CH 4 Motion in two and three Dimensions

Transcription

1 CH 4 Motion in two and three Dimensions I. Position and Displacement: A. Position: 1. The position of a particle can be described by a position vector, with respect to a reference origin. B. Displacement Vector: r 1. Displacement Vector: The displacement of a particle is the change of the position vector during a certain time. r Page 1

2 C. Example of 2D motion 1. Solution t = 15 sec SO: Page 2

3 2. Graphs II. Average Velocity and Instantaneous Velocity A. If a particle moves through a displacement of r in t time, then the average velocity is: B. In the limit that the t time shrinks to a single point in time, the average velocity is approaches instantaneous velocity. This velocity is the derivative of displacement with respect to time. Page 3

4 C. For the previous sample problem, let s find the velocity of the rabbit at 15 sec: 1. Solution t = 15 sec Page 4

5 2. Graph III. Average and Instantaneous Accelerations A. Following the same definition as in average velocity, B. If we shrink t to zero, then the average acceleration value approaches to the instant acceleration value, which is the derivative of velocity with respect to time: Page 5

6 C. For the previous sample problem, let s find the acceleration of the rabbit at 15 sec: 1. Solution t = 15 sec Page 6

7 2. Graph IV. Example Problem A. A Particle moves in the XY plane. Its coordinates vary with time as m 3 m x( t) 1.00 ( t ) 32.0 ( t) 3 s s and m 2 m yt ( ) 5.00 ( t ) 12.0 ( t) 2 s s. Find the position, velocity, and acceleration at t=3sec. 1. Solution Page 7

8 V. Next class (Friday), PROJECTILE MOTION. Read pages Page 8

9 VI. PROJECTILE MOTION. A. A projectile is 1. A particle moving in the vertical plane 2. With some initial velocity 3. Whose acceleration is always free fall acceleration (g) B. The movement of a projectile is projectile motion, with the only acceleration equal to the free fall acceleration, g. Examples in sports: Tennis Baseball Football Lacrosse Racquetball Soccer. C. In projectile motion, the horizontal motion and the vertical motion are ; that is neither motion is affected by the other. D. The initial velocity of the projectile is: v0 1. Here, v0x and v0 y Page 9

10 E. Projectile motion analyzed, assuming no external forces other than the weight: V fy = V 0y + DV V 0y + a y t 1. Follow my format and it will make these problems simpler. 2. Start by always drawing the following grid Vertical a y = g = 9.8 m/s 2 a x =0 Dy (height) = Dx (range) = V oy = V o Sinq= V ox = V o Cosq= V fy = t = t = Horizontal 3. First rule: Time is the ONLY Interloper! Nothing else can cross between Vertical and Horizontal 4. Then let s fill it in with our big 5 equations on the left above the word vertical. 5. After we have the big 5 equations on the left, we write the first equation on the right with x subscripts. Notice that a x = 0 so draw a line through the + ½ a x t 2 because this term goes to zero. 6. Now will fill in our given data from the word problem. 7. Then we circle which variable they want us to find. 8. Then we look at our equations and see which ones we will use to navigate through this problem. We will do a sample problem shortly to make sure we understand this process. However, what if we wanted to plot the projectiles position or know its range. Let s derive two equations which could help us determine these values more rapidly. Page 10

11 F. Projectile s Path 1. From a) Then solve Equation 1 for time. 2. And since 3. Plug that value of t back into Equation above: 4. Finally we get the Equation of Path. Page 11

12 G. Horizontal Range, assuming no external forces: [SHIVOK SP211] September 2, The horizontal range of a projectile is the horizontal distance when it returns to its launching height 2. The distance equations in the x and y directions respectively: 3. Eliminating t: 4. Thus R= 5. Caution: 6. Page 12

13 H. Example, projectile motion: Pleasant Hill 40m 200 m/s 30 ± April 9, 1864 During the civil war, a cannon placed on top of a 40 m high hill is fired with a velocity of 200 m/s at an angle of 30 from horizontal. How close did the cannon operators have to the enemy get in order to ensure a hit? 1. Solution: Vertical a y = g = 9.8 m/s 2 a x =0 Dy (height) = Dx (range) = V oy = V o Sinq= V ox = V o Cosq= V fy = t = t = Horizontal Page 13

14 Page 14 [SHIVOK SP211] September 2, 2015

15 I. Example II, projectile motion: [SHIVOK SP211] September 2, Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Fig. below). Although the fish sees the insect along a straight line path at angle and distance d, a drop must be launched at a different angle 0 if its parabolic path is to intersect the insect. If = 36 and d=0.900m, what launch angle 0 is required for the drop to be at the top of the parabolic path when it reaches the insect? a) Solution: 2 1/2 0 and thus 2 thus Tuesday s lesson (Monday s a Holiday) will be Uniform Circular Motion read pages Page 15

16 VII. Uniform Circular Motion A. A particle is in Uniform Circular Motion if it travels around a circle or a circular arc at a constant (uniform) speed. Note: although the speed is not changing, the particle is accelerating because the velocity is changing direction. B. Drawing C. Centripetal acceleration (center seeking) 1. Proof is in the book on page 77 (proof does not need to be memorized). Page 16

17 D. Period (or Period of revolution) Symbol T. 1. T is defined as the time to make one revolution. 2. T = 3. a c = E. Side notes (will be useful when we get to chapter 10, but only interesting right now) 1. Angular Velocity not meters per second, but number of radians per second. Symbol w. w = 2p/T 2. Velocity = w*r 3. a c = w 2 *r 4. Frequency = revolutions per second = 1/T. Units Hertz or rev/sec or cycles per sec. F. Sample Problem: Uniform Circular Motion 1. A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what distance does the tip move in one revolution? What are (b) the tip's speed and (c) the magnitude of its acceleration? (d) What is the period of the motion? a) Solution Page 17

18 VIII. Relative motion in one dimension [SHIVOK SP211] September 2, 2015 A. The velocity of a particle depends on the reference frame of whoever is observing the velocity. 1. Suppose Alex (A) is at the origin of frame A (as in Fig. 4 18), watching car P (the particle ) speed past. 2. Suppose Barbara (B) is at the origin of frame B, and is driving along the highway at constant speed, also watching car P. Suppose that they both measure the position of the car at a given moment. Then: where x PA is the position of P as measured by A. 3. Graph 4. Consequently, 5. Also, a) Since v BA is constant, the last term is zero and we have a PA = a PB. Page 18

19 6. Example, relative motion, 1 D: a) Page 19

20 Watch the following video about frames of reference it is very interesting and informative (it is a classic): IX. Relative motion in two dimensions A. A and B, the two observers, are watching P, the moving particle, from their origins of reference. B moves at a constant velocity with respect to A, while the corresponding axes of the two frames remain parallel. rpa refers to the position of P as observed by A, and so on. From the situation, it is concluded: Page 20