Centripetal Force
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Centripetal Force

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Explains centripetal and centrifugal force and acceleration.

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

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    Centripetal Force Centripetal Force Presentation Transcript

    • Centripetal Force Copyright Sautter 2003
    • The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www. wsautter .com
    • Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
    • 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.
    • 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 !
      The Second Law F = MA ALL CIRCULAR MOTION IS ACCELERATED MOTION. THE ACCELERATION IS ALWAYS TOWARDS THE CENTER OF THE CIRCULAR PATH.
    • THE INSTANTANEOUS VELOCITY VECTOR IN CIRCULAR MOTION IS TANGENTIAL TO THE CIRCULAR PATH 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 Deriving the Centripetal Acceleration Equation The smaller  gets, the better a c is approximated by a c = V 2 / r V 1 V 2 r r  r r  Distance traveled Over angle  S V 1  V V 2  Vector difference
    • Displacement Vector (Radius) Velocity Vector (Tangent to the path) Acceleration Vector (Towards the center) Vectors in Circular Motion
    • Circular Motion with Velocity Vector The velocity vector is always tangential to the circular path
    • Circular Motion with Velocity & Acceleration Vectors The acceleration vector is always towards the center of the circular path
    • Summary of Rotational Motion Equations  AVERAGE =  /  t = (  2 +  1 ) / 2  =  o t + ½  t 2  i =  o +  t  i = ½ (  i 2 -  o 2 ) /  s =  R V linear =  R a linear =  R f = 1/ T, T = 1 / f 1 revolution = 360 degrees = 2  radians  = 2  f ,  = 2  / T a centripetal = V 2 / r F centripetal = m V 2 / r
    • 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) a centripetal = V 2 / r = 9 2 / 30 = 2.7 m /s 2
      • (b) F centripetal = m V 2 / r = m x a c = 1000 x 2.7 = 2700 nt
      r = 30 m v = 9 m/s
    • 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.
      • F f =  F n when the car is on level ground the normal force is the car’s weight w = mg
      • Centripetal force is given by mv 2 /r
      • F c = F f ,  mg = mv 2 /r , canceling mass from both sides leaves  g = v 2 /r and rearranging the equation, r = v 2 / (  g)
      • R = (29.4) 2 / ( 0.80 x 32) = 34 ft
      r = ? ft v = 20mph  = 0.80
    • 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.
    • tan  = F c / W F centripetal = W x tan  Banked Curves & Centripetal Force reaction to normal force normal force weight centripetal force  F c W F n  Vector diagram
    • Centripetal Force & Acceleration Problems What is the angle of banking which would allow vehicles to round a curve of 1000 foot radius at 50 mph without friction ?
      • 50 mph x 5280 ft / 3600 sec = 73.5 ft/sec
      • The centripetal force supplied by the banking.
      • F banking = w x tan  = mg tan  , F c = m v 2 /r
      • F banking = F c , mg tan  = m v 2 /r , canceling mass gives g tan  = v 2 /r and tan  = v 2 / (g x r)
      • tan  = 73.5 2 / (32 x 1000) = 0.169
      • tan -1 (0.169) = 9.6 0
      r = 1000 ft v =50 mph 
    • Centripetal Force & Acceleration Problems
      • Centripetal force problems often involve objects whirled in a circle. An example is a rock tied to a string. The circular path may be in the horizontal plane or the vertical plane.
      • The rock moves in a circular path because of the pull of the string to which it is tied. The string supplies the centripetal force necessary for the rock to be moved from its inertial straight line path into a circular path. The pull of the string is referred to as tension in the string.
      • When moving in a horizontal path the forces acting (the tension, the weight and centripetal force) remain constant in magnitude
      • When moving in a vertical path the forces acting vary from point to point during the travel.
