Kinematics in Two Dimensions

Document Sample
Kinematics in Two Dimensions Powered By Docstoc
					Kinematics in Two

   Chapter 3

After Chapter 3, students will:
 generalize the concepts of displacement, velocity,
  and acceleration from one to two dimensions
 use the projectile-motion principle to analyze
  two-dimensional constant-acceleration problems
 apply the kinematic equations to solve constant-
  acceleration problems in two dimensions
 add relative velocities in two dimensions
Two-Dimensional Displacement

Sologdin the Turtle moves in     y
  the X-Y plane from the
  point (x0, y0) to the point         (x0, y0)
  (x1, y1).
R1 and R2 are vectors from the
  origin to his starting and
  finishing point,
  respectively.                           R1
                                                 (x1, y1)
Two-Dimensional Displacement

Sologdin’s displacement          y
  vector D is the vector that
  extends from the starting of        (x0, y0)
  his motion to its endpoint.
In keeping with our Chapter 2                  D
   definition of displacement,
   we will define Sologdin’s
   displacement by                        R1
                                                   (x1, y1)
         
       D  R1  R0                                            x
Two-Dimensional Displacement

We can see the truth of this by
 solving for R1:                  y
                     
  D  R1  R0  R1  R0  D            (x0, y0)

Our picture shows this
  addition being done,                R0
  graphically.                                  D

                                                    (x1, y1)
Two-Dimensional Displacement

Since both R0 and R1 start at
   the origin, it is very easy to
   write down the magnitudes
   of their X and Y                      (x0, y0)
   R0 X  x0       R0Y  y0                       D

   R1 X  x1      R1Y  y1
                                                      (x1, y1)
Two-Dimensional Displacement

Now we perform the                 y
  subtraction by subtracting
  the component magnitudes:             (x0, y0)
     
   D  R1  R0
   DX  R1 X  R0 X  x1  x0
   DY  R1Y  R0Y  y1  y0
The magnitude of D:
                                                     (x1, y1)
  D  DX  DY
           2         2

  D    x1  x0    y1  y0 
                 2             2
Two-Dimensional Displacement
                                              x1 – x0
The direction of D is more clearly
  seen if we translate D so that its                              +X
  starting point is at the origin.
  (Remember that vectors are not
  changed by translating them.)                         y1 – y0

We can always write
                       DY   y1  y0
               tan      
                       DX   x1  x0
                          y1  y0   
                 arctan
                          x x
                          1 0       
Two-Dimensional Displacement
                  y1  y0   
         arctan
                  x x
                                       x1 – x0
                  1 0                                    +X
We used the absolute value signs           
 here because we want simply the
 lengths of the two legs of the                  y1 – y0
 right triangle. The sense of the
 angle is apparent from the
 drawing. We would describe the
 direction of D as “60° below the
 positive X axis,” or “60°        -Y
 clockwise from +X.”
Two-Dimensional Velocity
                                           t = t0
Let’s return to the travels of
  Sologdin … and this time,           (x0, y0)
  we’ll use our stopwatch.

Solgdin’s journey begins at
  time t0, and ends at a later
                                                         t = t1
  time t1.
                                          R1        (x1, y1)
Two-Dimensional Velocity
Sologdin’s average velocity is               t = t0
  a vector. Its magnitude is
  the magnitude of his
  displacement, D, divided by                    V
  the elapsed time:                    R0
         v                                           t = t1
            t1  t0
Its direction is the same as the
    direction of the
    displacement vector.
Two-Dimensional Velocity
Sologdin’s average speed is             t = t0
  not merely the magnitude
  of his average velocity. Let
  S be the curvilinear length
  of his meandering path.
  Then his average speed is                 D
                                                 t = t1
     ave. speed 
                  t1  t0              R1
Note that his average speed
  exceeds his average
Two-Dimensional Velocity

The critical difference between speed and velocity:
Consider the race car that wins the Indianapolis 500.
It completes 200 laps at 2.5 miles each in 2.75 hours. Its average
    speed is 500 mi / 2.75 hr = 182 mi/hr.
What is its average velocity?
Two-Dimensional Velocity

What is its average velocity? Zero magnitude; pick a direction.

Its displacement is zero (the race both starts and ends at the start-
    finish line). And zero divided by any amount of time yields a
    magnitude of zero for the average velocity.
Two-Dimensional Acceleration

A truck travels east at 12 m/s. Following a bend in the road, the
   truck turns so that it is headed north, still at 12 m/s. The truck
   spends 8.0 s negotiating the curve. What is its average
   acceleration?                        y

From the definition of acceleration:                    -V0
              
           v1  v0
          a                                  V1 – V0

Here, we have graphically added
-v0 to v1.                                                           x
Two-Dimensional Acceleration

Having obtained v1 - v0, we need to multiply that vector by the
  scalar 1/t to obtain a:

      v1  v0
      2    2

                  12 m/s2  12 m/s2        y
         t               8.0 s            (north)

a  2.1 m/s2                                                  -V0

                                                               a        V1
                                                    V1 – V0         


Two-Dimensional Acceleration

The direction of a is the same as the direction of v1 - v0. Since the
  magnitudes of v1 and v0 are equal, we can write
                v0                 (north)
         tan    1
                v1                                     -V0

           arctan1  45
                                                        a        V1
                                             V1 – V0         
So, the average acceleration is
2.1 m/s2, 45° west of north.                           V0

Note that the speed did not change – only the direction.
Projectile Motion

A projectile is something
  that is launched or
In the general case, it has a
   nonzero horizontal
   component of initial                    v0x
Projectile Motion

In most cases, the projectile experiences a negative vertical
   acceleration (g, due to gravity).

