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Gerak dan Perubahan (Tinjauan Fisika) Elok Sudibyo P. Sains – FMIPA Unesa

1. Linear Kinematics.pptx

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Page 1: 1. Linear Kinematics.pptx

Gerak dan Perubahan(Tinjauan Fisika)

Elok SudibyoP. Sains – FMIPA Unesa

Page 2: 1. Linear Kinematics.pptx

Materi Perkuliahan:Gerak dan Perubahan (Tinjauan Fisika)

Pertemuan ke:11. Kinematika Linier (Linear Kinematics)12. Kinetika Linier (Linear Kinetics)13. Ujian Subsumatif-114. Kinematika Anguler (Angular Kinematics)15. Kinetika Anguler (Angular Kinetics)16. Ujian Subsumatif-2

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LINEAR KINEMATICS

Objectives:When you finish this chapter, you should be able to do the following:• Distinguish between linear, angular, and general

motion.• Define distance traveled and displacement, and

distinguish between the two.• Define average speed and average velocity, and

distinguish between the two.

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• Define instantaneous speed and instantaneous velocity.

• Define average acceleration.• Define instantaneous acceleration.• Calculate the distance traveled and displacement,

speed and velocity, and acceleration.

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• Use the equations of projectile motion to determine the vertical or horizontal position of a projectile given the initial velocities and time.

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Types of Motion

Motion

Linear Motion (Translation)

Rectilinear Motion

Curvilinear Motion

Angular Motion

(Rotation)

General Motion

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Rectilinear motion

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• Linear Kinematics: describing objects in linear motion (position, distance traveled, displacement, time, speed, velocity, acceleration)

LINEAR KINEMATICS

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• Mechanically, position is defined as location an object in space at any particular time.

• Example: A 100-meter swimming race in a 50-meter pool.

Position

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• a swimmer’s location at any particular time.

Position

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Distance Traveled & Displacement

• Distance:– Length of path which a body covers during motion– Units (SI): meter (m)– Scalar quantity

• Displacement:– The change in position of a body during motion– Units (SI): meter (m)– Vector quantity

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Distance & Displacement

• Distance: represented by BLUE colour line• Displacement: represented by YELLOW colour line

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Displacement as a Vector

• Vector has:– Magnitude– Direction– Point of origin

• Vector represented graphically by: – Line of action

• Magnitude and Direction quantified using:– Pythagorean Theorem– Trigonometry

Standing Broad Jump take-off

+-

+

-

P2P1

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Calculation of Displacement

• Calculation of Magnitude:Resultant displacement (dR)

==

= 0.63 m

• Calculation of Direction:Angle to horizontal (θ)

Tan θ = Opposite / AdjacentTan θ = dV / dH = 0.2 / 0.6

θ = Tan-1 (0.2 / 0.6) θ = 18.8º

2

V

2

H d+d

22 )2.0(+)6.0(

Vertical displacement (dV) = 0.2 m

Horizontal displacement (dH) = 0.6 m

Resultant displacement (dR)

P1

P2

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Speed and Velocity

• Average Speed (scalar):– Length of path (distance)

divided by change in time (∆t)

• Average velocity (vector):– Change in position (∆p)

divided by change in time (∆t)

– Displacement (d) divided by change in time (∆t)

If displacement = 50 m

Δt

d=

Δt

Δp=v

If t = 5 s

v = 50 / 5

= 10 m·s-1

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Speed and Velocity

• We can think of instantaneous speed as distance traveled divided by the time it took to travel that distance if the time interval used in the measurement is very small.

• If the motion of the object under analysis is in a straight line and rectilinear, with no change in direction, average speed and average velocity will be identical in magnitude.

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Velocity as a vector

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Average Velocity

• Average velocity not necessarily equal to instantaneous velocity.

