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Elementary Analysis of the Gyroscope

ERNEST . B A R K E RUniversity oj Michigan, Ann Arbor, Michigan

(Received March 3 , 1960)

The simple gyroscope is an excellent subject for a lecture tab le demonstration to classes inelementary physics. The only observable force acting upon the precessing top is a downwardpull due to grav ity, yet, instead of falling, i t moves with a continuous horizontal displacement.An adequat e and convincing explanation of t his curious behavior is essential, an d it mus t bestated in language familiar to the student. One possible approach to the problem, using a verysimple model, is given here. The in ternal reactions are described and their values are computed.Because of some difficulty in visualizing motion in three dimensions, it is recommended th at amodel be constructed. I t may be very crude an d yet quite adequate.

TE gyroscopic top provides one of the most

intr iguing of al'l lecture demonstrations inelementary physics. I t never fails to commandattention, and to arouse the keenest interest.When such a top, having been properly set inmotion, is supported at one end of i ts axis, i texecutes a simple precession about the vertical.Although the only obvious force acting upon itis the downward pull of gravity, the motion ishorizontal. Almost invariably the students ask,"What keeps it from falling, and what makesi t precess?" I t is of the utmost importance thatthe instructor be prepared to answer these ques-tions in a convincing way, and in familiar lan-guage. One approach, which experience hasproved t o be helpful, is presented here. I t isneither as economical nor as elegant as the ex-planation in terms of axial vectors given in manytextbooks, but it utilizes no ideas more compli-cated tha n the addition and subtract ion of veloc-ities, and yields specific information about in-

ternal forces which produce the gyroscopic effects.The problem is much simplified by the adop-

tion of a model for the rotor , like th at illustratedin Fig. 1. I t consists of four small bodies, eachof mass m , which may be considered as pointmasses. They are mounted symmetrically at D,E, G, and H by means of two perpendicular cross-bars passing through th e axis A B a t it s midpointC, which is the center of mass of t he movingsystem. For present purposes no complicatedbearings a re necessary.

Initially friction may be neglected, and theframe assumed to be weightless. Each particleis distant r from C, and rotates as indicatedabout the axis A B with an angular velocity w.The plane containing the four masses will bereferred to a s the rotor plane. I ts distance fromA , the poin t of suppor t, is R, and it precesseswith an angular velocity In. Th e linear velocityof the center of mass is V=OR. Each mass fol-lows a path which is continually changing indirection, and, therefore, the motion is acceler-ated. The accelerations arise from forces exertedupon the individual masses by the framework.

Each mass in turn reacts upon the frameworkwith a force proportional to its instantaneousacceleration but opposite in direction. Some ofthese reactions are simply centrifugal, due eitherto the rotation or to the precession, but theseneed not be considered, since they contributenothing to t he gyroscopic effect. I t is clear thatthere must be some equilibrating torque whichneutralizes the torque due to gravity, and itmust arise from interactions within the system

FIG. 1. Simple model of gyroscopic top. itself, i.e., the influence of precession upon the

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rotational velocities, and of rotation upon theprecessional velocities.

Assume that A B is horizontal (sine= 1) an dD E is vertical. The interactions will cause ac-celerations of the masses a t D and E , but produceno effects on those a t G an d H, since at thesepoints the velocities due t o rotat ion are parallelto the axis of precession; although displaced bythe precession, the rotational velocities of G andH suffer no changes either in direction or in mag-nitude. When the mass D is at its highest pointi t has a horizontal velocity v = w r due t o rotation,directed outward from the page. At the sameinstant the mass at E is moving into the page

with an equal velocity. These velocities, lying inthe rotor plane, will be changing in direction be-

FIG. . View looking downward along DE showing changein direction of the r otatio nal velocity v, because of pre-cession. The precessional velocity V , turns counterclock-wise and the rotational velocities v a t D and E lie in therotor plane which turns with the precession.

cause the rotor plane itself is turning abou t theaxis DE. Both of the velocities v will be deflectedas shown in Fig. 2. These are views looking down-ward along the line DE.

During a time At the linear displacement ofeach v due to precession is VAt, and the anglethrough which it turns is VAt/R. This is equalto the angle Av/v. Hence the acceleration is

The mass D is accelerated away from the axisZ (Fig. I) , while E is accelerated toward the axis,by forces each equal t o mal. The inertial reactionsin each case are forces in the opposite directionsexerted by th e masses upon the framework. MassD

pushes the framework inward (towardsZ)

FIG. 3. Change in velocity V due to rotation. Point Dmoves to D' an d V, which is perpendicular to the radiusvector from A , tur ns out of t he rotor plane. The samething happens at point E which is displaced to E' , the samedistance but in the opposite direction, causing V to swingto the left.

while mass E pushes it outward. Th e forces arein t he directions of Fl and Fz in Fig. 1. The tworesulting torques add together, giving a totaltorque due to the precession

This is incomplete, however, since i t does nottake into account an y effects due to rotation, sincethe line DE was assumed t o be vertical.

