Rutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 4. Home Page. Title Page. Page 1 of 35.

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1 Rutgers Uniersit Department of Phsics & Astronom 01:750:271 Honors Phsics I Fall 2015 Lecture 4 Page 1 of 35

2 4. Motion in two and three dimensions Goals: To stud position, elocit, and acceleration ectors To appl position, elocit, and acceleration insights to projectile motion To extend our linear inestigations to uniform circular motion To inestigate relatie elocit Page 2 of 35 in two and three dimensions.

3 Position and Displacement In 3D the position of a particle is gien b a position ector r = xî + ĵ + z k starting at the origin O. x,, z are the components of r, also called the coordinates of the particle. The path of the particle is generall a cure. Page 3 of 35

4 Displacement: the change of the position ector r oer a time interal t. r 1 D r r 2 x r 1 = x 1 î + 1 ĵ + z 1 k r 2 = x 2 î + 2 ĵ + z 2 k at time t1 at time t2 Page 4 of 35 z r = (x 2 î + 2 ĵ + z 2 k) (x1 î + 1 ĵ + z 1 k) = (x 2 x 1 )î + ( 2 1 )ĵ + (z 2 z 1 ) k. = x î + ĵ + z k.

5 i-clicker A particle traels along a 3D path as shown in the figure starting from the initial position r 1 = (1 m)î and ending with the final position r 2 = (0.5 m)î. What is the displacement ector? 0.7m 0.5m 1m x 0.3m z A) r = ( 1 m)î + (0.7 m)ĵ B) r = (0.7 m)ĵ (0.3 m) k C) r = ( 0.5 m)î D) r = 0 E) r = (0.5 m)î Page 5 of 35

6 Answer A particle traels along a 3D path as shown in the figure starting from the initial position r 1 = (1 m)î and ending with the final position r 2 = (0.5 m)î. What is the displacement ector? 0.7m z 0.3m 0.5m Dr=(-0.5 1m m) i ^ x A) r = ( 1 m)î + (0.7 m)ĵ B) r = (0.7 m)ĵ (0.3 m) k C) r = ( 0.5 m)î D) r = 0 E) r = (0.5 m)î Page 6 of 35

7 Aerage and Instantaneous Velocit Aerage elocit oer a time interal t Displacement ector ag = = r time interal t r 1 D r r 2 x ag Note: Since t > 0 ag is alwas parallel with r and points in the same direction. Page 7 of 35 z

8 Instantaneous elocit at time t r = lim t 0 t = d r ( ) ( ) dx d dt = î + ĵ + dt dt ( dz dt ) k r 1 Dr r 2 tangent to path x Note: is alwas tangent to the path of the particle at its current position. Page 8 of 35 z

9 Components of : x = dx dt, = d dt, z = dz dt Page 9 of 35

10 Aerage and Instantaneous Acceleration Aerage acceleration oer a time interal t a ag = t 1 a ag Note: Page 10 of 35 2 D x Since t > 0 a ag is alwas parallel with and points in the same direction. z

11 Instantaneous acceleration at time t a = lim t 0 t = d ( ) ( ) ( ) dt = dx d dz î + ĵ + k dt dt dt z 1 a 2 D x Acceleration is a ector which measures the change of the elocit ector. Components of a: a x = d x dt, a = d dt, a z = d z dt Page 11 of 35

12 The acceleration is non-zero if either the magnitude of changes. or the direction of changes. Page 12 of 35 The acceleration is 0 if and onl if both the magnitude and the direction of are constant.

13 i-clicker A particle moes on a path as shown below such that the magnitude of its elocit ector is constant. Where is the instantaneous acceleration a zero? z A A B B C x C A) Eerwhere B) At A C) At C D) At B E) Nowhere Page 13 of 35

14 Answer A particle moes on a path as shown below such that the magnitude of its elocit ector is constant. Where is the instantaneous acceleration a zero? z A A B B linear segment C x C A) Eerwhere B) At A C) At C D) At B E) Nowhere constant constant on linear segment; not constant on cured segments; changes direction. Page 14 of 35

15 i-clicker A particle moes on a helix as shown below such that the -component of its elocit is constant. Which of the following statements is false? j ^ constant A) a x 0 B) a z 0 Page 15 of 35 C) a = 0 D) a 0 E) None of the aboe. z x

16 Answer A particle moes on a helix as shown below such that the -component of its elocit is constant. Which of the following statements is false? constant A) a x 0 B) a z 0 Page 16 of 35 C) a = 0 ^ j D) a 0 E) None of the aboe. z x

17 Projectile Motion Projectile Motion: motion in a ertical plane with constant acceleration equal to the free fall acceleration. Page 17 of 35

18 Examples: tennis, baseball, basketball, football... How hard and at what angle should the quarterback throw? Page 18 of 35 How should the gun shoot in order to hit the target?

19 Set up coordinate sstem a=-g j^ a x a =0 =-g 0 0 Acceleration x 0x x Initial elocit Page 19 of 35 Note: no correlation between 0x, 0 ; can take an alues independentl from each other.

