Lecture 3 Physics Fundamentals 1 PDF

Lecture 3 (Chapter 3 Motion in Two or Three Dimensions) 1  Objectives a. To use vectors in describing the position and velocity of a particle in two or three dimensions b. To discuss the acceleration of a body in two or three...

Lecture 3 (Chapter 3 Motion in Two or Three Dimensions) 1  Objectives a. To use vectors in describing the position and velocity of a particle in two or three dimensions b. To discuss the acceleration of a body in two or three dimensions c. To describe and solve problems involving motions in curved and circular paths d. To discuss the concept of relative velocities 2 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Position vector 𝒓 The position vector 𝒓 of a particle at an instant is the vector from the origin 𝑂 to point 𝑃 given by 𝒓 = 𝑥 𝒊Ƹ + 𝑦𝒋Ƹ + 𝑧𝒌. ෡ 3.1 The displacement (change in position) can now be written as ∆𝒓 = 𝑥2 − 𝑥1 𝒊Ƹ + 𝑦2 − 𝑦1 𝒋Ƹ + 𝑧2 − 𝑧1 𝒌.෡ 3 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Average velocity 𝒗𝑎𝑣 The average velocity 𝒗𝑎𝑣 can be defined as ∆𝒓 𝒓2 − 𝒓1 𝒗𝑎𝑣 = = 3.2 ∆𝑡 𝑡2 − 𝑡1 where 𝒓2 and 𝒓1 are the final and initial positions, respectively. The average velocity is the change in position multiplied by by 1Τ∆𝑡. 4 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Instantaneous velocity 𝒗 The instantaneous velocity 𝒗 is ∆𝒓 𝑑𝒓 𝒗 = lim =. 3.3 ∆𝑡→0 ∆𝑡 𝑑𝑡 The magnitude of 𝒗 is the speed 𝑣 of the particle at a point along the path. The direction of 𝒗 is the direction where the particle is moving at an instant along the path. 5 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Components of 𝒗 The instantaneous velocity 𝒗 in three dimensions is ෡ 𝒗 = 𝑣𝑥 𝒊Ƹ + 𝑣𝑦 𝒋Ƹ + 𝑣𝑧 𝒌. where 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑣𝑥 = , 𝑣𝑦 = , 𝑣𝑧 =. 3.4 𝑑𝑡 𝑑𝑡 𝑑𝑡 We can generalize 𝒗 into 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝒗= 𝒊Ƹ + 𝒋Ƹ + 𝒌. ෡ 3.5 𝑑𝑡 𝑑𝑡 𝑑𝑡 6 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education. Magnitude and direction of 𝒗 The magnitude of 𝒗 is the speed 𝑣 of the particle at a point along the path given by 𝒗 =𝑣= 𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2. 3.6 The speed 𝑣 in the 𝑥𝑦-plane (two dimensions) and the direction are 𝑣= 𝑣𝑥2 + 𝑣𝑦2 3.7 𝑣𝑦 tan 𝛼 = , ▪ Young, D. and Freedman, R. (2016). 𝑣𝑥 University Physics with Modern Physics (15th ed). Pearson Education. respectively. 7  8 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  9 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Average acceleration 𝒂𝑎𝑣 The average acceleration 𝒂𝑎𝑣 is ∆𝒗 𝒗2 − 𝒗1 𝒂𝑎𝑣 = =. 3.8 ∆𝑡 𝑡2 − 𝑡1 10 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Instantaneous acceleration 𝒂 The instantaneous acceleration 𝒂 is ∆𝒗 𝑑𝒗 𝒂 = lim =. 3.9 ∆𝑡→0 ∆𝑡 𝑑𝑡 11 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Components of 𝒂 The instantaneous acceleration 𝒂 in three dimensions is ෡ 𝒂 = 𝑎𝑥 𝒊Ƹ + 𝑎𝑦 𝒋Ƹ + 𝑎𝑧 𝒌. where 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 𝑎𝑥 = , 𝑎𝑦 = , 𝑎𝑧 =. 3.10 𝑑𝑡 𝑑𝑡 𝑑𝑡 We can generalize 𝒂 into The magnitude and direction 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 ෡ 3.11 of 𝒂 in 2D are given by 𝒂= 𝒊Ƹ + 𝒋Ƹ + 𝒌. 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝒂 =𝑎= 𝑎𝑥2 + 𝑎𝑦2 As a result, 2 2 2 𝑎𝑦 𝑑 𝑥 𝑑 𝑦 𝑑 𝑧 tan 𝛼 = , 𝑎𝑥 = 2 , 𝑎𝑦 = 2 , 𝑎𝑧 = 2. 3.12 𝑎𝑥 𝑑𝑡 𝑑𝑡 𝑑𝑡 ▪ Young, D. and Freedman, R. (2016). University Physics with respectively. 12 Modern Physics (15th ed). Pearson Education.  13 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Parallel and perpendicular components of acceleration The acceleration 𝒂 has component that is parallel to the path, 𝑎∥ , and a component that is perpendicular to the path, 𝑎⊥. 𝑎∥ shows changes in speed. 𝑎⊥ shows changes in direction of motion. 14 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Parallel and perpendicular components of acceleration Figure 3.11(a) shows that 𝑎∥ ≠ 0 and 𝑎⊥ = 0 for cases where only the speed is changing 𝑣2 > 𝑣1. The path is a straight line Figure 3.11(b) shows that 𝑎∥ = 0 and 𝑎⊥ ≠ 0 for cases where only the direction is changing (speed is constant). The path is curved. 