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Physics 231

Physics 231. Topic 3: Vectors and two dimensional motion. Alex Brown September 8-12 201. What’s up? (Monday Sept 8) 1) Homework set 01 due Tuesday Sept 9 th 10 pm 2) Learning Resource Center (LRC) in 1248 BPS Monday 9 am to 8:30 pm and Tuesday 9 am to 8:30 pm

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Physics 231

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  1. Physics 231 Topic 3: Vectors and two dimensional motion Alex Brown September 8-12 201

  2. What’s up? (Monday Sept 8) 1) Homework set 01 due Tuesday Sept 9th 10 pm 2) Learning Resource Center (LRC) in 1248 BPS Monday 9 am to 8:30 pm and Tuesday 9 am to 8:30 pm Friday 9-12:30 3) My office hours in 1248 BPS Monday 9-10 am Wednesday 9-10 am

  3. Key Concepts: 2D Motion • Vectors and Scalars • Two Dimensional Motion • Velocity in 2D • Acceleration in 2D • Projectile motion • Throwing a ball or cannon fire • Covers chapter 3 in Rex & Wolfson

  4. Extra Credit Quiz • You throw a ball straight up (positive direction) and catch it again at the same location where you released it. At the highest point the acceleration is: • Impossible to tell • -9.8 m/s2 • 9.8 m/s2 • 0

  5. Trigonometry SOH – CAH - TOA: sin =opposite/hypotenuse =a/c cos =adjacent/hypotenuse =b/c tan =opposite/adjacent =a/b c Pythagorean theorem: a Note that sin,cos,tan are dimensionless. 2 radians corresponds to 360o  b

  6. Be Careful…. h y y   y= h sin() = hcos() Always check carefully which angle is given

  7. Vectors and Scalars • Scalar: A quantity specified by its magnitude only • Vector: A quantity specified both by its magnitude and direction. • To distinguish a vector from a scalar quantity, it is usually written with an arrow above it, or in bold to distinguish it from a scalar. • Scalar: A • Vector: A or A (bold face)

  8. Vectors and Scalars

  9. Question No Yes • Are these two vectors the same? • Are the lengths of these two vectors the same? Two vectors are equal if both their lengths and directions are the same!

  10. A+B B+A B A Vector Addition B A A+B=B+A

  11. Vector operations in equations (xa+b,ya+b) y (xb,yb) A+B B (xa,ya) A x Example: A+B

  12. A-B = A + (-B) Vector Subtraction B A -B

  13. A-B=A+(-B) -B Vector Subtraction A B

  14. b a More on Coordinate Systems y Cartesian Coordinates: (x,y)=(a,b) Also, r = a î + b ĵ î unit vector in x direction ĵ unit vector in y direction r  x Plane Polar Coordinates (r,)

  15. Vector length and its components Y (xa,ya)  x Length of vector (use pythagorean theorem):

  16. Vector operations in equations

  17. Vector operations in equations ff

  18. Clicker Question begin Which route is shorter? • Red • Black • The same • Don’t know end

  19. Question A) 3 km 1 km =-45o Y2=2.5 sin()=-1.77 2.5 km x2=2.5 cos()=1.77 A man walks 5 km/h. He travels 12 minutes to the east, 30 minutes to the south-east and 36 minutes to the north. A) What is the displacement of the man when he’s done? B) What is the total distance he walked? B) 1 + 2.5 + 3 = 6.5 km

  20. Key Concepts: 2D Motion • Vectors and Scalars • Two Dimensional Motion • Velocity in 2D • Acceleration in 2D • Projectile motion • Throwing a ball or cannon fire • Uniform Circular Motion • Centripetal acceleration • Covers chapter 3 in Rex & Wolfson

  21. Boat crossing the river Flow

  22. Question A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing (south) with 5 km/h. 1) At what angle  does the captain have to steer the boat the go straight across? 2) How long does it take for the boat to cross the river? 

  23. Answer A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. At what angle  does the captain have to steer the boat the go straight across? 2) How long does it take the boat to cross the river? v=10 km/h  sin = opposite/hypotenuse = 5/10 = 0.5  = sin-1(0.5) = 30o Flow=5km/h tan = opposite/adjacent tan30o = 0.577 = 5/velocityhor velocityhor= 8.66 km/h Time = (1 km)/(8.66 km/h) = 0.115 h = 6.9 min

  24. 3) If it doesn’t matter at what point the boat reaches the other side, at what angle should the captain steer to cross in the fastest way? 0o:the horizontal component of the velocity is then maximum.

