Physics 1: Mechanics - PDF Free Download

Physics 1: Mechanics Đào Ngọc Hạnh Tâm Office: A1.53, Email: dnhtam@hcmiu.edu.n HCMIU, Vietnam National Uniersity Acknowledgment: Most of these slides are supported by Prof. Phan Bao Ngoc

credits (3 teaching hours) [ 5/6 11/8/18: 7 weeks (includes midterm test)] Textbook: Principles of Physics, 9th edition, Halliday/Resnick/Walker (11),John Willey & Sons, Inc. Course Requirements Attendance + Discussion/homework + Assignment: 3% Mid-term exam: 3% Final: 4% No cellphone & laptop in class Absence rate >% > not allowed to take the final exam. Notice: Finish homework & Read text ahead of time

Contents of Physics 1 Part A Dynamics of Mass Point Chapter 1 Bases of Kinematics Chapter Force and Motion (Newton s Laws) Group Assignment Part B Laws of Conseration Chapter 3 Work and Mechanical Energy Midterm exam 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 Group Assignment Chapter 6 Equilibrium and Elasticity Chapter 7 Graitation Final exam

Part A Dynamics of Mass Point Chapter 1: Bases of Kinematics 1. 1. Motion in One Dimension 1.1.1. Position, Velocity, and Acceleration 1.1.. One-Dimensional Motion with Constant Acceleration 1.1.3. Freely Falling Objects 1.. Motion in Two Dimensions 1..1. Position, Velocity, and Acceleration Vectors 1... Two-Dimensional Motion with Constant Acceleration. Projectile Motion 1..3. Circular Motion. Tangential and Radial Acceleration 1..4. Relatie Velocity and Relatie Acceleration

Measurements Use laws of Physics to describe our understanding of nature Test laws by experiments Need Units to measure physical quantities Three SI Base Quantities : Length meter [m] Mass kilogram [kg] Time second [s] Systems: SI: Système International [m kg s] CGS: [cm gram second]

1.1. Motion in one dimension Kinematics Kinematics: describes motion Dynamics: concerns causes of motion F ma dynamic kinematic To describe motion, we need to measure: Displacement: x x t x (measured in m or cm) Time interal: t t t (measured in s)

1.1.1. Position, Velocity and Acceleration A. Position: determined in a reference frame. Motion of an armadillo t s: x-5 m t3 s: x m x-(-5)5 m Space s. time graph Two features of displacement: - its direction (a ector) - its magnitude

B. Velocity: describing how fast an object moes B.1. Aerage elocity: Unit: m/s or cm/s B.. Aerage speed: The ν ag of the armadillo: 6m ag 3s m/s s ag totaldistance Δt Note: aerage speed does not include direction

If a motorcycle traels m in s, then its aerage elocity is: If an antique car traels 45 km in 3 h, then its aerage elocity is:

Sample Problem (aerage elocity s aerage speed): A car traels on a straight road for 4 km at 4 km/h. It then continues in the opposite direction for another km at 4 km/h. (a) What is the aerage elocity of the car during this 6 km trip? (b) What is the aerage speed? (a) (b)

B.3. Instantaneous Velocity and Speed The aerage elocity at a gien instant ( t ), which approaches a limiting alue, is the elocity: (t) lim Δt Δx(t) Δt dx(t) dt The slope (tanθ) of the tangent line gies (t) Speed is the magnitude of elocity, ex: ±4 km/h, so s4 km/h.

