Ppt circular motion

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Indicator 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

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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)

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Uniform Circular MotionObjects moving in a circle is said irregular if the object is moving with constant angular velocity on a circular trajectoryThe position angle can be calculated by the equation:  = ω. T, with t = time (s)  Angular position versus time graph on the GMB:    

t

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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

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LINEAR VELOCITY (v):The distance traveled divided by the object latency

V = 2Лr / T

V = speed of the linear (m /s)r = radius of the circleT = period (second)

A

v

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ANGULAR VELOCITY (ω):The angles that have been taken within a certain time interval

ωω = = 2Л/T

With

ω = angular velocity (rad / s)T = period (s)

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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!

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Unknown:a.d = 800 m r = 400 mb. t = 1 jam = 3600 sc. n = 120 lap

Asked: a. T = …?b. f = …?c. v = …?d. ω = …?

Answer:

x

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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.

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Centripetal acceleration is mathematically written

as = V2/rWith as = centripetal acceleration (m/s2) V = velocity (m / s) r = radius of circle (m)

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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 : as = .....?

Answer :as = v2/r =(10)2/20

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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

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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 ωo on GMBB : ωt = ωo + t t

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The position angle traveled by the moving body uniformly accelerated circular mathematically written:

Θ = ωo . t + ½. .t2

With Θ = angle position of the object (rad)ωo = initial angular velocity (rad/s) = angular acceleration (rad/s2)

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Relations Wheelsa. Concentric relationship: On the relationship of two concentric wheels direction of rotation and angular velocity both same wheel. So: ω1 = ω2b. relationships intersect In this connection both the direction of rotation linear speed of the two opposite and equal. So v1 = v2c. Relationship of two wheels with a rope, then the direction of rotation and linear speed of the same. So v1 = v2

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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 velocity2. wheel linear speed of c

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Solution

Given:Ra = 4 cm = 4x10 -2 mRb = 6 cm = 6x10 -2 mRc = 8 cm = 8x10 -2 mωb = 8 rad / sasked:1. ωa2. vc

Answer:1. va = vb            ωaRa = ωc.Rc ωa. (4x10 -2) = 8. (6x10 -2)                ωa = 12 rad / s.                                    2. ωa = ωc             vc = ωa.Rc                 vc = 12. (8x10 -2)                 vc = 0.96 m / s

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ReferensiReferensi

1. Supriyanto.2006. Fisika SMA kelas X2. Marthin Kanginan. 2006. Fisika SMA kelas X3. Karyono.2007.Fisika SMA dan MA kelas XII4. Sutejo.2007.Fisika X

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Irna Ilfiana(09330105)