PHY183-Lecture11 - M o v in g R e fe r e n c e F r a m e s...

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Unformatted text preview: M o v in g R e fe r e n c e F r a m e s ! So far: we moved the origin of the coordinate system to a location that is convenient for calculations Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 11 • Example: shifting x0 so that x0 = 0 at the start of a projectile’s trajectory ! Until now we have always kept the coordinate system in the same place during the motion that we want to describe. ! But there are some situations for which a moving coordinate system is necessary • Example: Airplane landing on a moving aircraft carrier September 8, 2006 Physics for Scientists&Engineers 1 1 September 8, 2006 Physics for Scientists&Engineers 1 2 E x a m p le : A ir p o r t W a lk w a y ! Person walking with a velocity vw, as measured by an observer moving along with him on the walkway. ! Walkway surface moves with vwt relative to terminal. ! Two velocities add as vectors ! Velocity of person as measured by someone standing in the terminal: M o v in g R e fe r e n c e F r a m e s ! Demand that reference frame (= coordinate system) moves with constant velocity relative to a coordinate system that is at rest ! Then accelerations measured in both reference frames are the same ! Airport walkway example, again: • If vwt = const . ! dvwt / dt = 0 • From vt = vwt + vw we then obtain: dvt d (vwt + vw ) dvwt dvw dv = = + =0+ w dt dt dt dt dt ! at = aw 3 September 8, 2006 Physics for Scientists&Engineers 1 4 vt = vw t + v w September 8, 2006 Physics for Scientists&Engineers 1 3 D im e n s io n s ! Two coordinate systems x,y,z and x’,y’,z’ that have their axes parallel to each other and coincide at t=0. ! Origin of x’,y’,z’ moves with ! constant velocity vT relative to x,y,z ! After some time t, the origin of x’,y’,z’ is located at ! Vector addition gives us the transformation between frames 3 D im e n s io n s ( 2 ) ! Velocities: !!!!! r = r '+ rT = r '+ vT t d! d ! ! d! ! ! r = (r '+ vT t ) = r '+ vT dt dt dt !!! v = v '+ vT ! Accelerations: d! d ! ! d! v = (v '+ vT ) = v '+ 0 dt dt dt !! rT = vT t !!!!! r = r '+ rT = r '+ vT t September 8, 2006 Physics for Scientists&Engineers 1 5 !! a = a' September 8, 2006 Physics for Scientists&Engineers 1 6 A ir p la n e in C r o s s w in d Airplane moves with a speed of 160. m/s in direction NE. The wind is blowing with 32.0 m/s in direction W. A ir p la n e in C r o s s w in d ( 2 ) Airplane moves with a speed of 160. m/s in direction NE. The wind is blowing with 32.0 m/s in direction W. Question: ! What is the velocity vector – speed and direction – of the airplane relative to the ground? Answer: ! In coordinate system that moves with wind, airplane has velocity components: vx = v cos! = 160 m/s " cos45°=113 m/s vy = v sin ! = 160 m/s " sin 45°=113 m/s ! Wind velocity components: vx ,T = !32 m/s vy , T = 0 September 8, 2006 Physics for Scientists&Engineers 1 7 September 8, 2006 Physics for Scientists&Engineers 1 8 A ir p la n e in C r o s s w in d ( 3 ) ! Add them up to find plane’s velocity relative to ground: vx ' = vx + vx ,T = 113 m/s - 32 m/s = 81 m/s vy ' = vy + vy,T = 113 m/s A ir p la n e in C r o s s w in d ( 3 ) Question: ! What is the course change due to the wind? Answer: ! Easier: realize that the course change due to the wind is the wind velocity (towards the west) times 7,200 s: ! Absolute magnitude and direction: 2 2 v ' = vx '+ vy ' = 139 m/s rT = vT t = 32.0 m/s ! 7200 s = 230.4 km ! ' = arctan(vy '/ vx ') = 54.4° (as compared to 160 m/s and 45°, a reduction of 21 m/s and deflection by 9.4° from original course) September 8, 2006 Physics for Scientists&Engineers 1 9 September 8, 2006 Physics for Scientists&Engineers 1 10 D r iv in g T h r o u g h R a in ! When driving through the rain, one notices that the rain almost always comes straight at us. Why? ! Again, this is a consequence of moving reference frames. V ie w o f o b s e r v e r s ta n d in g o n s tr e e t V ie w fr o m th e c a r E x a m p le : D r iv in g T h r o u g h t h e R a in ! It is raining and there is no wind to speak of. Rain drops of 0.08 inches diameter (typical) are falling with a terminal velocity of 14 mph (6.26 m/s) ! A car is driving through the rain (direction does not matter!) with a speed of 25 mph (11.2 m/s). With which angle relative to the horizontal does the rain hit the car? ! vrain -vcar is the velocity of reference frame September 8, 2006 Physics for Scientists&Engineers 1 11 ! September 8, 2006 ! ! vcar ! = tan "1 vrain 14 = tan "1 = 29.2° vcar 25 12 Physics for Scientists&Engineers 1 P r o je c t ile s in M o v in g R e fe r e n c e F r a m e s ! Bow hunting: target is 25 m away, shoot an arrow with initial velocity 90 m/s, horizontal direction, exactly aimed at bull’s eye. Target moves with 3 m/s from left to right. ! Question: Where does the arrow hit? P r o je c t ile s in M o v in g R e fe r e n c e F r a m e s ! Answer: • It takes t = 25m/(90 m/s) = 0.28 s for arrow to arrive at target • Arrow falls by 0.5·9.81·0.282 m = 0.38 m • Target movement is moving reference frame => sideward deflection of arrow by vt = 3 m/s·0.28 s = 0.83 m Aim here! September 8, 2006 Physics for Scientists&Engineers 1 13 September 8, 2006 Physics for Scientists&Engineers 1 14 ...
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