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Systems with Multiple Accelerations in 2-Dimensions

This lecture discusses the various systems with multiple accelerations in 2-dimensions, including linear, projectile, and circular motion. It covers discerning different reference frames, identifying forces, and drawing free body diagrams. Homework assignment based on Chapters 2 and 3 is also mentioned.

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Systems with Multiple Accelerations in 2-Dimensions

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  1. Lecture 5 • Goals: • Address systems with multiple accelerations in 2-dimensions (including linear, projectile and circular motion) • Discern different reference frames and understand how they relate to particle motion in stationary and moving frames • Recognize different types of forces and know how they act on an object in a particle representation • Identify forces and draw a Free Body Diagram Assignment: HW2, (Chapters 2 & 3, due Wednesday) Read through Chapter 6, Sections 1-4

  2. Kinematics in 2 D • The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as: r = x i + y j v = vx i + vy j a = ax i + ay j Special Cases: 1. ax=0 ay= -g 2. Uniform Circular Motion

  3. xvst x 4 0 t Special Case 1: Freefall x and y motion are separate and Dt is common to both Now: Let g act in the –y direction, v0x= v0 and v0y= 0 xvsy yvst t = 0 y 4 y x 4 t 0

  4. xvsy t = 0 4 x y Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! Acceleration may or may not be! Example Problem Given How far does the knife travel (if no air resistance)?

  5. 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

  6. 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

  7. 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

  8. 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

  9. Exercise 2Trajectories with acceleration • A • B • C • D • None of these C C After C ? A B C C C D

  10. Exercise 3Relative Trajectories: Monkey and Hunter All free objects, if acted on by gravity, accelerate similarly. A hunter sees a monkey in a tree, aims his gun at the monkey and fires. At the same instant the monkey lets go. Does the bullet … • go over the monkey. • hit the monkey. • go under the monkey.

  11. Schematic of the problem • xB(Dt) = d = v0cosqDt • yB(Dt) = hf = v0sinqDt – ½ g Dt2 • xM(Dt) = d • yM(Dt) = h – ½ g Dt2 • Does yM(Dt) = yB(Dt) = hf? (x,y) = (d,h) Monkey Does anyone want to change their answer ? What happens if g=0 ? How does introducing g change things? hf g v0 q Bullet (x0,y0) = (0 ,0) (vx,vy) = (v0cosq, v0sinq)

  12. Case 2: Uniform Circular MotionCircular motion has been around a long time

  13. = + Changes only in the direction of = 0 a a a v a Generalized motion with only radial acceleration Uniform Circular Motion A particle doesn’t speed up or slow down!

  14. Uniform Circular Motion (UCM) is common so we have specialized terms • Arc traversed s = q r • Tangential velocity vt • Period, T, and frequency, f • Angular position, q • Angular velocity, w s vt r q Period (T): The time required to do one full revolution, 360° or 2p radians Frequency (f): 1/T, number of cycles per unit time Angular velocity or speed w = 2pf = 2p/T, number of radians traced out per unit time (in UCM average and instantaneous will be the same)

  15. s vt r q Angular displacement and velocity • Arc traversed s = q r in time Dt then Ds = Dq r so Ds / Dt = (Dq / Dt) r in the limit Dt  0 one gets ds / dt = dq / dt r vt = w r w≡ dq / dt if w is constant, integrating w = dq / dt, we obtain: q = qo + wDt Counter-clockwise is positive, clockwise is negative

  16. vt ar r Circular motion also has a radial (perpendicular) component Uniform circular motion involves only changes in the direction of the velocity vector, thus acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction. Quantitatively (see text) Centripetal Acceleration ar = vt2/r Circular motion involves continuous radial acceleration

  17. 1 2 at at r r What if w is linearly increasing … • Then angular velocity is no longer constant so dw/dt ≠ 0 • Define tangential acceleration as at = dvt/dt = r dw/dt • So s = s0 + (ds/dt)0Dt + ½ atDt2 and s = q r • We can relate at to dw/dt q = qo + woDt + Dt2 w = wo + Dt • Many analogies to linear motion but it isn’t one-to-one • Note: Even if the angular velocity is constant, there is always a radial acceleration.

  18. v2 ar = r d| | v at = dt Non-uniform Circular Motion For an object moving along a curved trajectory, with non-uniform speed a = ar + at(radial and tangential) at ar

  19. Angular motion, signs • If angular displacement velocity accelerations are counter clockwise then sign is positive. • If clockwise then negative

  20. Circular Motion • UCM enables high accelerations (g’s) in a small space • Comment: In automobile accidents involving rotation severe injury or death can occur even at modest speeds. [In physics speed doesn’t kill….acceleration does (i.e., the sudden change in velocity).]

  21. Mass-based separation with a centrifuge Before After ar=vt2 / r and f = 104 rpm is typical with r = 0.1 m and vt = wr = 2pf r How many g’s? ca. 10000 g’s bb5

  22. Relative motion and frames of reference • Reference frame S is stationary • Reference frame S’ is moving at vo This also means that S moves at – vo relative to S’ • Define time t = 0 as that time when the origins coincide

  23. Relative Velocity • The positions, r and r’, as seen from the two reference frames are related through the velocity, vo, where vo is velocity of the r’ reference frame relative to r • r’ = r – voDt • The derivative of the position equation will give the velocity equation • v’ = v – vo • These are called the Galilean transformation equations • Reference frames that move with “constant velocity” (i.e., at constant speed in a straight line) are defined to be inertial reference frames (IRF); anyone in an IRF sees the same acceleration of a particle moving along a trajectory. • a’= a (dvo/ dt = 0)

  24. y(t) motion governed by 1) a = -gy 2) vy = v0y – gDt 3) y = y0 + v0y – g Dt2/2 Reference frame on the moving cart. x motion: x = vxt Reference frame on the ground. Net motion: R = x(t) i + y(t) j(vector) Central concept for problem solving: “x” and “y” components of motion treated independently. • Example: Man on cart tosses a ball straight up in the air. • You can view the trajectory from two reference frames:

  25. Recap • Assignment: HW2, (Chapters 2 & 3, due Wednesday) • Read through Chapter 6, Sections 1-4

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