Sample Problem : The position of an object described by: x 4-1t+3t (x: meters; t: seconds) (1) What is its elocity at t 1 s? () Is it moing in the positie or negatie direction of x just then? negatie (3) What is its speed just then? dx/dt-1+6t-6 (m/s) S6 (m/s) (4) Is the speed increasing or decreasing just then? <t<: decreasing; <t: increasing (5) Is there eer an instant when the elocity is zero? If so, gie the time t; if not answer no. t s (6) Is there a time after t 3 s when the object is moing in the negatie direction of x? if so, gie t; if not, answer no. no

C. Acceleration: C1. Aerage acceleration: The rate of change of elocity: Unit: m/s (SI) or cm/s (CGS) C. Instantaneous acceleration: At any instant: a ag Δ Δt The deriatie of the elocity (or the second one of the position) with respect to time. t t Δ(t) d(t) d dx a(t) lim Δt Δt dt dt dt 1 1 d x dt

1.1.. Constant acceleration: If t : (1) dx dt t t x x + dt x + [ + a(t t)]dt d a dt t If t : () t a t const + adt + a(t t ) x t x x + + x (t t + t ) + at a(t t 1 + at )

Specialized equations: From Equations (1) & (): 1 x ( + ) t x

Example: An electron has a3. m/s At t (s): 9.6 m/s Find: at t 1 t-.5 (s) and t t+.5 (s)? Key equation: +at ( is the elocity at s) At time t: +at At t 1 : 1 + at 1 1 + a(t 1 -t) 9.6+3.x(-.5) 1.6 (m/s) At t : + at +a (t -t) 9.6+3.(.5) 17.6 (m/s)

1.1.3. Freely falling objects: Free-fall is the state of an object moing solely under the influence of graity. The acceleration of graity near the Earth s surface is a constant, g9.8 m/s toward the center of the Earth. Free-fall on the Moon Free-fall in acuum

Example (must do): y A ball is initially thrown upward along a y axis, with a elocity of. m/s at the edge of a 5-meters high building. (1) How long does the ball reach its maximum height? () What is the ball s maximum height? (3) How long does the ball take to return to its release point? And its elocity at that point? (4) What are the elocity and position of the ball at t5 s? (5) How long does the ball take to hit the ground? and what is its elocity when it strikes the ground? + at Using two equations: 1 y y + t + at

We choose the positie direction is upward. m/s, y, a -9.8 m/s (1) How long does the ball reach its maximum height? + at gt y At its maximum height, : t g 9.8.4 (s) () What is the ball s maximum height? 1 + t at y y + max +.4 (-9.8)(.4) y + 1 y max.4 (m)

We can use: y a( y y) At the ball s maximum height: (3) How long does the ball take to return to its release point? And its elocity at that point? So: 9.8 y y max.4 (m) 1 max + t at y y + At the release point: y t 1 + t 9.8t t or t 4. 8 (s) 4.8 (s)

+ at You can also use: 9.8(4.8) a( y y) gt (m/s) y : downward (4) What are the elocity and position of the ball at t5 s? gt -9.8 5 1 y t 9.8t.5 (m) 9. (m/s)

(5) How long does the ball take to hit the ground? and what is its elocity when it strikes the ground? y When the ball strikes the ground, y -5 m so t 1 y t 9.8t 5.83 t 1. 5 (s); 75 (s) t 5.83 (s) gt - 9.8 (5.83) 37.1(m/s)

Reiew lesson 1: Motion in one Dimension 1. Displacement (m): x x t x. Velocity (m/s): x/ t ; speed (m/s): 3. Acceleration (m/s ): a / t s total (x) Δt 4. Instantaneous elocity and acceleration: dx(t) (t) dt 5. Constant acceleration: d(t) d dx a(t) dt dt dt + 1 at x x + t + at d x dt 6. Free falling: affected by only graity (g9.8 m/s )

Homework: (1) Read Session -1, page 7 () Problems 1, 3, 4, 7, 1, 1, 3, 34, 48, 5 (Page 3-35)

Part A Dynamics of Mass Point Chapter 1 Bases of Kinematics 1.. Motion in Two Dimensions 1..1. Position, Velocity, and Acceleration Vectors 1... Two-Dimensional Motion with Constant Acceleration. Projectile Motion 1..3. Circular Motion. Tangential and Radial Acceleration 1..4. Relatie Velocity and Relatie Acceleration