20 i-clicker Recall that a = d dt Then what is the time dependence of x? 0 0 a=-g j ax =0 a =-g ^ A) x = 0 B) x = 0x C) x = 0x gt D) x = 0x + gt Page 20 of 35 0x x E) x = 0

21 Answer Recall that a = d dt Then what is the time dependence of x? A) x = 0 B) a=-g ^ x = 0x j 0 ax =0 C) x = 0x gt a =-g 0 D) x = 0x + gt Page 21 of 35 0x x E) x = 0 d x dt = a x = 0 x = constant = 0x

22 i-clicker Recall that a = d dt Then what is the time dependence of? 0 0 a=-g j ax =0 a =-g ^ A) = 0 B) = 0 C) = 0 gt D) = 0 + gt Page 22 of 35 0x x E) = 0x

23 Recall that a = d dt Then what is the time dependence of? 0 0 a=-g j ax =0 a =-g ^ A) = 0 B) = 0 C) = 0 gt D) = 0 + gt Page 23 of 35 0x x E) = 0x d dt = a = g = 0 gt

24 Is there an correlation between ertical and horizontal motion? x = 0x, No! = 0 gt 0x, 0 are completel independent. Can decompose projectile motion into ertical and horizontal parts. Page 24 of 35

25 i-clicker A ball is thrown straight up with elocit ector 2 from a truck moing with constant elocit 1. For which alues of 1, 2 will the ball land back in the truck at a later time? A) 1 < B) 1 > 2 C) 1 = 2 Page 25 of 35 D) It will not happen for an alues of 1, 2. E) For an alues of 1, 2

26 Answer A ball is thrown straight up with elocit ector 2 from a truck moing with constant elocit 1. For which alues of 1, 2 will the ball land back in the truck at a later time? A) 1 < B) 1 > 2 C) 1 = 2 D) It will not happen for an alues of 1, 2. Page 26 of 35 E) For an alues of 1, 2 The horizontal component 0x for the ball is equal to 1. Since a x = 0, x = 1 at an time t > 0.

27 Projectile motion analsed Page 27 of 35 Initial elocit 0 = 0x î + 0 ĵ, 0x = 0 cos θ 0, 0 = 0 sin θ 0

28 Horizontal motion Vertical motion a x = 0 a = g x = 0x = 0 cos θ 0 = 0 gt = 0 sin θ 0 gt x x 0 = 0x t = ( 0 cos θ 0 ) t 0 = 0 t gt 2 /2 = ( 0 sin θ 0 ) t gt 2 /2 Page 28 of 35 2 = ( 0 sin θ 0 ) 2 2g( 0 )

29 Trajector (θ 0 π/2) x x 0 = ( 0 cos θ 0 ) t, 0 = ( 0 sin θ 0 ) t gt 2 /2 Eliminate t : t = x x 0 0 cos θ 0 = 0 + ( tan θ 0 ) (x x0 ) g(x x 0) 2 2 ( 0 cos θ 0 ) 2 Page 29 of 35 Parabola

30 Horizontal range The horizontal distance R traelled until returning to initial height. 0 = ( tan θ 0 ) (x x0 ) g(x x 0) 2 = 0 ( tan θ 0 ) R gr 2 2 ( 0 cos θ 0 ) 2 2 ( 0 cos θ 0 ) 2 Page 30 of 35 R = 2 0 g sin (2θ 0) ( ) using 2 sinθ 0 cosθ 0 = sin(2θ 0 )

31 Maximum horizontal range For fixed 0, when is R maximum? 0 sin(2θ 0 ) 1, sin(2θ 0 ) = 1 2θ 0 = π/2 Hence R is maximum for θ 0 = π/4 and Page 31 of 35 R max = 2 0 g

32 i-clicker A projectile is launched from the origin as shown below. How long will it take for it to reach maximum height? A) ( 0 /g) sin θ 0 B) ( 0 /g) cos θ 0 C) 0 /g Page 32 of 35 D ( 0 /g) tan θ 0 E) None of the aboe.

33 Answer A projectile is launched from the origin as shown below. How long will it take for it to reach maximum height? A) ( 0 /g) sin θ 0 B) ( 0 /g) cos θ 0 C) 0 /g Page 33 of 35 D ( 0 /g) tan θ 0 E) None of the aboe. = 0 sin θ 0 gt. Max height: = 0 t = ( 0 /g) sin θ 0

34 i-clicker A projectile is launched from the origin as shown below. What is its maximum height? A) ( 2 0 /g) sin θ 0 B) (0 2/2g) cos θ 0 C) 0 2/2g D ( ) 2/2g 0 sin θ 0 E) None of the aboe. Page 34 of 35

35 Answer A projectile is launched from the origin as shown below. What is its maximum height? A) ( 2 0 /g) sin θ 0 B) (0 2/2g) cos θ 0 C) 0 2/2g D ( ) 2/2g 0 sin θ 0 E) None of the aboe. 2 = ( 0sin θ 0 ) 2 2g; = 0 = ( ) 2/2g 0 sin θ 0 Page 35 of 35