Chapter 2 MOTION IN TWO DIMENSIONS

1. Umm Al-Qura University جا معة أم القرى Faculty of Applied Scienceكـلية العلوم التطبيقية Department of Physics قسم الفيزياء General Physics (For medical students): 403104-3 403104-3 : المدخل للفيزياء الطبية Chapter 2 MOTION IN TWO DIMENSIONS A. Prof. Hamid NEBDI hbnebdi@uqu.edu.sa Faculty of Applied Science. Department of Physics. Room: 315 second floor. Phone: 3192

2. Course’s Topics 2.1- An Introduction to Vectors 2.2- The Velocity in Two Dimensions 2.3- The Acceleration in Two Dimensions 2.5- Projectiles 2.6- Projectiles in Biomechanics

3. Introduction • The definitions of position, velocity, and acceleration and their interrelations can be extended from the one-dimensional case to the more general situation if these quantities are represented by vectors. • Vectors are mathematical objects having both a magnitude and a direction and used to represent many other physical quantities in addition to the ones needed to describe motion. • By contrast, some physical quantities, such as temperature and time, have no direction and are presented by ordinary numbers or scalars. • Motion in a plane is equivalent to a pair of one-dimensional motions.

4. 2.1- An Introduction to Vectors • While vectors are written with an arrow over the symbol , in print they are denoted with boldface type, as in A, s, and v. • The magnitude of the vector A is written as A or |A|. • A vector is a pictured in a diagram by an arrow with a length proportional to its magnitude and oriented to indicate its direction. • For example, in the above Figure, A, B and C each have different directions, and C is longer or greater in magnitude than A.

5. Addition of Vectors - The concept of vector addition is illustrated by an example involving two vectors representing displacements. - The net change in position or displacement will depend on the magnitudes and directions of the two displacements. • Let us denote the first displacement by A and the second by B. The net displacement C is the sum of A and B: C = A + B • The procedure for finding C is shown in Fig.2.1. • Note the order of the vectors in the sum is irrelevant: A + B and B + A are the same thing.

6. Addition of Vectors • When there are three or more vectors to be summed, two vectors are added first, then the next, and so on. Again, in order of the vectors is irrelevant; for example, in Fig.2.2, A + B + C = B + A + C. • The following exercise shows how the magnitude of A + B depends on the relative directions of A and B.

7. Example 2.1: A person walks 1 km due east. If the person then walks a second kilometer, what is the final distance from the starting point if the second kilometer is walked : (a) due east; (b) due west; (c) due south? We will call the first displacement A and the second B.

8. Solution 2.1: Using the procedures outlined above, we construct the sum C = A + B for the three cases (Fig. 2.3). (a) Since A and B are in the same direction, C = A + B = 2A = 2km. C is directed due east. (b) Here, the vectors are opposite, so C = A – B = 0. (c) From the Pythagorean theorem, C2 = A2 + B2 = 2 A2, so C = A = km From Fig. 2.3c, C points toward the southeast.

9. Multiplication of a Vector by a Scalar • Multiplication of a vector by a scalar is defined so that the usual rules of algebra apply. • If 2A = A + A is to hold, then we must interpret 2A as a vector in the same direction as A but with twice the length. •  For example, if A is a 2 km displacement due north, 2A is a 4 km displacement due north, and 5A is a 10 km displacement due north.

10. Vector Subtraction • Subtraction of vectors is also defined so that the usual algebraic rules apply. • In order that A – A = 0 holds, -A must be interpreted as a vector equal in magnitude to A but opposite in direction. Thus if A is a 2 km displacement due north, -A is a 2 km -A displacement due south. • The vector difference C = B – A = B + (-A) is evaluated by adding the vector –A to the vector B as in Fig. 2.4.

11. Vector Components - A vector in a given plane is specified by two pieces of information, its magnitude and its direction. • Equivalently, the vector can be specified by two other quantities, its components along a pair of perpendicular axes. These components are often useful in vector calculations. • The procedure for finding the components of a vector is shown in Fig.2.5. The vector A is redrawn from the origin of the x-y axes, and the components of A, denoted by Ax and Ay, are constructed by drawing the dashed lines at right angles to the axes.

