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Kinematika 18Sept 2015

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9/22/2015
Mekanika
1
Mekanika

Mekanika


Kinematika


Cabang dari ilmu fisika yang mengkaji tentang materi,
gaya, dan gerakannya
Terfokus pada gerakannya saja
Dinamika

Gerakan dan penyebabnya
2i
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Gerak satu dimensi

Gerak 1 dimensi
Gerakan pada lintasan garis lurus
 Menganggap yang bergerak adalah partikle
bebas mengabaikan bentuk dan ukuran
 Setiap geraknya dapat direpresentasikan
secara vektor

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Perpindahan, Waktu, Kecepatan
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Posisi dan perpindahan

Bergeraknya benda adalah murni
translasional,
tanpa melibatkan gerak rotasi

Perpindahan
adalah sebuah vektor jarak dari satu
titik acuan ke titik tujuan yang masing-masing
merupakan koordinat dari partikel.
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Perpindahan
x  x2  x1
Vektor
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Jarak Tempuh
Skalar
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Posisi dan perpindahan

Dikatakan bergerak bila
Perubahan posisi dan waktu
Terjadi perpindahan bila
Perubahan posisi koordinat sistem
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Kecepatan rata-rata

Kecepatan rata-rata selama interval waktu adalah
kuantitas vektor yang mengalami perubahan komponen
koordinat x dibagi interval waktunya
x  x2  x1
t  t 2  t1
vav x 
x2  x1 x

t 2  t1
t
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Average Velocity


Average velocity is positive when during the time interval
coordinate x increased and particle moved in the positive
direction
If particle moves in negative x-direction during time
interval, average velocity is negative
x  x2  x1  19 m  277 m  258 m
t  t 2  t1  25.0 s  16.0 s  9.0 s
vav x  x   258 m
 29 m / s
t
9.0s
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X-t Graph

Kecepatan rata-rata hanya
bergantung pada
perubahan jarak terhadap
interval waktu, tidak
mengkaji secara detail pada
setiap perubahan waktunya
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Beberapa kuantitas kecepatan rata-rata
Mobil tercepat 341 m/s
Pejalan cepat 2 m/s
Gerakan random molekul2 udara
500 m/s
Kecepatan orbit satelit komunikasi
siput
10-3
3000 m/s
m/s
Kecepatan cahaya di ruang hampa
3 x 108 m/s
Pesawat tercepat 1000 m/s
Kecepatan orbit elektron dalam atom hidrogen 2 x 108 m/s
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Kecepatan sesaat

Kecepatan sesaat
sebuah
vektor yang
menyatakan kecepatan rata-rata yang dibatasi waktu yang
sangat singkat (mendekati 0)
x dx
v x  lim

t 0 t
dt
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Kecepatan sesaat
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Kecepatan sesaat
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Kecepatan sesaat
Concept Question
The graph shows position
versus time for a particle
undergoing 1-D motion.




At which point(s) is the
velocity vx positive?
At which point(s) is the
velocity negative?
At which point(s) is the
velocity zero?
At which point is speed
the greatest?
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Acceleration
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Acceleration
If the velocity of an object is changing with time, then the
object is undergoing an acceleration.
 Acceleration is a measure of the rate of change of velocity
with respect to time.
 Acceleration is a vector quantity.
 In straight-line motion its only non-zero component is along
the axis along which the motion takes place.

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Average Acceleration

Average Acceleration over a given time interval is defined
as the change in velocity divided by the change in time.

In SI units acceleration has units of m/s2.
aav x 
v2 x  v1x vx

t2  t1
t
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Instantaneous Acceleration

Instantaneous acceleration of an object is obtained by
letting the time interval in the above definition of average
acceleration become very small. Specifically, the
instantaneous acceleration is the limit of the average
acceleration as the time interval approaches zero:
v x dvx

t 0 t
dt
a x  lim
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Acceleration of Graphs
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Arah Percepatan vs Kecepatan
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Acceleration of Graphs
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aav x 
ax 
vxf  vxi
t 2  t1

vx
t
v xf  v xi
t 0
vxf  vxi  axt
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Constant Acceleration Motion
Position of a particle moving with constant acceleration
vav x 
v0 x  vx
2
vx  v0 x  axt
vav x 
x x  x0

t
t 0
 vav x  v0 x  v0 x  a x t   v0 x  a x t
1
2
x  x0
1
v0 x  a x t 
2
t

1
2
1
x  x0  v0 x t  a x t 2
2
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Constant Acceleration Motion

Relationship between position of a particle moving with
constant acceleration, and velocity and acceleration
itself:
vx  v0 x  axt  t 
vx  v0 x
ax
v v  1 v v 
1
x  x0  v0 x t  a x t 2  x  x0  v0 x  x 0 x   a x  x 0 x 
2
 ax  2  ax 
2
vx2  v02x  2ax ( x  x0 )
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Gerak Jatuh Bebas
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Gerak Jatuh Bebas

Example of motion with
constant acceleration is
acceleration of a body falling
under influence of the earth’s
gravitation

All bodies at a particular location
fall with the same downward
acceleration, regardless of
their size and weight

Aristotle
384 - 322 B.C.E.
Idealized motion free fall: we
neglect earth rotation, decrease
of acceleration with decreasing
altitude, air effects
Galileo Galilei
1564 - 1642
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Freely Falling Bodies

The constant acceleration of a freely falling
body is called acceleration due to
gravity, g

Approximate value near earth’s surface g =
9.8 m/s2 = 980 cm/s2 = 32 ft/s2

g is the magnitude of a vector, it is always
positive number

Exact g value varies with location
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Motion in 2 or 3 Dimensions
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Projectile Motion

A projectile is any object that is given an initial velocity and
then follows a path (trajectory) determined solely by gravity
and air resistance.

The motion of a projectile will take place in a plane (so, it is 2D motion).
For projectile motion we can analyze the x- and y-components
of the motion separately.




The horizontal motion (along the x-axis) will have zero
acceleration and thus have constant velocity.
The vertical motion (along the y-axis) will have constant
downward acceleration of magnitude g = 9.80 m/s2.
The initial velocity components, vox and voy, can be expressed
in terms of the magnitude vo and direction ao of the initial
velocity.
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
Kita kombinasikan gerak dalam 2 arah
Projectile
Motion

Arah Horisontal
a x  0  vx  v0 x

x  x0  v0 xt
Arah Vertikal
ay  g
v y  v0 y  gt
y  y0  v0 y t  1 2 gt 2

Kecepatan arah x dan y pada posisi
pusat koordinat :
v0 x  v0 cosa 0
v0 y  v0 sin a 0
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Projectile
Motion
 Lintasan (koord.)
benda pada suatu titik
x  (v0 cosa 0 )t
y  (v0 sin a 0 )t  1 2 gt 2
vx  v0 cosa 0
v y  v0 sin a 0  gt
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Koordinat horisontal max
Koord. Pada titik R (horisontal maximum)
Untuk arah gerak ke atas
y  y0  v0 y t 
1
2
at
2
nilai y=0 dan y0 = 0
t
2v0 y
g
0  0  v0 y t  1 2 gt 2
R  v0 xt  v0 x

g
2v 20 sin a cosa
g
2v 0 sin 2a
2

2v0 y
g
v0 x  v0 cosa 0
v0 y  v0 sin a 0
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