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Physics 111: Lecture 1, Pg 1

Fisika Dasar IWAHIDIN ABBAS

FT Mesin [email protected]

Pengukuran dan SatuanSatuan dasarSistem SatuanKonversi Sistem SatuanAnalisis Dimensional

Kinematika PartikelKecepatan dan percepatan rata-rata & sesaatGerak dengan percepatan konstan


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 kastiOrbit planet-planetdll...


Bagaimana mengukur dimensi? Semua ukuran di dalam mekanika klasik dapat dinyatakan

dengan satuan dasar:

Length L PanjangMass M MassaTime T Waktu

Contoh:Kecepatan mempunyai satuan L / T (kilometer per jam).Gaya mempunyai satuan ML / T2 .

Units


Panjang:

Jarak Panjang (m)Jari-jari alam semesta 1 x 1026

Ke galaksi Andromeda 2 x 1022

Ke bintang terdekat 4 x 1016

Bumi - matahari 1.5 x 1011

Jari-jari bumi 6.4 x 106

Sears Tower 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


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


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


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.


Converting between different systems of units

Useful Conversion factors:1 inch = 2.54 cm1 m = 3.28 ft1 mile = 5280 ft 1 mile = 1.61 km

Example: convert miles per hour to meters per second:

s

m4470

s

hr

3600

1

ft

m

283

1

mi

ft5280

hr

mi1

hr

mi 1 .

.


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= LRuas kanan = L / T x T2 = L x T

Dimensi ruas kiri tidak sama dengan dimensi ruas kanan, dengan demikian, jawaban di atas pasti salah!!

Analisis Dimensional


Lecture 1, Act 1Dimensional 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 ?

Pdg

2Pdg

2(a) (b) (c)

Given: d has units of length (L) and g has units of (L / T 2).

P = 2 (dg)2


Lecture 1, Act 1 Solution

Realize that the left hand side P has units of time (T ) Try the first equation

P dg2 2(a) (b) (c)

(a) LL

T

L

TT

2

2 4

4 Not Right !!

Pdg

2Pdg

2


LL

T

T T

2

2

P dg2 2(a) (b) (c)

(b) Not Right !!

Try the second equation


Pdg

2Pdg

2


TT

TLL 2

2

P dg2 2(a) (b) (c)

(c) This has the correct units!!

This must be the answer!!

Try the third equation


Pdg

2Pdg

2


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

t

x

t1 t2

x

t

x1

x2some particle’s trajectory

in 1-D


1-D kinematics

tx

tt)t(x)t(x

v12

12av

t

x

t1 t2

x

x1

x2trajectory

Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:

t

Vav = slope of line connecting x1 and x2.


Consider limit t1 t2

Instantaneous velocity v is defined as:

1-D kinematics...

dt)t(dx

)t(v

t

x

t1 t2

x

x1

x2

t

so v(t2) = slope of line tangent to path at t2.


1-D kinematics...

tv

tt)t(v)t(v

a12

12av

Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:

And instantaneous acceleration a is defined as:

2

2

dt)t(xd

dt)t(dv

)t(a

dt)t(dx

)t(v using


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!

adv

dt

d x

dt

2

2

vdx

dt

x x t ( )

x

a

vt

t

t


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:

2

1

t

t12 dttvtxtx )()()(

Graphically, this is adding up lots of small rectangles:

v(t)

t

+ +...+

= displacement


High-school calculus:

Also recall that

Since a is constant, we can integrate this using the above rule to find:

Similarly, since we can integrate again to get:

1-D Motion with constant acceleration

constt1n

1dtt 1nn

adv

dt

vdx

dt

0vatdtadtav

002

0 xtvat21

dt)vat(dtvx


Recap So for constant acceleration we find:

atvv 0

200 at

2

1tvxx

a const

x

a

v t

t

t

Planew/ lights


Lecture 1, Act 2Motion 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.

(c) v = 0, but a 0.

y


Lecture 1, Act 2Solution

x

a

vt

t

t

Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero.

Since the velocity is continually changing there must be some acceleration.

In fact the acceleration is caused by gravity (g = 9.81 m/s2).

(more on gravity in a few lectures)

The answer is (c) v = 0, but a 0.


Derivation:

Plugging in for t:

atvv 0 200 at

21

tvxx

Solving for t:

avv

t 0

200

00 avv

a21

avv

vxx

)xx(a2vv 02

02


Average Velocity

Remember that atvv 0

v

t

t

v

vav

v0

vv2

1v 0av


Recap: For constant acceleration:

From which we know:

v)(v21

v

)x2a(xvv

0av

02

02

Washers

atvv 0

200 at

2

1tvxx

a const


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

x = 0, t = 0ab

vo


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?

x = xf , t = tf

v = 0

x = 0, t = 0ab

v0


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


Problem 1...

To find stopping distance we use:

In this case v = vf = 0, x0 = 0 and x = xf

fb2

0 x)a(2v

b

20

f a2v

x

)x2a(xvv 02

02


Problem 1...

So we found that

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

b

20

fb

0f a

v

2

1x ,

a

vt


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).


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