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An analysis of gyroscopic couples

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Introduction

Whilst Gyroscopes are used extensively in aircraft instrumentation and have been utilised in monorail trains, the everyday impact of gyroscopic forces on our lives is unappreciated and significant. The simple example is a child's top which would not work but for the gyroscopic couple which keeps it upright. On a slightly different level, the gyroscopic couple helps us to keep a bicycle upright. It is interesting and instructive to remove a bicycle wheel from its frame, hold it by the axle, spin the wheel and then try to change the orientation of the axle. The force required to do so is considerable! However these gyroscopic forces are not always beneficial and it will be shown that the effect on the wheels of a car rounding a corner are to increase the tendency for the vehicle to turn over.

Gyroscopic Couple

Without an understanding of Angular movement it is difficult to understand Gyroscopic Couples. For this reason the Paragraph on Angular Displacement; Velocity and Acceleration shown in The Theory of Machines - Mechanisms, has been reproduced here.

Angular Displacement, velocity and Acceleration

Let:-

The line OP in the diagram rotates around O

Its inclination relative to OX be $\phi$ radians.

Then if after a short period of time the line has moved to lie along OQ, then the angle $\delta\theta$ is The Angular Displacement of the line.

Angular Displacement is a vector quantity since it has both magnitude and direction.

Angular Displacement

In order to completely specify and angular displacement by a vector, the vector must fix:-

The direction of the axis of rotation in space.

The sense of the angular displacement. i.e. whether clockwise or anti-clockwise.

The magnitude of the angular displacement.

In order to fix the vector can be drawn at right angles to the plane in which the angular displacement takes place, say along the axis of rotation and its length will be , to a convenient scale, the magnitude of the displacement.

The conventional way of representing the sense of the vector , is to use the right-hand screw rule. i.e,

The arrow head points along the vector in the same direction as a right handed screw would move, relative to a fixed nut.

Using the above convention, the angular displacement $\delta\theta$ shown in the diagram would be represented by a vector perpendicular to the plane of the screen and the arrow head would point away from the screen.

Angular Velocity

This is defined as the rate of change of angular displacement with respect to time.

As angular velocity has both magnitude and direction it is avector quantity and may be represented in the same way as angular displacement.

If the direction of the angular displacement vector is constant. i.e.The plane of the angular displacement does not change its direction,. Then the angular velocity is merely the change in magnitude of the angular displacement with respect to time.

Angular Acceleration

Defined as the rate of change of angular velocity with respect to time.

A Vector quantity.

The direction of the acceleration vector is not necessarily the same as the displacement and velocity vectors.

Assume that a given instant a disc is spinning with an angular velocity of $\omega$ in a plane at right angles to the screen and that after a short interval of $\delta t$ its speed has increased to $\omega + \delta\omega$ .

Then applying the right-hand rule:-

The angular velocities at the two instants are represented by the vectors oa and ob

The change of angular velocity in a time of $\delta t$ is represented by the vector ab. This can be resolved into two components ac and cb which are respectively parallel and perpendicular to oa

Hence.

The component parallel to oa is given by:- $\displaystyle \alpha_T=\frac{d\omega}{dt}$

The component perpendicular to oa is given by $\displaystyle \alpha_C=\omega\;\frac{d\theta}{dt}=\omega\times \omega_P$

Note

$\omegaP$ is the rate of change of direction of the vector oa

$\alpha_T is the rate of change of the magnitude of the velocity \omega$ of the disc.

$\alpha_C$ is the rate at which the direction of $\omega$ and therefore the plane of the rotation of the disc is changing.

The total angular acceleration of the disc is the vector sum of $\alpha_T$ and $\alpha_C$

Two particular cases should be noted:-

If the plane of rotation of the disc is constant in direction, then $\omega_P$ is zero and the component of acceleration $\alpha_C$ is zero.

If the angular acceleration of the disc is constant in magnitude but the plane of rotation changes direction at the rate $\omega_P$ radians per second, then the angular acceleration of the disc is given by:-

$\alpha_C=\omega\;\frac{d\theta}{dt}=\omega\times \omega_P$

(1)

The direction of this acceleration vector is at right angles to the angular velocity vector and lies in the plane of motion of the velocity vector.

Precessional Motion and Gyroscopic Acceleration.

The change in direction of the plane of rotation of the disc is known as Precessional motion and $\omega_P$ is known as the angular velocity of precession.

The angular acceleration $\alpha_C$ is called the Gyroscopic acceleration.

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Gyroscopic Couple

If a uniform disc of polar moment of inertia I is rotated about its axis with an angular velocity $\omega$ , its Angular Momentum $I\;\omega$ is a vector and can be represented in diagram (c), by the line op which is drawn in the direction of the axis of rotation. The sense of the rotation is clockwise when looking in the direction of the arrow.

