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Home , Gyrocompass, Gyroscope

UM-P-89/79

Coriolis and Magnetic Forces:

The and Magnetic as Analogues

by

Geoffrey I. Opat

School of Physics The University i.f Melbourne Parkville, Victoria 3052 Australia

Abstract:

The on a particle (acting within a rotating frame) and the magnetic force on a charge, are both proportional to and act perpendicular to the velocity. As a consequence, the action of the Coriolis force on rotating mass, and the action of the magnetic force on rotating charge are formally identical. Just as the action of a is to align the axis of the rotating charge distribution (magnetic dipole) with itself, so the Coriolis force aligns the axis of a rotating mass distribution

() with the of the rotating frame. This enables us to understand the gyrocompass by analogy with the magnetic compass.

A simple way to demonstrate the gyrocompass to a class is also presented. 2

1. Introduction

Many students of physics have a naive view of how a gyrocompass works. Starting from the notion of angular momentum conservation, they infer correctly that the axis of a fast-spinning rotor, not subject to external , will eternally point in a single direction. From this they conclude that this is how a gyrocompass must work whereas, in reality, it is the of the with respect to an inertial frame which is responsible for the pointing of the gyrocompass.

The gyrocompass is in fact a rotor, whose axis of is constrained to turn in a horizontal plane (see Fig. 1). The constraining mechanism

(gimbals) exerts a on the rotor, and this torque is directly or indirectly responsible for keeping it aligned. The gyrocompass can be analysed from the viewpoint of an inertial observer, for whom the gyrocompass is carried around by a rotating earth. Alternatively, it can be analysed from the more natural vantage point of an observer stationed at some point on the earth's surface. Such a non-inertial observer, must include the Coriolis force as well as the centrifugal force when applying

Newton's laws of motion. It is in this frame of reference, i.e. in that of an earth-bound observer, that the magnetic analogy holds.

A rotating mass is characterised by its angular momentum, L, which is the analogue of the magnetic moment, \i, which characterises a rotating charge distribution. In a similar way the earth's sidcrial angular velocity, ft, is the analogue of the magnetic field B. The torque T, on the rotating mass and charge distributions will be shown to be T = LxQ, and

T = jixB respectively.

To make the Coriolis force concept more concrete it is useful in lectures to identify it with a "quasi magnetic field", ft, which lies in the N-S meridian plane making an angle X upward (dc vnward) with respect to the

floor, towards (away from) the north (south) celestial pole in the northern 3

(southern) hemisphere. The angle X is the latitude, (sec Fig. 2). It is the vertical component of this field which causes the of the plane of the Foucault , whereas it is the horizontal component which is responsible for gyrocompass alignment.

In the following section we present an elementary account of the gyrocompass, and its analogy with the magnetic compass. In a final section we present a simple classroom demonstration of the gyrocompass. A more exact treatment of the gyrocompass is presented in an appendix.

2. Elementary Theory of the Gyrocompass and its Magnetic

Analogy.

Consider a circular ring of matter of radius a and mass M, rotating about its symmetry axis, e, with angular velocity to . Let A be the mass per unit length of the ring. Applying the general expression for the Coriolis force- on a mass m of velocity v F,, = 2mvxfl , (1) to this element we find

dFc = 2(Adr aco0)/.v; . (2)

In Eq. (2), r is a vector from the centre to a point on the ring. The term dr is a length clement of the ring taken in the sense of the rotation.

Using the fact that r«dr = 0, we find that the torque exerted by the Coriolis force on this element is given by

dTc , rxdFc

= Aaco0((rxdr)xQ + d(rr«n» . (3)

On integrating around the ring to obtain the total torque the second term, being a perfect differential, contributes nothing. We find 4

Tc = Lxfl , (4) where the orbital angular momentum is given by, L E E rxmv (sum over mass particles) = y rx(Adrao) )

A = I0o>0e . (5)

In obtaining Eq.(5) we have ussd the expression for the vector area, S | ^rxdr s S = na2e (6) and the moment of

2 IQ = Ma - (7)

In fact, with an appropriate I , Eq.(5) holds for an arbitrary symmetric rotor.

The force of a magnetic field, B, on a charge, q, is given by

Fq=q(v/c)xB. (8)

To calculate the torque T on a circulating cnarge, we replace qv in Eq.(8) by the current element idr. We find

Tq = | rx{(i/c)drxB} . (9)

Making the same step as in Eq.(3), we find

Tq - nxB (10)

where the magnetic moment is given by u. = (i/c)| ^"xdr

= iS/c as usual. (11)

Again we have made use of Eq.(6) for the vector area.

