 Phyllis Fleming Physics Physics 104

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.

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

4. 6. 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).

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

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

5. Example (Fig. 3) A = Axi + Axj, and 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)
See Fig. 3 7. Practice Problems in 104 Problem Set for Two Dimensional Motion: 1-4.

2. Motion In Two-Dimensions

1. Projectile Motion

Sample problem:

An object is given an initial velocity of 25 m/s at an angle of 53o with the horizontal. Find the initial position of the object along (a) the X-axis, xo, and (b) the Y axis, yo. Find 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. Approach to solution:

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. 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.

 X Y (a) xo = 0 (b) yo =25 m (c) vox = 25 m/s cos 53o   =15 m/s (d) voy = 25 m/s sin 530    = 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 Q = vy/vx = -30/15 = - 2.            Q = -63o

2. Uniform Circular Motion

1. An object has uniform circular motion when it moves in a circle with constant speed. The speed v is constant, but the velocity v is not constant, because v is always tangent to the path so it continually changes direction. In Fig. 5 above, | v1|= |v2|= |v3|= |v4|= v,  but this is not a case of constant velocity because the direction of v changes.

2. In Fig. 6 below, the two angles labeled Q are equal because v2 and v1 are perpendicular to their respective radii. Since |v1|= |v2|= v,  the triangle with the radii and the triangle with the velocities are similar because they are isosceles triangles and they have angles that are equal.

Thus  Dv/v = Dr/r  or  Dv = Dr(v/r).

Divide both sides by Dt and take limit as Dt approaches zero: Since a = dv/dt,  a has the same direction as Dv, or into the center of the circle.

3. The magnitude of the acceleration is constant:
|a1|= |a2|= |a3|= |a4|= a = v2/r,
but this is not a case of constant acceleration because the direction of a is not constant.

4. Descriptive Terms

1. Period T = time for one complete rotation = 2pr/v.

2. Frequency f = number of revolutions per second = the reciprocal of period = 1/T = v/2pr.

3. The speed of the object = 2pr/T = 2prf is a constant.

3. Practice problems in 104 Problem Set for Two Dimensional Motion: 6-14.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming September 25, 2002 April 1, 2003