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# Centripetal Force

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Explains centripetal and centrifugal force and acceleration.
**More good stuff available at:
www.wsautter.com

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### Centripetal Force

1. 1. Copyright Sautter 2015
2. 2. Centripetal Force • Newton’s First Law tells us that only a force can cause a body to move out of a straight line path. In circular motion the direction of the body is continually changing at every instant. Therefore a force must be acting. That force is called centripetal (central) force since it acts toward the circle of the circular path. “Objects in motion tend to remain in motion, at the same rate, And in the same direction, unless acted on by an outside force” ALL CIRCULAR MOTION REQUIRES A CENTRIPETAL FORCE, OTHERWISE THE BODY CONTINUES IN A STRAIGHT LINE PATH. 2
3. 3. Centripetal Force • All circular motion requires a centripetal force. Newton’s Second Law of Motion tells us that force equal mass times acceleration. Therefore, centripetal force must produce an acceleration (centripetal acceleration). Since the force acts towards the center of the circular path, the acceleration must also be towards the center ! ALL CIRCULAR MOTION IS ACCELERATED MOTION. THE ACCELERATION IS ALWAYS TOWARDS THE CENTER OF THE CIRCULAR PATH. 3
4. 4. THE INSTANTANEOUS VELOCITY VECTOR IN CIRCULAR MOTION IS TANGENTIAL TO THE CIRCULAR PATH V1 V2 r r r r Distance traveled Over angle  S Similar triangles give S/ r = V / V Distance traveled (S) = V t Therefore V t / r = V / V Rearranging the equation gives V / t = V x V / r a = V / t = V 2 / r V1 V V2  Vector difference The smaller  gets, the better ac is approximated by ac= V2 / r 4
5. 5. Displacement Vector (Radius) Velocity Vector (Tangent to the path) Acceleration Vector (Towards the center) 5
6. 6. The velocity vector is always tangential to the circular path 6
7. 7. The acceleration vector is always towards the center of the circular path 7
8. 8. AVERAGE =  /  t = (2 + 1) / 2  = o t + ½ t2 i = o + t i = ½ (i 2 - o 2) /  s =  R Vlinear =  R alinear =  R f = 1/ T, T = 1 / f 1 revolution = 360 degrees = 2  radians  = 2  f ,  = 2  / T acentripetal = V2 / r Fcentripetal = m V2 / r 8
9. 9. Centripetal Force & Acceleration Problems A 1000 kg car rounds a turn of 30 meter radius at 9 m/s. (a) What is its acceleration ? (b) What is the centripetal force ? • (a) acentripetal = V2 / r = 9 2 / 30 = 2.7 m /s2 • (b) Fcentripetal = m V2 / r = m x ac = 1000 x 2.7 = 2700 nt r = 30 mv = 9 m/s 9
10. 10. Centripetal Force & Acceleration Problems A car is traveling at 20 mph on a level road with a coefficient of friction of 0.80. What is the maximum curve radius ? • In the English system velocity must be in ft/sec. • 20 mph x 5280ft / 3600 sec = 29.4 ft/sec. • The centripetal force which allows the car to round the curve is supplied by friction. • Ff =  Fn when the car is on level ground the normal force is the car’s weight w = mg • Centripetal force is given by mv2/r • Fc = Ff ,  mg = mv2/r , canceling mass from both sides leaves  g = v2/r and rearranging the equation, r = v2/ ( g) • R = (29.4)2/ ( 0.80 x 32) = 34 ft r = ? ftv = 20mph  = 0.80 10
11. 11. Centripetal Force & Acceleration Problems • Roadways are often banked to help supply centripetal force thereby allowing cars to execute curves more readily. • Without banking, the friction between the tires and the road, are the only source of centripetal force. • As the banking angle increases the amount of centripetal force the roadway supplies increases. • The following slide analyzes the relationship between the angle of banking and the centripetal force. 11