Status: Unit 2, Chapter 3

Transcription

1 1 Status: Unit, Chapter 3 Vectors and Scalars Addition of Vectors Graphical Methods Subtraction of Vectors, and Multiplication by a Scalar Adding Vectors by Components Unit Vectors Vector Kinematics Projectile Motion Soling Problems in Projectile Motion Relatie Velocity /3/7 Physics 53

2 Section Two Problem Assignment Q3.4, P3.6, P3.9, P3.11, P3.14, P3.73 Q3.1, P3.4, P3.3, P3.43, P3.65, P3.88 /3/7 Physics 53

3 Kinematic Equations for Projectile Motion (+y up, a x, a y -g -9.8m/s ) 3 x x y y gt x x + xo t y y y + y yo t 1 gy gt /3/7 Physics 53

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6 6 What angle gies the most range? This can be cast as a nifty calculus question! And the results may surprise you. What we want is an expression for x in terms of all the other ariables. Then we take the deriatie with respect to the angle to find the maxima. So let s s get to it, we really just follow the illustratie problem from last lesson but keep results general. /3/7 Physics 53

7 7 The range is just the trael time multiplied by the initial x but time was deried from the equation 1 y yo + yot gt which reduced to + t ( 1 cosθ ) t cosθ ) t sinθ g Inserting this into the first equation we see x g xo yo o t ( yo o o gt o o cosθ sinθ g o g cosθ sinθ elocity /3/7 Physics 53

8 8 From the calculus you will recall that finding a maximum or minimum of a function is done by setting the deriatie with respect to the independent ariable to zero. This corresponds to finding those points where the slope of the function is zero the ery definition of an extreme point. /3/7 Physics 53

9 dx dθ d(cosθ sinθ) dθ Notice that the maximum range depends only on the angle with no dependence on the speed or acceleration of sinθ sinθ + cosθ cosθ sin sinθ g d(cosθ sinθ) dθ o cosθ d(cosθ sinθ) dθ graity! which only occurs for θ Now doing the deriatie d(cosθ) d(sinθ) sinθ + cosθ dθ dθ 45 degrees!!!! θ + cos θ 9 /3/7 Physics 53

10 Let's we'll see the maximum range : R R max max Ok, occurs at 45 degrees, goes as and inersely with the magnitude of graity. backtrack cosθ sinθ g ( g )( now, ) we know that at 45 degrees g cos 45sin 45 g there's some physics here!the maximum range the square of the elocity the acceleration of So doubling elocity will quadruple the range. And on the moon the range with be six times longer! 1 /3/7 Physics 53

11 There's more to be learned, starting with the basic result and using the trig identity cosθ sinθ R we can ask for the angle that corresponds to a specific range by recasting the equation as sin θ We're looking for an angle The plot shows that for a gien alue of y there are two alues of cosθ sinθ g Rg θ that is, if sin θ g < 18 θ is a solution so is18 - θ. < θ < 9 or < Which means there are two angles that gie a specific range! Let's explore with the little applet. sin θ θ 11 /3/7 Physics 53

12 1 capa.org/~mmp/kap3/cd6.htm /3/7 Physics 53

13 13 Properties of Trajectories w/ x y Maximum Range At θ 45 o Goes as elocity squared Goes inersely with acceleration For all other ranges Two initial angles. /3/7 Physics 53

14 14 Problem Soling Information can be propagated forward and backwards using the equations of motion as long as a minimum amount of information is aailable. To keep yourself organized follow the usual steps: 1) Draw a figure with a thoughtful origin and xy coordinate system ) Analyze the x and y motion separately, remember they share the same time interal 3) Make the known and unknown table with a x, a y -g, x is constant, and y at the apex. 4) Select the equations with some forethought. /3/7 Physics 53

15 Finding Initial Variables Gien Final Variables 15 The examples last lesson used information about initial position & elocity to find final location. The problem can be reersed. In the next example we use final information to derie initial parameters. Let s s consider that ery familiar situation of a hit softball or baseball A ball player hits a homer and the ball lands in the seats 7.5 m aboe the point at which the ball was hit. The ball lands with V36m/s, 8 degrees below the horizontal. Find the initial elocity of the ball. /3/7 Physics 53

16 Well at first glance this seems difficult. What are we after? Well we need V ox and V oy - from that we can find V o and the direction of the initial elocity ector using V o V x + V y and tanθ V oy /V ox. Remembering to keep the dimensions independent, let s work with the horizontal or x dimension first. Since there is no acceleration in the horizontal direction: V ox V x (36 m/s) ) (cos8 o ) 3 m/s. Now we are halfway there and can focus on y-direction. y Known y o y7.5m g9.8m/s y? y? This looks dire but we actually do hae V y -(36 m/s) ) (sin8 o ) -17 m/s Known y o y7.5m g9.8m/s y -17m/s Unknown t? Unknown t? y? Which is more than enough to proceed. 16 /3/7 Physics 53

17 17 y y y y 38m / s θ ( 17m / 1m / s (we choose only the postie root) y x tan y 1 gy + gy + oy ox y s) + (9.8m / (3m / tan 1 s) s 1m / s 3m / s )(7.5m) To finish up we just need to get the initial elocity ector magnitude and direction. + (1m / 33. s) /3/7 Physics 53

18 A Problem with Initial and final Information: The Air-mail Drop 18 A plane traeling at an altitude of m and a speed of 69m/s makes an air drop. At what distance x must the drop be made so that the package lands near the recipients? This is a mixed problem we know the initial speed and the final distance of the drop, from this we can get missing information. /3/7 Physics 53

