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Chapter 4

Chapter 4. Motion in Two Dimensions. 2-D position vector. Consider a particle moving in a flat 2-D plane (usually the x - y plane) Its position is described by the 2-D position vector r , which has two components: r describes a ‘ curve ’ (or ‘ path ’ ) on the x-y plane.

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Chapter 4

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  1. Chapter 4 Motion in Two Dimensions

  2. 2-D position vector • Consider a particle moving in a flat 2-D plane (usually the x-y plane) • Its position is described by the 2-D position vector r, which has two components: • rdescribes a ‘curve’ (or ‘path’) on the x-y plane. • The notation for the unit vectors is used interchangeably with

  3. 2-D displacement vector • The displacement of the particle can be defined as the change in the position: Δr = rf - ri

  4. 2-D velocity vector • The curve shown represents the ‘path’ of a particle moving in 2-D • Its instantaneous velocity vector is represented by

  5. Real life example of motion in 2-D space

  6. 2-D average velocity in vectorial form • Consider a particle tracing along the red path as shown. • The average velocity between the point A and any other point B along the path is defined as the ratio of displacement to the time interval between these two point, • The direction of the mean velocity is in the direction of Δr • For example Δr1/Δt1 is the average velocity between the point A to point B

  7. 2-D instantaneous velocity in vectorial form • The instantaneous velocity at a point A is the limit of the average velocity when Δt 0 (equivalent to the statement that BA) • In this limit Δr tends to the green straight line, which slope defines the DIRECTION of the instantaneous velocity at A.

  8. Average acceleration • Definition: ratio of change in instantaneous velocity to the time interval during which such change occurs • The mean acceleration is a vector which direction is the same as that of Δv

  9. Instantaneous acceleration • Definition: The limit of average acceleration when Δt  0

  10. Acceleration occurs when… • There is a change in the magnitude of the velocity vector • There is a change in the direction of the velocity vector • There is a change in the direction and the magnitude of the velocity vector

  11. Quick Quiz 4.1 Which of the following cannot possibly be accelerating? (a) An object moving with a constant speed (b) An object moving with a constant velocity (c) An object moving along a curve Answer: (b). An object moving with constant velocity has Δv = 0, so, according to the definition of acceleration, a = Δv/Δt = 0. Choice (a) is not correct becauase a particle can move at a constant speed and change direction. This possibility also makes (c) an incorrect choice.

  12. 2-D motion comprised of two modes of motion • In then absence of constraints, motion of constant acceleration in 2-D is equivalent to two modes of motion that are independent from each other. • One mode of motion is in the x-direction, and the other the y-direction. • If constraint presents, the mode modes are coupled and these modes are not independent from each others (e.g., in circular motion).

  13. Kinematic vectors in 2-D with constant acceleration • Position vector • Velocity • Acceleration c.f. for the 1-D case c.f. for the 1-D case

  14. Kinematic vectors in 2-D with constant acceleration 2D 1D • vf = vi + at • rf - ri = ½(vf + vi)t • rf - ri = vit + ½ at2 • (vf)2 = (vi)2 + 2a(rf - ri)

  15. Resolving vf = vi + at into two sets of components • From the graph we can see that the vectorial equation can be resolved into two sets of components: x-component: vxf = vxi + axt y-component: vyf = vyi + ayt

  16. Resolving rf = ri + vit + ½at2into two sets of components • From the graph we can see that the vectorial equation can be resolved into two sets of components: • xf = xi + vxit + ½axt2 • yf = yi + vyit + ½ayt2

  17. Contributions to rf • The final position vector is a sum of three contributions: • The initial position vector ri • Displacement due to the initial velocity vector vit • Displacement due to the acceleration, ½at2

  18. Projecting out the components • The 2-D vectorial kinematic equation can be easily resolved into there components by taking the dot project with the relevant unit vector (we speak of ‘projecting out the components from the vectorial equation’). E.g., • Taking the dot product of vf = vi + at with i projects out the x-component : (vf = vi + at)i vxf = vxi + axt • Taking the dot product of vf = vi + at with j projects out the x-component : (vf = vi + at)j vyf = vyi + ayt

