International University, Vietnam National University, HCMC Physics 1: Mechanics Tran Nguyen Lan, Ph.D Department of Physics, HCMIU-VNU Phone: 0905 623 462, email: lantrann@gmail.503 Week 01 FUNDAMENTALS OF Halliday & Resnick 1n edition EXTENDED https://drive.com/file/d/1aSR7PEIH6kfWOApoPtP8k4xLH_i6A8VA/view?usp=sharing * No of credits: 02 (30 teaching hours) * Textbook: Halliday/Resnick/Walker (2011) entitled Fundamentals of Physics, 9th edition, John Willey & Sons, Inc. Course Requirements ¢ Attendance + Discussion + Homework: 15% * Assignment: 15% * Mid-term exam: 30% * Final: 40% Preparation for each class * Read text ahead of time ¢ Finish homework Content Part A Dynamics of Mass Point - Chapter 1 Bases of Kinematics - Chapter 2 Force and Motion (Newton’s Laws) Midterm exam Part B Laws of Conservation - Chapter 3 Work and Mechanical Energy - Chapter 4 Linear Momentum and Collisions Part C Dynamics and Statics of Rigid Body - Chapter 5 Rotation of a Rigid Body About a Fixed Axis - Chapter 6 Equilibrium and Elasticity - Chapter 7 Gravitation Final exam Chapter 1 Bases of Kinematics 1. Motion in One Dimension 1. Position, Velocity, and Acceleration 1.
One-Dimensional Motion with Constant Acceleration 1. Freely Falling Objects 1. Motion in Two Dimensions 1. The Position, Velocity, and Acceleration Vectors 1.
Two-Dimensional Motion with Constant Acceleration. Tangential and Radial Acceleration 1. Relative Velocity and Relative Acceleration Measurement ° Why do we need physics? — Explain the nature (develop modern technologies!) ¢ How to know the laws of physics are correct? — Do experiment ¢ Physical quantities have their own units length meter (m) centimeter (cm) mass kilogram (kg) gram (g) time second (s) second (s) 1. Motion in one dimension ¢ Dynamics and kinematics: Động học — Kinematics: describing motion (how objects move) Động lực học — Dynamics: concerning causes of motion (why objects move) ¢ Two basis quantities of motion: — Displacement: Ax = x; — x; — Time interval: At =t,-t; ¢ Restriction for this chapter: — Along the straight line only (vertical, horizontal, slanted) — Will not discuss the cause of motion (force) — Consider particles or particle-like objects only A.
Position, velocity, and acceleration Position defined in terms of a frame of reference. x (m) Positive direction => < Negative direction Q2 Q9 mà | | | | | L x (m) 3 -2 -l O 1 2 3 Origin/ Scalar number: a val Two features of displacement: - its direction (a vector) f\Posidon ati=0| ||| - its magnitude a For positive direction: Ax>0 a_i er, For opposite direction: Ax <0 a (>) 4 (s) t Question Xo =0 tạ=1.5h x, = 40 km xX (origin) ổ To ” > x x; = 20 km (a) What is the displacement of the car after 1.5 h? 2" (b) What is the distance the car travelled after 1. Position-time graph Position x(m) © 2002 Brooks Cole Publishing - a division of Thomson Learning 100 80 60 @) 40 20 | | | | O 1.0 Time ¢(s) Note: position-time graph is not necessarily a straight line, even though the motion is along one dimension! A. Position, velocity, and acceleration Average velocity Unit: m/s, km/h, or cm/s - Magnitude: the slope of the straight Vave = AX _ Xo 7X1 line that connects two particular At t,-t, points on the x(t) curve.
- Sign: the sign of Ax x(m) 0-4 Xf ‘lone = \ Average speed ø0-L * ope = mh “0 i _ totaldistance xi a Note: average speed does | + “‡ + † (Ss) :. : O}u , 2 3 Yas 6 not include direction Question tạ =0 t,=1h Xa ty=15h — Xị=40km ° (origin) a ” > x x, = 20 km What is (a) the average velocity so: (b) the average speed 40 40 over < of the car during the total trip of 60 km? Insta nta neous velocity when delta t approach 0 (t2-t1 approx The instantaneous a particle at a particular time. Position x(m) © 2002 Brooks Cole Publishing - a division of Thomson Learning 100 80 60% 40 20 0 10 2.0 40 ao ume) Average velocity velocity corresponds to the velocity of Ax dx Ar>o At dt Position ©2002 Brooks Cole Pubishing- a division of Thomson Learning x(m) Tangent line đường tiép tuyér 190 : ma © 100 = 80 © 60 20 : 20 Time f(s) 0.0 Instantaneous velocity Speed (not average speed) is the magnitude of velocity: |v| - When move from 0 to t1, (1) the slope=tang = velocity increases From t1 to t2, the graph is approx. a line -> the velocity does not, From t2 to t3, the slope = tang.= velocity decreases x Oo position - ~ * ~ t tị I> lỳ tang.