If DE rotates through a small angle in the di-rection indicated in Fig. 1, the precessional veloc-it y V moves out of the rotor plane. This is indi-cated in Fig. 3, which is a view looking downwardalong DE. During a time At the rotation causesmass D to move outward from the page to apoint Df , with a new radius vector R'. Since theprecessional velocity V must be perpendicularto the radius vector, it will be turned throughan angle A V / V = DD1/R = vAt/R, and its acceler-ation is

a z = A V/At = Vv/R, (3 )

directed outward from the page. This Is exactlyequal to al and in the same direction. I t measuresthe effect of rotat ion upon precession. Th e cor-responding acceleration of E is in the oppositedirection.

Combining these four accelerations, since theyare simultaneous, and multiplying by m an d rgives the final torque

24 being the sum of the four masses. The torquedepends upon both w an d Q. If the magnitudeand direction of w are given, the magnitude anddirection of Q for steady motion would be de-termined 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 howits value is automatically selected.

Suppose that the rotor is supported at bothA and B, and is already spinning a t a fairlyhigh speed, but without precession. If the sup-port at A is removed, the rotor will start tofall. Both masses G and H will move downwardbecause of gravi ty, along circular arcs of radiusR with centers on the X axis (see Fig. 1). At thesame time G will be moving downward and Hwill be moving upward, because of t he rotation,while both will be turning inward, toward thepage. Consequently, G will be accelerated inwardtoward the X Z plane, while H will be accelerated

outward. Thus the framework will be subjectedto horizontal forces outward a t G and inwarda t H, producing a torque about a vertical axis,e.g., the Z axis, which initiates the precession.

As long as the system continues to fall, theangular momentum of precession will be acceler-ated. This precession, in turn, generates thetorque which opposes the fall, as has alreadybeen shown. When fi reaches the proper valueQ will equal the gravitational torque. At thisinstant, however, the downward motion will notcease since the downward momentum has justreached it s maximum. As this point of equilibriumis passed, an upward retarding torque is gener-ated. This brings the falling motion to a stop,and causes it t o return upward, to pass the pointof equilibrium again. Thu s there is an oscillationdownward and upward which, in the absence offriction, will continue indefinitely. This motionis called nutation. In an actual top there will befriction, of course, and thi s rapidly damps ou tthe nutation. In fact, this effect may be difficultto observe in a demonstration model, and themotion may appear to be a steady precessionfrom the beginning. The center of gravity willmove downward gradually, however, since po-tential energy is being transformed into heat.

Th e system always has some dead weight be-cause of framework and mountings. These con-tribute to the gravitational torque, and makenecessary an increased angular momentum tomaintain t he motion.

The accelerations derived previously apply

only to t he case where DE is vertical. As D movesoutward from this position, its velocity v has

both horizontal and vertical components, thelatt er being parallel to the axis Z. T he effectivecomponent v, decreases as cos4, 4 being th e angleof rotation, measured from the initial position.At the same time the radius r turns through thesame angle, and its effective component is verti-cal with the value cos4. Thus the contributionof D to the equilibrating torque Q decreases ascos24, and the same holds for th e contributionof mass E. The situation is saved, however, bythe fact that G an d H develop horizontal com-ponents which produce additions to Q, propor-

tional t o sin24, and the tota l value of Q remainsunchanged.

Since the foregoing solution holds for anyvalue of 4 , it must also hold if many addi tionalsimilar rotors were added to form a continuousring. Th e addition of other hoops of equal radius,with centers distributed along the rotational axis,would be allowed if a proper value of R wereintroduced. Similarly, hoops with different radiicould be added, if appropriate average values ofr are computed. In other words, the rotor mayhave a ny mass distribution, as long as i t is sym-metrical about the rotational axis. The generalcase may be expressed very simply: Q= Io8,where I is the moment of inertia abo ut the axis.This is exactly the result obtained in Eq. (4).

The student's questions have now been an-swered. The mass particles which form the rotorare moving with velocities which change con-tinuously in direction. These changes give riseto accelerations and hence to internal torquesexerted upon the moving system by the massparticles. I n steady motion these internal torquesexactly counterbalance th e gravitational torque.If th e spinning rotor has no precession, i t willbegin to fall, causing internal torques whichinit iate the precession.

Th e simple model makes it relatively easy tosee how these effects arise. Perhaps the greatestdifficulty is the problem of visualizing the three-dimensional motions. This can be relieved byconstructing a model, which will be quite ade-quate, even if it is very crude.