Fisika Dasar I
WAHIDIN ABBAS
FT Mesin UNY
abbas@uny.ac.id


Pengukuran dan Satuan
Satuan dasar
Sistem Satuan
Konversi Sistem Satuan
Analisis Dimensional
Kinematika Partikel
Kecepatan dan percepatan rata-rata & sesaat
Gerak dengan percepatan konstan
Physics 111: Lecture 1, Pg 1
Physics 111: Lecture 1, Pg 2
Physics 111: Lecture 1, Pg 3


Mekanika Klasik (Newton):
Mekanika: Bagaimana dan mengapa benda-benda
dapat bergerak
Klasik:
» Kecepatan tidak terlalu cepat (v << c)
» Ukuran tidak terlalu kecil
(d >> atom)
Pengalaman sehari-hari banyak yang terjadi berdasarkan
aturan-aturan mekanika klasik.
Lintasan bola kasti
Orbit planet-planet
dll...
Physics 111: Lecture 1, Pg 4
Units


Bagaimana mengukur dimensi?
Semua ukuran di dalam mekanika klasik dapat dinyatakan
dengan satuan dasar:
Length
Mass
Time

L
M
T
Panjang
Massa
Waktu
Contoh:
Kecepatan mempunyai satuan L / T (kilometer per
jam).
Gaya mempunyai satuan ML / T2 .
Physics 111: Lecture 1, Pg 5
Panjang:
Jarak
Jari-jari alam semesta
Ke galaksi Andromeda
Ke bintang terdekat
Bumi - matahari
Jari-jari bumi
Sears Tower
Lapangan sepak bola
Tinggi manusia
Ketebalan kertas
Panjang gelombang sinar biru
Diameter atom Hidrogen
Diameter proton
Panjang (m)
1 x 1026
2 x 1022
4 x 1016
1.5 x 1011
6.4 x 106
4.5 x 102
1.0 x 102
2 x 100
1 x 10-4
4 x 10-7
1 x 10-10
1 x 10-15
Physics 111: Lecture 1, Pg 6
Waktu:
Interval
Umur alam semesta
Umur Grand Canyon
32 tahun
1 tahun
1 jam
Perjalanan cahaya dari mh ke bumi
Satu kali putaran senar gitar
Satu putaran gel. Radio FM
Umur meson pi netral
Umur quark top
Time (s)
5 x 1017
3 x 1014
1 x 109
3.2 x 107
3.6 x 103
1.3 x 100
2 x 10-3
6 x 10-8
1 x 10-16
4 x 10-25
Physics 111: Lecture 1, Pg 7
Massa:
Object
Galaksi Bima Sakti
Matahari
Bumi
Pesawat Boeing 747
Mobil
Mahasiswa
Partikel debu
Quark top
Proton
Electron
Neutrino
Mass (kg)
4 x 1041
2 x 1030
6 x 1024
4 x 105
1 x 103
7 x 101
1 x 10-9
3 x 10-25
2 x 10-27
9 x 10-31
1 x 10-38
Physics 111: Lecture 1, Pg 8
Satuan ...

Satuan Internasional, SI (Système International) :
mks: L = meters (m), M = kilograms (kg), T = seconds (s)
cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)

Satuan Inggris:
Inci (Inches, In), kaki (feet, ft), mil (miles, mi), pon (pounds)

Pada umumnya kita menggunakan SI, tetapi dalam masalah
tertentu dapat dijumpai satuan Inggris. Mahasiswa harus dapat
melakukan konversi dari SI ke Satuan Inggris, atau sebaliknya.
Physics 111: Lecture 1, Pg 9
Converting between different systems of units

Useful Conversion factors:
1 inch
= 2.54 cm
1 m
= 3.28 ft
1 mile
= 5280 ft
1 mile
= 1.61 km

Example: convert miles per hour to meters per second:
1
mi
mi
ft
1 m
1 hr
m
1
 5280


 0.447
hr
hr
mi 3.28 ft 3600 s
s
Physics 111: Lecture 1, Pg 10
Analisis Dimensional

Analisis dimensional merupakan perangkat yang sangat
berguna untuk memeriksa hasil perhitungan dalam sebuah
soal.
Sangat mudah dilakukan!

