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Physics 107

Outline - Two Dimensional Motion

  1. Vector and Scalar Quantities

    1. Scalar quantity possesses magnitude only.  Examples: speed, mass and temperature.  When you state a magnitude you give a number, e.g. 50 miles/hr

    2. Vector quantities possess both magnitude and direction. Examples: displacement, velocity, acceleration, and force. Magnitude and direction means, for example, 50 miles/hr, south

    3. Two vectors are equal if they have the same magnitude and direction.  Moving a vector to a different location does not change the vector as long as its direction and magnitude remain constant.

    4. Usefulness of Vectors

      1. Principle of superposition: If an object is subjected to two separate influences, each producing a characteristic type of motion, it responds to each without modifying its response to the other.

      2. A vector analysis allows you to separate two-dimensional motion into two one-dimensional motions and then combine them at the end of the problem.  Example: projectile motion

  2. Rules for vector addition

    1. For vectors along a straight line use a positive sign for a vector to the right and a negative sign for a vector to the left.

    2. For vectors at an angle to each other use:

      1. The polygon method.  In Fig. 1a below, the vectors are placed head to tail.  The resultant vector goes from the tail of the first vector A to the head of the second vector B to give C = A + B

      2. The parallelogram method.  Draw the vectors with their tails at one point.  Complete the parallelogram and draw the diagonal to find the resultant. See Fig. 1b below.

    3. Cartesian coordinates

      1. Unit vectors.  The unit vector i, j, and k are unit vectors along the X, Y, and Z-axes, respectively.  When multiplied by a number or a symbol that represents a quantity it becomes a vector with the magnitude of the quantity (or symbol).  For example, 5m i is a vector of length 5 m along the X-axis.

      2. Components of vectors

        1. If a vector lies in the X-Y plane, it can be written as a component in the X-direction added vectorally to a component in the Y-direction. A = Axi + Ayj, where Ax and Ay are the X and Y components of the vector A (Fig. 2a below).

        2. From Fig. 2b above, we see that Ax = A cos Θ Ay = A sin Θ
          and A = (Ax2  + Ay2)1/2

        3. Utility of components.  If two vectors are equal, their components along any chosen axis are equal.   If C = D,  then C= Dx
          and C= Dy

        4. Example (Fig. 3 below)

          A = Axi + Axj     B = Bxi + Bxj
          A + B = (Axi + Ayj) + (Bxi + Byj)
                    = (Ax + Bx))i + (Ay+ By)j
          C = Cxi + Cxj.
            If C =A + B, Cx =(Ax + Bx)), and
          Cy  =(Ay+ By) [Fig. 3 above]

    4. Sample problems in 107 Problem Set for Two Dimensional Motion: 1-4.

  3. Constant Acceleration

    1. For one dimensional motion

      1. In the X-direction
        vx = vox + axt   and  x = xo + vox t + 1/2 axt2

      2. In the Y-direction
        vy = voy + ayt   and  y = yo + voy t + 1/2 ayt2

    2. For two dimensional motion
      v = vo + at   and  r = ro  + vot + 1/2 at2,
      where ro  = xo i + yoj, vo  = voxi  + voy j, and a = (axi + ayj)
      r  = xi + yj.  From A we see we may write:
      r = (xo + vox t + 1/2 axt2)i + (yo + voy t + 1/2 ayt2)j,  or
      r = (xoi  + yoj) + (voxi  + voy j)t + 1/2(axi + ayj)t2,  or
      r =      ro             +       vot           +       1/2 at2

  4. Motion In Two-Dimensions

    1. Projectile Motion

      1. Sample problem.   An object is given an initial velocity of 25 m/s at an angle of 530 with the horizontal.  Find the initial position of the object along (a) the X-axis, xo,  (b) the Y axis, yo,  the initial velocities along  (c) the X-axis, vox, and (d) the Y-axis, voy.  (e) Indicate the directions of vox and voy on the figure. Find (f) the maximum vertical height ymax, (g) the time for object to hit the ground, (h) the distance x the projectile travels in a horizontal direction, (i) vx, vy, and v just before the object hits the ground, and (j) the angle v makes with the horizontal. Take g = 10 m/s2.