19 Let s s use the usual up+y, right +x coordinate system with the origin on the initial position of the plane. What we need is the time of the projectile motion which as usual is obtained from the ertical fall. Then our table takes the form: Known y o m y-m y g9.8m/s Unknown t? y? And we can derie the missing time using the second projectile motion formula for the y direction So y y y t y 1 + gt y g gt gt ( m) t 9.8m / s t 6.39s And the drop distance is just gien by the horizontal distance corresponding to that time. yo + t 1 1 x x t + 69m / s 6.39s 44m 19 /3/7 Physics 53

20 Let s s make this a bit more difficult and for external reasons require that the drop occur 4 m before the recipients. This means some ertical elocity down (since 4 m is less that 44m) will be required to ensure the time taken is correct. What will that elocity be? /3/7 Physics 53

21 1 Well the table is clearly different now: Known y o m y-m g9.8m/s Unknown t? y? y? And doesn t t hae enough information. Howeer we can get the time t from the horizontal information. And the table now is Known y o m y-m t5.8s g9.8m/s Unknown y? y? And the initial ertical elocity can be found from the nd equation x t x x x t 4m 69m / s 5.8s y y + t yo 1 gt /3/7 Physics 53

22 y y yo y gt y + m + 6.1m / yo 1 y + gt y Since the initial s t 1 t 1 gt yo 1 + t gt yo (9.8m / 5.8s t postion is s )(5.8s) As we expected the package required a downward push. /3/7 Physics 53

23 Projectile Motion Parabolic Motion 3 It turns out in the absence of air resistance all projectile motion is simple parabolic motion. This can be show with combination of the x and y equations of motion through substitution for time. Setting x y the equations are considerable simplified x y x x x / 1 x 1 y yot gt We can sole the first for t and substitute into the second to get : x yo g y x x x x which is of the form of y yo t t x Ax Bx g x a parabola /3/7 Physics 53

24 4 Parabolic Mirrors /3/7 Physics 53

25 Relatie Velocity 5 As you can see from past discussions we need to moe easily between reference frames. The tyke tossing a ball from the wagon was a good example. From the ground s frame of reference the ball clearly has two elocity components. From the girl s reference frame there is only the ertical component. /3/7 Physics 53

26 Relatie Velocity 6 Actually moing between frames gets confusing. This can be addressed with a rather prescriptie approach. Let s s set it up with the rather easy example of two trains meeting each other and the extending to the more general situation. +X V A-E V B-E Gloucestershire Warwickshire Railway Cotswolds, England. /3/7 Physics 53

27 Suppose: Train A has elocity m/s relatie to the earth: V A-E m/s Train B has elocity -1m/s relatie to the earth: V B-E -1m/s Then from the trains points of iew: The earth has elocity of -m/s relatie to train A: V E-A -m/s The earth has elocity of 1m/s relatie to train B: V E-B 1m/s And the following equations are true: V B-A V B-E+ V E-A -1m/s+ 1m/s+-m/s m/s-3m/s V A-B V A-E+ V E-B m/s+1m/s+3m/s Notice the nifty way the notation behaes, we hae a sort of cancellation of inner subscripts Also V B-A -V B-A which is true for any ectors. +X V A-E V B-E The axes points in the same direction for all three frames. 7 /3/7 Physics 53

28 Lets consider a person walking in a train at a single instant From ector addition it s s pretty clear that r PE PT +r TE PE r PT TE If we simply take the time deriatie of this equation: PE PT + TE PE PT TE In words, this simply says that the elocity of the person with respect to earth equals the sum of the elocity of the person with respect to the train and the elocity of the train with respect to the earth. Because of commutatiity we could hae written this as PE PT + TE PE PT TE but the canceling of the inner products lends some preference to the former eq. 8 /3/7 Physics 53

29 Let s s consider the more complicated case of a boat in a current. Here we ll see the alue of the subscripts. If we want to go straight across the rier the boat will need to push at an angle with respect to water along the ector V BW. The water moes with respect to the shore according to V WS As a result the boat with respect to the shore will trael along V BS. Vectorially then, V BS V BW + V WS Note for this equation The final ector subscripts correspond to the outer subscripts on the right side The inner subscripts are the same. This notational rule demonstrated aboe, and proen in the preious slide helps us moe quickly and correctly between reference frames. 9 /3/7 Physics 53

30 3 Example: Relatie Velocity between Two Moing Objects Both cars shown are traeling at 11m/s. What is the relatie elocity of car 1 in the reference frame of car? /3/7 Physics 53

31 31 Let the elocity of car 1 in the reference frame of car be denoted V 1. We can also denote the elocity of car 1 relatie to the earth as V 1E. Likewise the elocity of the earth relatie to care is V E. Using our rules we propose V 1 V 1E + V E The last ector is not something we hae but can get with the identity V E -V E Or V 1 V 1E V E Which makes sense as it describes a ector moing at 45 degrees at car. The magnitude is simply gien by the Pythagorean Theorem as: or r 1 1E + (11m / s) 15.5m / s 11m / E + (11m / s) if + y is up, and + x to the right r r si + 11m / sj /3/7 Physics 53

32 3 Schedule Wednesday: We ll reiew the most important information and discus the test Start on Chapter 4: Dynamics! Friday: No class, use the time to study for the test Monday Feb 1 th th : Quiz #1 Problem sets for Units 1 and due. /3/7 Physics 53