Following Ohanian2 or Rcsnick and Halliday^ Eqs.(4) and (10) may be

obtained by more elementary force diagram methods. By comparing

Eqs.(4) and (5) with Eqs.(10) and (11) the analogy of the gyro and magnetic

is established.

To complete the analysis, we consider motion of the gyrocompass about

the vertical axis, k, i.e. an axis oithogonal to its spin axis, about which the mounting system exerts no torque. Let L be the of the 5 entire gyrocompass assembly about this vertical axis. The equation of motion becomes,

2 2 Itd 9/dt = £«T

= -0owoflcosA.)sin9 - (12)

The angle of deviation from north 9, and the latitude X, are defined in figs. (1) and (2), we see that for small deviations the gyrocompass hunts around true north with period t given by 1/2 t = 2Tt{Il/Io(0oQcosX} . (13)

For (o = 104 sec'1, at the earth's equator, this period is typically about

5 sec.

3. Demonstration

To demonstrate the gyrocompass, sit on a rotatablc stool holding a as shown in Fig.(3). (With many , a degree of freedom associated with one of the gimbal rings has to be "frozen out" to constrain the rotor axis to be as shown).

With the stool not rotating, set the rotor spin axis to be horizontal

(east-west). With a good grip on the gyroscope and a gentle rotation of the stool, the rotor axis will flip into the north-south direction. Fig.(3) has the earth's globe drawn in around the physicist to indicate the gyroscope's disposition on earth. On reversing the sense of rotation of the stool, the north and south poles are interchanged. The gyroscope will then flip into the reverse direction.

Acknowledgements

The author is indebted to Graham Sargood and Tony Klein for their valuable discussions and advice. 6

Appendix

Rotating Mass Distributions

In dealing with motion referred to a frame of reference rotating with a constant angular velocity, CI, with respect to an inertial frame, Newton's Law of motion becomes1;

md2R/dt2 = F + 2m(dR/dt)xQ - mQx(QxR) (Al)

where the rate of change, d/dt, is that perceived within the rotating frame.

The position vector. R, has its origin in a point on the axis of rotation. (For our later applications, this point will be the centre of the earth). The second term on the right of Eq. (Al) is the Coriolis force, and the third, the centrifugal force. In deriving this equation we employed the well known relation for rotating frames1

dV/dt ! = dV/dt + flxV (A2) inertial where V is an arbitrary vector.

We now apply Eq. (Al) to a rotating rigid body which has its centre-of- mass at R , a fixed point on the earth. The coordinate vector, r , is the position of a mass point with respect to R . Thus

R = RQ + r. (A3)

The equation of motion now reads

2 2 md r/dt = F + 2m(dr/dt)xQ - mQx(flx(R0 + r)) . (A4)

By way of comparison, the equation of motion of a particle of charge, q, in a i constant uniform magnetic field B is given by

md2r/dt2 = F + q(dr/cdt)xB . (A5)

F is any other non-magnetic force. In the following section we apply

Eq. (A5) to charge distributions.

Let L be the angular momentum of the rigid body about its centre-of-

mass

L s Z rxindr/dt, (A6)

the summation being over all particles in the body. With Eq. (A4) we find

dL/dt = E rxF + E rx(2mdr/dtxO) - £ rx(mflx(flx(R0 + r))) .(A7)

The torque due tc the non-inertial forces, F, is defined by

T =1 rxF. (A8)

Owing to the cancellation of the internal forces, T contains the external

torques only. The second term un the right of Eq. (A7) is developed as follows:

£ rx(2mdr/dtxfi) = E m{(rxdr/dt)xQ + d(rr»£2 - r?Q)/dt}

= LxQ - d(I»0)/dt (A9)

where, as usual, the inertia dyadic is given by

I ^ E m(r2E - rr) . (AlOa)

The components of this dyadic arc the inertia tensor Ijj - Z m(r2 6jj - r-jrj) . (A 10b)

The unit dyadic, E, has tensor components 8-. The steps required to obtain 8

Eq.(A9) become apparent on resolving ^(dr-Mt) into its symmetric and antisymmetric components with respect to ij.

