PHY183-Lecture09 - R e v ie w : Y e s t e r d a y ( 1 ) ! H...

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Unformatted text preview: R e v ie w : Y e s t e r d a y ( 1 ) ! H o r i z o n t a l m o t i o n : c o n s t a n t v e lo c i t y (1) x = x0 + vx 0 t vx = vx 0 Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 9 (2 ) ! V e r t i c a l m o t i o n : f r e e f a ll (3) (4) (5) (6) ! U se n o t at io n co n v e n t io n : y = y0 + v y 0t ! 1 gt 2 2 v y = v y 0 ! gt y = y0 + v y t v y = 1 (v y + v y 0 ) 2 2 2 v y = v y 0 ! 2 g ( y ! y0 ) (7 ) vx 0 ! vx (t = 0 ); vy 0 ! vy (t = 0 ) September 8, 2006 Physics for Scientists&Engineers 1 1 September 8, 2006 Physics for Scientists&Engineers 1 2 R e v ie w : Y e s t e r d a y ( 2 ) ! Trajectory is a parabola in space (xy-plane) for x0 =0 g y = y0 + x ! 2 x2 vx 0 2 vx 0 vy 0 v (t) ! Look at x- and y-components separately ! Horizontal component of the velocity stays constant v( t ) in time -> horizontal line ! Vertical component falls in vy 0 time, with slope -g vx 0 ! Note: if vertical velocity starts positive, it will reach t = vy0/g a point at which it is 0. y or: y = y0 + x tan ! 0 " T h a n k s to vx ( t ) Sl op e g x2 2 2 v cos ! 0 2 0 t = x vx 0 = v0 cos ! 0 v y 0 = v0 sin ! 0 (1) (2) 3 September 8, 2006 vx (t ) = vx 0 v y (t ) = v y 0 ! gt Physics for Scientists&Engineers 1 vy ( t ) -g September 8, 2006 Physics for Scientists&Engineers 1 4 v ( t ) a n d T r a je c t o r y ! Superimpose snapshots of velocity vectors on trajectory at different times • Green arrows: horizontal v-component • Red arrows: vertical v-component • Blue arrows: velocity vector Note: at apex of trajectory, vy changes sign D e p e n d e n c e s o f V e lo c it y ( 1 ) ! What is the dependence of the magnitude of the velocity vector on time and the coordinates? v = vx 2 + v y 2 ! Coordinate dependence: 2 2 v 2 x = v 2 x 0 ; v y = v y 0 ! 2 g ( y ! y0 ) ! Magnitude of the velocity vector: 2 2 v = vx + v y 2 2 = vx 0 + v y 0 ! 2 g ( y ! y0 ) 2 = v0 ! 2 g ( y ! y0 ) 5 September 8, 2006 Physics for Scientists&Engineers 1 6 ! Important: velocity vector forms tangent at every point of trajectory September 8, 2006 Physics for Scientists&Engineers 1 D e p e n d e n c e s o f V e lo c it y ( 2 ) ! U s e p r e v i o u s r e s u lt a n d i n s e r t d e p e n d e n c e o f y -c o o r d i n a t e o n t i m e , y = y0 + vy 0t ! 1 gt 2 2 " y ! y0 = vy 0t ! 1 gt 2 2 M a x im u m H e ig h t o f T r a je c t o r y ! Starting point: trajectory g y = y0 + x tan ! 0 " 2 x2 2 v0 cos 2 ! 0 ! Take derivative of y with respect to x: R e s u lt : 2 v = v0 " 2 g ( y " y0 ) 2 = v0 " 2 g (v y 0t " 1 gt 2 ) 2 2 = v0 " 2 gv y 0t + g 2t 2 & dy d # g g = % y0 + x tan ! 0 " 2 x 2 ( = tan ! 0 " 2 x 2 dx dx $ 2 v0 cos ! 0 ' v0 cos 2 ! 0 ! Find root of this derivative to determine extremum: 0 = tan ! 0 " # xH = 7 September 8, 2006 g xH v cos 2 ! 0 2 0 = v " 2 gtv0 sin ! 0 + g t 2 0 September 8, 2006 Physics for Scientists&Engineers 1 22 2 2 v0 cos 2 ! 0 tan ! 0 v0 v2 = sin ! 0 cos! 0 = 0 sin 2! 0 g g 2g Physics for Scientists&Engineers 1 8 M a x im u m H e ig h t o f T r a je c t o r y ! Starting point: trajectory g y = y0 + x tan ! 0 " 2 x2 2 v0 cos 2 ! 0 ! Take derivative of y with respect to x: ! Result: M a x im u m H e ig h t o f T r a je c t o r y ( 2 ) 2 v0 xH = sin 2! 0 2g & dy d # g g = % y0 + x tan ! 0 " 2 x 2 ( = tan ! 0 " 2 x 2 dx dx $ 2 v0 cos ! 0 ' v0 cos 2 ! 