    • Motion in a Vertical Circle Centripetal force (F c ) weight weight Centripetal force (F c ) Centripetal force (F c ) Centripetal force (F c ) weight weight tension tension tension tension At the top of the path Weight & F c are Opposite in direction At the bottom of the path Weight & F c are in the same direction
    • Circular Motion in a Vertical Plane A string 0.50 meters long is used to whirl a 1 kg rock in a vertical circle at 5.0 m/s. (a) Find the tension at the top of the circle (b) Find the tension at the bottom.
      • (a) T + w = F c , T + mg = mv 2 /r, T + (1 x 9.8) = 1 (5.0) 2 / 0.50
      • T + 9.8 = 25 / 0.50, T = 40.2 nt (downward) or – 40.2 nt
      • (b) T = F c + w , T = mv 2 /r + mg = T = 1 (5.0) 2 / 0.50 + 1 (9.8)
      • T = 59.8 nt (upward) or + 59.8 nt
      At the top T + w = F c At the bottom T = F c + w Downward vectors are – Upward vectors are + F c = m v 2 / r , w = mg weight weight tension tension Centripetal force (F c ) Centripetal force (F c )
    • Circular Motion in a Vertical Plane A 2 lb ball on the end of a string is spun in a vertical circle at 10 ft/s. The radius of the circle is 3.0 ft. What is the maximum tension in the string ?
      • T = F c + w = mv 2 /r + w, w = mg, m =w/g
      • m = 2/32 slugs
      • T = ((2/32) x 10 2 ) / 3) + 2 = 4.08 lbs.
      The maximum tension Occurs at the bottom Of the circle & equals The weight + centripetal force weight = 2 lbs tension Centripetal force (F c ) v = 10 ft/s
    • Circular Motion & Satellites
      • Satellites orbiting the earth are held in their circular path by the force of gravity. Gravity supplies the centripetal force required for satellite motion.
      • Gravity at the Earth’s surface is, on average, 9.8 meters per second 2 or 32 feet per second 2 . As we move away from the Earth’s surface, the force of gravity decreases.
      • Since the force of gravity is the centripetal force acting on a satellite, the velocity of the satellite depends on the distance of the satellite from the Earth.
    • Center of Earth Radius of Earth = 4000 miles scale 150 lbs 8000 miles scale 37.5 lbs 12000 miles scale 16.7 lbs ¼ g 1 / 9 g Normal gravity Distance , Gravity
    • Circular Motion & Satellites (a) What is the velocity of moon orbiting the Earth ? The moon is 242,000 miles from earth. (b) What is the period of the moon’s rotation about the Earth ?(g at the moon’s location = 8.74 x 10 -3 ft/s 2 )
      • 242,000 miles = 242,000 x 5280 = 1.28 x 10 9 ft
      • v = (r x g ) 1/2 = ((1.28 x 10 9 ) x (0.00874)) 1/2 = 3.34 x 10 3 ft/sec
      • (3.34 x 10 3 ) x 3600 x (1/5280) = 2,280 mph
      • (b) Circumference of moon’s orbit = 2  1.28 x 10 9 = 8.04 x 10 9 ft
      • Distance / velocity = time, 8.04 x 10 9 / 3,340 = 2.41 x 10 6 seconds
      • 2.41 x 10 6 / 3600 = 669 hours or 27.9 days
      F centripetal = F gravity mv 2 /r = mg V 2 /r = g v = (r x g ) 1/2 Earth F c g moon
    • Now it's time for you to try some problems on your own ! The problems are similar to the ones which have been solved so look back and review the appropriate problem if you get stuck !
    • A 1200 kg car rounds a turn of 30 meter radius at 6.0 m/s What is the centripetal acting on the car ? (A) 48 nt (B) 147 nt (C) 240 nt (D) 1440 nt A car just rounds a turn at 60 km/hr. The coefficient of friction between the and the road is 0.65. What is the turning radius of the car ? (A) 9.4 m (B) 22 m (C) 44 m (D) 565 m A 0.500 kg rock is whirled in a vertical circle of radius 60 cm. The velocity of the rock at the bottom of the swing is 4.0 m/s. What is the string tension ? (A) 4.9 nt (B) 8.4 nt (C) 13.3 nt (D) 18.2 nt A satellite orbits the Earth at 8000 km from the Earth’s center. Gravity At this location is 6.2 m/s 2 . What is the velocity of the satelliter ? (A) 0.90 km /s (B) 7.0 km /s (C) 8.9 km /s (D) cannot be determined A road is banked at 14 degrees and has a radius of 250 meters. What is maximum speed of the cars rounding the turn without considering friction ? (A) 2.47 m/s (B) 9.8 m /s (C) 32 m /s (D) 14 m/s Click here for answers
    • The End