PHYS 1050 Announcements:

Transcription

1 PHYS 050 Announcements: Net week: Tutorial covers chapters and 3.5 hr problem solving with instruction Short quiz (multiple choice) Formula sheet provided! PHYS 050 Lecture 7 Constant acceleration in 3d Sept., 0 -- HRW, Ch. 4 in chapter 3, we saw how to add vectors, e.g. C A+ B = C B A in Lecture 6: `everthing we wrote down about -d motion (ch. ) can be etended to 3-d using vector notation just replace with r. in -d, we had, for constant acceleration: = o + vo t + a t

2 Constant acceleration with vectors: then we can generalize to (3d): r = r o + v o t + a t WHAT DOES THIS LOOK LIKE? à let s draw the case r o = 0. Earlier time: t. Later time: t r(t ) a t r(t ) a t t t 3 Constant acceleration cont d. The path of the particle is traced out b the tip of the position vector If we want to include a position offset at t = 0, we need to add all 3 vectors together: particle is here at time t r = r o + v o t + a t r t ( ) a t r(t) the origin is here r o t Particle was here at t = 0 4

3 Equations in 3d for constant acceleration The vector formalism is ver efficient: r = r o + v o t + a t means: o o = + v t + a t o o = + v t + a t o oz z z = z + v t + a t All the relations we worked out in d (HRW table.) appl independentl to each component of r Wow! 5 Projectile Motion ver important case stud: motion under the influence of gravit, a.k.a. projectile motion fortunatel, we onl need d: (horizontal) and (vertical) acceleration has a constant value in the direction: a = -g g = 9.8 m/s a = g ĵ (earth) 6 3

4 Projectile Motion: path Starting from the origin: vo t at Path is a parabola r (t) Slope at t=0 d d t =0 = vo vo θo = tan θo what does this look like in real time? 7 What does it look like? Tr the concept simulation in Wile Plus: 8 4

5 Or, tr the java applet at this web link: Launch speed: Launch angle: θ o θ o R h Ma. height: h Range: R For given speed, greatest distance is for ~45 launch angle. For given angle, greater speed goes farther. à ANYONE who has ever plaed a game of catch KNOWS these basic features of the motion! We just need the equations to go with it J 9 Projectile Motion analsis (start from the origin!) We have two independent motions:. Horizontal (constant v ). Vertical: (constant a = -g ) = t = t g t Idea: use as a measure of time: t = v / o θ o Plug this into the equation for : (This is the eq n of a parabola, with = a + b!)! $ = # " v & o % = tanθ o g! $ # " v & o % g v o cos θ o 0 5

6 Path of versus (starting from the origin) θ o h g = tanθo v cos θ o o Question: How do we find the maimum height h of the projectile? A. Use the fact that B. Use the fact that 0 at C. Either A or B will work D. Wait for Dr. Page to eplain this d / d = 0 at = h v = = h h = g Path of versus : Range v R = g o sin θ o g = tanθo v cos θ o o θ o R Question: How do we find the Range R of the projectile? = 0 at = R v = v at = R A. Use the fact that B. Use the fact that C. Either A or B will work o D. Wait for Dr. Page to eplain this 6

7 Practice Question (HRW Q9) The figure shows three paths for a football kicked from ground level. Rank the paths in order of initial horizontal velocit component, greatest first, and indicate how the horizontal velocit component changes with time: A.,, 3; it is constant B.,, 3; it increases with acceleration g C. 3,, ; it is constant D. 3,, ; it increases with acceleration g 3 Practice Question (HRW Q9) The figure shows three paths for a football kicked from ground level. Rank the paths in order of initial vertical velocit component, greatest first. A.,, 3 B. 3,, C. The are all the same 4 7

8 Chapter 4, Problem 8 In the figure, a stone is projected at a cliff of height h with an initial speed of 4 m/s directed at angle θ o = 60 above the horizontal. The stone lands at point A, 5.5 sec after being launched. Find: a) The height h of the cliff (Ans: 5.8 m) b) The speed of the stone just before it lands at point A (Ans: 7.4 m/s) c) The maimum height H reached above the ground (Ans: 67.5 m) 5 Summar and Outlook: We ve covered the vector equations and graphical analsis for motion in 3d Special case: constant acceleration due to gravit = projectile motion Read section 4.8 relative motion for Monda Tr these problems for toda s class:, 3, 7,