1. Linear Kinematics.pptx

Gerak dan Perubahan(Tinjauan Fisika)

Elok SudibyoP. Sains – FMIPA Unesa

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

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.

• Define instantaneous speed and instantaneous velocity.

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

speed and velocity, and acceleration.

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

Types of Motion

Motion

Linear Motion (Translation)

Rectilinear Motion

Curvilinear Motion

Angular Motion

(Rotation)

General Motion

Rectilinear motion

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

LINEAR KINEMATICS

• 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

• a swimmer’s location at any particular time.

Position

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

Distance & Displacement

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

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

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

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

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.

Velocity as a vector

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

Kinematic analysis of 100 m sprint

Kinematic analysis of 100 m sprint

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

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

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.

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

Positive/Negative Acceleration

Velocity Curve for Sprinting

Velocity Curves for Two Sprinters

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

• Angle of projection

• Projection speed• Relative height of

projection

Factors Influencing Projectile Trajectory

• Perfectly vertical• Parabolic• Perfectly horizontal

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

Angles of Projection

Relative Projection Height

• 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

Range at Various Angles

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.

General Kinematic Equations for Constant Acceleration in Two Dimensions

• Horizontal (x component):

• Vertical (y component):

Kinematic Equations for Projectile Motion

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)

• 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

THANKS....

WISH FOR GOOD LUCK....