Chapter 4: Motion in Two and Three Dimensions

Chia sẻ: Kieu Cong Viet | Ngày: | Loại File: PPT | Số trang:16

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Chapter 4: Motion in Two and Three Dimensions

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In this chapter we will continue to study the motion of objects without the restriction we put in chapter 2 to move along a straight line. Instead we will consider motion in a plane (two dimensional motion) and motion in space (three dimensional motion)

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Nội dung Text: Chapter 4: Motion in Two and Three Dimensions

  1. Chapter 4 Motion in Two and Three Dimensions In this chapter we will continue to study the motion of objects without the restriction we put in chapter 2 to move along a straight line. Instead we will consider motion in a plane (two dimensional motion) and motion in space (three dimensional motion) The following vectors will be defined for two- and three- dimensional motion: Displacement Average and instantaneous velocity Average and instantaneous acceleration We will consider in detail projectile motion and uniform circular motion as examples of motion in two dimensions Finally we will consider relative motion, i.e. the transformation of velocities between two reference systems which move with respect to each other with constant velocity (4 -1)
  2. Position Vector r The position vector r of a particle is defined as a vector whose tail is at a reference point (usually the origin O) and its tip is at the particle at point P. Example: The position vector in the figure is: r ˆ ˆ ˆ r = xi + yj + zk r (ˆ j ˆ r = −3i + 2 ˆ + 5k m ) P (4 -2)
  3. Displacement Vector r r For a particle that changes postion vector from r1 to r2 we define the displacement r r r r vector ∆r as follows: ∆r = r2 − r1 r r The position vectors r1 and r2 are written in terms of components as: r ˆ ˆ + y1 ˆ + z1k r ˆ r1 = x1i j r2 = x2i + y2 ˆ + z2 k ˆ j r The displacement ∆ r can then be written as: r ∆r = ( x2 − x1 ) i + ( y2 − y1 ) ˆ + ( z2 − z1 ) k = ∆xi + ∆yj + ∆zk ˆ j ˆ ˆ ˆ ˆ ∆x = x2 − x1 ∆y = y2 − y1 ∆z = z2 − z1 1 t 2 (4 -3) t
  4. Average and Instantaneous Velocity Following the same approach as in chapter 2 we define the average velocity as: displacement average velocity = time interval r ˆ ˆ ˆ ˆ ˆ ˆ r ∆r ∆xi + ∆yj + ∆zk ∆xi ∆yj ∆zk vavg = = = + + ∆t ∆t ∆t ∆t ∆t We define as the instantaneous velocity (or more simply the velocity) as the limit: r r r ∆r dr v = lim = t ∆t dt t + Δt ∆t → 0 (4 - 4)
  5. If we allow the time interval ∆t to shrink to zero, the following things happen: r r r 1. Vector r2 moves towards vector r2 and ∆r → 0 r ∆r r 2. The direction of the ratio (and thus vavg )approaches the direction ∆t of the tangent to the path at position 1 r r 3. vavg → v r d ˆ ˆ v= dt ( ) ˆ dx ˆ dy ˆ + dz k = v i + v ˆ + v k xi + yj + zk = i + dt dt j dt ˆ x ˆ yj zˆ The three velocity components are given by the equations: t dx vx = t + Δt dt r r dr dy v= vy = dt dt dz (4 - 5) vz = dt
  6. Average and Instantaneous Acceleration The average acceleration is defined as: r r r change in velocity r v − v ∆v average acceleration = aavg = 2 1= time interval ∆t ∆t We define as the instantaneous acceleration as the limit: r r ( ) ∆v dv d dv r a = lim = = ˆ + v y ˆ + vz k = dvx i + y ˆ + dvz k = ax i + a y ˆ + az k vx i j ˆ ˆ j ˆ ˆ j ˆ ∆t dt dt dt dt dt ∆t → 0 Note: Unlike velocity, the acceleration vector does not have any specific relationship with the path. The three acceleration components are given by the equations: r dv dv y dvz r dv ax = x ay = az = a= dt dt dt dt (4 - 6)
  7. Projectile Motion The motion of an object in a vertical plane under the influence of gravitational force is known as “projectile motion” r The projectile is launched with an initial velocity vo The horizontal and vertical velocity components are: vox = vo cos θ o voy = vo sin θ o g Projectile motion will be analyzed in a horizontal and a vertical motion along the x- and y-axes, respectively. These two motions are independent of each other. Motion along the x- axis has zero acceleration. Motion along the y-axis has uniform acceleration ay = -g (4-7) (4 - 7)
  8. Horizontal Motion: ax = 0 The velocity along the x-axis does not change v x = v0 cos θ 0 (eqs.1) x = xo + ( vo cos θ o ) t (eqs.2) Vertical Motion: ay = − g Along the y-axis the projectile is in free fall gt 2 v y = v0 sin θ 0 − gt (eqs.3) y = yo + ( v0 sin θ 0 ) t − (eqs.4) 2 If we eliminate t between equations 3 and 4 we get: v y − ( v0 sin θ 0 ) = −2 g ( y − yo ) 2 2 g Here xo and yo are the coordinates of the launching point. For many problems the launching point is taken at the origin. In this case xo = 0 and yo = 0 Note: In this analysis of projectile motion we neglect the effects of (4-8) air resistance (4 - 8)