      2. Approach to solution

        1. Choose an appropriate X and Y axis.  For a projectile problem, the appropriate X-axis is the horizontal direction and the Y-axis the vertical direction.  Draw the figure, as shown in Fig. 4b below.

        2. Set up a chart for the X and Y components of the descriptive quantities. Identify the quantities asked for in (a) through (d) of the statement of the problem along with the components of the acceleration and enter them in the chart. Then write appropriate formulae for the components of motion.



          (a) xo = 0

          (b) yo =25 m

          (c) vox = 25 m/s cos 53o
            =15 m/s

          (d) voy = 25 m/s sin 53o
             = 20 m/s

          ax = 0

          ay = -10 m/s2

          x(t) = xo + voxt + 1/2 axt2

          vy2(y) = (voy)2 + 2ay(y - yo)    

          (e) x(t) = 0 + 15 m/s t + 0

          (f) At ymax, vy2(y) = 0 =
          (20m/s)2- 20m/s2(ymax-25m)
          (ymax- 25 m) = 20 m,  or
          ymax = 45 m


          (g) y(t) = yo + voyt + 1/2 ayt2 when object hits ground,
          0 = 25 m + 20 m/s t - 5 m/s2 t2, or  t2- 4 s t - 5 s2 = 0
          (t - 5 s)(t +1 s) = 0
          t = 5 s.  t cannot be negative.

          (h) x(5s) = (15m/s)(5s) = 75m


          (i) vx(t) = vox + axt
          vx(5 s) = vox = 15m/s

          (i) vy(t) = voy + ayt
          vy(5 s) = 20 m/s -(10 m/s)(5s)
          = -30 m/s

          v = (vx2 + vy2)1/2 = [152 + (-30)2]1/2 m/s = 33.5m/s

          (j) tan Θ = vy/vx = -30/15 = - 2.
                     Θ = - 63o

    2. Sample problems in 107 Problem Set for Two Dimensional Motion: 6-14.

  5. Uniform Circular Motion

    1. Definition:  A body travelling in a circle with a constant speed.
      In Fig. 6 below, |v| = |v + Δv|

      1. Instantaneous velocity tangent to the circle at any instant.

      2. Displacement vector sweeps through same angle as the velocity vector.

    2. Centripetal acceleration

      1. Look at the triangles of Fig. 6b above.

      2. Δv/v = Δr/r  or  Δv = v Δr/v.

      3. To find magnitude of acceleration consider:
        Δv/ Δt = (v/r)( Δr/ Δt)

        a = lim Δv/ Δt = (v/r)(v) = v2/r
              Δt o

      4. Since a = lim of Δv/ Δt as Δt 0, find the direction of the acceleration by looking at the direction of Δv as Δt approaches 0.   Δv and a point into the center of the circle.

  6. Descriptive Terms

    1. Period T is the time for one complete revolution

    2. Frequency f is the number of revolutions per second = 1/T

    3. The speed of the object = 2 πr/T = 2 πrf is a CONSTANT.  It's velocity is not constant because it changes direction.

    4. The magnitude of the acceleration  = v2/r is a constant, but the direction of the acceleration is not a constant since it continually changes direction.  The direction changes so this is not a case of constant acceleration and you cannot use the formulas for constant acceleration.

    5. Angular velocity ω is the time rate of change of angular displacement =

      1. The angle ΔΘ subtends an arc whose length is Δs.

      2. By definition, the angle in radians,  ΔΘ= Δs/r.

      3. Since v =  Δs/ Δt,  ΔΘ/ Δt = ( Δs/r)/ Δt = v/r.  The limit as Δt approaches zero of ΔΘ/ Δt is ω.  So ω = v/r.

      4. The direction of ω is given by the right hand rule:  When the fingers of your right hand are curled in the direction of the motion, the direction of your thumb points in the direction of ω.  For Fig. 6a above, ω is out of the page.

      5. While v is not constant for uniform circular motion,  ω  is constant.
        Θ = ωt.

    6. Sample problems in 107 Problem Set for Two Dimensional Motion: 15-18.

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Susan D. Kunk
Phyllis J. Fleming
October 8, 2002
April 15, 2003