Winner of the Men's 100 m at the 2004 Athens Olympics

in 9.85 sAverage velocity:

= 100/9.85= 10.15 m·s-1

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Kinematic analysis of 100 m sprint

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Kinematic analysis of 100 m sprint

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Velocity during 100 m

Average velocity: v = d / ∆t 0-10 m

= 10 / 2.2 = 4.5 m·s-1

10-20 m= 10 / 1.2 = 8.3 m·s-1

20-30 m= 10 / 0.8 = 12.5 m·s-1

30-40 m= 10 / 0.7 = 14.3 m·s-1

40-50 m= 10 / 0.8 = 12.5 m·s-1

50-60 m= 10 / 0.8 = 12.5 m·s-1

60-70 m= 10 / 0.7 = 14.3 m·s-1

70-80 m= 10 / 0.8 = 12.5 m·s-1

80-90 m= 10 / 0.9 = 11.1 m·s-1

90-100 m= 10 / 0.9 = 11.1 m·s-1

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Acceleration

• Average Acceleration (a):Change in velocity (∆v) divided by change in time (∆t)

• As with displacement & velocity, acceleration can be resolved into components using trigonometry & Pythagorean theorem

2 1(v - vva = =

t t

)

V1 = 4.5 m·s-1 V2 = 8.3 m·s-1

∆t = 1.2 s

a = (8.3 - 4.5) / 10 = 3.2 m·s-2

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Acceleration

• When an object speeds up, slows down, starts, stops, or changes direction, it is accelerating.

• The direction of motion does not indicate the direction of the acceleration.

• Instantaneous acceleration is the acceleration of an object at an instant in time.

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Acceleration during 100 m

Acceleration at start of racea = (v2 - v1) / ∆t

= (8.3 - 4.5) / 1.2 Positive Acceleration= 3.2 m·s-2

_____________________________________________________________________________________________________________________________________________

Acceleration during middle of racea = (v2 - v1) / ∆t

= (12.5 - 12.5) / 0.8 Constant Velocity= 0

_____________________________________________________________________________________________________________________________________________

Acceleration at end of racea = (v2 - v1) / ∆t

= (11.1 - 14.3) / 0.9 Negative Acceleration= -3.5 m·s-2

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Positive/Negative Acceleration

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Velocity Curve for Sprinting

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Velocity Curves for Two Sprinters

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Bodies projected into the air are projectiles.

Horizontal & Vertical Components:• Vertical is influenced by gravity.• No force (neglecting air resistance) affects the

horizontal.• Horizontal relates to distance.• Vertical relates to maximum height achieved.

Kinematics of Projectile Motion

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• Angle of projection

• Projection speed• Relative height of

projection

Factors Influencing Projectile Trajectory

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• Perfectly vertical• Parabolic• Perfectly horizontal

Air resistance may cause irregularities.In this chapter, neglecting air resistance.

Angles of Projection

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Relative Projection Height

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• Maximize the speed of projection• Maximize release height• Optimum angle of projection:

Release height = 0, then angle = 450

Release height, then angle Release height, then angle

Optimum Projection Conditions

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Range at Various Angles

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Analyzing Projectile Motion

Path of a projectile fired with initial velocity vo at angle qo to the horizontal. Path is shown in black, the velocity vectors are green arrows,

and velocity components are dashed.

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General Kinematic Equations for Constant Acceleration in Two Dimensions

• Horizontal (x component):

• Vertical (y component):

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Kinematic Equations for Projectile Motion

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v = projection speed (m/s)q = angle of projection (degree)h = relative height of projection (m)g = acceleration due to gravity (9,8 m/s2)

TUGAS: BUKTIKAN...!!!

Range of Projectile (R)

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• Kerapian pekerjaan (keseriusan dalam mengerjakan tugas).

• Sistematika (keruntutan) tahap-tahap penyelesaian atau pembuktian.

• Kelengkapan tahap-tahap penyelesaian.• Kebenaran pekerjaan.• Tugas ditulis dengan tinta hitam pada kertas double-

folio bergaris.• Ketepatan dalam penyerahan tugas.

Kriteria Penilaian Tugas

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THANKS....

WISH FOR GOOD LUCK....