aconstant speed of 10.0m/s. Neglectingthe effects of gravity, what is thecentripetal force acting on the rock?
26. 26. SOLUTION:Given:V=10.0 m/sR=1.0 mM=400g= 0.4 kgFind: Fc(0.4 kg)(10 m/s)2___________1.0 mmv2___r40 N40 kg m/s2
27. 27. PROBLEM 3B: Centripetal Force• How long does it take a 50 kg runner torun a circular track starting and endingat the same point, if the radius of thetrack is 30m and a force of 68 N keepshim running at constant speed in thecircular path?
28. 28. SOLUTION:Given:M=50 kgR=30 mFc=68 NFind: Tmv2___r29. 50 s2r__Tm( )2___________r___________T2___________T2________T
29. 29. • FORCES SUCH AS THE GRAVITATIONAL FORCE (w=mg), TENSIONFORCE (tied to a string; pushing/ pulling), FRICTIONAL FORCE (Ex. a carturning) and Normal force (on a surface) can be the centripetal forces• GRAVITATIONAL FORCE. For satellites in orbit around a planet, thecentripetal force is supplied by gravity.• TENSILE FORCE. For an object swinging around on the end of a rope in ahorizontal plane, the centripetal force on the object is supplied by thetension of the rope.
30. 30. DRAWING A FREE BODYDIAGRAM CAN HELP YOUUNDERSTAND THE PROBLEMBETTER.
31. 31. • A free body diagram, is a pictorial device, used to analyze the forcesand moments acting on a body.What is included:1. The body2. The external forces: These are indicated by labelled arrows.The forces acting on the object include friction, gravity, normalforce, drag, tension, or a human force due to pushing orpulling.FREE BODY DIAGRAMFORCE DIAGRAM/
32. 32. FG
33. 33. FNFG
34. 34. FNFGFT
35. 35. FNFGFTFf
36. 36. FREE BODY DIAGRAMFNFGFTFf
37. 37. PROBLEM 4: Friction as Centripetal ForceA car is going around in a circular road.Given R, Ff b/w the tires and the roadand m ; Find v.
38. 38. FREE BODY DIAGRAMFN(on asurface)Fg (weight)Ff (friction)Ff is the force going to thedirection of theacceleration, therefore:It is the one causing thecentripetal acceleration
39. 39. ANSWERV=10 m/s
40. 40. THE CENTRIFUGAL FORCE ACTS AWAY FROM THE CENTER.THE WORD ITSELF MEANS “FLEEING FROM THE CENTER” .THIS FORCE IS A FICTITIOUS FORCE. IT DOESN’T ACTON A BODY IN MOTION, BUT ONLY ON NON-INERTIALCOORDINATE SYSTEMS SUCH AS A ROTATING ONE.CENTRIFUGAL FORCE
41. 41. PROBLEM 5: Which way will the object go?• An object on a guideline is in uniform circularmotion. The object is symbolized by a dot, and atpoint O it is release suddenly from its circular path.• If the guideline is cut suddenly, will the object movealong OA or OP ?
42. 42. ANSWER:• the object would move along thestraight line between points Oand A, not on the circular arcbetween points O and P.
43. 43. SOLUTION:• NEWTON’S LAW OF MOTION:“An object continues in a state of rest/ motion at a constantspeed unless compelled to changes to its net force.” When the object was suddenly released from its path, there was no longer a netforce (i.e. centripetal force) being applied to the object. In the absence of a net force, the object will continue to move ata constant speed but, along a straight line in the direction it hadat the time of release.
44. 44. CENTRIFUGAL FORCE vs. INERTIAIf you let go of the rope (orthe rope breaks) the objectwill no longer be kept inthat circular path and it willbe free to fly off on atangent.
45. 45. 53Uniform circular motionemphasizes that1.The speed, or the magnitude of the velocity vector, is constant.2. Direction of the vector is not constant.3. Change in direction, means acceleration4. “Centripetal acceleration” , points toward the center of thecircle.5. “Centripetal Force” is the net force that causes centripetalacceleration
46. 46. (a)T= 1__f
47. 47. (b)Hertz (Hz) = 1 rev/sec
48. 48. (c)
49. 49. rvac2(d)
50. 50. 5g=0.005 kg(e)mv2___r
51. 51. AP PHYSICS Circular Motion Mrs. CoylePractical ad Explorational Physics Padua et alYou and The Natural world Physics Navaza, ValdesINTERNETYOUTUBEGOOGLE……..SOURCES:• Francis Marlon Cabredo• Mary Minette Geñorga• Mary Judith VerdejoAPPLICATIONSPROBLEMS