Materi PowerPoint Pembelajaran Mata Kuliah Fisika Dasar

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 > 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 L Panjang
 Mass M Massa
 Time T 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

Panjang (m)
1 x 1026
2 x 1022
4 x 1016
1.5 x 1011
6.4 x 106
4.5 x 102

Lapangan sepak bola
1.0 x 102
Tinggi manusia

2 x 100
Ketebalan kertas
1 x 10-4
Panjang gelombang sinar biru
4 x 10-7
Diameter atom Hidrogen
1 x 10-10
Diameter proton

1 x 10-15
Physics 111: Lecture 1, Pg 6

Waktu:
Interval
Time (s)
Umur alam semesta
5 x 1017
Umur Grand Canyon
3 x 1014
32 tahun

1 x 109
1 tahun
3.2 x 107
1 jam
3.6 x 103
Perjalanan cahaya dari mh ke bumi 1.3 x 100
Satu kali putaran senar gitar
2 x 10-3
Satu putaran gel. Radio FM
6 x 10-8
Umur meson pi netral
1 x 10-16
Umur quark top
4 x 10-25

Physics 111: Lecture 1, Pg 7

Massa:
Object
Mass (kg)

Galaksi Bima Sakti
4 x 1041
Matahari
2 x 1030
Bumi
6 x 1024
Pesawat Boeing 747
4 x 105
Mobil
1 x 103
Mahasiswa
7 x 101
Partikel debu
1 x 10-9
Quark top
3 x 10-25
Proton
2 x 10-27
Electron
9 x 10-31

Neutrino
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
 1m
= 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


(a)

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 ?

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

(a)

(a)

L

L



 T2 

2

L4
 4 T
T

2

P 2  dg  (b)

Not Right !!

d
P 2
g

(c)

d
P 2
g

Physics 111: Lecture 1, Pg 13

Lecture 1, Act 1
Solution


(b)

(a)

Try the second equation

L
T 2 T
L
T2

2

P 2  dg  (b)

Not Right !!

d
P 2
g

(c)

d
P 2
g

Physics 111: Lecture 1, Pg 14

Lecture 1, Act 1
Solution


(c)

(a)

Try the third equation

L
 T 2 T
L
T2

2

P 2  dg  (b)

This has the correct units!!
This must be the answer!!

d
P 2
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

x2
x1
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

x2

Vav = slope of line connecting x1 and x2.

x1
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.
x2
x1
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 
 2
dt
dt

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


n
High-school calculus: t dt 



dv
a

Also recall that
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
a const

t

v

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
be some acceleration.
v
 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

1
x  x0  v 0 t  at 2
2


Plugging in for t:

 v  v0  1  v  v0 
x  x0  v 0 
  a

a
2
a





2

2

v 2  v 0 2 a( x  x0 )

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 = x f , 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

2

 v 0 2(  ab ) xf
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
(Text: 1-1)
 Converting between systems of units
(Text: 1-2)
 Dimensional Analysis
(Text: 1-3)
1-D Kinematics
(Chapter 2)
 Average & instantaneous velocity
and acceleration
(Text: 2-1, 2-2)
 Motion with constant acceleration
(Text: 2-3)



Example car problem

(Ex. 2-7)



Look at Text problems Chapter 2: # 6, 12, 56, 119

Physics 111: Lecture 1, Pg 35