Physics 1: Mechanics Đào Ngọc Hạnh Tâm WEEK 2-3, LESSON 2 Physics 1: Mechanics Đào Ngọc Hạnh Tâm

Review lesson 1: Motion in one Dimension 1. Displacement (m): x = xt – x0 2. Velocity (m/s): v = x/t speed (m/s): 3. Acceleration (m/s2): a = v/t 4. Instantaneous velocity and acceleration: 5. Constant acceleration: 6. Free falling: affect by only gravity (g=9.8 m/s2)

Part A Dynamics of Mass Point Chapter 1 Bases of Kinematics 1.1. Motion in One Dimension 1.2. Motion in Two Dimensions 1.2.1. The Position, Velocity, and Acceleration Vectors 1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion 1.2.3. Circular Motion. Tangential and Radial Acceleration 1.2.4. Relative Velocity and Relative Acceleration

Vectors (Recall) R1. Vectors and scalars: A vector has magnitude and direction; vectors follow certain rules of combination. Some physical quantities that are vector quantities are displacement, velocity, and acceleration. Some physical quantities that does not involve direction are temperature, pressure, energy, mass, time. We call them scalars. R2. Components of vectors: A component of a vector is the projection of the vector on an axis. If we know a vector in component notation (ax and ay), we determine it in magnitude-angle notation (a and θ):

θ R3. Adding vectors: R3.1. Adding vectors geometrically: Vector subtraction: R3.2. Adding vectors by components: θ

ϕ is the smaller of the two angles between and R4. Multiplying a vector by a vector: ϕ is the smaller of the two angles between and R4.1. The scalar product (the dot product): R4.2. The vector 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:

y x 1.2. Motion in Two Dimensions 1.2.1. The Position, Velocity, and Acceleration Vectors Position: y M Y A particle is located by a position vector: O X x and are vector components of x and y are scalar components of

y M Y1 Displacement:  Y2 M O X1 X2 x Three dimensions:

Average Velocity and Instantaneous Velocity: Instantaneous Velocity, t0: The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle position.

The scalar components of Three dimensions:

Average Acceleration and Instantaneous Acceleration: Instantaneous Acceleration, t0: The scalar components of Three dimensions:

1. 2. 2. Two-Dimensional Motion with Constant Acceleration 1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion Key point: To determine velocity and position, we need to determine x and y components of velocity and position Along the x axis: Along the y axis:

y x Sample Problem: A particle with velocity (m/s) at t=0 undergoes a constant acceleration of magnitude a = 3.0 m/s2 at an angle = 1300 from the positive direction of the x axis. What is the particle’s velocity at t=5.0 s, in unit-vector notation and in magnitude-angle notation? Key issues: This is a two-dimensional motion, we must apply equations of straight-line motion separately for motion parallel v0x=-2.0 (m/s) and v0y=4.0 (m/s) y At t= 5 s: x The magnitude and angle of :

Projectile motion θ0: launch angle R: horizontal range

Projectile Motion: A particle moves in a vertical plane with some initial velocity but its acceleration is always the free-fall acceleration. Ox, horizontal motion (no acceleration, ax = 0): Oy, vertical motion (free fall, ay = -g if the positive y direction is upward): The equation of the path: Horizontal range:

y x Example: A projectile is shot from the edge of a cliff 115m above ground level with an initial speed of 65.0 m/s at an angle of 350 with the horizontal (see the figure below). Determine: the maximum height of the projectile above the cliff; the projectile velocity when it strikes the ground (point P); point P from the base of the cliff (distance X). (a) At its maximum height: y Hmax x

(b) its velocity: Hmax x (c) Calculate X:

Sample Problem (page 70): Figure below shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0= 82 m/s. (a) At what angle 0 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)

Vo Vo 1 2 R1 R2 However 1>2  t2 < t1:  It is likely answer 4 However when the shells hit the ships, 1>2  t2 < t1: the answer is B the farther ship gets hit first

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

1.2.3. Circular Motion. Tangential and Radial Acceleration If the speed is not constant, means velocity vector changes both in magnitude and in direction at every point then the acceleration includes radial and tangential components. The path of a particle’s motion Radial (centripetal) acceleration Tangential acceleration

The tangential acceleration causes the change in the speed of the particle: parallel to the instantaneous velocity: The radial acceleration arises from the change in direction of the velocity vector : perpendicular to the path (R : radius of curvature of the path at the point)

The magnitude of the acceleration vector : Uniform circular motion (v is constant ) : aT = 0  the acceleration is always completely radial

1.2.4. Relative Velocity and Relative Acceleration A. In one dimension: A is parked, watching a moving car P B is driving at constant speed and also watching P:  The velocity vPA of P as measured by A is equal to the velocity vPB of P as measured by B plus the velocity vBA of B as measured by A. If car P is moving with an acceleration:  Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving object.

B. In two dimensions: Note:

Example: A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. What is the resultant velocity of the motorboat? If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? What distance downstream does the boat reach the opposite shore?

time to cross the river: distance downstream:

From Page 78, in the Chapter 4, Principle of Physics Homework: 6, 7, 11, 20, 27, 29, 54, 58, 66 70, 76 From Page 78, in the Chapter 4, Principle of Physics Acknowledgment: Some slides and images used in this lecture are referred from Lectures of Prof. Phan Bao Ngoc and Dr. Do Xuan Hoi in Department of Physics, HCM International University.