  19. Example ax = 4.0 m/s2 0 • A particle is set to motion from the origin at t=0. It’s x- and y-initial velocity component are respectively 20 m/s and -15 m/s. The particle move in the x-y plane and experience an acceleration in the x-direction only, given by ax = 4.0 m/s2. x vx = 20 m/s vy = -15 m/s -y

  20. (a) Determine the components of its velocities as a function of time • Use vf = vi + at, with

  21. vx = 40 m/s q vy = -15 m/s vf (b) Calculate the velocity and position at t=5s. Determine also the direction of it. • =tan-1(vy / vx) = tan-1 (-15/40) = - 20.6 o The direction of vf is as shown in the figure.

  22. Checking for consistency

  23. (c) Determine the position vector at time t = 5s rx(t=5s)= 20(5)+2(5)2 m = 150m • =tan-1(ry / rx) = tan-1 (-75/150) = - 26.6 o The direction of the position vector can be inferred from the figure q ry(t=5s) = -15(5)m = -75m rf

  24. Projectile motion • Defined as the 2-D motion of an object under the influence of gravity

  25. The free-fall acceleration g is a constant, and is always pointing downwards. Air friction is ignored Under these assumptions, the trajectory of the projectile will trace out a parabola in the x-y plane. (y is in the vertical direction and x is in the horizontal direction, both are with respect to the Earth’s surface) Assumptions Acitve figure 4.7: trajectory of a projectile

  26. y x y>0, x>0 y>0, x<0 0 j y<0, x<0 y<0, x>0 i Preparing the coordinate system • First, choose the coordinate system as done earlier: • Choose a origin as reference point. • The position to the left of O has x < 0, to the right has x>0 • The position above O has y < 0, below has y>0 • +j is defined as pointing in the upward direction; +i is defined as pointing in the right direction

  27. Initial conditions Acceleration: • a= ayj + axi = -gj + 0i; g = 9.81 m/s2. Initial position: r(t=0)=ri=0 Initial velocity: • vi= vyij + vxii = vi cosqj + vi sinq i; • vi = (vxi2 + vyi2)½ is the magnitude of the initial velocity, • vxi =vi cosq , vyi = vi sinq are the components of the initial velocity

  28. The projectile’s trajectory ax = 0 ay = -g vyi = vi sin qi vxi = vi cos qi

  29. The trajectory is parabolic: verification • The displacement in the x- and y- directions xf = vxit = (vi cos q)t yf = vyit + ½ay t2 = (vi sin q)t - ½gt2 • Combining both equation by eliminating the parameter t: • This is in the form of y = ax–bx2, i.e. the standard expression for a parabolic curve, where atan qi, b  are the parameters that characterise shape of the parabola.

  30. How to analyse a projectile motion • Assume that the motion is a superposition of the motions in the x- and y- direction • The projectile moves with a constant velocity (= the initial vix) in the x-direction; ax = 0 • The projectile undergoes a free fall motion in the y-direction with ay = -g • The position vector at time t is: rf = ri+ vit + ½gt2 • Note that the initial position vector riin general is not 0; its value depends on the choice of your coordinate system set-up. For convenience it can be chosen to be zero in most cases.

  31. The position vector of a projectile in a generic coordinate system rf = ri+ vit + ½gt2 Projectile’s trajectory ½gt2 Initial position from which the projectile is launched vit Note: In some scenario we will choose ri = 0 so that the launching point of the projectile coincides with the origin for convenience. Generically this is not always the case. ri y x The chosen origin. One can choose any point as the origin as the reference point. 0

  32. Range and maximal height • Two important characteristics of a projectile: • Range R, the horizontal distance when it crosses the same height level with the launching point, i.e. when yf = yi. • Maximal height h: As in the case of a 1-D free-fall, the y-component of the velocity is zero at the maximum height of the trajectory: vf = 0 when y–yi = h

