Harold A. Daw and M. Ramona Daw Pomeroy
Physics Department, New Mexico State University
Las Cruces, NM 88003
Phone 505 646 3831
email hadaw@nmsu.edu



Lecture-Demonstration or Advanced Laboratory

The Free Rotator Demonstrator

Support required for apparatus (i.e., special needs in order to set up the equipment): compressed air

Approximate size: The device itself is approximately 1 foot by 1 foot. An area of 3 feet by 3 feet would be adequate.

Does this device require electrical power? No, but if compressed air is not available, I will need to provide a pump which will require electricity.

Will you be present to set up your apparatus? Yes

Abstract

The free rotator device consists of a ball floated on an air-bearing surface ( 4.4 inch and 8.58 inch diameter balls are used). In operation the ball is set in rotation by hand and the path of the rotation vector is directly recorded on the spinning ball surface with a marker. The principal moments of inertia can be changed. The path may be displayed using a video camera and projector.

Introduction

The free rotator is almost always discussed in mechanics classes. Much of what is discussed is verbally descriptive but lacks a live, clear display of the behavior. This demonstration provides a means of clearly demonstrating the behavior. First we give a brief outline of the theory involved, then a description of the apparatus, and conclude with how the apparatus is used. References , figures, and an appendix of physical properties are included.

The theory of the free rotator behavior is well covered in a number of texts. [1,2,3] . The following is a brief description of the theory. (Due to the limitations of HTML, the reader will have to exercise some judgement in deciding what was intended in the way of greek letters, superscripts and subscripts)

The equations of motion of the rigid rotator are:

L = angular momentum, N = torque, omega = angular velocity

L = m r x v = m r x (omega x r) = m omega (r.r) -m r (omega.r) = II.omega .....................(1)

where II is the inertia tensor: (in equation (2), I is the identity matrix)

II = m (r^2 I - r r) ............................................................................................................(2)

N = dL/dt = d/dt (II.omega) ...............................................................................................(3)

If one changes to a system anchored in the body, then

N = d'/dt(II.omega) + omega x (II.omega) = II. d/dt omega + omega x (II.omega) ..................(4)

where the time derivative in the body frame is designated by d'/dt.

In component form Eq.(4) is:

I1 d/dt'(omega1) + (I3 - I2) omega3 omega2 = N1...............................................................................(5a)

I2 d/dt'(omega2)+ (I1 - I3) omega3 omega1 =  N2...............................................................................(5b)

I3 d/dt'(omega3)+ (I2 - I1) omega2 omega1 = N3, ..............................................................................(5c)

Euler's equations for the rigid body.

In these equations omega1, omega2, and omega3 are the components of the rotation vector omega in the body system. The I's are the principal moments of inertia.

If there is no torque, the free asymmetric rotator, then N is zero. Eq.(5) become:

I1 d/dt'(omega1) + (I3 - I2) omega3 omega2 = 0...............................................................................(6a)

I2 d/dt'(omega2)+ (I1 - I3) omega3 omega1 = 0...............................................................................(6b)

I3 d/dt'(omega3)+ (I2 - I1) omega2 omega1 = 0, ...............................................................................(6c)

For the free symmetric rotator, I1 = I2, and Eq.(6) become:

I1 d/dt'(omega1) + (I3 - I1) omega3 omega2 = 0.............................................................................(7a)

I1 d/dt'(omega2)+ (I1 - I3) omega3 omega1 = 0...............................................................................(7b)

I3 d/dt'(omega3) = 0, ....................................................................................................................(7c)

Since by Eq.(7c) I3 d/dt'(omega3) = 0, omega3 is constant. Energy E is also constant and therefore:

1/2 I1 omega1^2 + 1/2 I1 omega2^2 = E - 1/2 I3 omega3^2 ............................................................(8)

From (8) the magnitude of the component of omega in the 12 plane, omega12, is constant and given by

omega12 = sqrt(omega1^2+omega2^2)=(2/I1)( E -1/2 I3 omega3^2). .............................................(8a)

The instaneous omega axis traces out a circle on the surface of the ball. This circle has a radius

r = R sin (alpha) .............................................................................................................................(9)

where R is the radius of the ball, and alpha is the angle between the omega vector and the 3-axis of the ball.

The sine of the angle alpha is given by

sin(alpha) = sqrt(omega1^2 + omega2^2)/sqrt( omega1^2 + omega2^2+omega3^2). .............................(10)

The magnitude of the spin vector is set by the hand spin, and the direction of the spin vector relative to the symmetry axis is set by the initial spin direction.

