Barker, Elementary Analysis Of The Gyroscope

Elementary Analysis of the Gyroscope ERNEST F. BARKER University oj Michigan, Ann Arbor, Michigan (Received March 3, 1960) The simple gyroscope is an excellent subject for a lecture table demonstration to classes in elementary physics. The only observable force acting upon the precessing top is a downward pull due to gravity, yet, instead of falling, it moves with a continuous horizontal displacement. An adequate and convincing explanation of this curious behavior is essential, and it must be stated in language familiar to the student. One possible approach to the problem, using a very simple model, is given here. The internal reactions are described and their values are computed. Because of some difficulty in visualizing motion in three dimensions, it is recommended that a model be constructed. It may be very crude and yet quite adequate.

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H E gyroscopic top provides one of the most intriguing of al'l lecture demonstrations in elementary physics. I t never fails to command attention, and to arouse the keenest interest. When such a top, having been properly set in motion, is supported at one end of its axis, i t executes a simple precession about the vertical. Although the only obvious force acting upon it is the downward pull of gravity, the motion is horizontal. Almost invariably the students ask, "What keeps it from falling, and what makes i t precess?" I t is of the utmost importance that the instructor be prepared to answer these questions in a convincing way, and in familiar language. One approach, which experience has proved t o be helpful, is presented here. I t is neither as economical nor as elegant as the explanation in terms of axial vectors given in many textbooks, but i t utilizes no ideas more complicated than the addition and subtraction of velocities, and yields specific information about in-

FIG.1. Simple model of gyroscopic top.

ternal forces which produce the gyroscopic effects. The problem is much simplified by the adoption of a model for the rotor, like that illustrated in Fig. 1. I t consists of four small bodies, each of mass m, which may be considered as point masses. They are mounted symmetrically a t D, E, G, and H by means of two perpendicular crossbars passing through the axis A B a t its midpoint C, which is the center of mass of the moving system. For present purposes no complicated bearings are necessary. Initially friction may be neglected, and the frame assumed to be weightless. Each particle is distant r from C, and rotates as indicated about the axis A B with an angular velocity w. The plane containing the four masses will be referred to as the rotor plane. Its distance from A , the point of support, is R, and i t precesses with an angular velocity In. The linear velocity of the center of mass is V=OR. Each mass follows a path which is continually changing in direction, and, therefore, the motion is accelerated. The accelerations arise from forces exerted upon the individual masses by the framework. Each mass in turn reacts upon the framework with a force proportional to its instantaneous acceleration but opposite in direction. Some of these reactions are simply centrifugal, due either to the rotation or to the precession, but these need not be considered, since they contribute nothing to the gyroscopic effect. I t is clear that there must be some equilibrating torque which neutralizes the torque due to gravity, and i t must arise from interactions within the system itself, i.e., the influence of precession upon the 808

GYROSCOPE ANALYSIS

rotational velocities, and of rotation upon the precessional velocities. Assume that A B is horizontal (sine= 1) and DE is vertical. The interactions will cause accelerations of the masses a t D and E , but produce no effects on those a t G and H, since a t these points the velocities due to rotation are parallel to the axis of precession; although displaced by the precession, the rotational velocities of G and H suffer no changes either in direction or in magnitude. When the mass D is a t its highest point it has a horizontal velocity v = wr due to rotation, directed outward from the page. At the same instant the mass a t E is moving into the page with an equal velocity. These velocities, lying in the rotor plane, will be changing in direction be-

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FIG.3. Change in velocity V due t o rotation. Point D moves t o D' and V , which is perpendicular to the radius vector from A, turns out of the rotor plane. The same thing happens a t point E which is displaced to E', the same distance but in the opposite direction, causing V t o swing to the left.

while mass E pushes it outward. The forces are in the directions of Fl and Fz in Fig. 1. The two resulting torques add together, giving a total torque due to the precession

FIG.2. View looking downward along DE showing change in direction of the rotational velocity v, because of precession. The precessional velocity V, turns counterclockwise and the rotational velocities v a t D and E lie in the rotor plane which turns with the precession.

