Principles of Physics Chapter 1 PDF

Principles of Physics Dr. Mohaned Mohmmed School of Biotechnology, Badr University in Assiut Introduction Lectures will be delivered face-to-face. Textbooks: Physics for Scientists and Engineers, R.A. Serway and J.W. Jewett, 8th Edition, Thomson Books/Cole 2013...

Principles of Physics Dr. Mohaned Mohmmed School of Biotechnology, Badr University in Assiut Introduction Lectures will be delivered face-to-face. Textbooks: Physics for Scientists and Engineers, R.A. Serway and J.W. Jewett, 8th Edition, Thomson Books/Cole 2013. College Physics, PAUL PETER URONE, ROGER HINRICHS, OpenStax Evaluation: Mid-term exam 10 Marks Quizzes/ assignments 10 Marks Practical exam 40 Marks Final exam 40 Marks Principles of Physics Dr. Mohaned Mohammed 2 Course Topics & Timeline Topic No. of Weeks Chapter (1) Physics and Measurement 2 Chapter (2) Statics and Torque 4 MID Term 1 Chapter (3) Temperature 2 Chapter (4) Heat and Thermodynamics 4 Final EXAM Principles of Physics Dr. Mohaned Mohammed 3  CHAPTER 1 Physics and Measurements Topics: 1. Standards of length, mass and time 2. Dimensional Analysis 3. Conversion of units Principles of Physics Dr. Mohaned Mohammed 4  1. Standards of length, mass and time The laws of physics are expressed in terms of basic quantities that require a clear definition. Physical Quantities Mechanics Basic Quantities Force Velocity Volume ……etc. Length Mass Time L M T Any physics quantity in mechanics can be expressed in terms of these three basic quantities. Principles of Physics Dr. Mohaned Mohammed 5  Standards of length: Standard of length is based on speed of light in vacuum. Since 1983: Meter: distance travelled by light in vacuum during a time of 1/299,792,458 sec Standards of mass: Standard of mass is based on the mass of an alloy. Since 1887: Kilogram: mass of a Pt-Ir alloy cylinder. (Pt-Ir alloy is very stable ) Standards of time: Since1960: is based on the atomic clock sec: is based on the characteristic frequency (period of vibration ) of radiation from the Cs137atom. 1 sec = 9 192 631 770 x T T Time Principles of Physics Dr. Mohaned Mohammed 6 International System (SI) of units : In 1960, an International committee established a set of standards for the three basic quantities as follows: Quantity Unit Mass Kilogram (kg) Mechanics Length Meter (m) Time Second (s) Electric current Ampere (Amp) Other Temperature Kelvin (K) Quantities Luminous intensity Candela Amount of substances mol cgs mks British Eng. System Other Systems Length Mass Time Length Mass Time Length Mass Time cm gm s m kg s Foot (ft) Pound (lb) s Closely related to SI Principles of Physics Dr. Mohaned Mohammed 7  2. Dimensional Analysis Dimension: denotes the physical nature of a quantity. Basic quantities in mechanics: Length L Mass M Time T Dimension of some physics quantities: Area [A] =L.L = L2 Volume [V] =L.L.L = L3 Velocity [ν] =L / T = LT-1 Acceleration [a] =L / T2 = LT-2 Momentum [P] = [m]. [ν] = MLT-1 Kinetic Energy [K.E] = [1/2].[m]. [ν2] = M((LT-1)2) = ML2T-2 Potential Energy [P.E] = [m]. [g]. [h] = M(LT-2) L = ML2T-2 Note that [K.E] = [P.E] Principles of Physics Dr. Mohaned Mohammed 8  Dimensional Analysis: A method that can be used to assist in Checking the validity of a specific formula Deriving some formulas Example 2 : A car started from rest and moves with a constant acceleration a. What is the distance x travelled in a time t. 1) If we know the formula and want to check it. x=½ a t2 [L.H.S] = [x ] = L [R.H.S] = [ ½] [a]. [t2] [R.H.S] = LT-2. T2 = L [R.H.S]= [L.H.S] Since dimensions of both side are the same: The formula is correct. Principles of Physics Dr. Mohaned Mohammed 9  (2) If we wish to derive the formula We assume the factors that affect x which are the Acceleration and the Time  x  t n am  x  tn [ x]  [t ] n [ a ]m  x  am L = T n  ( LT − 2 ) m L = T n  LmT −2 m m =1 L = Lm  T n − 2 m n − 2m = 0  n = 2m = 2  x  t 2 a1 x  at 2 x = k at2 k : can not be determined by this method Principles of Physics Dr. Mohaned Mohammed 10  Ex. 3 Analysis of a power law: Suppose we are told that the acceleration of a particle moving with a uniform speed v in a circular orbit of radius r is proportional to some power of r say rn and some power of v say vm. How can we determine the power of r and v? a  r n ( LT -2 ) = (L) n.(LT -1 )m a  v m LT -2 =L nL m T -m LT - 2 = L n + m T -m a  r n v m 1= n+ m  -2 = -m a=kr n v m m=2 2 n = 1 − 2 = −1 a =k r a =kr −1 2 Principles of Physics Dr. Mohaned Mohammed 11  3. Conversion of units Sometimes it is necessary to convert units from one system to another. Examples : 1 mile = 1609 m=1.609 km 1 ft= 0.3048 m = 30.48 cm 1 inch =0.0254 m = 2.54 cm 1m= 100 cm 1kg =1000 gm Ex. 4: The mass of a solid cube is 856 g, and each edge has a length of 5.35 cm. Determine the density  of the cube in basic SI units? 1g = 10-3 kg 1cm = 10-2 m Volume = L3=(5.35x10-2 )3 =1.53x10-4 m3 Mass=856x10-3= 0.856 kg Density = 5.59x103 kg /m3 Principles of Physics Dr. Mohaned Mohammed 12  Quick Quizzes Principles of Physics Dr. Mohaned Mohammed 13  Answer to Quick Quizzes Principles of Physics Dr. Mohaned Mohammed 14 HW- 1 Serway Version 6 P. 15 and 17 15. Which of the following equations are dimensionally correct? (a) vf = vi + ax (b) y = (2 m)cos(kx), where k = 2 m–1. 17. Newton’s law of universal gravitation is represented by Here F is the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance. Force has the SI units kg·m/s2. What are the SI units of the proportionality constant G? Mm F =G r2 Principles of Physics Dr. Mohaned Mohammed 15  𝑀𝑚 Prob.17: Newton’s Law of universal gravitation is 𝐹 = 𝐺 2 𝑟 where F is the force of gravity and M,m are masses; and r is a length. If the Force have the SI units of kg m/s2. What are the units of G? Mm F =G 2 M m r r kg m kg kg 2 = [G ]. 2 s m 𝑚3 [𝐺] = 𝑘𝑔. 𝑠 2 Principles of Physics Dr. Mohaned Mohammed 16 Problem :The period of a pendulum t is measured in time units and is  t = 2 g  = length of the pendelium g = free - fall acceleration measured in length divided by square of time Show that the equation is dimensionally correct.  t = 2 g 1/ 2 R.H.S = 2 1/ 2 g [ L]1/ 2 [ R.H.S] = [ LT − 2 ]1/ 2 1/ 2 [R.H.S] = T = [L.H.S] L 1 [ R.H.S] = = =T L1/ 2T − 2 T −1 Equation is dimensionally correct. Principles of Physics Dr. Mohaned Mohammed 17 End of Lecture Principles of Physics Dr. Mohaned Mohammed 18