PHYS 1441 Section 501 Lecture 4 Monday June 14 2004 Dr Jaehoon Yu Chapter three Motion in two dimension Two dimensional equation of motion Projectile motion Chapter four Newton s Laws of Motion Force and Mass Newton s Laws of Motion Solving problems using Newton s Laws Friction Forces First term exam at 6pm next Wednesday June 23 Monday June 14 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 1 Announcements E mail distribution list 36 of you have registered This Wednesday is the last day of e mail registration 5 extra points if you don t register by next Wednesday June 23 A test message was sent out today Term exam Date Next Wednesday June 26 Time In the class 6 7 50pm Covers Chapters 1 6 5 and Appendix A Location Classroom SH 125 Mixture of multiple choices and numeric calculations There will be a total of 3 exams of which two best will be chose for your grades Missing an exam is not permitted unless approved prior to the exam by me I will only be in e mail contact for this exam Class Schedule Dr White will teach you on June 16 and 21 Mr Kaushik Rm 133 x25700 will proctor your exam on June 23 Monday June 14 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 2 Displacement Velocity and Acceleration in 2 dim Quantity One Dimension Two Dimensions Displacement x xf xi r r f r i Average Velocity xf xi x vx tf ti t Instantaneous Velocity Dx vx lim t 0 Dt r r f ri v t t f ti r r Dr v lim Dt 0 Dt Average Acceleration Instantaneous Acceleration Monday June 14 2004 vxf vxi Dvx a v v f v i ax t t f ti tf t i Dt Dvx ax lim t 0 Dt PHYS 1441 501 Summer 2004 Dr Jaehoon Yu r r Dv a lim Dt 0 Dt 3 2 dim Motion Under Constant Acceleration r r r r i xi i yi j r r r v i vxi i v yi j Position vectors in x y plane Velocity vectors in x y plane Velocity vectors in terms of acceleration vector r r r r f x f i y f j r r r v f vxf i v yf j vxf vxi axt v yf v yi a y t r r r r r v f vxi ax t i v yi a y t j vi at How are the position vectors written in acceleration vectors 1 1 x f xi vxi t axt 2 y f yi v yi t a y t 2 2 2 r r r r r 1 1 2 2 xi vxi t a x t i yi v yi t a y t j r f x f i y f j 2 2 r r r r r 2 u r r 1 r 1r2 xi i yi j vxi i v yi j t ax i a y j t ri v i t at 2 2 Monday June 14 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 4 Example for 2 D Kinematic Equations A particle starts at origin when t 0 with an initial velocity v 20i 15j m s The particle moves in the xy plane with ax 4 0m s2 Determine the components of velocity vector at any time t vxf vxi ax t 20 4 0t m s v yf v yi a y t 15 0t 15 m s Velocity vector r r r r r v t vx t i v y t j 20 4 0t i 15 j m s Compute the velocity and speed of the particle at t 5 0 s r r r r r r r vt 5 vx t 5 i v y t 5 j 20 4 0 5 0 i 15 j 40i 15 j m s r 2 2 speed v vx v y Monday June 14 2004 40 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 2 15 43m s 2 5 Example for 2 D Kinematic Eq Cnt d Angle of the Velocity vector v y q tan vx 1 15 3 1 1 o tan tan 21 40 8 Determine the x and y components of the particle at t 5 0 s 1 2 1 x f vxi t ax t 20 5 4 52 150 m 2 2 y f v yi t 15 5 75 m Can you write down the position vector at t 5 0s r r r r r r f x f i y f j 150i 75 j m Monday June 14 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 6 Projectile Motion Monday June 14 2004 The only acceleration in this PHYS 1441 501 Summer 2004 Dr Jaehoon motion It isYua constant 7 Projectile Motion A 2 dim motion of an object under the gravitational acceleration with the assumptions Gravitational acceleration g is constant over the range of the motion Air resistance and other effects are negligible A motion under constant acceleration Superposition of two motions Horizontal motion with constant velocity and Vertical motion under constant acceleration ow that a projectile motion is a vxi vi cos qi r r r r a ax i a y j g j 1 y f v yi t g t 2 2 1 2 vi sin qi t gt 2 Monday June 14 2004 In a projectile motion the only acceleration is gravitational one whose direction is always toward the center of parabola the earth downward v yi vi sin qi ax 0 Plug in the t above x f vxi t vi cos qi t t xf vi cos qi xf 1 xf g y f vi sin i vi cos i 2 vi cos i 2 g x f y f x f tan i 2 2 2vi cos i PHYS 1441 501 Summer 2004 What kind of parabola is this Dr Jaehoon Yu 8 2 Example for Projectile Motion A ball is thrown with an initial velocity v 20i 40j m s Estimate the time of flight and the distance the ball is from the original position when landed Which component determines the flight time and the distance 1 y f 40t g t 2 0m 2 Flight time is determined by y component because t 80 gt 0 the ball stops moving when it is on the ground after the flight Distance is determined by x component in 2 dim because the ball is at y 0 position when it completed it s flight Monday June 14 2004 80 t 0 or t 8 sec g t 8sec x f vxi t 20 8 160 m PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 9 Horizontal Range and Max Height Based on what we have learned in the previous pages one can analyze a projectile motion in more detail Maximum height an object can reach What happens at the maximum height Maximum range At the maximum height the object s vertical motion stops …
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