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 / rFc = mac = m v2 /rTorque T = radius r X ForceTorque T = Radius X Force sinθ
3 Homework : P 255- #48,49,50,55 P 295 - # 22, 24, 26,27,28 Chapter 7
4 Read and Summarize Read p 219-220, p 226 Read p 96 Princeton Review VocabularyUniform Circular MotionCentripetal acceleration –3. Equation for centripetal accelerationDiscuss Centripetal force p229
5 1. Uniform MotionType 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 motionCenter seeking accelerationConstant speed but changing direction constantlymust be directed radially inwardAcceleration vector points toward the centerDepends on the object’s speed, radius r
7 3. 3. Equation of Centripetal Acceleration : ac= v2 / r
8 4. Centripetal Force Fc = mac = m v2 /r Net inward force which causes an inward accelerationFc = mac = m v2 /rDirected toward the center of the circular pathAlways perpendicular to the direction of motionCould be theA. tension of the string ,B. friction orC. 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 Examples6. 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 = m/ s2Fc = 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 /rm = Fw / g = 800 N / 10 m/s2 = 80 kgFc = 80 kg ( 5m/s) 2 / 5m = 400 NFriction provides the centripetal force .The athlete will slip .
13 9. A roller coaster car enters the circular loop portion of the ride 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 =40mWeight provides v = 15m/s r = 20mCentripetal forceFc = FN + Fw = mac = m v2 / r = 1200 kg ( 15m/s)2/ 20m = 13,500 NFN = Fc – Fw = 13,500 N – mg= 13,500 N- 1200kg (10m/s2) = 1500 N
14 Centripetal Force is the combination of Normal force and weight . 10. How would the normal force change in example 9 if the car was at the bottom of the circle ?Centripetal Force is thecombination of Normal force andweight .Fc = FN- FwFN = Fc + Fw = 13, 500 N + 12,000N = 25,500 NFNFw
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 m/sπ2r/s23. FT = 44 N , 2.24 m/s
18 A. Fc = FT – FW = mV2/rFc = 40 N – 2kg ( 10m/s2)= 40 N -20N =20Nv = √(Fc r /m )v = √(20 N X 0.5 m) / 2kgV = m/sB. Fc = FT + FW = m v2 / rFc = kg (10m/s2) = N = 20 Nv = √( Fc r / m)v = √( 20 N X 0.5 m ) / 2 kgv = m/s
19 ac = V2 / r =1 revolution = 2πrac= ( 2 X 2πr/ s) 2 /r = 16 π2r2 /s2 r = 16π2r/s23. Fc = FT – FW = mv2 /rFT = mV2/r + FwFT = ( 2kg ( 4m/s)2 / 0.5 m ) + 2kg ( 10m/s2)FT = 84 NFc = mg FT = 0Fc = 2 kg ( 10m/s2)= 20 Nv =√ Fc r /mv = √ 20 N X 0.5 m / 2kg = 2.24 m/s
20 Experiment : Roller Coaster PURPOSETo 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 trackTo calculate the potential energy and the kinetic energy of the marble for each trackTo calculate the centripetal acceleration and centripetal force for 1 loop and 2 loop track.
25 ANALYSIS AND CONCLUSIONS COMPARE THE 1 LOOP AND 2 LOOP TRACK ROLLER COASTERCOMPARE 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 Torque13. Moment arm or lever arm14. Equation of torque15. Symbol of torque16. Unit of Torque
27 11. TorqueIs the effectiveness of a force in producing rotational acceleration.
28 12. Factors affecting Torque A. Magnitude of the forceB. perpendicular distance from the axis of rotation
29 13. Moment or lever armPerpendicular distance, m
30 15. Torque is tau τ 16 . Unit of Torque is N-m 14. Equation of Torque Torque T = radius r X ForceTorque T = Radius X Force sinθTorque is tau τ16 . Unit of Torque is N-m
31 EXAMPLES17. 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 Nτ= r X F = m X 40 N = 2 Nm5 cm
32 Counterclockwise + F1 =100N Clockwise - F2 = 80 N τ= τ1 –τ2 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 =100NClockwise F2 = 80 Nτ= τ1 –τ2τ= 100N (0 .12m)- 80N ( 0.08 m)τ= +5.6 N8cm12cm
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 xIdentify all Forces along ySet up the equation EFx = 0 at equilibriumSet up the equation EFy = 0 at equilibriumChoose a pivot point to set up equation for ET= 0Torque at the pivot point is zero.Torque = perpendicular Force X lever armClockwise Torque is negativeCounterclockwise Torque is positive .
35 Examples on Torque24. 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 .55om 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 = Sin kg(10m/s2) + 8kg (10m/s2) = N Fc = √ (-80.2 )2 = N
37 Choose a pivot point or fulcrum ; Torque on the pivot point =0 Στ=0 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 =0Στ=0+τboy – τgirl = 0Fwboy ( 0.6 m ) - Fwgirl (X) = 045 kg(10m/s2) (0.6m) = 60kg(10m/s2) X X = 270 Nm/ 600kgX = 0.45 m0.6mX
38 Pivot point = center of the bar ; torque of the weight of the bar=0 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=040 cm cmΣτ=00.030 kg (10m/s2) .4m kg (10m/s2) .2m kg(10m/s2)0.05m – 0.01kg(10m/s2) X – 0.05kg(10m/s2)(.4m)= .01 X x = 0.1m or 10 cmX20 cm5cm30g40g20g10g50g
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 : TORQUE1. 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 =3NF2= 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 + τ2Στ = - F1 ( 0.75m) +F2 ( 0.35m)= - 3N ( 0.75m) + 5N (0.35m)= Nm Nm= Nmτ= 20 N X .2m = - 4 Nm3kg(10m/s2) (0.5m) – M(10m/s2)(0.3m )=0M= 5 kg
44 Newton’s Law of Gravitation 27. gravitation force -Any two objects in the universe exert an attractive force on each other28. G – is the Universal Gravitational Constant =6.67 X Nm2/kg229. 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 .FG = Gm1m2 / r2
46 Examples32. 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) X 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 1024 kg )Let m – satellite’s massM- the mass of the earthR – distance between the center of the earth and the center of the satelliteFG = Fc = mv2 / r
48 M mRFcFG = G Mm/ R ( 2πR/T )2 = GM/RFG = Fc = mv2 / R π2R2/T2 = GM/Rmv2 / R = G Mm/ R π2R3 /T2 = GMV2 = GM /R R = √ GMT2 / 4π2
49 Cube rootR = √ ( 6.67 X 10-11)( 5.98 X 1024) (24 X 60 X 60 ) 2 / 4π2R = X 107mIf the radius of the earth is 6.37 X 106 m , the altitude of the satellite above the earth isH = X 107m-6.37 X 106 m = 3.59 X 10 7 m
50 CW1. 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?
51 The centripetal force on the satellite is provided by the Earth’s gravitational pull. mV2/R = G mM/R2V = √ G M /R 2
52 2. A moon of Jupiter has a nearly circular orbit of radius R and an orbit period of T. What is the mass of Jupiter ?
53 mV2/R = G mM/R2v2 = GM/RM = v2 R /GM = (2πR/T)2 R / GM = 4 π2 R3 /G T2
Your consent to our cookies if you continue to use this website.