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

 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.

 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/ T}2 .

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

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

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

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)}

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:

 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!!

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) (c)

P

d

g



2



P

d

Lecture 1,

Act 1

Solution

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

(a) (b) (c)

(a) Not Right !!





P



2



dg

2

L

L

T

L

T

T



















2 2 ₄ 4

P

d

g



2



P

d

g

(b) _{Not Right !!}

 Try the second equation

Lecture 1,

Act 1

Solution

L

L

T

T

T

2

2





(a) (b) (c)

(c) This has the correct units!!

This must be the answer!!  Try the third equation

Lecture 1,

Act 1

Solution

T

T

T

L

L

₂

2









P



2



dg

2

P

d

g



2



P

d

g

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

x₁

x₂ some particle’s trajectory

1-D kinematics

t x

t₁ t₂

x

x₁

x₂ trajectory

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

V_av = slope of line connecting x₁ and x₂.

 Consider limit t₁ t₂

 Instantaneous velocity v is defined as:

1-D kinematics...

t x

t t

x

x₁ x₂

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

) t ( dx )

t (

1-D kinematics...

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

 And instantaneous acceleration a is defined as:

using t v t t ) t ( v ) t ( v a 1 2 1 2 av       2 2 dt ) t ( x d dt ) t ( dv ) t (

a  

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

a

v t

t

a dv

dt

d x dt

 

2 2

v dx

dt



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:

 Graphically, this is adding up lots of small rectangles:

v(t)

t

+ +...+

= displacement







2

1

t

t 1

2

x

t

v

t

dt

t

 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

const t 1 n 1 dt

tn n 1 

    a dv dt  v dx dt      

 a dt a dt at v₀

v

0 0

2

0 at v t x

2 1 dt ) v at ( dt v

Recap

 So for constant acceleration we find:

x

a

v t

t t

Plane w/ lights

at v

v  ₀ 

2 0

0 v t ₂1 at

x

x   

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.

Lecture 1,

Act 2

Solution

x

a

v t

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)}

Derivation:

 Plugging in for t:  Solving for t:

at v

v  ₀  _{0 0} at2

2 1 t v x

x   

a v v

t _  0

2 0 0 0 0 a v v a 2 1 a v v v x x _                 ) x x ( a 2 v

Average Velocity

 Remember that

v

t t

v v_av v₀

at v

v  ₀ 



v v



2 1

Recap:

 For constant acceleration:

 From which we know:

Washers v) (v 1 v ) x 2a(x v

v2 ₀2 ₀

     at v

v  ₀ 

2 0

0 v t ₂1 at

x

x   

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

a_b

x = 0, t = 0 a_b

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

a_b. At what time t_f does the car stop, and how much farther

x_f does it travel?

x = x , t = t v = 0

x = 0, t = 0 a_b

Problem 1...

 Above, we derived: v = v₀ + at

 Realize that a = -a_b

 Also realizing that v = 0 at t = t_f : find 0 = v₀ - a_b t_f or

Problem 1...

 To find stopping distance we use:

 In this case v = v_f = 0, x₀ = 0 and x = x_f

f b 2

0 2( a )x

v  



2 0 f

v x 

) x 2a(x v

Problem 1...

 So we found that

 Suppose that v_o = 65 mi/hr = 29 m/s  Suppose also that a_b = g = 9.81 m/s2

 Find that t_f = 3 s and x_f = 43 m

b 2 0 f

b 0

f

a

v

2

1

x

,

a

v

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 !

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)