Lecture Ch 03

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Lecture Ch 03

  1. 1. Chapter 3 Kinematics in Two Dimensions; Vectors
  2. 2. Units of Chapter 3 <ul><li>Vectors and Scalars </li></ul><ul><li>Addition of Vectors – Graphical Methods </li></ul><ul><li>Subtraction of Vectors, and Multiplication of a Vector by a Scalar </li></ul><ul><li>Adding Vectors by Components </li></ul><ul><li>Projectile Motion </li></ul><ul><li>Solving Problems Involving Projectile Motion </li></ul>
  3. 3. 3-1 Vectors and Scalars A vector has magnitude as well as direction . Some vector quantities: displacement , velocity , force , momentum A scalar has only a magnitude . Some scalar quantities: mass , time , temperature
  4. 4. 3-2 Addition of Vectors – Graphical Methods For vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs , as the figure indicates. 8 km + 6 km = 14 km East 8 km - 6 km = 2 km East
  5. 5. 3-2 Addition of Vectors – Graphical Methods If the motion is in two dimensions, the situation is somewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem .
  6. 6. 3-2 Addition of Vectors – Graphical Methods Adding the vectors in the opposite order gives the same result:
  7. 7. 3-2 Addition of Vectors – Graphical Methods <ul><li>Even if the vectors are not at right angles, they can be added graphically by using the “ tail-to-tip ” method: </li></ul><ul><ul><li>1. Draw V1 & V2 to scale. </li></ul></ul><ul><ul><li>2. Place tail of V2 at tip of V1 </li></ul></ul><ul><ul><li>3. Draw arrow from tail of V1 to tip of V2 </li></ul></ul><ul><ul><li>This arrow is the resultant V (measure length and the angle it makes with the x -axis ) </li></ul></ul><ul><ul><li>Same for any number of vectors involved. </li></ul></ul>
  8. 8. 3-2 Addition of Vectors – Graphical Methods The parallelogram method may also be used; <ul><ul><li>1. Draw V1 & V2 to scale from common origin. </li></ul></ul><ul><ul><li>2. Construct parallelogram using V1 & V2 as 2 of the 4 sides. </li></ul></ul><ul><ul><li>Resultant V = diagonal of parallelogram from common origin (measure length and the angle it makes with the x -axis) </li></ul></ul>
  9. 9. 3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction. Then we add the negative vector:
  10. 10. 3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar A vector V can be multiplied by a scalar c ; the result is a vector c V that has the same direction but a magnitude c V . If c is negative , the resultant vector points in the opposite direction.
  11. 11. 3-4 Adding Vectors by Components Any vector can be expressed as the sum of two other vectors, which are called its components . Usually the other vectors are chosen so that they are perpendicular to each other.
  12. 12. 3-4 Adding Vectors by Components If the components are perpendicular , they can be found using trigonometric functions.
  13. 13. 3-4 Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically :
  14. 14. 3-4 Adding Vectors by Components <ul><li>Adding vectors: </li></ul><ul><li>Draw a diagram ; add the vectors graphically. </li></ul><ul><li>Choose x and y axes. </li></ul><ul><li>Resolve each vector into x and y components . </li></ul><ul><li>Calculate each component using sines and cosines. </li></ul><ul><li>Add the components in each direction. </li></ul><ul><li>To find the length and direction of the vector, use: </li></ul>
  15. 15. Example <ul><li>V = displacement 500 m, 30º N of E </li></ul>
  16. 16. Problem 3-14
  17. 17. Note that since both components are negative, the vector is in the 3 rd quadrant. Problem 3-14
  18. 18. 3-5 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity ; its path is a parabola . Of the form A parabola in the x-y plane !!
  19. 19. It can be understood by analyzing the horizontal and vertical motions separately. 3-5 Projectile Motion
  20. 20. 3-5 Projectile Motion The speed in the x -direction is constant; in the y -direction the object moves with constant acceleration g . This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x -direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly. Show clip
