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Kinematics in Dimensions

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					          Chapter 4
Kinematics in 2 & 3 Dimensions

              Kinematics in 2 & 3 Dim
              Projectile Motion
              Circular Motion
Position and Displacement in 3-D
Average Velocity

   Displacement per unit time
                       Dr Dx ˆ Dy ˆ Dz ˆ
                vavg =   = i+     j+ k
                       Dt Dt   Dt   Dt

   Instantaneous velocity

              dr dx ˆ dy ˆ dz ˆ
     v(t) =     = i+ j+ k
              dt dt   dt   dt
Kinematics in 2 & 3 Dimensions

   Kinematic equations must be written in
    vector form
    –   Unit vector form is most advantageous
    –   Takes advantage of properties of Cartesian coordinates; i.e.,
        independence of coordinates from one another

                 r ( t ) = x ( t ) iˆ + y ( t ) j
                                                ˆ
                         dr ( t ) dx ( t ) ˆ dy ( t ) ˆ
                 v (t) =         =         i+         j = vx ( t ) iˆ + vy ( t ) j
                                                                                 ˆ
                          dt       dt          dt
                         dv ( t ) dvx ( t ) ˆ dvy ( t ) ˆ
                 a (t) =         =          i+          j = ax ( t ) iˆ + ay ( t ) j
                                                                                   ˆ
                          dt        dt           dt
In projectile motion the x and y
components of motion are independent

   Use vector notation for position and
    velocity                     v  v cos
                                x0          0

                               v y 0  v0 sin 

                                                   ˆ
                                v0 = v0 xiˆ + v0 y j
                                v ( t ) = vx ( t ) iˆ + vy ( t ) j
                                                                 ˆ
What is rabbit’s position at t=15 sec


                   x(t )  0.31t 2  7.2t  28
                   y (t )  0.22 t 2  9.1t  30
                                 ˆ
                   r (t )  x(t ) i  y (t ) ˆ
                                             j

                  a) What is position at t=15
                  b) Plot the path for 0<t<25 sec
Rabbit’s Velocity at t=15 sec

         
       dr dx ˆ dy ˆ
v (t)      i    j
        dt dt   dt
      Acceleration



   Here the acceleration is constant


            
          dv dv x ˆ dv y ˆ
    a(t)         i     j
           dt   dt    dt
A freely falling object and a object with
a horizontal velocity fall at the same rate



                   The motion in the x and
                    y directions are
                    independent
                   So we know the EOMs!
Projectile Motion Problems

   Conclusions from previous slide:
    –   Two sets of kinematic equations used
            One in the x direction
            Another in the y direction
    –   Special Conditions apply
            If no air resistance, ax = 0
            If no air resistance, ay = g
The Range equation – A special
case

   The projectile lands at the same elevation it
    starts
   There is no air resistance
   g is a constant

        y  yo  0   v0 sin   t  gt
                                     1
                                     2
                                           2


       R  x  x0   v0 cos   t
The Range equation continued

   Eliminating time between the two previous
    equations yields
                                   v 2 cos sin 
                                   2
                    x f  x0  R  0
                                         g
                             v0 sin 2
                              2
                           
                                 g
Graphical Behavior of the Range
equation

   Note that each value of range has two
    solutions for a give v0 except 0o,90o and 45o
   Example
Prob 30


          a) ywall=y(tf)
          b) vx(tf), vy(tf)
          c) Has it passed apogee
Uniform Circular Motion
   Motion in a circle at
    constant speed
     –   Magnitude of radius is
         constant
     –   Magnitude of velocity is
         constant
     –   Direction of velocity
         CHANGES
   Velocity vector is tangent
    to circle
   Radius if perpendicular to
    tangent to circle (and
    velocity)
A change in velocity requires an
acceleration



                      Assume D is small
                                 Ds
                            D 
                                 r
                                 Dv
                            D 
                                 v
     Determining the acceleration

      Using similar triangles
       Ds      Dv                                    SO!
            
        r      v
        1 Ds      1 Dv
                
       Dt r       Dt v
       v     a
          
       r     v
                                                    v2
              v2                                 a r ˆ
           a                                       r
              r
   Result is called radial or centripetal acceleration
Uniform circular Motion     
                          r (t), v (t), a (t)




                                                θ
Example 3

   A wheel 10 cm in radius rotates at 150 rpm.
    What is the centripetal acceleration at a
    point on the edge of the wheel?