Vectors (Recall) R1. Vectors and scalars: A ector has magnitude and direction; ectors follow certain rules of combination. Some physical quantities that are ector quantities are displacement, elocity, and acceleration. Some physical quantities that does not inole direction are temperature, pressure, energy, mass, time. We call them scalars. R. Components of ectors: A component of a ector is the projection of the ector on an axis. If we know a ector in component notation (a x and a y ), we determine it in magnitude-angle notation (a and θ):

R3. Adding ectors: R3.1. Adding ectors geometrically: θ Vector subtraction: R3.. Adding ectors by components:

R4. Multiplying a ector by a ector: ϕ is the smaller of the two angles between and R4.1. The scalar product (the dot product): R4.. The ector product (the cross product):

The direction of is determined by using the right-hand rule: Your fingers (right-hand) sweep into through the smaller angle between them, your outstretched thumb points in the direction of In the right-handed xyz coordinate system:

1.. Motion in Two Dimensions 1..1. The Position, Velocity, and Acceleration Vectors Position: A particle is located by a position ector: y Y M x î r xî + yĵ and yĵ are ector components of r x and y are scalar components of r O X x

Displacement: r r r 1 Δr (x î + y ĵ) (x1î + y1ĵ) r (x - x )î + (y - y )ĵ xî Δ 1 1 + Three dimensions: y Y 1 r xî + yĵ + Y O zkˆ X 1 yĵ M M X x Δr (x - x1)î + (y - y1)ĵ + (z - z1)kˆ xî + yĵ + zkˆ

Aerage Velocity and Instantaneous Velocity: aerage elocity ag r t Instantaneous Velocity, t : lim Δt r dr t dt displacement time interal The direction of the instantaneous elocity of a particle is always tangent to the particle s path at the particle position.

d dt (xî x î + + y yĵ) ĵ dx dt The scalar components of dx x, y dt Three dimensions: î ĵ x + y + î + dy dt dy dt z kˆ ĵ z dz dt

Aerage Acceleration and Instantaneous Acceleration: aerage acceleration a ag Instantaneous Acceleration, t : a lim Δt t a a î The scalar components of a Three dimensions: a x + t change in elocity time interal d dt a ĵ y d d a x x, ay dt dt a î + a ĵ+ x y a z y kˆ;, a z d dt z

1... Two-Dimensional Motion with Constant Acceleration. Projectile Motion Key point: To determine elocity and position, we need to determine x and y components of elocity and position Along the x axis: Along the y axis:

Sample Problem: A particle with elocity.î + 4.ĵ (m/s) at t undergoes a constant acceleration a of magnitude a 3. m/s at an ngle α 13 from the positie direction of the x axis. What is the particle s elocity at t5. s, in unit-ector notation and in magnitude-angle notation? Key issues: This is a two-dimensional motion, we can apply equations of straight-line motions separately x x + a xt; y y + a yt x -. (m/s) and y 4. (m/s) a y α a a x y x a a cosα sinα At t 5 s: 3. cos(13 3. sin(13 ) ) -1.93 (m/s.3 (m/s x -11.7 (m/s); y 15.5 (m/s) 11.7î + 15.5ĵ The magnitude and angle of : y tan( θ) 1.33 θ 17 x + x y ) ) 19.4 (m/s)

Projectile motion θ : launch angle R: horizontal range

Projectile Motion: A particle moes in a ertical plane with some initial elocity but its acceleration is always the freefall acceleration. Ox, horizontal motion (no acceleration, a x ): Oy, ertical motion (free fall, a y -g if the positie y direction is upward): y y + 1 sinθt - gt The equation of the path (trajectory): y ( tanθ ) Horizontal range: x - R ( cosθ ) gx sinθ g