  21. 21. 3-5 Projectile Motion If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.
  22. 22. 3-6 Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.
  23. 23. 3-6 Solving Problems Involving Projectile Motion <ul><li>Read the problem carefully, and choose the object(s) you are going to analyze. </li></ul><ul><li>Draw a diagram. </li></ul><ul><li>Choose an origin and a coordinate system. </li></ul><ul><li>Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g . </li></ul><ul><li>Examine the x and y motions separately. </li></ul>
  24. 24. 3-6 Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that v x never changes, and that v y = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.
  25. 25. Projectile Motion <ul><li>Simplest example : Ball rolls across table, to the edge & falls off edge to floor. Leaves table at time t = 0 . Analyze y part of motion & x part of motion separately . </li></ul><ul><li>y part of motion : Down is positive & origin is at table top: </li></ul><ul><li>y 0 = 0 . Initially, no y component of velocity: v y0 = 0 </li></ul><ul><li> v y = gt and y = ( ½) g t 2 </li></ul><ul><li>x part of motion : Origin is at table top: x 0 = 0 . No x component of acceleration(!) : a = 0 . Initially x component of velocity is: v x0 </li></ul><ul><li> v x = v x0 and x = v x0 t </li></ul>
  26. 26. Ball Rolls Across Table & Falls Off <ul><li>Vector velocity v has </li></ul><ul><li>2 components : </li></ul><ul><li>v x = v x0 , v y = gt </li></ul><ul><li>Vector displacement D has </li></ul><ul><li>2 components : </li></ul><ul><li>x = v x0 t , y = (½)g t 2 </li></ul>
  27. 27. General case
  28. 28. <ul><li>General Case : Take y positive upward & origin at the point where it is shot: x 0 = y 0 = 0 </li></ul><ul><li>v x0 = v 0 cos θ 0 v y0 = v 0 sin θ 0 </li></ul><ul><li>Horizontal motion: </li></ul><ul><li>NO ACCELERATION IN THE x DIRECTION ! </li></ul><ul><li>v x = v x0 x = v x0 t </li></ul><ul><li>Vertical motion : </li></ul><ul><li>v y = v y0 - gt y = v y0 t - ( ½) g t 2 </li></ul><ul><li>(v y ) 2 = (v y0 ) 2 - 2gy </li></ul><ul><ul><li>If y is positive downward, the - signs become + signs. </li></ul></ul><ul><ul><li>a x = 0, a y = -g = -9.8 m/s 2 </li></ul></ul>
  29. 29. Example 3-8 <ul><li>Range ( R ) of projectile  Maximum horizontal distance before returning to ground. Derive a formula for R . </li></ul>
  30. 30. <ul><li>Range R  the x where y = 0 ! </li></ul><ul><li>Use v x = v x0 , x = v x0 t , v y = v y0 - gt </li></ul><ul><li> y = v y0 t – ( ½) g t 2 , (v y ) 2 = (v y0 ) 2 - 2gy </li></ul><ul><li>First , find the time t when y = 0 </li></ul><ul><li>0 = v y0 t - ( ½) g t 2 </li></ul><ul><li> t = 0 (of course!) and t = (2v y0 )/g </li></ul><ul><li>Put this t in the x formula: x = v x0 (2v y0 )/g  R </li></ul><ul><li>R = 2(v x0 v y0 )/g, v x0 = v 0 cos( θ 0 ), v y0 = v 0 sin( θ 0 ) </li></ul><ul><li>R = (v 0 ) 2 [2 sin( θ 0 )cos( θ 0 )]/g </li></ul><ul><li> </li></ul><ul><li>R = (v 0 ) 2 sin(2 θ 0 )/g (by a trig identity) </li></ul>
  31. 31. Example 3-4: Driving off a cliff!! <ul><li>How fast must the motorcycle leave the cliff to land at </li></ul><ul><li>x = 90 m, y = -50 m ? v x0 = ? </li></ul>y is positive upward y 0 = 0 at top v y0 = 0 v x = v x0 = ? v y = -gt x = v x0 t y = - ( ½)gt 2 Time to Bottom: t = √ 2y/(-g) = 3.19 s v x0 = (x/t) = 28.2 m/s
  32. 32. Example 3-9, A punt! <ul><li>v 0 = 20 m/s, θ 0 = 37º </li></ul><ul><li>v x0 = v 0 cos( θ 0 ) = 16 m/s </li></ul><ul><li>v y0 = v 0 sin( θ 0 ) = 12 m/s </li></ul><ul><li>y = y 0 + v y0 t – ( ½) g t 2 </li></ul><ul><li>x = v x0 t </li></ul>
  33. 33. Summary of Chapter 3 <ul><li>A quantity with magnitude and direction is a vector. </li></ul><ul><li>A quantity with magnitude but no direction is a scalar. </li></ul><ul><li>Vector addition can be done either graphically or using components. </li></ul><ul><li>The sum is called the resultant vector. </li></ul><ul><li>Projectile motion is the motion of an object near the Earth’s surface under the influence of gravity. </li></ul>

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