Chapter 11: Combined Parametric Motions

Introduction

Sometimes we are interested in the motion of a point P which has a simple motion relative to some other point , which, in turn, also is moving in a particular way. It is not hard to write parametric formulas , for the point , if we proceed carefully in a step-by-step manner.

Suppose we want to describe the motion of a tip of a baton as it rotates after being thrown into the air. More precisely, suppose that at time a 2-foot long baton is situated vertically with the tip we are interested in at height ft above ground level and the other tip at height ft. At that moment the baton is tossed at ft/sec at an angle of above the horizontal, and it is also set spinning at rev/sec clockwise in the plane of its trajectory. (It could have been set spinning in a different plane, but that would have made this into a three-dimensional problem, and that would be too complicated for our present purposes.

To set up the parametric equations for the motion of the tip of the baton, the basic procedure is to divide the motion into two parts: the motion of the center of the baton and the rotation of the tip around the center . The first of these is a falling body problem, and the second is a circular motion problem. We shall take the -axis to be the vertical line along which the baton is situated at time , and we shall take the -axis to be the line on the ground under the baton's trajectory.

According to a principle of physics, the center of the baton follows the same path as a small object (say, a ball) that is thrown with the same initial velocity. In other words, the first half of our procedure --- the motion of the center of the baton --- is like Example 10.6 of the last chapter. Let ( ) denote the - and -coordinates of the center of the baton at time . Repeating the procedure used in Example 10.6 --- with ft/ , = 14 ft/sec (rounded to the nearest ft/sec), v[0, vert] = ft/sec --- we obtain:

, .

(Notice that ft, because the initial location of the center is half-way between and .)

Now let ( ) denote the - and -coordinates of the tip of the baton at time . The second half of our procedure consists in determining how much differs from and how much differs from , i.e., how far the tip is from the center in the x- and y-directions. Let be the angle in radians through which the baton has spun at time t. We have the following diagram:

code for diagram

code for diagrams

Example 11.2: The Cycloid

We shall express the coordinates of the center ( ) and the coordinates of the pebble ( ) in terms of the following parameter: the angle in radians through which the wheel has rotated after picking up the pebble at the origin. We shall also introduce another letter s to denote the linear distance the wheel has traveled. is the distance along the x-axis from the origin to the point of contact of the wheel, and is also the x-coordinate of the center of the wheel.

There is a simple relation between the distance the wheel has traveled and the angle in radians through which it has rotated. Namely, the distance traveled along the x-axis is also equal to the distance measured around the circumference through which the wheel has turned (here we are assuming that the wheel rolls without slipping or sliding). For example, every time the wheel rotates through 1 revolution (i.e., radians), the wheel moves a distance equal to the circumference . More generally, if the wheel rotates through an angle of radians, the wheel moves a distance equal to . This is because the distance along the circumference of the unit circle of the part of the circumference corresponding to an angle of radians is simply , by the definition of the measure of an angle in radians. In a circle of radius --- i.e., the unit circle magnified by a factor of --- the distance along the circumference will be times as great:

code for diagram

code for cycloid function and diagrams

code for sequence of images of rolling wheel

The same as in Example 11.2, except that when the wheel rolls over the origin, instead of getting stuck in the tread on the outside of the tire, the pebble flies up and gets stuck in a spoke at a distance a directly below the center (where a<r). Thus, the pebble again rotates around the center, but at a distance a rather than r from the center. The path of this pebble is called a trochoid.

In this problem everything is just like in Example 11.2, except that in the circular motion part the hypotenuse of the triangle (which is the line from the center to the pebble after the wheel has rotated through an angle of theta radians) is now a rather than r. This leads to the formulas:

= - =

- =

Here is a picture of the trochoid path in the case when :

code for diagram