Example: A projectile is shot from the edge of a cliff 115m aboe ground leel with an initial speed of 65. m/s at an angle of 35 with the horizontal (see the figure below). Determine: (a) the maximum height of the projectile aboe the cliff; (b) the projectile elocity when it strikes the ground (point P); (c) point P from the base of the cliff (distance X). (a) At its maximum height: y sinθ - gt sinθ t g 1 y y + sinθt - gt y ( sinθ ) g H max H max 7.9 (m) y H max x

(b) its elocity: cosθ x y a y y 53.5 (m/s) y sinθ 37.8 (m/s) a 9.8 (m/s ) y y P y 115 (m) y 6.36 (m/s) 53.5(m/s) î 6.36 (m/s) ĵ H max x VV VV xx + VV yy 6666. 5555 ( mm ss ) (c) Calculate X: t 9.96 (s) y sin θ + at t X t cosθt x sin θ - y g 53.37 (m)

Sample Problem (page 7): Figure below shows a pirate ship 56 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea leel, fires balls at initial speed 8 m/s. (a) At what angle θ from the horizontal must a ball be fired to hit the ship? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannon balls? (a) (b) θ R θ 7 sinθ g sin 1 or θ gr 63 8 R max sinθ sin( 45) g 9.8 686 (m)

V o θ 1 θ V o R 1 R x t x 1 + Howeer y t cosθ R1 cosθ sin θ g 1 t ; t ; t R cosθ sin θ g It is likely answer 4 sin θ gt when the shells hit the ships, y sin θ 1 1 the farther ship gets hit first θ 1 >θ t < t 1 : the answer is B

1..3. Circular Motion. Tangential and Radial Acceleration Uniform Circular Motion: A particle moes around a circle or a circular arc at constant speed. The particle is accelerating with a centripetal acceleration: a r Where r is the radius of the circle the speed of the particle T r π (T: period)

1..3. Circular Motion. Tangential and Radial Acceleration If the speed is not constant, means elocity ector changes both in magnitude and in direction at eery point then the acceleration includes radial and tangential components. The path of a particle s motion a r a a a t r a a t a a r + a t Radial (centripetal) acceleration Tangential acceleration

The tangential acceleration causes the change in the speed of the particle: parallel to the instantaneous elocity: The radial acceleration arises from the change in direction of the elocity ector: perpendicular to the path a T a R d dt R (R : radius of curature of the path at the point) The magnitude of the acceleration ector : a a + a T R Uniform circular motion ( is constant ) : a T a a R

1..4. Relatie Velocity and Relatie Acceleration A. In one dimension: An is parked, watching a moing car P. Bao is driing at constant speed and also watching P: x x + PA PB x d(x ) d(xpb) dt dt + PA + BA d(x dt PA PB BA The elocity PA of P as measured by A is equal to the elocity PB of P as measured by B plus the elocity BA of B as measured by A. If car P is moing with an acceleration: BA a constant : a PA d( dt ) d( dt ) PA PB + Obserers on different frames of reference that moe at constant elocity relatie to each other will measure the same acceleration for a moing object. a PB d( dt BA BA ) )

B. In two dimensions: r PA PA r PB + PB r BA + BA d( PA ) d(pb) d(ba + dt dt dt a a PA PB ) Note:

Example: A motorboat traeling 4 m/s, East encounters a current traeling 3. m/s, North. What is the resultant elocity of the motorboat? If the width of the rier is 8 meters wide, then how much time does it take the boat to trael shore to shore? What distance downstream does the boat reach the opposite shore? boat/shore boat/shore boat/rier + boat/rier + rier/shore 3 R 5 (m/s); tanθ θ 4 rier/shore 36.9 R

Time to cross the rier: t distancea boat/rier distance downstream: t distanceb distancea boat/rier distanceb rier/shore 8 4 distancec (s) rier/shor et 3 boat/shore 6 (m)

Homework: 6, 11,, 7, 9, 54, 66, 7, 76 From Page 78, in the Chapter 4, Principle of Physics