0 ! Find root of this derivative to determine extremum: ! Convince ourselves that 2nd derivative is smaller than 0: d2y d # g = % tan ! 0 " 2 2 dx dx $ v0 cos 2 ! 0 & g x( = " 2 <0 v0 cos 2 ! 0 ' 0 = tan ! 0 " # xH = September 8, 2006 g xH v cos 2 ! 0 2 0 2 2 v0 cos 2 ! 0 tan ! 0 v0 v2 = sin ! 0 cos! 0 = 0 sin 2! 0 g g 2g Physics for Scientists&Engineers 1 9 ! xH is indeed the point where maximum height is reached September 8, 2006 Physics for Scientists&Engineers 1 10 M a x im u m H e ig h t o f T r a je c t o r y ( 3 ) ! What is the maximum height? ! Answer: Illu s t r a t io n ! Plot maximum height H as a function of the launch angle (for a given v0) H = y ( xH ) = y0 + xH tan ! 0 ' g 2 xH 2v cos 2 ! 0 2 0 2 2 " v0 # v2 g = y0 + 0 sin 2! 0 tan ! 0 ' 2 sin 2! 0 % $ 2 2g 2v0 cos ! 0 & 2 g ( = y0 + 2 v0 v2 v2 sin 2 ! 0 ' 0 sin 2 ! 0 = y0 + 0 sin 2 ! 0 g 2g 2g H " y0 = 2 v0 si n 2 ! 0 2g ! Since vy 0 = v0 sin ! 0 , we finally find: H = y0 + 2 vy 0 2g September 8, 2006 Physics for Scientists&Engineers 1 11 September 8, 2006 Physics for Scientists&Engineers 1 12 R ange ! The range, R, of a projectile is defined as the horizontal distance between the launching point and the point where the projectile reaches the same height again from which it started, y(R)=y0: R ange ! The range, R, of a projectile is defined as the horizontal distance between the launching point and the point where the projectile reaches the same height again from which it started, y(R)=y0: y0 = y0 + R tan ! 0 " # R tan ! 0 = # tan ! 0 = #R= September 8, 2006 g R2 2v cos 2 ! 0 2 0 y0 = y0 + R tan ! 0 " # R tan ! 0 = # tan ! 0 = #R= 13 September 8, 2006 g R2 2v cos 2 ! 0 2 0 g R2 2v cos 2 ! 0 2 0 2 0 g R2 2v cos 2 ! 0 2 0 2 0 g R 2v cos 2 ! 0 g R 2v cos 2 ! 0 2 2v0 v2 sin ! 0 cos ! 0 = 0 sin 2! 0 g g Physics for Scientists&Engineers 1 2 2v0 v2 sin ! 0 cos ! 0 = 0 sin 2! 0 g g Physics for Scientists&Engineers 1 14 M a x im u m R a n g e ! For which angle is the range at maximum (for a given v0)? ! Take derivative with of R respect to angle: 2 % dR d " v0 v2 = sin 2! 0 ' = 2 0 cos 2! 0 d! 0 d! 0 $ g g # & Illu s t r a t io n ! Plot range as a function of launch angle ! This is 0 when cos 2! 0 = 0 ! In other words, the launch angle has to be 45 degrees. ! Maximum range for an ideal projectile is then: 2 v0 = g 15 September 8, 2006 Physics for Scientists&Engineers 1 16 Rmax September 8, 2006 Physics for Scientists&Engineers 1 E x a m p le : B a s e b a ll T h r o w ! “frozen rope” = expression for a particularly strong throw from second or third base to first base Question: ! we know that you cannot throw a ball on a straight line, but it instead must follow a parabola; how much does the trajectory deviate from a straight line? R1 Answer: R2 = 2 R1 ! Assume a throw with 90 mph = 40.2 m/s ! Distance from second to first base = 27.4 m = R1 2 ! Range formula # d12 g & v0 1 R= g sin 2! 0 " ! 0 = 2 arcsin % 2 ( = 4.79° $ v0 ' September 8, 2006 Physics for Scientists&Engineers 1 17 E x a m p le : B a s e b a ll T h r o w ( 2 ) ! Now use this angle to obtain the height: 2 v0 sin 2 ! 0 H = y0 + = 1.83 m + 0.57 m = 2.4 m 2g ! Here we have assumed a height of 6 ft = 1.83 m from which the ball is thrown. The middle of the trajectory is 57 cm height (~ 2 ft). ! Throw from third to first base with the same velocity: launch angle 6.81° and maximum height 1.16 m (~ 4 ft) above release point. September 8, 2006 Physics for Scientists&Engineers 1 18 ...
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