12. - Expressions relating the components to the magnitude of A and its angle  (theta) with the x axis can be obtained using some properties of right triangles. - Figure 2.6 shows a right triangle with sides a, b, and c. These satisfy the Pythagorean relation: a2 + b2 = c2 - The sine, cosine, and tangent of the angle  are defined by: - Thus, in Fig.2.5b, the components Ax and Ay satisfy: Also, or Ax is positive when it is directed in the +x direction and negative when it is directed toward the –x direction. Similarly for Ay.

13. Example 2.2: Find the components of the vectors A and B in Fig.2.7, if A = 2 and B = 3.

14. Solution 2.2: Using trigonometric tables or an electronic calculator, we find that cos 30° = 0.866 and sin 30° = 0.500 Thus, Ax = A cos = 2 cos 30° = 2(0.866) = 1.73 Ay = A sin  = 2 sin 30° = 2 (0.500) = 1.00 From Fig.2.7b, Bx is positive and By is negative. With cos 45° = sin 45° = 0.707, Bx= 3 cos 45° = 3 (0.707) = 2.12 By = -3 sin 45° = -3 (0.707) = -2.12

15. Sum of Vectors using Components • The sum of two more vectors can be calculated conveniently in terms of components. To do this, we define the unit vector (read “x-hat”), a vector of length one in the +x direction. Similarly, is a unit vector in the +y direction. - With these unit vectors, a vector A with components Ax and Ay can be written as: If a vector B is also written as: The sum C = A + B is : - The components of C are the sums of the components of A and B(Fig. 2.8).

16. Example 2.3: In Fig.2.9, (a) Find the components of C = A + B. (b) Find the magnitude of C and its angle  with the x axis.

17. Solution 2.3: (a) Using the following Eq.: we can write : Thus Cx = 6, and Cy = 8. (b) From the Pythagorean theorem: so C = 10. From Fig. 2.9, we see that the angle  satisfies. Using a calculator, we find  = 53.1.

18. 2.2- The Velocity in Two Dimensions • In two dimensions, position, velocity, and acceleration are presented by vectors. • Their definitions are very similar to those for motion along a straight line, and their x components are related to each other in the same way as are x, v, and a for a straight line motion. • Since this is also true for the y components, a problem involving motion in a plane is effectively a pair of one-dimensional motion problems.

19. Figure 2.10 shows an object moving in a plane labeled with x and y axes. The object could be a car, an animal, or a red blood cell, and it is represented symbolically by a point. • If the displacement in a time interval ∆t is denoted by the vector ∆s, then the average velocity of the object is parallel to ∆sand is given by: We can write ∆s in terms of its components in the x and y direction: • The average velocity can also be written in terms of its components: Thus the components of are: • Each component of the average velocity looks like a one-dimensional average velocity.

20. The instantaneous velocity v is the average velocity for an extremely short time interval. In component form: • Here vx and vy are the instantaneous rates of change of the x and y coordinates of the object. • At any instant, v is directed along the tangent to the path or trajectory of the object (Fig. 2.10b).

21. Example 2.4: A car travels halfway around an oval racetrack at a constant speed of the 30 m s-1 (Fig. 2.11). (a) What are its instantaneous velocities at points 1 and 2? (b) It takes 40 s to go from 1 to 2, and these points are 300 m apart. What is the average velocity of the car during this time interval?

22. Solution 2.4: (a) The instantaneous velocity is tangent to the path of the car, and its magnitude is equal to the speed. Thus, at point 1 the velocity is directed in the +y direction and v1 = 30 ms-1 . Similarly, at point 2 the velocity is long the –y direction, and v2 = (30 m s-1)(- ) = -30 m s-1 (b) The average velocity is the displacement divided by the elapsed time. The displacement is entirely along the x direction, so ∆s = 30m . . Since ∆t = 40 s, The average velocity during this time interval is directed along the +x axis. Its magnitude is less than the speed of 30 m s-1 because the car does not travel in a straight line.

