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Chapter 4: Motion in Two Dimensions Chapter 4 Goals: • To define and understand the ideas of vector position,, displacemen, velocity and acceleration • To understand what is meant by the path (trajectory) of an object as it moves in space • To work out the implications of motion with a constant acceleration: projectile motion • To work out the implications of motion on a circle: circular motion • To semi-quantitatively understand ‘curvy’ motion • To discuss how motion can be described from a moving frame of reference Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ I: position and displacement as vector • position is now a vector: r : x ˆi y ˆj z kˆ • the three components of the position: r:= <x, y, z> • if the arena is 2d, drop the z-stuff • displacement is change in position: Dr = rB – rA or Dr = rf – ri or, best of all, Dr = r(t+Dt) – r(t) • it is a vector with tail at r(t) and tip at r(t+Dt) • displacement vector not usually drawn in standard position but may be, especially if you are adding a second displacement to the first and you don’t really care about ‘initial’ position, or you take that to be zero • the displacement vector is origin-independent! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’II • how do we get velocity from this? • start with average velocity: of course, vavg := Dr/Dt Dr(t) y r(t) path of object r(t+Dt) x • vavg is a vector: it’s a vector (Dr) times a number (1/Dt) • magnitude: |vavg | = = |Dr /Dt| = |Dr |/|Dt| • direction: same as direction of Dr Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ III: the (instantaneous) velocity vector • velocity is now a vector: d Dr v(t ) r (t ) : lim [drop ' t ' argument] Dt 0 Dt dt dx ˆ dˆi dy ˆ dˆj dz ˆ dkˆ i x j y k z dt dt dt dt dt dt dx ˆ dy ˆ dz ˆ i 0 j 0 k 0 dt dt dt • here we used the fact that the unit vectors do not vary with time as the body moves around • the three components of the velocity v:= <vx, vy , vz > where vx(t)=dx(t)/dt and similarly for y and z Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ IV • we have let Dt get really small (Dt 0) and called it dt • as that shrunk, so did Dr shrink : call it dr dr(t) Dr(t) y r(t) path of object r(t+dt) r(t+Dt) x • magnitude: |v| = |dr /dt| = speed v v x2 v 2y v z2 • direction: tangent to the path in space!!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ V: the acceleration vector • acceleration is now a vector: d Dv a(t ) v(t ) : lim a avg : lim Dt 0 Dt 0 Dt dt dvx ˆ dˆi dv y ˆ dˆj dvz ˆ dkˆ i vx j vy k vz dt dt dt dt dt dt : a ˆi 0 a ˆj 0 a kˆ 0 x y z • three components of the acceleration a:= <ax, ay, az > where ax(t)= dvx(t)/dt2 = d2x(t)/dt2 and similarly for y and z Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ VI • precisely the same procedure gives us the acceleration • however, the direction of the acceleration is not tangent to the path: it will have components both tangent to (at), and normal to (ar for ‘radial’), the path a(t) v(t) y path of object Dv(t) v(t+Dt) v(t+dt) dv(t) x a is parallel to dv and rescaled by 1/dt dv 2 2 2 2 2 a : a : a a a a a x y z t r • magnitude: dt Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What kind of motion can occur near the Earth? • gravity is down: a = – g j [g = 9.81 m/s2 ≈ 10 m/s2] • object is projected with an initial velocity vi • g and vi define a plane in space: the xy plane • vertical component vyi horizontal component vxi • let projection angle be qi so vi = {vi ,qi} • vyi = vi sin qi and vxi = vi cos qi so vi = < vxi , vyi > • assume projectile starts at the origin: <xi, yi> = <0,0> •now allow x and y motion to unfold by components v y t v yi gt v x t v xi (constant!!) 1 y(t ) v yit 2 gt 2 xt v xit (steady!!) • We still don’t know the ‘shape’ of the path, but we see that the motion remains in the xy plane forever Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Some questions about projectile motion • what is the time to maximum height? (call it T) v y T 0 v0 y v0 sin q 0 gT T g g voy • clearly, the time to return to starting height is 2T • what is the maximum height ? (call it ym) v0 sin q 0 1 v0 sin q 0 g ym y (T ) v0 yT gT v0 sin q 0 2 g g 1 2 2 v02 sin 2 q 0 v02 sin 2 q 0 v02 sin 2 q 0 g 2g 2g Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 2 One more question about projectile motion, on level ground (ending y is same as yi = 0) • what is the range – the maximum distance along the ground? (call it R) . . . This is the x value at t = 2T 2v0 sin q 0 2vo2 cosq 0 sin q 0 R : x(2T ) v0 x 2T v0 cosq 0 g this can be rewritten using double - angle identity : sin 2 2 sin cos v02 sin 2q 0 R g • note that R = 0 for qi = 0° and 90° • note that the range for qi is the same as the range for 90° qi • the maximum