If now the axis of rotation is precessing with a uniform angular velocity $\Omega$ about an axis perpendicular to that of $\omega$ , then after a time $\delta t$ , the axis of rotation will have turned through an angle $\delta \theta=\Omega t$ and the momentum vector will be oq. The Gyroscopic Couple $\tau$ is given by:-

$\tau=\text{The rate of change of angular momentum}=\frac{pq}{\delta t}$

(2)

$= \frac{I\,\omega\,\delta\theta}{\delta t}=I\,\omega\,\Omega\;\;\;\text{In the limit}$

(3)

The direction of the couple acting on the gyroscope is that of a clockwise rotation when looking in the direction pq.

In the limit the direction of the couple is perpendicular to the axe of both $\omega$ and $\Omega$

The reaction couple exerted by the gyroscope on its frame is in the reverse sense( It is advisable to draw the vector triangle opq in each case.

Worked Examples

The solutions to the following questions have been hidden. They can be seen by clicking the red buttons.

Example 1

The diagram shows the Gyro unit of an aircraft instrument in which the rotor is carried in a closed casing mounted in bearings so that its axis is normally vertical but free to take up any direction.

The rotor speed is maintained constant in an anti-clockwise direction, when seen from the top, by air jets inside the casing which impinge onto its slotted periphery. The used air may then leave the casing by the four orifices indicated, these are uncovered as required by gravity controlled vanes. If the rotor axis is tilted through an angle $\theta$ in the XOZ plane, explain which orifice must be uncovered so that the reaction from the jet tends to restore the axis to the vertical.

If this force is 0.0004 lb. at a distance of 0.825 in. from O and the wheel is equivalent to a uniform solid disc 1.75 in. in diameter weighing 0.3 lb. and running at 9000 r.p.m., find the time required for the axle to return to the vertical is $\theta$ is $1^0$

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Example 2

A pair of locomotive driving wheels with the axle have a moment of inertia of $4 tons-ft^2$ . The diameter of the wheels is 6 ft.3 in. and the distance between the wheel centres is 5 ft. When the locomotive is traveling at 70 m.p.h. on a level track, defective ballasting causes one wheel to fall $frac{3}{8}\;in.$ and to rise again in a total time of 0.1 sec. If the displacement of the wheel takes place with simple harmonic motion, find the gyroscopic reaction on the locomotive.

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Example 3

A pair of flanged wheels 4 ft. in diameter, mounted on an axle, roll along rails spaced 5 ft. apart and at the same level and around a curve of 500 ft. mean radius, at a speed of 40 m.p.h. Each wheel is assumed to be a thin disc weighing 600 lb. Determine the vertical force between each wheel and the rail allowing for both centrifugal and gyroscopic effects and also the horizontal force assuming this to be at the outer rail.

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Example 4

The turbine rotor of a ship weighs 6 tons and has a radius of gyration of 19.5 in. It rotates at 1800 r.p.m. clockwise when looking forward from the stern. Determine the gyroscopic effects set up if:-

(a) The ship , when traveling at 15 knots, steers to the left in a curve of 200 ft. radius.

(b) The ship is pitching and the bow is descending with its maximum velocity. The pitching is simple harmonic, the periodic time is 20 sec. and the total angular movement between extreme positions is $10^0$

(c) The ship is rolling and at a certain instant has an angular velocity of 0.02 radians per second clockwise when looking from the stern.

In each case explain clearly how you determine the direction in which the ship tends to move as a result of the gyroscopic action. (U.L.)

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Note One ton = 2240 lb. : One Knot = one nautical mile per hour = 6080 ft. per hour.

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Example 5

(a) A motor car takes a right hand bend of 100 ft. radius at a speed of 40 m.p.h. . Determine the magnitude of the centrifugal and gyroscopic couples acting on the vehicle and state the effect that each of these has on the road reactions of the road wheels. Assume that:-

Each road wheel has a moment of inertia of $50\;lb.ft.^2$ and an effective radius of 15 in.

The rotating parts of the engine and transmission are equivalent to a flywheel weighing 140 lb. with a radius of gyration of 4 in.

The engine turns in a clockwise direction when viewed from the front.

The back axle ratio is 4 to 1, the drive through the gear box is direct. Also the gyroscopic effects of the half shafts at the back axle are to be ignored.

The car weighs 2500 lb. and its centre of gravity is 24 in. above the road.

(b) If the turn had been in a left-hand direction, and all other details had remained the same, which answers, if any, would need to be modified. (U.L.)

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Example 6

A uniform beam of weight 4 lb and length 2 ft. rotates in a vertical plane about its mid point with an angular velocity of 65 rad./sec. Its axis of rotation is precessed in the horizontal plane with an angular velocity of 10 rad./sec.

(a) Calculate the value of the gyroscopic torque when the beam is at an angle $\theta$ to the horizontal.

(b) Sketch the graph connecting gyroscopic torque with $\theta$ for angles from 0 to $360^0$ (U.L.)