Ppt circular motion
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Ppt circular motion Ppt circular motion Presentation Transcript

  • Allow Sad,   But Do not Despair
  • Allow Sad,   But Do not Despair
  • Allow Sad,   But Do not Despair
    • I ndicator
    • Identify the amount of frequency, angular frequency, period, and the travel angle contained in a circular motion with constant speed
    • Applying the principle of the wheels which are interconnected in a qualitative
    • Analyzing the amount of conduction-related linear motion and circular motion on a rolling motion with constant speed
  • Allow Sad,   But Do not Despair Circular motion is the motion of objects on the circular trajectory, speed of object speed and direction remain changeable. Circular motion can be divided into: Uniform circular motion (GMB) Circular motion uniformly accelerated (GMBB)
  • Allow Sad,   But Do not Despair Uniform Circular Motion Objects moving in a circle is said irregular if the object is moving with constant angular velocity on a circular trajectory The position angle can be calculated by the equation:    = ω. T, with t = time (s)   Angular position versus time graph on the GMB:      t
  • Allow Sad,   But Do not Despair MAGNITUDES IN CIRCULAR MOTION PERIOD (T): time needed by the body to take the path to a full circle. Where is the second unit (s). T = 1/f FREQUENCY (f): Number of walks taken a full circle objects in one second. Where is the Hertz unit (Hz). f = 1/T
  • Allow Sad,   But Do not Despair LINEAR VELOCITY (v): The distance traveled divided by the object latency V = 2 Л r / T V = speed of the linear (m /s) r = radius of the circle T = period (second) A v
  • Allow Sad,   But Do not Despair ANGULAR VELOCITY (ω): The angles that have been taken within a certain time interval ω = 2 Л /T With ω = angular velocity (rad / s) T = period (s)
  • Allow Sad,   But Do not Despair Example Problem: Bakri spur bike on a circular trajectory within 1 hour. In that time, Bakri has done 120 laps. Determine the period, frequency, kecepatn linear and angular velocity Bakri if the track has a diameter of 800 m!
  • Allow Sad,   But Do not Despair
    • Unknown :
    • d = 800 m r = 400 m
    • t = 1 jam = 3600 s
    • n = 120 lap
    Asked : a. T = …? b. f = …? c. v = …? d. ω = …? Answer : x
  • Allow Sad,   But Do not Despair Centripetal Acceleration
    • Centripetal acceleration is the acceleration of the direction toward the center of the circle
    • Acceleration occurs because the linear speed of the object that keeps changing.
  • Allow Sad,   But Do not Despair
    • Centripetal acceleration is mathematically written
      • a s = V 2 /r
      • With a s = centripetal acceleration (m/s 2 )              V = velocity (m / s)               r = radius of circle (m)
  • Allow Sad,   But Do not Despair Example Problem : Bambang riding a motorcycle through a bend circle 20 m fingered finger when going to school. If Bambang motor speed 10 m / s, determine the acceleration Bambang trajectory leading to the center! Unknown : Asked : a s = .....? Answer : a s = v 2 /r =(10) 2 /20
  • Allow Sad,   But Do not Despair Position Angle
    • The position angle (  ) is the position of particle moving along the arc of linear displacement (s) within r from the center of the circle. Mathematically calculated angular position:
    •  = s / r
    • Where the unit  in radians                         s in meters                         r in meters
    • Since the circumference of a circle = 2Пr then  = 2П rad
  • Allow Sad,   But Do not Despair Regular Changed Circular Motion (GMBB)
    • Said to be irregular objects moving in a circle if the angle velocity uniformly accelerated so that a constant angular acceleration .
    • Graph velocity function of time at GMBB:
    • ω Mathematically,
    • angular velocity
    • ω o on GMBB :
    • ω t = ω o +  t
    • t
  • Allow Sad,   But Do not Despair
    • The position angle traveled by the moving body uniformly accelerated circular mathematically written:
    • Θ = ω o . t + ½.  .t 2
    • With
      • Θ = angle position of the object (rad)
      • ω o = initial angular velocity (rad/s)
      •  = angular acceleration (rad/s 2 )
  • Allow Sad,   But Do not Despair Relations Wheels a. Concentric relationship:           On the relationship of two concentric wheels           direction of rotation and angular velocity both           same wheel. So: ω1 = ω2 b. relationships intersect           In this connection both the direction of rotation           linear speed of the two opposite and equal.           So v1 = v2 c. Relationship of two wheels with a rope, then the direction of rotation and linear speed of the same. So v1 = v2
  • Allow Sad,   But Do not Despair Look at pictures of the three wheels are in a relationship as follows: If Rc = 4 cm, 6 cm and Rb = Ra = 8cm, and the wheel angular velocity w = 8 rad / s. discussion: 1. a wheel angular velocity 2. wheel linear speed of c
  • Allow Sad,   But Do not Despair Solution Given: Ra = 4 cm = 4x 10 -2 m Rb = 6 cm = 6x 10 -2 m Rc = 8 cm = 8x 10 -2 m ωb = 8 rad / s asked: 1. ωa 2. vc Answer: 1. va = vb             ωaRa = ωc.Rc ωa. (4x 10 -2 ) = 8. (6x 10 -2 )                 ωa = 12 rad / s.                                      2. ωa = ωc              vc = ωa.Rc                  vc = 12. (8x 10 -2 )                  vc = 0.96 m / s
  • Allow Sad,   But Do not Despair
  • Allow Sad,   But Do not Despair Referensi
    • Supriyanto . 2006. Fisika SMA kelas X
    • Marthin Kanginan. 2006. Fisika SMA kelas X
    • Karyono . 2007 . Fisika SMA dan MA kelas XII
    • Sutejo . 2007 . Fisika X
  • Allow Sad,   But Do not Despair Irna Ilfiana (09330105)