This is incomplete, however, since it does not take into account any effects due to rotation, since the line DE was assumed to be vertical. If DE rotates through a small angle in the direction indicated in Fig. 1, the precessional velocity V moves out of the rotor plane. This is indicated in Fig. 3, which is a view looking downward along DE. During a time At the rotation causes mass D to move outward from the page to a point D f , with a new radius vector R'. Since the precessional velocity V must be perpendicular to the radius vector, it will be turned through an angle A V / V = DD1/R= vAt/R, and its acceleration is a z= A V/At = Vv/R, (3)

cause the rotor plane itself is turning about the axis DE. Both of the velocities v will be deflected as shown in Fig. 2. These are views looking downward along the line DE. During a time At the linear displacement of each v due to precession is VAt, and the angle through which i t turns is VAt/R. This is equal to the angle Av/v. Hence the acceleration is

directed outward from the page. This Is exactly equal to al and in the same direction. I t measures the effect of rotation upon precession. The corresponding acceleration of E is in the opposite direction. Combining these four accelerations, since they are simultaneous, and multiplying by m and r gives the final torque

The mass D is accelerated away from the axis Z (Fig. I), while E is accelerated toward the axis, by forces each equal to mal. The inertial reactions in each case are forces in the opposite directions exerted by the masses upon the framework. Mass D pushes the framework inward (towards Z)

24 being the sum of the four masses. The torque depends upon both w and Q. If the magnitude and direction of w are given, the magnitude and direction of Q for steady motion would be determined by the requirement that Q must be equal to the gravitational torque MgR. I t remains

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to show how the precession is set up, and how its value is automatically selected. Suppose that the rotor is supported a t both A and B, and is already spinning a t a fairly high speed, but without precession. If the support a t A is removed, the rotor will start to fall. Both masses G and H will move downward because of gravity, along circular arcs of radius R with centers on the X axis (see Fig. 1).At the same time G will be moving downward and H will be moving upward, because of the rotation, while both will be turning inward, toward the page. Consequently, G will be accelerated inward toward the X Z plane, while H will be accelerated outward. Thus the framework will be subjected to horizontal forces outward a t G and inward a t H, producing a torque about a vertical axis, e.g., the Z axis, which initiates the precession. As long as the system continues t o fall, the angular momentum of precession will be accelerated. This precession, in turn, generates the torque which opposes the fall, as has already been shown. When fi reaches the proper value Q will equal the gravitational torque. At this instant, however, the downward motion will not cease since the downward momentum has just reached its maximum. As this point of equilibrium is passed, an upward retarding torque is generated. This brings the falling motion to a stop, and causes it to return upward, to pass the point of equilibrium again. Thus there is a n oscillation downward and upward which, in the absence of friction, will continue indefinitely. This motion is called nutation. In an actual top there will be friction, of course, and this rapidly damps out the nutation. In fact, this effect may be difficult to observe in a demonstration model, and the motion may appear to be a steady precession from the beginning. The center of gravity will move downward gradually, however, since potential energy is being transformed into heat. The system always has some dead weight because of framework and mountings. These contribute to the gravitational torque, and make necessary an increased angular momentum to maintain the motion. The accelerations derived previously apply

only to the case where DE is vertical. As D moves outward from this position, its velocity v has both horizontal and vertical components, the latter being parallel to the axis Z. The effective component v, decreases as cos4, 4 being the angle of rotation, measured from the initial position. At the same time the radius r turns through the same angle, and its effective component is vertical with the value cos4. Thus the contribution of D to the equilibrating torque Q decreases as cos24, and the same holds for the contribution of mass E. The situation is saved, however, by the fact that G and H develop horizontal components which produce additions to Q, proportional to sin24, and the total value of Q remains unchanged. Since the foregoing solution holds for any value of 4, it must also hold if many additional similar rotors were added to form a continuous ring. The addition of other hoops of equal radius, with centers distributed along the rotational axis, would be allowed if a proper value of R were introduced. Similarly, hoops with different radii could be added, if appropriate average values of r are computed. In other words, the rotor may have any mass distribution, as long as it is symmetrical about the rotational axis. The general case may be expressed very simply: Q= Io8, where I is the moment of inertia about the axis. This is exactly the result obtained in Eq. (4). The student's questions have now been answered. The mass particles which form the rotor are moving with velocities which change continuously in direction. These changes give rise to accelerations and hence to internal torques exerted upon the moving system by the mass particles. In steady motion these internal torques exactly counterbalance the gravitational torque. If the spinning rotor has no precession, it will begin to fall, causing internal torques which initiate the precession. The simple model makes i t relatively easy to see how these effects arise. Perhaps the greatest difficulty is the problem of visualizing the threedimensional motions. This can be relieved by constructing a model, which will be quite adequate, even if it is very crude.