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Fisika Dasar I WAHIDIN ABBAS FT Mesin UNY [email protected] Pengukuran dan Satuan Satuan dasar Sistem Satuan Konversi Sistem Satuan Analisis Dimensional Kinematika Partikel Kecepatan dan percepatan rata-rata & sesaat Gerak dengan percepatan konstan. Mekanika Klasik (Newton) : - PowerPoint PPT Presentation

### Transcript of Fisika Dasar I WAHIDIN ABBAS FT Mesin UNY [email protected]

Physics 106P: Lecture 1 Notes

Fisika Dasar IWAHIDIN ABBASFT Mesin [email protected] dan SatuanSatuan dasarSistem SatuanKonversi Sistem SatuanAnalisis DimensionalKinematika PartikelKecepatan dan percepatan rata-rata & sesaatGerak dengan percepatan konstanPhysics 111: Lecture 1, Pg #1Page #

Physics 111: Lecture 1, Pg #

Physics 111: Lecture 1, Pg #Mekanika Klasik (Newton):

Mekanika: Bagaimana dan mengapa benda-benda dapat bergerakKlasik: Kecepatan tidak terlalu cepat(v > atom)Pengalaman sehari-hari banyak yang terjadi berdasarkan aturan-aturan mekanika klasik.Lintasan bola kastiOrbit planet-planetdll...

Physics 111: Lecture 1, Pg #Bagaimana mengukur dimensi?Semua ukuran di dalam mekanika klasik dapat dinyatakan dengan satuan dasar:

Length LPanjangMass MMassaTime TWaktu

Contoh:Kecepatan mempunyai satuan L / T (kilometer per jam).Gaya mempunyai satuan ML / T2 .UnitsPhysics 111: Lecture 1, Pg #Panjang:JarakPanjang (m)Jari-jari alam semesta 1 x 1026Ke galaksi Andromeda2 x 1022Ke bintang terdekat4 x 1016Bumi - matahari1.5 x 1011Jari-jari bumi6.4 x 106Sears Tower 4.5 x 102Lapangan sepak bola1.0 x 102Tinggi manusia 2 x 100Ketebalan kertas 1 x 10-4Panjang gelombang sinar biru 4 x 10-7Diameter atom Hidrogen 1 x 10-10Diameter proton 1 x 10-15

Physics 111: Lecture 1, Pg #Waktu:IntervalTime (s)Umur alam semesta 5 x 1017Umur Grand Canyon 3 x 101432 tahun 1 x 1091 tahun3.2 x 1071 jam3.6 x 103Perjalanan cahaya dari mh ke bumi1.3 x 100Satu kali putaran senar gitar 2 x 10-3Satu putaran gel. Radio FM 6 x 10-8Umur meson pi netral 1 x 10-16Umur quark top 4 x 10-25

Physics 111: Lecture 1, Pg #Massa:ObjectMass (kg)Galaksi Bima Sakti 4 x 1041Matahari 2 x 1030Bumi 6 x 1024Pesawat Boeing 747 4 x 105Mobil 1 x 103Mahasiswa 7 x 101Partikel debu 1 x 10-9Quark top 3 x 10-25Proton 2 x 10-27Electron 9 x 10-31Neutrino 1 x 10-38

Physics 111: Lecture 1, Pg #Satuan ...Satuan Internasional, SI (Systme 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 #9Page #Converting between different systems of unitsUseful 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:

Physics 111: Lecture 1, Pg #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 TDimensi ruas kiri tidak sama dengan dimensi ruas kanan, dengan demikian, jawaban di atas pasti salah!!Analisis DimensionalPhysics 111: Lecture 1, Pg #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 ?

(a)(b)(c)Given: d has units of length (L) and g has units of (L / T 2).P = 2 (dg)2Physics 111: Lecture 1, Pg #Lecture 1, Act 1 SolutionRealize that the left hand side P has units of time (T )Try the first equation

(a)(b)(c)(a)

Not Right !!

Physics 111: Lecture 1, Pg #Page #

(a)(b)(c)(b)Not Right !!Try the second equation

Lecture 1, Act 1 Solution

Physics 111: Lecture 1, Pg #Page #

(a)(b)(c)(c)This has the correct units!!This must be the answer!!Try the third equation

Lecture 1, Act 1 Solution

Physics 111: Lecture 1, Pg #Page #Motion in 1 dimension In 1-D, we usually write position as x(t1 ).

Since its 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

txt1t2xtx1x2some particles trajectoryin 1-DPhysics 111: Lecture 1, Pg #1-D kinematics

txt1t2xx1x2trajectoryVelocity v is the rate of change of positionAverage velocity vav in the time t = t2 - t1 is: tVav = slope of line connecting x1 and x2.Physics 111: Lecture 1, Pg #Consider limit t1 t2Instantaneous velocity v is defined as:1-D kinematics...

txt1t2xx1x2tso v(t2) = slope of line tangent to path at t2.Physics 111: Lecture 1, Pg #1-D kinematics...

Acceleration a is the rate of change of velocityAverage acceleration aav in the time t = t2 - t1 is: And instantaneous acceleration a is defined as:

usingPhysics 111: Lecture 1, Pg #RecapIf 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!

xavtttPhysics 111: Lecture 1, Pg #More 1-D kinematicsWe 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++...+= displacementPhysics 111: Lecture 1, Pg #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

Physics 111: Lecture 1, Pg #RecapSo for constant acceleration we find:

xavttt

Planew/ lightsPhysics 111: Lecture 1, Pg #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.yPhysics 111: Lecture 1, Pg #Lecture 1, Act 2Solution xavtttGoing 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.

Physics 111: Lecture 1, Pg #Derivation:Plugging in for t:

Solving for t:

Physics 111: Lecture 1, Pg #Average VelocityRemember that

vttvvavv0

Physics 111: Lecture 1, Pg #Recap:For constant acceleration:From which we know:

Washers

Physics 111: Lecture 1, Pg #Problem 1A 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 abx = 0, t = 0abvo

Physics 111: Lecture 1, Pg #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 = tfv = 0

x = 0, t = 0abv0Physics 111: Lecture 1, Pg #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 #Problem 1...To find stopping distance we use:

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

Physics 111: Lecture 1, Pg #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

Physics 111: Lecture 1, Pg #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.