Cut a piece of pipe that is three times as long as its diameter. (Err on the long side; this experiment will still work with tubes
up to 3.15 diameters long.)
Mark an X on the side of one end of the pipe, and an O at the other end.
(See Figure 1.)
After the first experiment, you will need a few more pieces of pipe. Cut them so that you have a set of pipes with lengths that are two, three, four, and five times their diameters.
Place your finger on the X on the piece of pipe that is three times as long as its diameter, and rapidly push your finger down while at the same time pulling it toward you. See Figure 2.Press your finger down rapidly to make the cylinder spin.
Next place your finger on the O and spin the rod.
The cylinder will spin and rotate, making a blurred circle in which X's can be seen. Notice that as the spinning cylinder stabilizes, you can see three X's which mark the vertices of a triangle. Notice that the O does not appear.
Next place your finger on the O and spin the cylinder. Notice that as the motion stabilizes, three O's appear, each at the vertex of a triangle.
Notice that the X does not appear
Try some experiments to figure out what is going on. Here are some suggestions:
TO GET THE MOST OUT OF THIS ACTIVITY, DON'T READ THIS SECTION UNTIL YOU HAVE DONE SOME EXPERIMENTS YOURSELF!
When you launch the cylinder, it spins about its long axis and rotates about a line perpendicular to this axis. As it rotates about its center, the cylinder forms a blurry circle on the table top. As the cylinder spins, the top of one end moves in the same direction as the end that is rotating, while the top of the other end moves opposite the rotation. The arrows in Figure 3. show these relationships.
The two arrows within the cylinder show how it spins. The two arrows outside the cylinder show how it rotates. On the right end the two motions cancel each other, and when the mark on the spinning cylinder is at the top, it actually stops momentarily. (See Etc. below for the mathematical explanation). On the left end the two motions add, and when the mark on the spinning cylinder comes to the top, it moves twice as fast as it would with either motion alone. (See Etc. below for the mathematical explanation).
Human eyes can see the stopped mark easily, while the extra-fast moving mark is a blur. Thus only the mark on one end is visible. Since we see three marks around the blurred circle, we know that the cylinder is making three spins for every rotation. (See Etc. below for the mathematical explanation). Cylinders that are cut so that their length is four diameters have a stable square with four markings. Those cut to two diamters create a stable pattern of two marks. At first with the cylinder that is three diameters long, the marks on one end appear but they do not form a stable pattern. After a few seconds, however, the marks settle into a stable triangular pattern which persists until the cylinder slows to a stop. To understand this behavior, notice that the cylinder spins and rotates with one end on the table, and one end in the air. The cylinder makes a stable pattern when the end touching the table rolls without slipping. Usually the cylinder is launched so that it is spinning faster than it is rotating. This means that the end touching the table rubs against the table, dissipating energy and slowing down until it reaches a speed where it rolls without slipping. This is why the pattern is not stable at first, but then stabilizes.
The discussion below gives the "Math Root" of some of the behavior noted previously. References to the LEFT end and the RIGHT end of the cylinder refer to Figure 3.
The circumference of the circle that the RIGHT end moves through as it goes around its blurry circle is pi times the length of the cylinder, or pi times three diameters of the cylinder itself.
The circumference of the cylinder is pi times its diameter.
C = pL = 3pd
The circumference of the cylinder itself is pi times its diameter.
c = pd
So the number of times that the cylinder spins in one rotation is C/c = 3. This is why there are three markings!
The linear speed due to ROTATION at either end of the cylinder is Vrotation = distance/time = 3pd/T , where T is the time for one whole ROTATION.
The linear speed due to SPIN at either end of the cylinder is
Vspin = distance/time = pd/t , where t is the time for one whole
But there are three spins for every rotation, so T = 3t, and therefore
Vrotation = 3pd/T = 3pd/3t = pd/t = Vspin
This shows that the two speeds are equal (see arrows in figure 3). This means that for the instant that the marks are facing upward, the mark at the RIGHT end actually comes to a stop, since the speeds are in opposite directions, and the mark at the LEFT end moves doubly fast, since the speeds are in the same direction.