2D Motion and Vector Algebra Lecture

1. Lecture 4 • Goals for Chapter 3 & 4 • Perform vector algebra • (addition & subtraction) graphically or by xyz components • Interconvert between Cartesian and Polar coordinates • Work with 2D motion • Distinguish position-time graphs from particle trajectory plots • Trajectories • Obtain velocities • Acceleration: Deduce components parallel and perpendicular to the trajectory path • Solve classic problems with acceleration(s) in 2D (including linear, projectile and circular motion) • Discern different reference frames and understand how they relate to motion in stationary and moving frames Assignment: Read thru Chapter 5.4 MP Problem Set 2 due this Wednesday

2. Example of a 1D motion problem • A cart is initially traveling East at a constant speed of 20 m/s. When it is halfway (in distance) to its destination its speed suddenly increases and thereafter remains constant. All told the cart spends a total of 10 s in transit with an average speed of 25 m/s. • What is the speed of the cart during the 2nd half of the trip? • Dynamical relationships (only if constant acceleration): And

3. The picture v1 ( > v0 ) a1=0 m/s2 v0 a0=0 m/s2 • Plus the average velocity • Knowns: • x0 = 0 m • t0 = 0 s • v0 = 20 m/s • t2 = 10 s • vavg = 25 m/s • relationship between x1 and x2 • Four unknowns x1 v1 t1 & x2 and must find v1 in terms of knowns x0 t0 x1 t1 x2 t2

4. v1 ( > v0 ) a1=0 m/s2 v0 a0=0 m/s2 x0 t0 x1 t1 x2 Using • Four unknowns • Four relationships t2

5. Using v1 ( > v0 ) a1=0 m/s2 v0 a0=0 m/s2 • Eliminate unknowns first x1 next t1 then x2 x0 t0 x1 t1 x2 t2 1 2 3 4 2 & 3 1 Mult. by 2/ t2 4

6. v1 ( > v0 ) a1=0 m/s2 v0 a0=0 m/s2 x0 t0 x1 t1 x2 Fini • Plus the average velocity • Given: • v0 = 20 m/s • t2 = 10 s • vavg = 25 m/s t2

7. B B A C C Vectors and 2D vector addition • The sum of two vectors is another vector. A = B + C D = B + 2C ???

8. B B - C -C C B Different direction and magnitude ! A 2D Vector subtraction • Vector subtraction can be defined in terms of addition. = B + (-1)C B - C A = B + C

9. U = |U| û û Reference vectors: Unit Vectors • A Unit Vector is a vector having length 1 and no units • It is used to specify a direction. • Unit vector u points in the direction of U • Often denoted with a “hat”: u = û • Useful examples are the cartesian unit vectors [i, j, k] • Point in the direction of the x, yand z axes. R = rx i + ry j + rz k y j x i k z

10. By C B Bx A Ay Ax Vector addition using components: • Consider, in 2D, C=A + B. (a) C = (Ax i + Ayj ) + (Bxi + Byj ) = (Ax + Bx )i + (Ay + By ) (b)C = (Cx i+ Cy j ) • Comparing components of (a) and (b): • Cx = Ax + Bx • Cy = Ay + By • |C| =[(Cx)2+ (Cy)2 ]1/2

11. ExampleVector Addition • {3,-4,2} • {4,-2,5} • {5,-2,4} • None of the above • Vector A = {0,2,1} • Vector B = {3,0,2} • Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C?

12. ExampleVector Addition • {3,-4,2} • {4,-2,5} • {5,-2,4} • None of the above • Vector A = {0,2,1} • Vector B = {3,0,2} • Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C?

13. tan-1 ( y / x ) Converting Coordinate Systems • In polar coordinates the vector R= (r,q) • In Cartesian the vectorR= (rx,ry) = (x,y) • We can convert between the two as follows: y (x,y) r ry  rx x • In 3D cylindrical coordinates (r,q,z), r is the same as the magnitude of the vector in the x-y plane [sqrt(x2 +y2)]

14. Resolving vectors into componentsA mass on a frictionless inclined plane • A block of mass m slides down a frictionless ramp that makes angle  with respect to horizontal. What is its acceleration a ? m a 

15. y’ x’ g q q y x Resolving vectors, little g & the inclined plane • g (bold face, vector) can be resolved into its x,yorx’,y’ components • g = - g j • g = - g cos qj’+ g sin qi’ • The bigger the tilt the faster the acceleration….. along the incline

16. Dynamics II: Motion along a line but with a twist(2D dimensional motion, magnitude and directions) • Particle motions involve a path or trajectory • Recall instantaneous velocity and acceleration • These are vector expressions reflecting x, y & z motion r = r(t) v = dr / dt a = d2r / dt2

17. v Instantaneous Velocity • But how we think about requires knowledge of the path. • The direction of the instantaneous velocity is alonga line that is tangent to the path of the particle’s direction of motion. • The magnitude of the instantaneous velocity vector is the speed, s. (Knight uses v) s = (vx2 + vy2 + vz )1/2

18. Average Acceleration • The average acceleration of particle motion reflects changes in the instantaneous velocity vector (divided by the time interval during which that change occurs). • The average acceleration is a vector quantity directed along ∆v ( a vector! )

19. Instantaneous Acceleration • The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero • The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path • Changes in a particle’s path may produce an acceleration • The magnitude of the velocity vector may change • The direction of the velocity vector may change (Even if the magnitude remains constant) • Both may change simultaneously (depends: path vs time)

20. = + = 0 = 0 a a a a v v a Generalized motion with non-zero acceleration: need bothpath&time Two possible options: Change in the magnitude of Change in the direction of Animation

21. Kinematics • The position, velocity, and acceleration of a particle in 3-dimensions can be expressed as: r = x i + y j + z k v = vx i + vy j + vz k(i,j,kunit vectors ) a = ax i + ay j + az k • All this complexity is hidden away in r = r(Dt) v = dr / dt a = d2r / dt2

22. xvst x 4 0 t Special Case Throwing an object with x along the horizontal and y along the vertical. x and y motion both coexist and t is common to both Let g act in the –y direction, v0x= v0 and v0y= 0 xvsy yvst t = 0 y 4 y x 4 t 0

23. Another trajectory Can you identify the dynamics in this picture? How many distinct regimes are there? Are vx or vy = 0 ? Is vx >,< or = vy ? xvsy t = 0 y t =10 x

24. Another trajectory Can you identify the dynamics in this picture? How many distinct regimes are there? 0 < t < 3 3 < t < 7 7 < t < 10 • I. vx = constant = v0 ; vy = 0 • II. vx = vy = v0 • III. vx = 0 ; vy = constant < v0 xvsy t = 0 What can you say about the acceleration? y t =10 x

25. Exercise 1 & 2Trajectories with acceleration • A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces. • Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C • Sketch the shape of the path from B to C. • At point C the engine is turned off. • Sketch the shape of the path after point C

26. B B B C C C B C Exercise 1Trajectories with acceleration • A • B • C • D • None of these From B to C ? A B C D

27. Exercise 3Trajectories with acceleration • A • B • C • D • None of these C C After C ? A B C C C D

28. Lecture 4 Assignment: Read through Chapter 5.4 MP Problem Set 2 due Wednesday