increas (1) velocity v = const Constant velocity The instantaneous velocities are always the same All the instantaneous velocities will also equal the average velocity v= const t t xX=X,+[vdt; x=x,+Vv[ at; by đ X =Xẹ +V(£ - fq) fạ=Ũ ——> x=xX,+VvV 7 tị f2 đã Sample Problem : The position of an object described by: x = 4-12t+3t2 (x: meters; t: seconds) (1) What is its velocity at t=1s? v=dx/dt=-12+6t=-6 (m/s) (2) Is it moving in the positive or negative direction of x just then? negative (3) What is its speed just then? S=6 (m/s) (4) Is the speed increasing or decreasing just then? cons6e' !Ð S22? “decreasing; 2«†: increasing (5) Is there ever an instant when the velocity is zero? If so, give the time t; if not answer no. t=2 s (6) Is there a time after t= 3 s when the object is moving in the negative direction of x? if so, give t: if not, Sample Problem : The position of an object described by: x = 4-12t+3t2 (x: meters; t: seconds) (1) What is its velocity at t=1s? v=dx/dt=-12+6t=-6 (m/s) (2) Is it moving in the positive or negative direction of x just then? negative (3) What is its speed just then? S=6 (m/s) (4) Is the speed increasing or decreasing just then? O<t<2: decreasing; 2<t: increasing (5) Is there ever an instant when the velocity is zero? If so, give the time t; if not answer no. t=2 s (6) Is there a time after t= 3 s when the object is moving in the negative direction of x? if so, give t: if not, A. Position, velocity, and acceleration Acceleration When a particle’s velocity changes, the particle is said to undergo acceleration Average acceleration = Av = Y2—Ÿ| Unit: m/s2 “es At t,-t, meter per second squ % derive: khai triér Instantaneous acceleration Av(t)_ dv(t)_ d.
dx\ d°x a(t) = lim =—/— |= At=0 At dt dt\dt) at? 1*t derivative 2"3 derivative ° Example 1=0 v2=-40 (moving the opposite of positive directi t=10s (1) t=0s speed = 40 m/s speed = 0 m/s a —allA = > 1=0 X v2=40 (2) t=40s t=0 a speed =40m/s speed= 0m/s E What is the average acceleration in each case? > X ° Example -4m/s^¿ t=10s (1) t=0s speed = 40 m/s speed = 0m/s ——#blA ean 4m/ > -4m/s^2 X (2) t=40s t=0 a speed =40m/s speed= 0m/s E What is the average acceleration in each case? > X Note on the sign of acceleration: If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases. Chapter 1 Bases of Kinematics 1. Motion in One Dimension 1.
Position, Velocity, and Acceleration 1. One-Dimensional Motion with Constant Acceleration 1. Freely Falling Objects 1. Motion in Two Dimensions 1.
The Position, Velocity, and Acceleration Vectors 1. Two-Dimensional Motion with Constant Acceleration. Tangential and Radial Acceleration 1. Relative Velocity and Relative Acceleration 1.
Constant acceleration ' position dv a=—=a const dt t — it “ *>v = vụ + ƒadt 3v = vọ +a(tT— tọ) t, t t to velocity If †az0: V=Vo tat! (1) _ dx ⁄ ` “at 3 2 xX= Xg+ frdt= xy + flvgetalt— tạ)]dt ——ø ° tọ tọ a t, t, ty a(t—f ¥ acceleration X=Xọ + Vọ(t~fạ)+ a —] TY tu: X =Xq + Vot+— 3 + (2) Missing From Eqs (1) and (2) Equation Quantity one can derive three specialized equations v= vo + au ; _- for the constant X ~ Xo = vol + zat ý acceleration ve = vo + 2a(x — Xo) x — #xg = (vụ + v)f a xX — Xo = vt — jal? Vo Checkpoint The following equations give the position x(t) of a particle in four situations: (1) x= 3t-4; (2) x = -5t? + 4t2+ 6; (3) x= 2/t? +4/t; (4) x= 5t - 3. To which of these situations do the equations of the above Table apply? Chapter 1 Bases of Kinematics 1. Motion in One Dimension 1. Position, Velocity, and Acceleration 1.
One-Dimensional Motion with Constant Acceleration 1. Freely Falling Objects 1. Motion in Two Dimensions 1. The Position, Velocity, and Acceleration Vectors 1.
Two-Dimensional Motion with Constant Acceleration. Tangential and Radial Acceleration 1. Relative Velocity and Relative Acceleration In ideal conditions (no air resistance, 1. Free falling objects Galileo Galilei (1563-1642) ‘/ / .com In ideal conditions (no air resistance, 1.
Free falling objects Galileo Galilei (1563-1642) / sclencephoto.com In ideal 1. Free falling objeCtS ,¿nsgonc (no air resistance, no wind, .) Le Galileo Galilei (1563-1642) © ‘/ / .com - “Free-fall” is the state of an object moving solely under the influence of gravity. - Free-fall acceleration is the same for all objects. - The acceleration of gravity near the Earth’s surface is a constant, g = 9.8 m/s? toward the center of the Earth.
Free-falling experiment at NASA (https://www.com/watch Pv=E 43-CfukEgs) Free-falling motion is a special case of the constant acceleration! v=vo-— gt 1 y— 0= 2 † vi)! L2 vot — —Øf 0 25 y—30 vÝ = vệ — 2ø — yo) Free-falling motion is a special case of the constant acceleration! Sample problem: v=vo— gt A ball is thrown directly upwards with an initial velocity of 15 m/s. On its way down, it y—yo= Tự +yo)t was caught at a distance of 1 m below the 2 point from where it was thrown. 1 › Determine: y— y= vot — 28! (a) the maximum height reached by the ball; (b) the time it takes the ball to reach that height; (c) the velocity of the ball when it is caught; v* = v6 — 2e(y — yo) (d) the total time elapsed from where the ball was thrown to where it was caught. Solution (a) First we take y = 0 at the position where the ball is thrown and positive y to be upwards.
At the maxi- mum height the velocity of the ball is zero, v? = v5 — 28(y — y0) 0 = (15 m/s)* — 2(9.6 m/s or if we take the initial position at y = 11. (d) v = vo — gf, substituting for v and vp we have (—15.