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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4. Kinematics in Two Dimensions Chapter Goal: To learn to solve.

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Presentation on theme: "Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4. Kinematics in Two Dimensions Chapter Goal: To learn to solve."— Presentation transcript:

1 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 4. Kinematics in Two Dimensions Chapter Goal: To learn to solve problems about motion in a plane.

2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Ch. 4 - Student Learning Objectives To identify the acceleration vector for curvilinear motion. To compute two-dimensional trajectories. To read and interpret graphs of projectile motion. To solve problems of projectile motion using kinematics equations. To understand the kinematics of uniform and non-uniform circular motion.

3 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration in two dimensions Is the particle speeding up, slowing down, or traveling at a constant speed? Draw a dot to indicate the position of this particle after 1 second.

4 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration in two dimensions Since there is a component of the acceleration vector both parallel and perpendicular to the velocity vector, the particle must change speed as well as direction. ∆v

5 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. 0–1 s B. 1–2 s C. 2–3 s D. 3–4 s During which time interval is the particle described by these position graphs at rest?

6 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. 0–1 s B. 1–2 s C. 2–3 s D. 3–4 s During which time interval is the particle described by these position graphs at rest?

7 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another Given the initial velocity vector and acceleration vector shown, construct a motion diagram for a by generating a series of velocity vectors. Assume a time interval of 1 second.

8 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another This looks like a parabolic trajectory. Is this mathematically true?

9 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another This looks like a parabolic trajectory Is this mathematically true?

10 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another A toy rocket ship is launched with a constant vertical acceleration of 8 m/s 2 from a rolling cart with an initial velocity of 2 m/s. Draw a graph of this particle’s trajectory for t = 0,1, 2,3,4 and 5 s. Label the x,y coordinates of each point. Is this a parabolic trajectory? How do you know?

11 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration is constant along one axis and zero along another A toy rocket ship is launched with a constant vertical acceleration of 8 m/s 2 from a rolling cart with an initial velocity of 2 m/s. The equation of this trajectory is y = (a/2v 0x 2 )x 2 – i.e. a parabolic trajectory This equation is only valid if all quantities are in SI units!

12 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A rocket-powered hockey puck moves on a frictionless horizontal table. It starts at the origin. At t=5s, v x = 40 cm/s. a.In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis. b.How far from the origin is the puck at t = 5s?

13 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

14 Projectile motion on another planet A student on planet Exidor throws a ball that follows a parabolic trajectory as shown. The ball’s position is shown at 1- second intervals until t = 3s. At t=1s, the ball’s velocity is the value shown. a.Determine the ball’s velocity at t = 0,2, and 3 seconds. b.Determine the value of g on Exidor. c.Determine the angle at which the ball was thrown.

15 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

16 Projectile motion on another planet |g| = 2 m/s 2 θ = tan -1 (v 0y /v 0x ) = 63 0

17 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion Period (T) – the time it takes to go around the circle once. The SI units of T are seconds

18 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion For a particle traveling in circular motion, we can describe it’s position by distance from origin, r, and angle θ, in radians: θ = s/r where s is the arc length (meters) and r is the radius of the circular motion (also in meters). The unit of radians is dimensionless. There are 6.28 (2π) radians in a circle. By convention, counterclockwise direction is positive.

19 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion

20 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moves cw around a circle at constant speed for 2.0 s. It then reverses direction and moves ccw at half the original speed until it has traveled through the same angle. Which is the particle’s angle-versus-time graph?

21 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moves cw around a circle at constant speed for 2.0 s. It then reverses direction and moves ccw at half the original speed until it has traveled through the same angle. Which is the particle’s angle-versus-time graph?

22 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Circular Motion – Velocity Vector Components Consider the velocity vector The components are v r = 0, v t = where r denotes radial component and t denotes tangential component. For tangential components, ccw = positive, cw = negative. For radial components, positive is into the center of the circle.

23 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion v t = ds/dt and s = rθ v t = r dθ/dt v t = ωr (with ω in rads/s)

24 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration for uniform circular motion Consider the acceleration vector, graphical vector subtraction shows that: a t = 0, a r = It can be shown that: a r = v 2 /r or a r = ω 2 r For uniform circular motion, the magnitude of a is constant, but direction is not, so kinematic equations are not valid.

25 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e.

26 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. (a r ) b > (a r ) e > (a r ) a = (a r ) c > (a r ) d Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e.

27 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Nonuniform Circular Motion If the object changes speed: a t = dv/dt a r = v 2 /r where v is the instantaneous speed. if a t is constant, arc length s and velocity v t can be found with kinematic equations.

28 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Nonuniform Circular Motion For non-uniform circular motion: the acceleration vector no longer points towards the center of the circle. a t can be positive (ccw) or negative (cw) but a r is always positive.

29 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration a t = dv/dt where v = v t v t = r ω a t = r dω/dt The quantity dω/dt is defined as the angular acceleration: α = dω/dt, in rad/s 2 a t = r α

30 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Relationship between angular and tangential quantities Point 1 and 2 on the rotating wheel have different radii. θ 1 = θ 2 but s 1 ≠ s 2 ω 1 =ω 2 but v 1 ≠ v 2 α 1 = α 2 but a t1 ≠ a t2

31 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration the graphical relationships for angular velocity and acceleration are the same as those for linear velocity and acceleration

32 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Acceleration if α is constant, θ and ω can be found with the angular equivalents of the kinematic equations.

33 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The fan blade is slowing down. What are the signs of ω and α?  is positive and  is positive. B.  is negative and  is positive. C.  is positive and  is negative. D.  is negative and  is negative.

34 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The fan blade is slowing down. What are the signs of ω and α?  is positive and  is positive. B.  is negative and  is positive. C.  is positive and  is negative. D.  is negative and  is negative.

35 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Numerical Problem (#36) A 3.0 cm diameter crankshaft that is rotating at 2500 rpm comes to a halt in 1.5 s. a.What is the tangential acceleration of a point on the surface? b.How many revolutions does the crankshaft make as it stops?

36 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Convert all non SI units Decide whether time is important

37 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Note negative sign on the value for a t is not solely because it is slowing down, but because direction of ω is positive


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