In most cases (assuming we can ignore resistance due to the
   air), the projectile experiences zero horizontal
   acceleration. Its horizontal velocity is constant.

The general approach to all problems of this kind: apply the
  kinematic equations separately to the vertical and
  horizontal motion.
Projectile Motion: Example

Illustrative example: the golfer in the picture has launched
   his ball with an initial velocity of 45 m/s, 31° above the
   horizontal. We will analyze the flight of the ball, in order
   to be able to say:

   How far will the ball “carry” over level ground?
   How high above the ground will the ball reach at its
    highest point?
   What is the total time the ball will be in the air?
Projectile Motion: Example
First step: resolve the
  ball’s initial velocity
  into horizontal and
  vertical components.

   v0 x  v0 cos                    v0

   v0 y  v0 sin                         


Projectile Motion: Example
Keep firmly in mind
  that v0y is not going
                                   v0 x  v0 cos
  to remain constant …             v0 y  v0 sin 
  but that v0x will.



Projectile Motion: Example
During the time the ball      y
  takes to rise vertically,            v0 x  v0 cos
  stop vertically, and fall
  vertically back to the               v0 y  v0 sin 
  ground, the ball moves
  horizontally with
  constant velocity.                     v0

We must determine just            30

 how long that time of                        v0x
 flight is.
Projectile Motion: Example
Vertically, the ball obeys the first kinematic equation:
                         v  v0  at
In our case, the initial vertical velocity is:   v0 y  v0 sin

The acceleration is downward: g, the acceleration due to
  gravity. If we define our coordinate system with +Y
  pointing up, we rewrite the first kinematic equation:

           vy  v0 y  gt  v0 sin   gt
Projectile Motion: Example
In terms of its vertical motion, the ball stops momentarily at
   the top of its trajectory. We also showed, in chapter 2,
   that its vertical travel is symmetric: it rises for half the
   time of flight, and falls for half the time. So, if t is the
   full time of flight, we can rewrite our kinematic equation
   for the upward travel by setting the final (y) velocity to
                     1                  1 
      v y  v0 y  g  t   v0 sin   g  t   0
                     2                  2 
Projectile Motion: Example
Solving for t, the total time of flight:

                   1               g
      v0 sin   g  t   0          t  v0 sin 
                   2               2
          2v0 sin  245 m/ssin 31
      t                          2
                                           4.730 s
              g             9.8 m/s
Projectile Motion: Example
The ball’s horizontal motion is governed by another
  kinematic equation:
                                   1 2
                          x  v0t  at

In the horizontal, however, the acceleration is zero, and the
   initial velocity is v  v cos
                        0x    0

so we can rewrite our equation:
   x  v0 cos  t  45 m/s cos 31 4.730 s   182 m
Projectile Motion: Example
How high does the ball go at the top of its trajectory?

Recall that we already calculated the total time of flight:
t = 4.730 s. Recall also that the ball spent half that time
falling (with zero initial vertical velocity) from its highest
point. How far did it fall?
Projectile Motion: Example
We can calculate this maximum height directly from the
 third kinematic equation:
                                             1 2
                                 y  v0 y t  gt
Notice that the initial y velocity (for the falling part of the
  trajectory) is zero, reducing the equation to:
         1 2    1
    y   gt   9.8 m/s
                                   4.730 
                                         s   27 m
         2      2                   2      

The time we used was half the time-of-flight, t.
Parametric Representation of a Trajectory

So far, we have written both x and y as independent
  functions of time, t. We can combine these descriptions
  to express y as a function of x: the equation of the
  projectile’s trajectory.

Suppose our object is launched from the origin: x = y = 0.
Then:       x  v cos  t
                   0                           (1)

             y  v0 sin  t  gt
                               1 2
Parametric Representation of a Trajectory

Solve eq’n (1) for t:
               x  v0 cos  t
                                          t
                                              v0 cos
Substitute this result for t in eq’n. 2:

                 v0 sin  x  1 g 
                                          x 
              y                             
                    v0 cos       2  v0 cos 
                                             
                                gx 2
              y  x tan  
                            2v0 cos2 
Parametric Representation of a Trajectory



 y, m




             0   20   40   60   80         100    120   140   160   180   200
                                          x, m
Projectile Motion: Things to Remember
   Basic principle: the vertical motion and the horizontal
    motion are treated separately.
   In the vertical motion, there is acceleration (usually, g).
   In the horizontal motion, acceleration is zero and velocity
    is constant.
   Resolve the initial velocity into horizontal and vertical
   Apply the kinematic equations to each motion,
   If “g” appears in an equation with “x”, slap your own
Relative Motion
Consider an airplane flying through air, when the air is in
  motion relative to the ground (wind).

The wind is blowing eastward. The pilot points his airplane
  straight north. In what direction does the airplane move
  over the ground? How fast does it go?

(In other words, what is the airplane’s velocity over the
   ground … magnitude and direction?)
 Relative Motion
Principle of relative
                                                  
   velocity:     vplane - ground    vplane - air  vair - ground
                                      y (north)


Suppose the airplane’s
  speed through the air is                        vp-a          vp-g
  55 m/s, and the air moves
  over the ground with a
  wind speed of 8.9 m/s.
                                                                       x (east)
 Relative Motion
Velocity of plane over ground                     y (north)
Magnitude (speed):
 vp  g  vp  a  va  g
                2           2

 vp  g    55 m/s 2  8.9 m/s 2    56 m/s

Direction:                                           vp-a
           va  g 
  arctan          arctan 8.9 m/s   9.2
                                      
                             55 m/s 
           vp  a 
    (9.2° east of north)                                                  x (east)