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Vectors

Sunrise and sunset can be very unusual from certain vantage points on the planet Mercury. Since the period of one orbit around the Sun (i.e. one Mercury ``year'') for Mercury is about 1.5 times the period for one rotation of Mercury on its axis (i.e. one Mercury ``day''), the apparent path of the Sun across the sky for an observer on the surface of Mercury can be quite complex. From certain points on the surface, an astronaut would not see a sunrise followed by smooth progression of the sun across the sky from east to west to sunset, but a something altogether more interesting. After sunrise in the east, the Sun travels up the sky to a certain point but stops before reaching the zenith, reverses and sets in the east! It then rises again in the east, moves quickly across the sky to set in the west, then rises in the west, travels a short distance eastward across the sky, then reverses and sets again in the west. Click on the mosaicked image of Mercury above or here to see a video from such a vantage point. The inset to the video shows a red dot which traces the apparent path of the Sun relative to a Mercury-centered coordinate system. Vectors, along with the mathematics of circular motion described in the next section, provide an easy way of quantitatively describing this rather complex celestial motion.

The description of more than one dimension demands the introduction of vectors to our discussion. Vectors are simply a mathematical short-hand for describing any quantity that has both a direction and a magnitude. As usual, we will start our discussion of this new concept of combining direction and magnitude by relating back to things we already know and understand (by the way, it is a common practice in physics to build on subjects covered before so it is crucial that you understand each topic as it is introduced. You are bound to see that topic again as the basis for yet another new topic).

The simplest description of a vector in action is the displacement. Up to this point, we have really only considered motion along one axis, but, obviously, its possible for a object to have a more complicated path of motion. Consider a particle which moves along the dashed path between points and as shown in the following figure.

For practical purposes, all that matters as far as displacement is concerned is how far your final position is from your original position. The fact that you took a long path to get there means your average velocity will be small when calculated between and , but the displacement is the same no matter what path you take. In other words, when we calculate displacement, we take into account only the distance between the final and initial positions. We lose all information about the path taken. In some sense, this seems like a loss of information, but, as you will see later, the independence of displacement from the path taken allows you to make very powerful statements about describing what the particle is doing when it reaches its final position. This is because nature itself, in many instances, cares only about the initial and final positions and not on the path taken, but we leave the exploration of that to later courses ...

Now that we've described in words what the displacement is, how do we write it down mathematically? From our knowledge of trigonometry, we know that one way of specifying the distance between points and is to form a right triangle as shown below.

One leg of the right triangle runs from to while the other runs from to . We can calculate the length of each of these legs trivially, as for the horizontal leg and for the vertical leg. The length of the displacement (let's call it r is then

by the Pythagorean theorem. Another way of looking at it is to say that the displacement, r, is really made of two components, and . Each of these components is also a displacement, in fact, these components are just the one-dimensional displacements we have been used to using all along. The displacement r is a vector which has two components, one along the x-direction and one along the y-direction. The magnitude or length of r is given by the equation we showed above while the direction of r is given, typically, by specifying its angle with respect to the horizontal. In this case, the angle is , where

The symbol used most often for indicating that a certain quantity behaves like a vector (i.e. needs to have a direction and magnitude to be completely specified) is to place a little arrow above the variable for that quantity, e.g. is the vector displacement shown in the above figures. We can note here that there are two ways of specifying a vector. One is to give its magnitude and its direction in terms of an angle with respect to an axis of the coordinate system. The other is to specify its components. Specifying the components is equivalent to specifying the vector since we can derive both the magnitude and direction of the vector from its components as we just showed. We can also derive the components from knowledge of the magnitude and direction by again resorting to trigonometry

Generally, the component specification for a vector is given by a set of symbols which may be new to you. The symbol stands for the unit vector in the y-direction while the symbol stands for the unit vector in the x-direction. A unit vector is simply a vector that has a magnitude of 1 (in whatever the appropriate units are for the coordinate system you are using). For the present case, the component along the horizontal would be while that along the vertical is . These horizontal and vertical vectors have magnitudes equal to the lengths of the components and directions along the x and y directions which is just what we want. To specify , then, we would write

We can easily generalize this scheme to three dimensions by adding in an z-direction component as shown below.

The unit vector notation for the z-direction is . The generalization from two to three dimensions is easy. We state, without proof, that the length of the vector in three dimesions is

Obviously, the vector notation for is

In addition, it's important to note that the components are at right angles to each other and are therefore completely independent of each other. That is to say, no component along any axis has itself a component along any other axis. This is the reason for the independence of motion along horizontal and vertical directions that we saw in the trajectory motion section. Since we have defined the displacement as a vector, it should be obvious to you that velocity and acceleration, which are derived from the displacement, are also vectors, i.e. they have direction and magnitude. Since the components of vectors are independent, we expect that the components of motion are independent. An object can be accelerating along one component's direction and completely motionless along the other component direction (e.g. a falling object can be accelerating downward while having no motion along the horizontal whatsoever).

Of course, it should be obvious that force acts like a vector in that it has a direction and a magnitude. All of the rules on getting components, combining them to make a complete vector, etc. apply to force.

With components comes the inevitable task of remembering our trigonometry, so if you have not refreshed your memory of it by now, you will with the following example and next set of exercises.

SOLUTION: We begin by drawing an accurate sketch of the situation so that we can define our coordinate system and keep straight the forces.

The figure above shows the golf club force at angle , the force of gravity, the coordinate system for y and x, and the trajectory of the ball due to the forces acting on it. We first get the components of the forces acting. For , we have

Thus, the force could be written as in unit vector notation. The weight has only one component in the direction. Therefore, . Although the weight continues to act on the ball throughout its trajectory, acts only for 0.001 s. During this time, which we will call , the net external force on the ball is

We can find the acceleration on the ball during this time as

We can find the final velocity attained over this time as

After losing contact with the club, the only force acting on the ball is gravity. We can take the final velocity just after losing contact with the club to be the initial velocity for free-fall flight of the ball. For the moment, we can use a shorter notation for the velocity by noting [ v_0 = v_0x i + v_0y j ] where and for the initial velocity just after contact with the club.

We can look at the x component of the ball's trajectory to find the time for the ball to travel from the ground to its highest point and back to the ground again since the beginning and end point are known [

Now we can find the horizontal distance (called the range R) traveled in this time. Note that during the ball's flight there are no forces acting in the y direction and hence no y acceleration.

Try exercise 4-4 to get some experience with trajectory problems.