Using the centre-of-mass identity

E mr=0 (All) the R term in the third right hand term of Eq.(A7) drops out, leaving the expression, £2xI«Q. In summary Eq.(A7) becomes:

dL/dt = T +Lxft - d(I«Q)/dt - OxI»Q (A12)

We now develop the inertia dyadic as follows. Let ej, e2» e^ be three orthogonal (time dependent) unit vectors which are imbedded in the rotating rigid body and move with it. In terms of these, the inertia dyadic reads:

I = El-e-ej in general (A13a)

= Z I- eje- referred to principle axes (A 13b)

= (I - I ) ee + I,E for an axially symmetric body (A13c)

In Eq. (A 13c), e is the axis of symmetry (or spin axis), and I is the moment- of-inertia about that axis. I, is the moment of inertia about any transverse axis through the centre-of-mass. Note that I-., I-, I , L are all time independent moments of inertia.

We now analyse the four terms on the right of Eq.(A12), for the specific case of an axially symmetric rotor (gyroscope) rotating at high speed around its symmetry axis. The angular velocity is designated by (0 where

0) = © e + co. (A14a)

with e«wt = 0 . (A14b)

All wc need in order to evaluate dl/dt is the single expression for the rate of exchange of e;

de/dt = coxe = cotxe . (A15) 9

In the case of gyroscope,

(i) T is the torque supplied by the rotor support system. Its actual nature

depends on the degrees of freedom of the mount.

(ii) Lxft the torque which aligns the angular momentum with the frame

angular velocity

(iii) The term d(NQ)/dt is a torque due to the rotation of the gyroscope

about an axis perpendicular to the spin axis. Eq.(AlS) shows this term

is of order I0cotft. As the spin angular velocity is much greater than

the angular velocities perpendicular to it, o>t « coQ , this third term is

very small compared with the second. We shall see below, that for the

normal gyrocompass mounting it plays no role at all for geometric

reasons. For later use we note that

V d(I«ft)/dt = (IQ - It )(ootxdee -Q + eojjxe.fl). (A16)

(iv) The last term, ftxl ft, has nothing to do with the rotation of the

gyroscope at all. It is simply due to the fact that the centrifugal

force field being slightly non-uniform, docs not act through the

centrc-of-mass and thereby generates a torque. This term is of order

2 (I - It)ft , and is negligibly small.

In summary, for a system in which w » co, , ft

dL/dt - T + Lxft . (A17)

Except for the last (centrifugal) term, resolving Eq.(A12) along the zenith direction, k, yields Eq.(12), as does the approximate Eq.(AI7).

Although the centrifugal term in Eq.(A12) is very small, it too has a potential minimum when the gyrocompass points north.

If the matter distribution were continuous, and the matter current density was designated by J_, we would write instead of Eq.(A6),

3 L = J rx.Tmd r . (A18) 10

Rotating Charge Distributions

Taking moments of Eq. (A5) and developing it in the same manner as the previous section we find,

dL/dt = T + Erx(qdr/cdtxB) . (A 19) Introducing the electric current density J

£rx(qdr/cdtxB) = J d3r rx(J xB)/c. (A20)

Following Jackson4, using v"»J = 0

2 {rx(JqxB)}. = {j(rxJq)xB}. + -X B^.^r^) - j B; V-(Jqr ) (A21)

With a uniform magnetic field, and the application of Gauss' theorem, the last two terms of Eq.(A21) vanish to yield

dL/dt = T + nxB where the magnetic Jipolc moment is given by

V=\\ rxjqd3r . (A22) 11

References

^.Goldstein, Classical Mechanics (Addison-Wesley, 2nd ed. 1980), sections 4.9,

4.10.

2H.C.Ohanian, Physics (2nd expanded ed. 1989 - W.W.Norton), section

31.6,p.769.

3D.Halliday and R.Resnick, Fundamentals of Physics (3rd extended ed. - Wiley

1988), sections 30.8, 30.9, p701.

4J..D.Jackson, Classical Electrodynamics (Wiley, 2nd ed. 1980), section 5.6. 12

Figure Captions.

Fig. 1. A gyrocompass is depicted with its axis of spin, e, constrained to lie in

a horizontal plane. The gyrocompass is shown displaced by an angle

6 from the north-south meridian circle, its direction of equilibrium.

Fig. 2. The earth's angular velocity vector is shown as it appears to an

observer in a lecture theatre or laboratory at latitude X in the

northern or southern hemisphere.

Fig.3. The disposition of a physicist sitting on a rotatable stool holding a

gyroscope in a manner which permits a demonstration of the

gyrocompass. The degrees of freedom of the gyroscope arc carefully

depicted on the diagram and its detailed enlargement. The globe of

the earth is indicated, so that the orientation of an actual

gyrocompass is made clear. A magnetic cumpass needle of the same

general orientation is also indicated in the diagram. Fig 1 14

V77777T7777777777 N '///////////////7//// Northern Hemisphere Southern Hemisphere

Fig 2 15

Fig. 3