  33. Equation for the maximal height • The maximal height is determined by the y-component of the initial velocity, vyi =vi sinqi

  34. Equation for the range of projectile • The range is determined by • (i) the initial horizontal velocity, vxi, and • (ii) 2x the time of flight to the maximum height:

  35. The symmetry of the trajectory • In the previous derivation for R, we have assumed a left-right symmetry along the axis of maximal height h, so that • In realistic case this symmetry is broken by the air resistance. • However when air resistance is ignored, such symmetry is a valid assumption. Axis of the left-right symmetry

  36. The range varies with the launching angle (fixed vi) Active figure 4.11

  37. Complimentary angles • For a given projectile launched at a given initial speed vi, the range R fall in the range of 0  R  Rmax. • The lower limit occurs qi =90, the upper limit, Rmax occurs when qi = 45. • A projectile can land at a range lying between 0 < R <Rmax at two different initial angles with the same initial launching speed. • These angle are called complimentary angle. Rmax

  38. Complimentary projectile trajectories • A pair of complimentary trajectories with complimentary angles {qi,qi’}: • Has a common initial velocity vi • Has different maximal height • Has different t.o.f

  39. Complimentary angles {qi,qi’} • Given an initial angle qi how would you obtain the complimentary angle? • They are given by qi’ + qi = 90 • Can you prove this?

  40. Quick Quiz 4.6 Rank the launch angles for the five paths in the figure below with respect to time of flight, from the shortest time of flight to the longest. Answer: 15°, 30°, 45°, 60°, 75°. The greater the maximum height, the longer it takes the projectile to reach that altitude and then fall back down from it. So, as the launch angle increases, the time of flight increases.

  41. Example • Referring to the experiment demonstration in the picture, prove that if the gun points to the target at rest, the projectile launched by the gun will hit the target if it is shot at the same time the target is released.

  42. Solution • Initial condition: arbitrary launching speed vi. Initial angle qi is such that tan qi = H/xT; • H vertical displacement of the rest target from the gun. • Boundary condition for the target be hit by the projectile: The horizontal displacement of the projectile xP(t=t’)= xT, where t’ is the time when the hitting occurs. • What assumption is made in the above calculation? • ANS: xT must be within the range. Fig. 4.13b, p.88

  43. Solution • To prove that hitting occurs, we need to show that, when t=t’, yP(t’) = yT(t’). Fig. 4.13b, p.88

  44. Uniform circular motion Satellites are arranged in several planes of nearly-circular orbits

  45. Definition • Uniform circular motion (UCM) occurs when an object moves in a circular path with constant a speed. • Such object experiences an acceleration since its direction is constantly changing.

  46. Uniform circular motion in action • In a uniform circular motion: (i) The velocity vector is always tangential to the path of the object, i.e. vr ); (ii) The speed of the object |v| ( v),is constant; so is the radius of the circle, |r |( r) [Active figure 6.2]

  47. The relative orientation of r, v, a • Consider an object is making a UCM in clockwise direction. • When the object is displaced from AB in Dt, the corresponding angular change is Dq. • The change in velocity is denoted by Dv = vf–vi • In the limit Dt0, Dq  0, Dv v • By definition, acceleration is given by • From the diagram, we conclude that the direction of a (Dv) is anti-parallel to r

  48. v a Unit vectors • is the unit vector constantly pointing in the direction away from the origin and along the radius. • is the unit vector constantly tangential to the trajectory, pointing in the direction in which q increases. • Conventionally, if q increases in the anti-clockwise direction, the increase is defined as +ve.

  49. v a The relative orientation of r, v, a in terms of the unit vectors Dv,a

  50. Centripetal acceleration a has only radial component • The acceleration experienced by an object executing UCM, a, is called the centripetal acceleration. • In UCM a always points in the direction towards the center of the circular motion. • In other words, in UCM, a has only radial component but no tangential one. • Generically, for non UCM, the acceleration a has also a tangential component (discussed later)

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