Contoh:
Dalam menghitung suatu jarak yang ditanayakan di dalam
sebuah soal, diperoleh jawaban
d = vt 2 (kecepatan x waktu2)
Satuan untuk besaran pada ruas kiri= L
Ruas kanan = L / T x T2 = L x T
Dimensi ruas kiri tidak sama dengan dimensi ruas kanan,
dengan demikian, jawaban di atas pasti salah!!

Physics 111: Lecture 1, Pg 11
Lecture 1, Act 1
Dimensional Analysis

The period P of a swinging pendulum depends only on
the length of the pendulum d and the acceleration of
gravity g.
Which of the following formulas for P could be
correct ?
(a) P = 2
(dg)2
(b)
d
P  2
g
(c)
P  2
d
g
Given: d has units of length (L) and g has units of (L / T 2).
Physics 111: Lecture 1, Pg 12
Lecture 1, Act 1
Solution


Realize that the left hand side P has units of time (T )
Try the first equation
L
(a)  L 

 T2 
2
L4
 4 T
T
(a) P  2 dg 
2
(b)
Not Right !!
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 13
Lecture 1, Act 1
Solution

Try the second equation
(b)
L
 T2  T
L
T2
(a) P  2 dg 
2
Not Right !!
(b)
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 14
Lecture 1, Act 1
Solution

Try the third equation
(c)
L
 T2 T
L
T2
(a) P  2 dg 
2
(b)
This has the correct units!!
This must be the answer!!
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 15
Motion in 1 dimension

In 1-D, we usually write position as x(t1 ).

Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t2 - t1 is
x = x(t2) - x(t1) = x2 - x1
x
x
some particle’s trajectory
in 1-D
x
2
x
1
t1
t
t2
t
Physics 111: Lecture 1, Pg 16
1-D kinematics


Velocity v is the “rate of change of position”
Average velocity vav in the time t = t2 - t1 is:
v av 
x( t 2 )  x( t1 ) x

t 2  t1
t
x
x
trajectory
x
2
Vav = slope of line connecting x1 and x2.
x
1
t1
t2
t
t
Physics 111: Lecture 1, Pg 17
1-D kinematics...


Consider limit t1 t2
Instantaneous velocity v is defined as:
v( t ) 
x
x
dx( t )
dt
so v(t2) = slope of line tangent to path at t2.
x
2
x
1
t1
t2
t
t
Physics 111: Lecture 1, Pg 18
1-D kinematics...


Acceleration a is the “rate of change of velocity”
Average acceleration aav in the time t = t2 - t1 is:
aav 

v ( t 2 )  v ( t1 ) v

t 2  t1
t
And instantaneous acceleration a is defined as:
dv ( t ) d 2 x( t )
a( t ) 

dt
dt 2
using v ( t ) 
dx( t )
dt
Physics 111: Lecture 1, Pg 19
Recap

If the position x is known as a function of time, then we can
find both velocity v and acceleration a as a function of time!
x
x  x( t )
dx
v 
dt
dv
d 2x
a 

dt
dt 2
v
a
t
t
t
Physics 111: Lecture 1, Pg 20
More 1-D kinematics


We saw that v = dx / dt
In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
t2
x (t 2 )  x (t1 )   v (t )dt
t1

Graphically, this is adding up lots of small rectangles:
v(t)
+ +...+
= displacement
t
Physics 111: Lecture 1, Pg 21
1-D Motion with constant acceleration

High-school calculus:  t n dt 

Also recall that a 
dv
dt
1
t n 1  const
n 1

Since a is constant, we can integrate this using the above
rule to find:
v   a dt  a  dt  at  v 0

