Chapter 4 - Motion in 2D and 3D

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Britney Miles
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1 Never confuse motion with action. - Benjamin Franklin David J. Starling Penn State Hazleton PHYS 211

2 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) r(t) = x(t)î + y(t)ĵ + z(t)ˆk

3 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) x r(t) = x(t)î + y(t)ĵ + z(t)ˆk r = r 2 (t) r 1 (t)

4 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) x v avg (t) r(t) = x(t)î + y(t)ĵ + z(t)ˆk r = r 2 (t) r 1 (t) v avg (t) = r t

5 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) x v avg (t) v(t) r(t) = x(t)î + y(t)ĵ + z(t)ˆk r = r 2 (t) r 1 (t) v avg (t) = r t v(t) = d r dt = v x(t)î + v y (t)ĵ + v z (t)ˆk

6 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) x v avg (t) v(t) a avg (t) r(t) = x(t)î + y(t)ĵ + z(t)ˆk r = r 2 (t) r 1 (t) v avg (t) = r t v(t) = d r dt = v x(t)î + v y (t)ĵ + v z (t)ˆk a avg (t) = v t

7 Position, displacement, velocity and acceleration can be generalized to 3D using vectors. x(t) x v avg (t) v(t) a avg (t) a(t) r(t) = x(t)î + y(t)ĵ + z(t)ˆk r = r 2 (t) r 1 (t) v avg (t) = r t v(t) = d r dt = v x(t)î + v y (t)ĵ + v z (t)ˆk a avg (t) = v t a(t) = d v dt = a x(t)î + a y (t)ĵ + a z (t)ˆk

8 We can also generalize two of our constant acceleration equations. v(t) = v 0 + at x(t) = x 0 + v 0 t at2 v(t) = v 0 + at r(t) = r 0 + v 0 t at2

9 We can also generalize two of our constant acceleration equations. v(t) = v 0 + at x(t) = x 0 + v 0 t at2 v(t) = v 0 + at r(t) = r 0 + v 0 t at2 v 2 x = v 2 0,x + 2a x x v 2 y = v 2 0,y + 2a y y v 2 z = v 2 0,z + 2a z z

10 Objectives (Ch 4) Lecture Question 4.1 When an object is thrown (ignoring air drag), after it has left the thrower s hand, (a) v x and v y are constant. (b) v x and v y change with time. (c) v x changes with time but v y is constant. (d) v x is constant but v y changes with time.

11 Projectile motion is a very common example of 2D motion where objects move under the influence of gravity.

12 Projectile motion is a very common example of 2D motion where objects move under the influence of gravity. This ball is also rotating we ll get to that later (Ch 10).

13 In projectile motion, the acceleration in the horizontal direction is 0 m/s 2.

14 In projectile motion, the acceleration in the horizontal direction is 0 m/s 2. If we pick +x as right, a x = 0 m/s 2.

15 In projectile motion, the acceleration in the vertical direction is g = 9.81 m/s 2.

16 In projectile motion, the acceleration in the vertical direction is g = 9.81 m/s 2. If we pick +y as up, a y = 9.8 m/s 2.

17 In projectile motion, the horizontal and vertical motion are independent of each other.

18 In projectile motion, the horizontal and vertical motion are independent of each other. We use our standard equations: x(t) = x 0 + v 0,x t a xt 2 y(t) = y 0 + v 0,y t a yt 2

19 Lecture Question 4.2 A bullet is aimed at a target on the wall a distance L away from the firing position and the bullet strikes the wall a distance y below the mark. If the distance L was half as large, and the bullet had the same initial velocity, how would y change? (a) y 2 y (b) y 4 y (c) y y/2 (d) y y/4 (e) Need more information.

20 An object is in uniform circular motion when its speed is constant and it travels in a circle.

21 An object moving in a circle experiences acceleration (even if it s moving at constant speed!).

22 An object moving in a circle experiences acceleration (even if it s moving at constant speed!).

23 An object moving in a circle experiences acceleration (even if it s moving at constant speed!). If the object moves faster, should the acceleration be larger or smaller?

24 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus.

25 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus.

26 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus. v = v x î + v y ĵ

27 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus. v = v x î + v y ĵ = [ v sin(θ)]î + [v cos(θ)]ĵ

28 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus. v = v x î + v y ĵ = [ v sin(θ)]î + [v cos(θ)]ĵ ( = vy ) ( vx ) î + ĵ r r

29 For uniform circular motion, we can find the centripetal acceleration a r using geometry and calculus. v = v x î + v y ĵ = [ v sin(θ)]î + [v cos(θ)]ĵ ( = vy ) ( vx ) î + ĵ r r a = d v dt

30 v = v r ( yî + xĵ )

31 v = v ( ) yî + xĵ r a = d v dt = v ( vy î + v x ĵ ) r

32 v = v ( ) yî + xĵ r a = d v dt = v ( vy î + v x ĵ ) r a = a 2 x + a 2 y

33 v = v ( ) yî + xĵ r a = d v dt = v ( vy î + v x ĵ ) r a = a 2 x + a 2 y = v v r 2 y + v 2 x

34 v = v ( ) yî + xĵ r a = d v dt = v ( vy î + v x ĵ ) r a = a 2 x + a 2 y = v v r 2 y + v 2 x a = v2 r (uniform circular motion)

35 Lecture Question 4.3 A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball? (a) Centripetal acceleration increases. (b) Centripetal acceleration decrease. (c) Centripetal acceleration stays the same. (d) Not enough information.

36 The velocity of an object depends on the reference frame from which it is measured.

37 The velocity of an object depends on the reference frame from which it is measured. frame A (Alice) is stationary frame B (Bob) moves with some constant velocity object P (Parakeet) is measured

38 x BA : position of Bob relative to Alice x PB : position of Parakeet relative to Bob x PA : position of Parakeet relative to Alice

39 x BA : position of Bob relative to Alice x PB : position of Parakeet relative to Bob x PA : position of Parakeet relative to Alice x PA = x PB + x BA

40 x BA : position of Bob relative to Alice x PB : position of Parakeet relative to Bob x PA : position of Parakeet relative to Alice x PA = x PB + x BA v PA = v PB + v BA

41 x BA : position of Bob relative to Alice x PB : position of Parakeet relative to Bob x PA : position of Parakeet relative to Alice x PA = x PB + x BA v PA = v PB + v BA a PA = a PB + a BA

42 r BA : position of Bob relative to Alice r PB : position of Parakeet relative to Bob r PA : position of Parakeet relative to Alice

43 r BA : position of Bob relative to Alice r PB : position of Parakeet relative to Bob r PA : position of Parakeet relative to Alice r PA = r PB + r BA

44 r BA : position of Bob relative to Alice r PB : position of Parakeet relative to Bob r PA : position of Parakeet relative to Alice r PA = r PB + r BA v PA = v PB + v BA

45 r BA : position of Bob relative to Alice r PB : position of Parakeet relative to Bob r PA : position of Parakeet relative to Alice r PA = r PB + r BA v PA = v PB + v BA a PA = a PB + a BA

Motion in Two or Three Dimensions

Chapter 3 Motion in Two or Three Dimensions PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by Pui Lam 5_25_2011 Goals for