23. Velocities are vectors, and are added using the rules for vectors addition. • When a boat moves relative to a moving stream, or a person walks on a moving vehicle, the motion relative to the ground is found by adding the velocity vectors.

24. Example 2.5: A boat moves at 10 m s-1 relative to the water in a river. It is pointed toward the opposite shore. There is a 5 m s-1 current (Fig. 2.12a). Find the magnitude and direction of the boat’s velocity relative to the shore.

25. Solution 2.5: The boat is swept downstream by the current as it crosses the river. We must add the velocities vb and vw to find the velocity v of the boat relative to the shore. Since the velocities form a right triangle, = (10 m s-1)2 + (5 m s-1)2 = 125 m2 s-2 v = 11.18 m s-1 The angle  between the direction of the boat and the direction of the current is found from The boat does not move straight across the stream, and it will reach the opposite bank downstream of its starting point. For the boat to follow a direct line toward the opposite shore, the boat must be pointed somewhat upstream (Fig. 2.12b).

26. 2.3- The Acceleration in Two Dimensions • The acceleration of an object moving in a plane is defined such as in the one-dimensional case. • Suppose that in a time interval ∆t=t2-t1 the velocity changes by ∆v=v2-v1 . • Then the average acceleration is: - The instantaneous acceleration a is the average acceleration during an extremely short interval (Fig.2.13).In components: Here ax and ay are the rates of change of vx and vy.

27. The Acceleration in Two Dimensions

28. Example 2.6: In Exercise 4 the velocity of the car changed from to in 40s. What was the average acceleration of the car in that time interval?

29. Solution 2.6: The average acceleration is defined as the velocity change divided by elapsed time: Thus the average acceleration during the time the car goes from point 1 to point 2 is directed in the –y direction, or downward in Fig. 2.11b.

30. This exercise illustrates two important points: 1- If the velocity is constant, the acceleration is zero , since a is the rate of change of the velocity. However, when the speed is constant, the acceleration may or may not be zero. If an object moves at a constant speed along a curved path, its velocity is changing direction, and it is accelerating. We feel the effects of this acceleration when a car turns quickly. The acceleration is zero only when the speed and direction of motion are both constants. 2- The directions of the velocity and acceleration at any instant can be related in many ways. The magnitude and direction of a are determined by how v is changing. When a car moves along a straight road, the acceleration is parallel to the velocity if v is increasing and opposite if v is decreasing. When the motion is along a curved path, the acceleration is at some angle to the velocity.

31. 2.5- Projectiles

32. 2.5- Projectiles • Kicked or thrown balls, jumping animals, and objects dropped from windows are all examples of projectiles. • If air resistance is neglected, the motion of a projectile is influenced only by the constant gravitational acceleration. • The equations for the motion of a projectile are given by: • The x and vx equations show that the horizontal motion is uniform, and the y and vy equations show the vertical motion is that of an object acting only under the influence of gravity.

33. Example 2.9: A ball is kicked from ground level with a velocity of 25 m/s at an angle of 30 to the horizontal direction. (Fig. 2.16) a) when does it reach its greatest height? b) Where is it at that time? Solution 2.9: a) The initial velocity has components as follows: vox= v0cos 30° = (25 m/s) (0.866) = 21.7 m/s voy= v0 sin 30° = (25 m/s) (0.500) = 12.5 m/s The greatest height is reached when vy= 0. Using vy = v0y – g t, this occurs when t= (v0y - vy)/ g = 12.5 m.s-1 – 0)/ 9.8 m/s2 = 1.28 s b) With Eqs. 2.15, the displacement after 1.28 s is given by: x = v0x t= 27.8 m, y = v0y t – (g t2)/2= 7.97 m

34. 2.6- Projectiles in Biomechanics • Many applications of projectile motion occur in athletics and in animal motion. • In applications of projectile motion, it is convenient to have a formula for the horizontal distance traveled or range, R given by: • The maximum range on flat ground occurs when the launch angle is 45. • The projectile lands after an elapsed time t when y returns to zero: