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Circular Motion

Circular Motion. Focused Learning Target. Given Circular Motion and Torque Problems , I will be able to calculate the centripetal acceleration , centripetal force , rotational speed and torque using the following equations: a c = v 2 / r F c = ma c = m v 2 /r

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Circular Motion

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  1. Circular Motion

  2. Focused Learning Target • Given Circular Motion and Torque Problems , I will be able to calculate the centripetal acceleration , centripetal force , rotational speed and torque using the following equations: • ac= v2 / r • Fc = mac = m v2 /r • Torque T = radius r X Force • Torque T = Radius X Force sinθ

  3. Homework : Chapter 7 Chapter 8 P 295 - # 22, 24, 26,27,28 P 255- #48,49,50,55

  4. Read and Summarize • Read p 219-220, p 226 • Read p 96 Princeton Review Vocabulary • Uniform Circular Motion • Centripetal acceleration – 3. Equation for centripetal acceleration • Discuss Centripetal force p229

  5. 1. Uniform Motion • Type of circular motion which occurs when an object moves at constant speed in a circular path . • Circular motion which Object’s speed around its path remains constant with changing velocity or changing direction

  6. 2. Centripetal Acceleration • Acceleration in a uniform circular motion • Center seeking acceleration • Constant speed but changing direction constantly • must be directed radially inward • Acceleration vector points toward the center • Depends on the object’s speed, radius r

  7. 3. 3. Equation of Centripetal Acceleration : • ac= v2 / r

  8. 4. Centripetal Force • Net inward force which causes an inward acceleration • Fc = mac = m v2 /r • Directed toward the center of the circular path • Always perpendicular to the direction of motion • Could be the • A. tension of the string , • B. friction or • C. caused by gravity

  9. 5.Critical Point – CRITICAL VELOCITY • Critical point occurs at the highest point of the circular motion . • The velocity at the highest point of a vertical uniform circular motion is Critical Velocity. • Critical point or critical velocity happens when the tension on the string is 0 N . This is the speed necessary to maintain circular motion.

  10. Examples • 6. An object of mass 5 kg at a constant speed of 6 m/s in a circular path of radius 2m . Find the object’s acceleration and the net force responsible for this motion . • ac= v2 / r = ( 6 m /s)2 / 2m = 18 m/ s2 • Fc = mac = 5kg ( 18m/s2) = 90 N

  11. 7. A 10 kg mass is attached to a string that has a breaking point of 200 N . If the mass is whirled in a horizontal circle of radius 80 cm , what is the maximum speed can it have? Tension on the string provides the centripetal force FT = Fc = m ac = m v2 / r v = √ Fc r /m v = √ ( 200 N ) ( 0.8 m ) / 10 kg v = 4m/s

  12. 8. An athlete who weighs 800 N is running around a curve at a speed of 5m/s in an arc whose radius of curvature r, is 5 m . Find the centripetal force acting on him . What provides the centripetal force? What could happen to him if r were smaller ? Fc = m ac = m V2 /r m = Fw / g = 800 N / 10 m/s2 = 80 kg Fc = 80 kg ( 5m/s) 2 / 5m = 400 N Friction provides the centripetal force . The athlete will slip .

  13. 9. A roller coaster car enters the circular loop portion of the ride. At the very top of the circle( where the people in the car are upside down) the speed of the car is 15 m/s and the acceleration points straight down. If the diameter of the loop is 40m and the total mass of the car (plus the passengers) is 1200 kg , find the magnitude of the normal force exerted by the track on the car at this point. • Normal Force and m = 1200kg d =40m Weight provides v = 15m/s r = 20m Centripetal force Fc = FN + Fw = mac = m v2 / r = 1200 kg ( 15m/s)2/ 20m = 13,500 N FN = Fc – Fw = 13,500 N – mg= 13,500 N- 1200kg (10m/s2) = 1500 N

  14. 10. How would the normal force change in example 9 if the car was at the bottom of the circle ? Centripetal Force is the combination of Normal force and weight . • Fc = FN- Fw • FN = Fc + Fw= 13, 500 N + 12,000N = 25,500 N FN Fw

  15. CW: 1.A 50 cm rope is tied to the handle of the bucket which is then whirled in a vertical circle. The mass of the bucket is 2 kg. • What is the speed of the bucket if the tension of the rope at the lowest point of its path is 40 N ? b. What is the critical speed if the rope would become slack when the bucket reaches the highest point in the circle ?

  16. 2. An object moves at a constant speed in a circular path of radius r at a rate of 2 revolution per second . What is its acceleration in terms of r and π ? 3. An object m =2 kg at the end of the string is whirled around in the vertical circle. The circular motion has a radius of 0.5 m. If the speed of the object is 4m/s at the bottom of the circle a. What is the tension of the string at that point ? b. What is the minimum speed necessary for the object to obtain circular motion ?

  17. 1 a 2.24 m/s b. 3.16 m/s • 2. 16π2r/s2 • 3. FT = 44 N , 2.24 m/s

  18. A. Fc = FT – FW = mV2/r Fc = 40 N – 2kg ( 10m/s2)= 40 N -20N =20N v = √(Fc r /m ) v = √(20 N X 0.5 m) / 2kg V = 2.24 m/s B. Fc = FT + FW = m v2 / r Fc = 0 + 2kg (10m/s2) = 0 + 20 N = 20 N v = √( Fc r / m) v = √( 20 N X 0.5 m ) / 2 kg v = 2.24 m/s

  19. ac = V2 / r = 1 revolution = 2πr ac= ( 2 X 2πr/ s) 2 /r = 16 π2r2 /s2 r = 16π2r/s2 3. Fc = FT – FW = mv2 /r FT = mV2/r + Fw FT = ( 2kg ( 4m/s)2 / 0.5 m ) + 2kg ( 10m/s2) FT = 84 N Fc = mg FT = 0 Fc = 2 kg ( 10m/s2)= 20 N v =√ Fc r /m v = √ 20 N X 0.5 m / 2kg = 2.24 m/s

  20. Experiment : Roller Coaster • PURPOSE • To allow the marble to complete the 1 and 2 loop track from start to finish . • To calculate the speed of the marble form start to finish on 1 and 2 loop track • To calculate the potential energy and the kinetic energy of the marble for each track • To calculate the centripetal acceleration and centripetal force for 1 loop and 2 loop track.

  21. MATERIALS • foam track • Marbles • Stopwatch • Meterstick • Calculator • Triple beam balance

  22. PROCEDURE EQUATIONS USED

  23. Data Table

  24. CALCULATIONS

  25. ANALYSIS AND CONCLUSIONS • COMPARE THE 1 LOOP AND 2 LOOP TRACK ROLLER COASTER • COMPARE YOUR ROLLER COASTER WITH 2 OTHER GROUPs’ ROLER COASTER WITH DIFFERENT STARTING HEIGHT . COMPARE THEIR RESULTS AND YOURS. HOW DOES THE START HEIGHT AFFECTS ALL OF THE VARIABLES ?

  26. 11. Torque • Read 264 and p 95-96 11. Definition of Torque 12. Factors affecting Torque 13. Moment arm or lever arm 14. Equation of torque 15. Symbol of torque 16. Unit of Torque

  27. 11. Torque • Is the effectiveness of a force in producing rotational acceleration.

  28. 12. Factors affecting Torque • A. Magnitude of the force • B. perpendicular distance from the axis of rotation

  29. 13. Moment or lever arm • Perpendicular distance, m

  30. 14. Equation of Torque • Torque T = radius r X Force • Torque T = Radius X Force sinθ 15. Torque is tau τ 16 . Unit of Torque is N-m

  31. EXAMPLES • 17. A student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm . What is the torque of this force ? • F = 40 N • τ= r X F = 0.05 m X 40 N = 2 Nm 5 cm

  32. 18. What is the net torque on the cylinder which is pinned at the center of the two forces , F1 upward force , 12 cm rightward from the center and F2 upward force ,8 cm leftward from the center. • Counterclockwise + F1 =100N • Clockwise - F2 = 80 N • τ= τ1 –τ2 • τ= 100N (0 .12m) • - 80N ( 0.08 m) • τ= +5.6 N ° 8cm 12cm

  33. Equilibrium • Read p 265 -266 p 98 • What is translational Equilibrium ? 20. What is rotational equilibrium ? 21. What are concurrent forces ? 22. What is mechanical equilibrium ? 23. What is static equilibrium ?

  34. Solving Torque Problems at Equilibrium • Identify all Forces along x • Identify all Forces along y • Set up the equation EFx = 0 at equilibrium • Set up the equation EFy = 0 at equilibrium • Choose a pivot point to set up equation for ET= 0 • Torque at the pivot point is zero. • Torque = perpendicular Force X lever arm • Clockwise Torque is negative • Counterclockwise Torque is positive .

  35. Examples on Torque • 24. A uniform bar of mass m and the length L extends horizontally from a wall. A supporting wire connects the wall to the bar’s midpoint , making an angle of 55o with the bar. A sign of mass M hangs from the end of the bar . • If the system is in static equilibrium and the wall has friction , determine the tension in the wire and the strength of the force exerted on the bar by the wall if M =8kg and m=2 kg . • 55o • m M

  36. ΣFx =0 Fcx – FTX =0 Fcx – FT Cos 55o= 0 Σfy= 0 Fcy + FTY -mg – Mg Fcy + FT Sin 550 –mg-Mg =0 Στ=0 FTy ( L/2) – mg(L/2) – Mg(L) =0 Fty (L/2) = gL/2 (m + 2M) FTY = g(m+2M) FT Sin 55o = 10m/s2 ( 2kg + 2 (8kg)) FT = 180 N / 0.819= 220 N Fcx = FT Cos 550 = 220 N ( 0.57) = 126 N Fcy = - 220 Sin 55 + 2kg(10m/s2) + 8kg (10m/s2) = - 80.2 N Fc = √1262 + (-80.2 )2 = 149.36 N

  37. 25. A 45 kg boy is sitting on a seesaw 0.6 m from the balance point as shown below. How far , on the other side should a60 kg girl sit so that seesaw will remain in balance ? Counter clockwise + • Clockwise – • Choose a pivot point or fulcrum ; Torque on the pivot point =0 • Στ=0 • +τboy – τgirl= 0 • Fwboy ( 0.6 m ) - Fwgirl (X) = 0 • 45 kg(10m/s2) (0.6m) = 60kg(10m/s2) X X = 270 Nm/ 600kg • X = 0.45 m 0.6m X

  38. 26. A balanced stick is shown below . The distance from the fulcrum is shown for each mass except the 10 g mass. What is the approximate position of the 10g mass, based on the diagram ? • Pivot point = center of the bar ; torque of the weight of the bar=0 • 40 cm 40 cm • Στ=0 • 0.030 kg (10m/s2) .4m + 0.04kg (10m/s2) .2m + 0.02kg(10m/s2) • 0.05m – 0.01kg(10m/s2) X – 0.05kg(10m/s2)(.4m) • .012 + .008 +.001 = .01 X + .02 x = 0.1m or 10 cm X 20 cm 5cm 30g 40g 20g 10g 50g

  39. 27. What is the torque about the pendulum’s suspension point produced by the weight of the bob , given that the length of the pendulum ,L is 50 cm and m = 0.6kg? 50o L τ= F sinθ X r τ= mg sin 50o X L τ= 0.6 kg (10m/s2) sin 50o X 0.5 m τ= 2.3 Nm m m

  40. CW : TORQUE • 1. A solid cylinder consisting of an outer radius R1 and an inner radius R2 is pivoted on a frictionless axle as shown below. A string is wound around the outer radius and is pulled to the right with the force F1=3N. A second string is wound around the inner radius and is pulled down with the force F2=5N . If R1 = 0.75 m and R2= 0.35m , what is the net torque acting on the cylinder? • F1 =3N F2= 5N

  41. 2. In an effort to tighten a bolt , a force F is applied as shown in the figure above . If the distance from the end of the wrench to the center of the bolt is 20 cm and F = 20 N , • What is the magnitude of the torque produced by F? • F

  42. 3. A uniform meter stick of mass 1 kg is hanging from a thread attached at the stick’s midpoint. One block of mass m =3kg hangs from the left end of the stick and another block , of unknown mass M, hangs below the 80 cm mark on the meter stick. If the stick remains at rest in the horizontal position shown above, what is M ? • m M

  43. Στ = τ1 + τ2 Στ = - F1 ( 0.75m) +F2 ( 0.35m) = - 3N ( 0.75m) + 5N (0.35m) = -2.25Nm + 1.75 Nm = -0.5 Nm • τ= 20 N X .2m = - 4 Nm • 3kg(10m/s2) (0.5m) – M(10m/s2)(0.3m )=0 M= 5 kg

  44. Newton’s Law of Gravitation 27. gravitation force -Any two objects in the universe exert an attractive force on each other 28. G – is the Universal Gravitational Constant = • 6.67 X 10-11 Nm2/kg2 29. Newton’s Law of Gravitation : Any two objects in the universe exert an attractive force on each other called gravitational force whose strength is proportional to the product of the object’s masses and inversely proportional to the square of the distance between them as measured from center to center . 30. FG = Gm1m2 / r2

  45. 31. • m1 m2 • F1-2 F2-1 • r

  46. Examples 32. Given that the radius of the earth is 6.37 X 10 6 m , determine the mass of the earth. M m r FG = Gm1m2 / r2 FG= G Mm/r2 mg= GMm/r2 M= gr2/G = (10m/s2)(6.37 X 10 6 m)2 ---------------------------------- 6.67 X 10-11 Nm2/kg2 M=6.1X 1024kg

  47. 33. Communications satellites are often parked in geosynchoronous orbits above the Earth’s surface. These satellites have orbit periods that are equal to earth’s rotation period, so they remain above the same position on Earth’s surface. Determine the altitude that a satellite must have to be in geosynchoronous orbit above a fixed position on earth’s equator. ( the mass of the earth is 5.98 X 1024kg ) • Let m – satellite’s mass • M- the mass of the earth • R – distance between the center of the earth and the center of the satellite FG = Fc = mv2 / r

  48. M m • R • Fc • FG = G Mm/ R2 ( 2πR/T )2 = GM/R • FG = Fc = mv2 / R 4π2R2/T2 = GM/R • mv2 / R = G Mm/ R2 4π2R3 /T2 = GM • V2 = GM /R R = √ GMT2 / 4π2

  49. Cube root • R = √ ( 6.67 X 10-11)( 5.98 X 1024) (24 X 60 X 60 ) 2 / 4π2 • R = 4.23 X 107m • If the radius of the earth is 6.37 X 106 m , the altitude of the satellite above the earth is H = 4.23 X 107m-6.37 X 106 m = 3.59 X 10 7 m

  50. CW • 1. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius R around the earth (mass M). What is the speed of the satellite?

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