  9. The equation of the path: gt 2 x = ( vo cos θ o ) t (eqs.2) y = ( v0 sin θ 0 ) t − (eqs.4) 2 If we eliminate t between equations 2 and 4 we get: g y = ( tan θ o ) x − x2 This equation describes the path of the motion 2 ( vo cos θ o ) 2 The path equations has the form: y = ax + bx 2 This is the equation of a parabola Note: The equation of the path seems too complicated to be useful. Appearances can deceive: Complicated as it is, this equation can be used as a short cut in many projectile motion problems (4 - 9)
  10. x = ( vo cos θ o ) t sin v x = v0 cos θ 0 (eqs.1) (eqs.2) ϕ 3π /2 gt 2 ϕ v y = v0 sin θ 0 − gt (eqs.3) y = ( v0 sin θ 0 ) t − (eqs.4) O π /2 2 Horizontal Range: The distance OA is defined as the horizantal range R At point A we have: y = 0 From equation 4 we have: gt 2  gt  ( v0 sin θ 0 ) t − = 0 → t  v0 sin θ 0 −  = 0 This equation has two solutions: 2  2 Solution 1. t = 0 This solution correspond to point O and is of no interest gt Solution 2. v0 sin θ 0 − = 0 This solution correspond to point A 2 2v0 sin θ 0 From solution 2 we get: t = If we substitute t in eqs.2 we get: g 2 2 2vo vo R= sin θ o cos θ o = sin 2θ o t g g O A R has its maximum value when θ o = 45° 2 R vo Rmax = (4 -10) 2sin A cos A = sin 2 A g
  11. tA Maximum height H g H vo sin 2 θ o 2 H= 2g The y-component of the projectile velocity is: v y = v0 sin θ 0 − gt v0 sin θ 0 At point A: v y = 0 → v0 sin θ 0 − gt → t = g 2 gt2 v sin θ 0 g  v0 sin θ 0  H = y (t ) = ( v0 sin θ 0 ) t − = ( v0 sin θ 0 ) 0 −   → 2 g 2 g  vo sin 2 θ o 2 H= (4 -11) 2g
  12. tA Maximum height H (encore) g vo sin 2 θ o 2 H H= 2g We can calculate the maximum height using the third equation of kinematics for motion along the y-axis: v y − v yo = 2a ( y − yo ) 2 2 In our problem: yo = 0 , y = H , v yo = vo sin θ o , v y = 0 , and a = − g → 2 vo sin 2 θ o v yo 2 −v yo = −2 gH → H = 2 = 2g 2g (4 -12)
  13. Uniform circular Motion: A particles is in uniform circular motion it moves on a circular path of radius r with constant speed v. Even though the speed is constant, the velocity is not. The reason is that the direction of the velocity vector changes from point to point along the path. The fact that the velocity changes means that the acceleration is not zero. The acceleration in uniform circular motion has the following characteristics: 1. Its vector points towards the center C of the circular path, thus the name “centripetal” v2 2. Its magnitude a is given by the equation: a= r Q The time T it takes to complete a full revolution is r known as the “period”. It is given by the equation: C r P r R 2π r T= v (4 -13)
  14. r yP xP v = vx i + v y ˆ = ( −v sin θ ) i + ( v cos θ ) ˆ ˆ j ˆ sin θ = j cosθ = r r Here xP and yP are the coordinates of the rotating particle r r  y P  ˆ  xP  ˆ r dv  v dyP  ˆ  v dxP ˆ v =  −v  i +  v  j Acceleration a = = − i +  j  r   r  dt  r dt   r dt  dyP dxP We note that: = v y = v cos θ and = vx = −v sin θ dt dt r  v2  ˆ  v2 ˆ v2 v2 ( cos θ ) + ( sin θ ) 2 2 a =  − cos θ  i +  − sin θ  j a = ax + a y = 2 2 =  r   r  r r ay − ( v 2 / r ) sin θ r tan φ = = = tan θ → φ = θ → a points towards C ax − ( v 2 / r ) cos θ vx = −v sin θ v y = v cos θ P C A C ( cos θ ) + ( sin θ ) = 1 2 2 (4 -14)
  15. Relative Motion in One Dimension: The velocity of a particle P determined by two different observers A and B varies from observer to observer. Below we derive what is known as the “transformation equation” of velocities. This equation gives us the exact relationship between the velocities each observer perceives. Here we assume that observer B moves with a known constant velocity vBA with respect to observer A. Observer A and B determine the coordinates of particle P to be xPA and xPB , respectively. xPA = xPB + xBA Here xBA is the coordinate of B with respect to A d d d We take derivatives of the above equation: ( xPA ) = ( xPB ) + ( xBA ) → dt dt dt vPA = vPB + vBA If we take derivatives of the last equation and take dvBA aPA = aPB into account that =0→ dt Note: Even though observers A and B measure different velocities for P, they measure the same acceleration (4 -15)
  16. Relative Motion in Two Dimensions: Here we assume that observer B moves with a known constant velocity vBA with respect to observer A in the xy-plane. Observers A and B determine the position vector of particle P to be r r rPA and rPB , respectively. r r r rPA = rPB + rBA We take the time derivative of both sides of the equation d r d r d r r r r r r r rPA = rPB + rBA → vPA = vPB + vBA vPA = vPB + vBA dt dt dt If we take the time derivative of both sides of the last equation we have: r d r d r d r dvBA r r vPA = vPB + vBA If we take into account that = 0 → aPA = aPB dt dt dt dt Note: As in the one dimensional case, even though observers A and B measure different velocities for P, they measure the same acceleration (4 -16)
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