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Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Dec 14, 2015

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Page 1: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Circular Motion

Page 2: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 3: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Homework :

Chapter 7

P 255- #48,49,50,55

Chapter 8

• P 295 - # 22, 24, 26,27,28

Page 4: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Read and Summarize • Read p 219-220, p 226 • Read p 96 Princeton Review Vocabulary 1. Uniform Circular Motion 2. Centripetal acceleration – 3. Equation for centripetal acceleration 4. Discuss Centripetal force p229

Page 5: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 6: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 7: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

3.

3. Equation of Centripetal Acceleration :

• ac= v2 / r

Page 8: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 9: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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.

Page 10: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 11: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 12: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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 .

Page 13: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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 = 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

Page 14: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Fw

FN

Page 15: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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.

a. 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 ?

Page 16: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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 ?

Page 17: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 1 a 2.24 m/s b. 3.16 m/s• 2. 16π2r/s2

• 3. FT = 44 N , 2.24 m/s

Page 18: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

1. 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

Page 19: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

2. 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

Page 20: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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.

Page 21: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 22: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

PROCEDURE EQUATIONS USED

Page 23: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Data Table loop Mass,

Marble,kg

Highest height , m

Total Energy , Joules

V marble , m/s

Ac, m/s2 Fc , NewtonsAnd Torque , Nm

1

2

Page 24: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

CALCULATIONS

Page 25: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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 ?

Page 26: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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 torque15. Symbol of torque16. Unit of Torque

Page 27: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

acceleration.

Page 28: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 29: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

13. Moment or lever arm • Perpendicular distance, m

Page 30: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 31: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 32: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 33: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Equilibrium

• Read p 265 -266 p 98

19. What is translational Equilibrium ?

20. What is rotational equilibrium ?

21. What are concurrent forces ?

22. What is mechanical equilibrium ?

23. What is static equilibrium ?

Page 34: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

Solving Torque Problems at Equilibrium1. Identify all Forces along x2. Identify all Forces along y3. Set up the equation EFx = 0 at equilibrium4. Set up the equation EFy = 0 at equilibrium 5. Choose a pivot point to set up equation for ET= 06. Torque at the pivot point is zero. 7. Torque = perpendicular Force X lever arm 8. Clockwise Torque is negative 9. Counterclockwise Torque is positive .

Page 35: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 36: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 37: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 38: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

20 cm 5cm

30g 40g 20g 10g

X

50g

Page 39: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Lτ= F sinθ X rτ= mg sin 50o X Lτ= 0.6 kg (10m/s2) sin 50o X 0.5 mτ= 2.3 Nm

m m

Page 40: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 41: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 42: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 43: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

1. Στ = τ1 + τ2 Στ = - F1 ( 0.75m) +F2 ( 0.35m) = - 3N ( 0.75m) + 5N (0.35m) = -2.25Nm + 1.75 Nm = -0.5 Nm

2. τ= 20 N X .2m = - 4 Nm3. 3kg(10m/s2) (0.5m) – M(10m/s2)(0.3m )=0 M= 5 kg

Page 44: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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

Page 45: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

31.• m1 m2

• F1-2 F2-1

• r

Page 46: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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)2

---------------------------------- 6.67 X 10-11 Nm2/kg2

M=6.1X 1024kg

Page 47: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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 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

Page 48: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 49: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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

Page 50: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

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?

Page 51: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• The centripetal force on the satellite is provided by the Earth’s gravitational pull.

• mV2/R = G mM/R2

• V = √ G M /R 2

Page 52: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• 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 ?

Page 53: Circular Motion. Focused Learning Target Given Circular Motion and Torque Problems, I will be able to calculate the centripetal acceleration, centripetal.

• mV2/R = G mM/R2

• v2 = GM/R• M = v2 R /G• M = (2πR/T)2 R / G• M = 4 π2 R3 /G T2