15 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Dependence on speed The acceleration vector is normal to the path when the speed is constant. Otherwise, it points ahead or behind. 16 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  17 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  18 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Projectile motion Projectile is any body that is given an initial velocity and then follows a path determined by the effects of gravitational acceleration and air resistance. A projectile’s path is called trajectory. In this lecture, we ignore the effects of air resistance, and the curvature and rotation of the earth. 19 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Projectile motion Projectile motion is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Projectile motion is a two- dimensional motion. The horizontal motion of the yellow projectile has no effect on its vertical motion. Acceleration due to gravity is purely vertical. 20 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Equations of motion along the 𝑥-axis Since the acceleration is due to gravity only 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔. 3.13 Consider the motion along the 𝑥- axis, 𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 (2.8) 𝑣𝑥 = 𝑣0𝑥 (3.14) 1 𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 + 𝑎𝑥 𝑡 2 2.12 2 𝑥 = 𝑥0 + 𝑣0𝑥 𝑡. 3.15 21 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Equations of motion along the 𝑦-axis Since the acceleration is due to gravity only 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔. 3.13 Consider the motion along the 𝑦- axis, 𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡 (2.8) 𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡 (3.16) 1 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2 2.12 2 1 2 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 − 𝑔𝑡. 3.17 2 22 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Trajectory of a projectile Let 𝑥0 = 0 and 𝑦0 = 0 at 𝑡 = 0 𝑣𝑦 = 0 at the highest point 23 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Velocity in component form The initial velocity, 𝒗0 , can be written in component form. If it makes an angle 𝛼0 wrt the horizontal, then 𝑣0𝑦 = 𝑣0 sin 𝛼0 𝑣0𝑥 = 𝑣0 cos 𝛼0. 3.18 The magnitude 𝑣0 is the initial speed. 24 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Equations of motion The equations of motion become 𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 3.15 𝑥 = 𝑣0 cos 𝛼0 𝑡 3.19 1 2 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 − 𝑔𝑡. 3.17 2 1 2 𝑦 = 𝑣0 sin 𝛼0 𝑡 − 𝑔𝑡. 3.20 2 𝑣𝑥 = 𝑣0𝑥 (3.14) 𝑣𝑥 = 𝑣0 cos 𝛼0 (3.21) 𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡 (3.16) 𝑣𝑦 = 𝑣0 sin 𝛼0 − 𝑔𝑡 (3.22) 25 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Distance, speed, direction of velocity The distance of the projectile 𝑟 from the origin at any time 𝑡 is 𝑟= 𝑥 2 + 𝑦 2. 3.23 The magnitude of velocity (projectile speed) at any time 𝑡 is 𝑣= 𝑣𝑥 2 + 𝑣𝑦 2. 3.24 The direction of velocity is 𝑣𝑦 tan 𝛼 =. 3.25 𝑣𝑥 26 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Shape of trajectory From 3.19 and 3.20, we obtain 𝑥 𝑦 = 𝑣0 sin 𝛼0 𝑣0 cos 𝛼0 2 1 𝑥 − 𝑔 2 𝑣0 cos 𝛼0 𝑦 = tan 𝛼0 𝑥 2 1 𝑥 − 𝑔. (3.26) 2 𝑣0 cos 𝛼0 Its form is similar to the equation of a parabola Trajectory’s shape for 𝑦 = 𝑏𝑥 − 𝑐𝑥 2 projectile motion without air where 𝑏 and 𝑐 are constants. resistance is parabola. 27 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Effects of air resistance Acceleration is not constant. Trajectory is not parabolic in shape 28 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  29 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  30 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  31 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education. Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education. 32  33 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  34 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  35 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  36 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Uniform circular motion Motion in a circular path with constant speed. 𝑎∥ = 0 and 𝑎⊥ ≠ 0 Acceleration vector is normal to the path. 37 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Velocity change Comparing the Figs. 3.28(a) and 3.28(b) ∆𝒗 ∆𝑠 = 𝑣1 𝑅 ∆𝑠 ∆𝒗 = 𝑣1 𝑅 38 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Average and instantaneous acceleration The average acceleration is ∆𝒗 𝑎𝑎𝑣 = ∆𝑡 𝑣1 ∆𝑠 𝑎𝑎𝑣 = 𝑅 ∆𝑡 The instantaneous acceleration is 𝑣1 ∆𝑠 𝑣1 ∆𝑠 𝑎 = lim = lim ∆𝑡→0 𝑅 ∆𝑡 𝑅 ∆𝑡→0 ∆𝑡 39 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Radial acceleration From the instantaneous acceleration 𝑣1 ∆𝑠 𝑣1 ∆𝑠 𝑎 = lim = lim ∆𝑡→0 𝑅 ∆𝑡 𝑅 ∆𝑡→0 ∆𝑡 ∆𝑠 The lim ∆𝑡 is the speed of the ∆𝑡→0 particle at point 𝑃1 , 𝑣1. Therefore, 𝑣2 𝑎𝑟𝑎𝑑 = 3.27 𝑅 where 𝑎𝑟𝑎𝑑 is the radial acceleration. For uniform circular motion, this is also called centripetal acceleration. 40 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Radial acceleration and period The time it takes to complete a single revolution is the period 𝑇. The speed is 2𝜋𝑅 𝑣=. 3.28 𝑇 The radial acceleration is now 𝑣 2 4𝜋 2 𝑅 𝑎𝑟𝑎𝑑 = = 2. 3.29 𝑅 𝑇 41 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Uniform circular motion vs projectile motion 42 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  43 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  44 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Non-uniform circular motion The speed varies in non-uniform circular motion. As a result, 𝑎𝑟𝑎𝑑 is not constant. Recall from previous slides 14-15 that 𝑎∥ ≠ 0. The component 𝑎∥ is called tangential acceleration 𝑎𝑡𝑎𝑛 which is the rate of change of speed Also, 𝑑 ∆𝒗 𝑣2 𝑎𝑡𝑎𝑛 =. 3.30 𝑎𝑟𝑎𝑑 =. 3.30 𝑑𝑡 𝑅 45 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Non-uniform vs uniform circular motion 46 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Relative velocity in 1D Relative velocity is the velocity seen by a particular observer that is relative to that observer. Each observer forms a frame of reference. In Fig. 3.32, the change of position of the passenger from the cyclist’s frame of reference is 𝑥𝑃Τ𝐴 = 𝑥𝑃Τ𝐵 + 𝑥𝐵Τ𝐴. (3.31) The relative velocity is 𝑑𝑥𝑃Τ𝐴 𝑑𝑥𝑃Τ𝐵 𝑑𝑥𝐵Τ𝐴 = + 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑣𝑃Τ𝐴−𝑥 = 𝑣𝑃Τ𝐵−𝑥 + 𝑣𝐵Τ𝐴−𝑥. (3.32) 47 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Relative velocity in 1D From the frame of reference of the passenger, the cyclist is moving to the left. The relative velocity of the cyclist from the passenger’s frame of reference 𝑣𝑃Τ𝐴−𝑥 = −𝑣𝐴Τ𝑃−𝑥 In general, 𝑣𝐴Τ𝐵−𝑥 = −𝑣𝐵Τ𝐴−𝑥. (3.33) 48 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  49 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Relative velocity in 2D or 3D In Fig. 3.34(a), the change of The relative velocity is position of the passenger from 𝒗𝑃Τ𝐴 = 𝒗𝑃Τ𝐵 + 𝒗𝐵Τ𝐴. (3.35) the cyclist’s frame of reference In general, is 𝒗𝐴Τ𝐵 = −𝒗𝐵Τ𝐴. (3.36) 𝒓𝑃Τ𝐴 = 𝒓𝑃Τ𝐵 + 𝒓𝐵Τ𝐴. (3.34) 50 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  51 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  52 ▪ Young, D. and Freedman, R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Quiz 2 Quiz2 Problems: 3.78, 3.80, 3.81 Deadline (2024/11/7 at 11:58PM) 53 ▪ Young, D. and Freedman R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.  Seatwork2, Probset2 Seatwork2 Exercises: 3.3, 3.6, 3.10, 3.22, 3.37, 3.40 Probset2 Problems: 3.46, 3.47, 3.58, 3.68, 3.70 Deadline (2024/11/12 at 11:58PM) 54 ▪ Young, D. and Freedman R. (2016). University Physics with Modern Physics (15th ed). Pearson Education.