Similarly, since v 
dx
we can integrate again to get:
dt
1
x   v dt   ( at  v 0 )dt  at 2  v 0 t  x0
2
Physics 111: Lecture 1, Pg 22
Recap

So for constant acceleration we find:
Plane
w/ lights
x
1
x  x0  v 0 t  at 2
2
v  v 0  at
v
t
a  const
a
t
t
Physics 111: Lecture 1, Pg 23
Lecture 1, Act 2
Motion in One Dimension

When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the
highest point in its path?
(a) Both v = 0 and a = 0.
(b) v  0, but a = 0.
y
(c) v = 0, but a  0.
Physics 111: Lecture 1, Pg 24
Lecture 1, Act 2
Solution

Going up the ball has positive velocity, while coming down
it has negative velocity. At the top the velocity is
momentarily zero.
x

Since the velocity is
continually changing there must
v
be some acceleration.
In fact the acceleration is caused
by gravity (g = 9.81 m/s2).
(more on gravity in a few lectures) a

The answer is (c) v = 0, but a  0.
t
t
t
Physics 111: Lecture 1, Pg 25
Derivation:
v  v 0  at

Solving for t:
t
v  v0
a
x  x0  v 0 t 

1 2
at
2
Plugging in for t:
 v  v0  1  v  v0 
x  x0  v 0 
  a

 a  2  a 
2
v 2  v 0  2 a( x  x0 )
2
Physics 111: Lecture 1, Pg 26
Average Velocity

Remember that v  v 0  at
v
v
vav
v0
t
t
v av 
1
v 0  v 
2
Physics 111: Lecture 1, Pg 27
Recap:

For constant acceleration:
Washers
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const

From which we know:
v 2  v 02  2a(x  x0 )
1
v av  (v 0  v)
2
Physics 111: Lecture 1, Pg 28
Problem 1

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab
vo
ab
x = 0, t = 0
Physics 111: Lecture 1, Pg 29
Problem 1...

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab. At what time tf does the car stop, and how much farther
xf does it travel?
v0
ab
x = 0, t = 0
v=0
x = xf , t = tf
Physics 111: Lecture 1, Pg 30
Problem 1...

Above, we derived: v = v0 + at

Realize that a = -ab

Also realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or
tf = v0 /ab
Physics 111: Lecture 1, Pg 31
Problem 1...

To find stopping distance we use:
v 2  v 02  2a(x  x0 )

In this case v = vf = 0, x0 = 0 and x = xf
 v 0  2( ab )xf
2
2
v
xf  0
2 ab
Physics 111: Lecture 1, Pg 32
Problem 1...



So we found that
2
1 v0
tf 
, xf 
ab
2 ab
v0
Suppose that vo = 65 mi/hr = 29 m/s
Suppose also that ab = g = 9.81 m/s2
 Find that tf = 3 s and xf = 43 m
Physics 111: Lecture 1, Pg 33
Tips:

Read !
Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what
information is given, what is asked for, and the meaning
of all the terms used in stating the problem.

Watch your units !
Always check the units of your answer, and carry the
units along with your numbers during the calculation.

Understand the limits !
Many equations we use are special cases of more
general laws. Understanding how they are derived will
help you recognize their limitations (for example,
constant acceleration).
Physics 111: Lecture 1, Pg 34
Recap of today’s lecture

Scope of this course
Measurement and Units (Chapter 1)
Systems of units
Converting between systems of units
Dimensional Analysis
1-D Kinematics
(Chapter 2)
Average & instantaneous velocity
and acceleration
Motion with constant acceleration
Example car problem

Look at Text problems Chapter 2: # 6, 12, 56, 119



(Text: 1-1)
(Text: 1-2)
(Text: 1-3)
(Text: 2-1, 2-2)
(Text: 2-3)
(Ex. 2-7)
Physics 111: Lecture 1, Pg 35