range is at qi = 45° Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. g What is the actual Trajectory in Projectile Motion? What is y(x)? • solve the x(t) equation for t and insert into y(t) equation x 1 t into y (t ) vi sin q i t 2 gt 2 vi cosq i x y ( x) vi sin q 0 vi cosq i x tan q i 1 x g 2 vi cosq i 2 gx2 2vi2 cos 2 q i • this is a parabolic path in space!! • Gravity’s Rainbow… {show Active figure 04_07} Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Side view of the parabolic trajectory, as seen in the plane of g and vi r rm y m r • note that the position vector is pretty complicated • confusingly, we use q for the angle of the velocity Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The velocity vector is less complicated Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example of a problem involving projectile motion You have locked yourself out of the dorm and your roommate throws your keys to you from a window that is 12 m above the ground, with a launch speed of 4 m/s, at a launch angle of 50°. a) how much higher do the keys rise? b) where should you stand to catch the keys? vi2 sin 2 qi 4 m/s 2 sin 50 a) ym 2g 2 10 m/s 2 b) trajectory equation : y ( x) x tan q i 2 .47 m gx2 2vi2 cos 2 q i We know y 12.0 m and so putting all # s in we get 12.0 1.19 x .76 x 2 or .76 x 2 1.19 x 12.0 0 x 1.19 1.19 2 4.76 (12.0) 2.76 1.19 1.42 36.48 1.52 1.19 6.16 4.84 m out from the base of the wall 1.52 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Another example A projectile is launched at a certain angle with a certain speed. a) suppose qi corresponds to the maximum range (call it Rm). What is ym in terms of Rm? b) suppose the angle corresponds to the situation where the range is the same as the maximum height ym. What is qi? vi2 sin 2 45 vi2 a) The launch angle is qi 45 so Rm g g 1 vi2 sin 2 45 vi2 2 Rm ym 2g g 2 4 vi2 sin 2q i vi2 sin 2 q i b) equate expressions for R and ym : g 2g 2vi2 sin q i cosq i vi2 sin 2 q i use sin/cos form for R : g 2g tan q i 4 q i Arctan4 76 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path I • the car moves on a flat (2d) curve at speed v, of radius r • for now assume a circle: constant speed v = V; constant radius r = R… • more complicated: varying r and varying v… • goal is to understand the acceleration • assume a time Dt elapses • for now v and r change in direction only, both by angle Dq • note the triangle made by ri, rf and Dr, with Dq in there Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path II • the velocity vectors form a precisely similar triangle, using vi, vf and Dv, with Dq in there Dr Dv Dr : R :: Dv : V or but dr dt : V and dv dt : a so we have V 2 V result : ac R V Dr Dt Dv Dt divide both sides by Dt : R V dr dt dv dt take limit as Dt 0 : R V R for the magnitude of a c Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. R a V Kinematics of motion on a curved path III • Now take that same Dt 0 limit • Dv gets shorter and shorter… • in the limit, dv points exactly along a, and becomes exactly perpendicular to v itself • result: direction of a is perpendicular to the path (and recall that v was tangent to the path) • this all assumed constant speed V and constant radius R • Conclusion: the centripetal acceleration!! • direction is toward the circle’s center • magnitude is ac = V2/R = w2R since V = w R Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path IV • what if v and r are changing? • new unit vectors rhatand that which are path-dependent • rhat points radially out from the center and is ┴ to path • that points tangentially to the path and is ║to path that rhat rhat that Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path V • then we can say ac = – rhatV2/R = – rhatw2R • fact #1: if radius r varies but speed V does not, then there is only centripetal acceleration, as before (R r) • so, the acceleration is purely radial (centripetal) • BUT if the speed v is changing TOO, then the acceleration has a tangential component as well v v tˆ ( true generally) so a : dv dt v dtˆ dt tˆ dv dt ˆ it can be shown that dt define at dv dt 2 v v rˆ so first term is just a c rˆ dt r r so second term is just a t tˆat ˆa c tˆ dv thus we have a : tˆa t r with v wr and dw dt dt 2 v rˆ r we can get a : tˆr rˆw 2 r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path IV The concept of the centripetal force • if a body moves on a curve, there must be a centripetal force (toward the instantaneous center) acting on it • if the speed is a constant, then Fc is the net force • it is not a ‘new’ force, it is a new name: it might be N, or W, or T, or some combination • We can always say Fc = mac so Fc = mv2/r • OK, so what is the centrifugal force,huh?? Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.