  25. plane in the wind

  26. Displacement in 2D Often, we replace motion in 2D into horizontal and vertical components. r y In vector notation: r= x+y (r,x,y: vectors) We can then work on horizontal and vertical components separately. x

  27. 1d motion 2d motion; decompose into horizontal and vertical components

  28. Near the surface of the earth with the y-axis pointing up ax = 0 ay = -g So for motion in x-direction and for motion in y direction the same equations as before but changing x to y

  29. (A) (B) (C) (D) (E) with a = -g

  30. Pop and Drop

  31. Pop and Drop B A For A: vy = - ½gt2 vx = 0 y = y0- ½gt2 x= 0 For B: vy= - ½gt2vx= v0 y = y0 – ½gt2 x=x0t

  32. Clicker Quiz Galileo and Newton stand on top of the tower of Pisa. Galileodrops a stone of mass 2 kg straight down (no initial velocity). Newtonthrows a 2 kg stone with an initial horizontal velocity of 3 m/s. Which stone will hit the ground first? (ignore effects of friction) a) The stone thrown by Galileo b) The stone thrown by Newton c) Both stones arrive at the same time d) Not enough information to say.

  33. Clicker Quiz Galileo and Newton stand on top of the tower of Pisa. Galileodrops a stone of mass 2 kg straight down (no initial velocity). Newtonthrows a 2 kg stone with an initial horizontal velocity of 3 m/s. Which stone will hit the ground first? (ignore effects of friction) a) The stone thrown by Galileo b) The stone thrown by Newton c) Both stones arrive at the same time d) Not enough information to say.

  34. Clicker QuizFiring Balls I A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart

  35. Clicker QuizFiring Balls I when viewed from cart when viewed from ground A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart In the frame of reference of the cart, the ball only has a vertical component of velocity. So it goes up and comes back down. To a ground observer, both the cart and the ball have the same horizontal velocity, so the ball still returns into the cart.

  36. Relative motion Motion is relative to a reference frame! Motion of the ball in rest-frame of cart Resulting motion Motion of the cart

  37. While studying motion in 2D one almost always makes a decomposition into horizontal and vertical components of the motion, which are both described in 1D • Remember that the object can accelerate in one direction, but remain at the same speed in the other direction. • Remember that after decomposition of 2D motion into horizontal and vertical components, you should should investigate both components to understand the complete motion of a particle. • After decomposition into horizontal and vertical directions, treat the two directions independently.

  38. Parabolic motion

  39. Parabolic motion vx(t) = v0x vy (t) = v0y – g t vx=v0 cos vy=v0 sin-2g=0 vx=v0 cos vy=v0 sin-1g vx=v0 cos vy=v0 sin-3g vx=v0 cos vy=v0 sin-4g v = (vx, vy) vx=v0 cos vy=v0 sin Vx remains constant throughout the flight!  t=0 t=1 t=2 t=3 t=4

  40. Extra Credit Quiz • You throw a ball in an arch (“up” is the positive y direction), at the highest point in the arc the y component of the acceleration is: • Impossible to tell • -9.8 m/s2 • 9.8 m/s2 • 0

  41. Parabolic motion  • Where is the speed… • …highest? t=0 t=1 t=2 t=3 E C D A B t=5

  42. Parabolic motion  • Where is the speed… • …lowest ? t=0 t=1 t=2 t=3 E C D A B t=5

  43. Question A hunter aims at a bird that is some distance away and flying very high (i.e. consider the vertical position of the hunter to be 0), but he misses. If the bullet leaves the gun with a speed of v0 and friction by air is negligible, with what speed vf does the bullet hit the ground after completing its parabolic path? v0 vf

  44. Answer • First consider the horizontal direction: v0x = v0cos() Since there is no friction, there is no change in the horizontal component: vx(tf) = v0x Next the vertical direction: vy(t) = v0y -gty(t) = voyt - ½ gt2 (g = 9.8 m/s2) • When the bullet hits the ground y(tf)=0: • 0 = voytf - ½gtf2 tf = 0 or tf = 2v0y/g So, vy(tf) = v0y - g(2v0y/g) = - v0 Total speed = v0x2 + (-v0y)2 = v0!!!! The speed of the bullet has not changed, but the vertical component of the velocity has changed sign.

  45. Clicker QuizDropping a Package a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will:

  46. What does the motion of the object look like according to the pilot?

  47. Clicker QuizDropping a Package a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: Both the plane and the package have the samehorizontalvelocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned.

  48. Another example x(t) = v0 cos()t y(t) = v0 sin()t – ½gt2 (A) A football player throws a ball with initial velocity of 30 m/s at an angle of 30o degrees w.r.t. the ground. 1) How far will the ball fly before hitting the ground? 2) What about an angle of 60o? 3) At which angle is the distance thrown maximum? y(t) = 0 if… t( v0 sin() - ½gt )=0 tf=0or tf= 2 v0sin()/g (B) x(tf) = v0 cos() tf = 2 (v0 )2cos() sin() / g (C) if  = 30o x = 79.5 m if  = 60o x = 79.5 m !! (D) Maximum if cos() sin() is maximum, so =45o x(=45o) = 91.7 m (E)

  49. And another example…

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