1. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4: Motion in Two Dimensions 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 Chapter 4 Goals:

2. 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: the three components of the position: r:= if the arena is 2d, drop the z-stuff displacement is change in position:  r = r B – r A or  r = r f – r i or, best of all,  r = r(t+  t) – r(t) it is a vector with tail at r(t) and tip at r(t+  t) 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!

3. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’II x y r(t)r(t) r(t+t)r(t+t) r(t)r(t) path of object how do we get velocity from this? start with average velocity: of course, v avg :=  r/  t v avg is a vector: it’s a vector (  r) times a number (1/  t) magnitude: |v avg | = = |  r /  t| = |  r |/|  t| direction: same as direction of  r

4. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ III: the (instantaneous) velocity vector 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:= where v x (t)=dx(t)/dt and similarly for y and z velocity is now a vector:

5. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ IV x y r(t)r(t) r(t+t)r(t+t) r(t)r(t) path of object we have let  t get really small (  t  0) and called it dt as that shrunk, so did  r shrink  : call it dr r(t+dt) dr(t)dr(t) direction: tangent to the path in space!!! magnitude: |v| = |dr /dt| = speed

6. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics in ‘higher dimensions’ V: the acceleration vector three components of the acceleration a:= where a x (t)= dv x (t)/dt 2 = d 2 x(t)/dt 2 and similarly for y and z acceleration is now a vector:

7. 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 (a t ), and normal to (a r for ‘radial’), the path magnitude: x y v(t)v(t) v(t+t)v(t+t) v(t)v(t) path of object v(t+dt) dv(t)dv(t) a(t)a(t) a is parallel to dv and rescaled by 1/dt

8. 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/s 2 ≈ 10 m/s 2 ] object is projected with an initial velocity v i g and v i define a plane in space: the xy plane vertical component v yi horizontal component v xi let projection angle be  i so v i = { v i,  i } v yi = v i sin  i and v xi = v i cos  i so v i = assume projectile starts at the origin: = now allow x and y motion to unfold by components We still don’t know the ‘shape’ of the path, but we see that the motion remains in the xy plane forever

9. 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) clearly, the time to return to starting height is 2T what is the maximum height ? (call it y m )

10. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. One more question about projectile motion, on level ground (ending y is same as y i = 0) what is the range – the maximum distance along the ground? (call it R)... This is the x value at t = 2T note that R = 0 for  i = 0° and 90° note that the range for  i is the same as the range for 90°  i the maximum range is at  i = 45°

11. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 this is a parabolic path in space!! Gravity’s Rainbow… {show Active figure 04_07}

12. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Side view of the parabolic trajectory, as seen in the plane of g and v i ymym note that the position vector is pretty complicated confusingly, we use  for the angle of the velocity r r rmrm

13. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The velocity vector is less complicated

14. 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?

15. 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  i corresponds to the maximum range (call it R m ). What is y m in terms of R m ? b) suppose the angle corresponds to the situation where the range is the same as the maximum height y m. What is  i ?

16. 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  t elapses for now v and r change in direction only, both by angle  note the triangle made by r i, r f and  r, with  in there

17. 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 v i, v f and  v, with  in there

18. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Now take that same  t  0 limit  v 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 Kinematics of motion on a curved path III Conclusion: the centripetal acceleration!! direction is toward the circle’s center magnitude is a c = V 2 /R =  2 R since V =  R

19. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. what if v and r are changing? new unit vectors r hat and t hat which are path-dependent r hat points radially out from the center and is ┴ to path t hat points tangentially to the path and is ║to path Kinematics of motion on a curved path IV r hat t hat

20. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of motion on a curved path V then we can say a c = – r hat V 2 /R = – r hat  2 R 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

21. 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 F c 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 F c = ma c so F c = mv 2 /r OK, so what is the centrifugal force,huh??