The precession rate of the angular velocity vector about the symmetry axis is given by

omega_pr= omega3 (I3 - I1)/I1. ..........................................................................................................(11)

In this demonstration the rate of precession and the rate of rotation can be determined. The path of the angular velocity is plotted directly on the surface of the rotating sphere.

The path for the asymmetric rotator is much more complicated. The paths are discussed in the cited references but are given in terms of the polhod,the path of the angular momentum vector on the inertia ellipsoid, and the hyperhod, the path of the angular momentum vector on the invariable plain etc. This demonstration plots the path directly on the surface of a sphere, the ball, and not the inertia ellipsoid. Never-the-less all of the aspects of asymmetric rotators can be illustrated: stable precession about the larger and smaller moments of inertia, and inversion if set in rotation about the intermediate axis. Other settings results a variety of interesting paths.

While many physics texts cover the free rotator, there is a paucity of ways to make a truly free rotator. The reason for this is that most supports for the spinning rotator makes it not truly free^6. A gyroscope made of a spherical ball floated in a hemispherical bowl using compressed air as the support mechanism was described some years ago in the American Journal of Physics^4. The degree of interaction between the support and the rotating sphere is minimal. A number of uses have been made of the air supported bearing for gyroscopes. [8,9,10,11]

In the course of the preparation of the written description of this device we ran across a most interesting and very relevant article[7]. In this article Soodak and Tiersten discuss the path of the motion of a free rotator and then use the result to demonstrate the motion on a 2 3/8 inch ball with an air-bearing support. This demonstration is closely allied to their work. It differs in that larger balls are used, and more importantly, the path of the instantaneous rotation axis is directly recorded on the surface of the rotating body and is available for display directly or on a video screen.

The Apparatus

This apparatus consists of three balls. One is a 11.19 cm diameter Bocce Ball with an aluminum rod 1.43 cm in diameter inserted along a diameter of the ball. This ball is used as the symmetric rotator (see Fig. 1).


A second Bocce Ball has four holes drilled into it, 1/2 inch in diameter and 1 inch deep. Metal plugs are slid into these holes to allow one to set the relative magnitudes of the three principal moments of inertia. In this case two aluminum plugs are inserted 1800 apart, and two brass plugs are inserted in the remaining two holes. The plugs are held by friction. The ends of the plugs are machined to match the curvature of the ball surface. This ball is used as the non symmetrical free rotator (see Fig. 2).

A third ball is the size of a bowling ball. While the smaller balls are quite adequate for small classes, the larger ball is much better for larger classes. Of course a video camera and screen can be used in connection with both.


The third ball is made like the second Bocce Ball. However in this case the holes in the ball are drilled with a tapped hole in the bottom. The plug is inserted and then screwed into the tapped hole. This allows one to prepare plugs of a variety of configurations such that the ball may be used as a balanced ball i.e. the density of the plugs match the density of the ball, as a symmetric ball i.e. two of the plugs do not match the density of the ball, and as an asymmetric ball i.e. there are two sets of plugs which do not match and have a different density than the ball. One can prepare plugs containing liquid as suggested by Soodak and Tiersten[7]. This allows great flexibility. One of course can have a net off balance and therefore a torque on the ball. See Figs. 3,4 for typical construction of the insert plugs. This ball is a magnified version of the smaller ball with plugs 1 inch in diameter and 2 inches long (see Fig.5) (below, right). Fig.(6) [not included here] shows the ball holes being drilled.

How Used.

This apparatus can be used in two formats. It can be used as a demonstration where the larger ball is to be preferred, or a TV camera can be used to magnify the image. Or it can be used as a laboratory experiment.

1. For The Demonstration : The Symmetric Rotator

Spin the ball shown in Fig.(1). Be sure to spin the ball so that the axis of rotation faces the class. Mark the ball as shown in Fig. 7 (below). One should mark the ball about every 1 or 2 seconds, but adjust this time with the spin rate such that one has an adequate number of points to clearly show the path. If one wishes to quantify the results, one can measure the spin rate, the precession rate, and the radius of the circular path of the angular momentum vector. One can use formulas to see that they are consistent with the results.

Fig. 8 (above) shows a marked ball.

2. For The Demonstration: The Asymmetric Rotator

This procedure is the same as for the symmetric rotator. If one rotates the ball about an axis close to the axis of the larger moment of inertia, or an axis close to the axis of the smaller moment of inertia near circular paths similar to the symmetrc rotator are obtained. If one rotates about an axis close to the axis of the intermediate moment of inertia, the rotator will flip completely over. The paths are not circles, and the simple formulas above do not apply. Axis's of rotation that are some distance from the symmetry axis with result in a variety of shapes.

See Fig. 9 (above) for a typical path.

3. For The laboratory Experiment Spin the ball and make a plot of the decay curve of the spin rate. Then spin the ball and have three students do the following:

Now calculate the precession rate, the spin rate, and the radius of the circle of precession. Note one needs to estimate the time for a full revolution if only a part of a revolution has been marked. The angle between the body axis and the precession path is measured. The average spin rate should be determined from the spin rates measured and the decay curve. Compare the results with those given by Eq. 11.

Conclusions

This apparatus can be used to make a clear presentation of the motion of a symmetric free rotator, and or an asymmetric free rotator. It is relatively simple to make and to use. Using a variety of interchangeable inserts or plugs, one can demonstrate a wide variety of rotational phenomenon, including that of gyroscopic motion.

With choices made in this demonstration the value of (I1 - I3)/I1 is about 1.145 x 10^(-2) for the symmetric rotator. This means that for spin rates for the ball in the vicinity of 2 rev/sec, the precession rate is on the order of .2 revolutions per minute.

References

[1.] H. C. Corben and Philip Stehle, Classical Mechanics (John Wiley and Sons Inc. New York, 1960 2 nd edition), pp. 148-51.

[2.] Keith R. Symon , Mechanics Addison (Wesley Publishing Company, London, 1961), pp. 450-458.

[3.] Herbert Goldstein, Classical Mechanics (Addison Wesley Publishing Company, London, 1956), pp. 156-63.

[4.] Harold A. Daw "Two Air Supported Devices for Physics Laboratories and for Physics Demonstrations," Am. J. Phys. 33, 322 (1965).

[5.] Harry F. Meiners, Physics Demonstration Experiments (The Ronald Press Company, New York, 1970), pp. 266-267.

[6.] William B. Case and Michael A. Shay "On the interesting behavior of a gimbal-mounted gyroscope," Am. J. Phys. 60, 503-6 (1992)

[7.] Harry Soodak and Martin S. Tiersten, "Resolution Analysis of gyroscopic motion," Am. J. Phys. 62, 687-94 (1994)

[8.] Robert G. Marclay, "Air Suspension Gyroscope," Am. J. Phys. 28, 150 (1960)

[9.] Bernard H. Duane, "Air suspension gyroscope," 23, 147 (1955)

[10.] O. L. de Lange and J. Pierrus "Measurement of inertial and non inertial properties of an air suspension gyroscope," Am. J. Phys. 61, 974-81 (1993)

[11.] William G. Harter and Chong C. Kim "Singular motions of asymmetric rotators," Am. J. Phy. 44, 1080-3 (1976)

Appendix


SMALL BALL DATA - Bocce Ball

Ball diameter 4.407 inches or 11.19 cm.

Ball mass 1287 grams

Volume of the sphere 44.79 in^3 or 733.3 cm^3

Density of the sphere 1.753 g-/cm^3

Moment of inertia of the sphere 16115 g- cm^2


LARGE BALL DATA - Bowling Ball Size

Ball diameter 8.58 inches 21.79 cm

Ball mass 14.75 lbs. 6690 g.

Volume of the sphere 5421 cm^3

Density 1.234 g-/cm3

Moment of inertia of the sphere 317,936 g-cm^2


ROD AND PLUG DATA

Aluminum rod used in the symmetric bocce ball

Mass of the rod 49.0 g.

Rod length 11.19 cm.

Rod diameter 1.429 cm.

Density of the rod 2.733 g/cm^3

Volume of rod 17.93 cm^3.


Brass and Aluminum plugs for Bocce ball

size 1 inch - 2.54 cm long by ½ inch - 1.27 cm in diameter

weight al 8.5 g brass 26.8 g


Brass and Aluminum plugs for large ball

size 2 inches long by 1 inch in diameter

weight brass 220.1 g al 70.2 g


MOMENT OF INERTIA

1. Symmetric Bocce Ball

Iy = 16,301 g cm^2

Ix = 16,119 g cm^2


2. Bocce Ball

Ibrass = 16,235 g-cms^2

Ialuminum = 16,937 g-cm^2

Iperpendicular = 17,047 g-cm^2


3. Large Ball

Ibrass = 328,167 g-cm^2

Ialuminum = 349,057 g-cm^2

Iperpendicular = 358,836 g-cm^2