110   THE GYROSCOPIC PENDULUM

axis OA, is dO/dt. The component angular velocity of the earth about a horizontal axis DV, Fig. 88, in the meridian plane, is we cos X. This is the component angular velocity of the earth about ON, Fig. 87. The component of this about OH' or the parallel line N H is we cos X sin 0. Hence, the velocity of precession about a horizontal line 0 H' fixed in space is

dt + we cos X sin 0

The torque about the axis perpendicular to OH' and OA, that is about the axis DG, is zero. Consequently,

0 = hs (de + we cos A sin o)

Differentiating (87) we have

z

mgl dt0 h, dt2

Here are two equations, one obtained by considering precession about a vertical axis and the other by considering precession about a horizontal axis. Each equation involves both 0 and 0. Now we shall eliminate 0 and from the resulting equation study the motion of the spin-axis in the horizontal plane.

On combining the last two equations by eliminating dt , we find h, d20 _

mgl dt2   -We cos X sin ¢ which can be put into the form

dt2~ _ - ( hslWecos A) sin 0

In order to abbreviate the labor of repeatedly copying the quantity within the parenthesis, we will represent it by the symbol A. When 0 is small, sin 0 = 0 radians and the above equation assumes the form

d20 _ -A~

dt2

Multiplying both sides by 2 do and integrating, we obtain do) 2 _ A02+C

dt - -

GENERAL PROPERTIES   111 For convenience in subsequent integration, let C = AB 2. Then

do2

dt~ = A(B2 - 4,2)

Extracting the square root,

dot = VA VB2 - 02 or ~Bdo = dt VA

2_02

Measuring time from the instant when 4, = 0, that is, when the gyro-axle is in the meridian, and integrating the last equation, we obtain

sin'-=t'\//-A +C1

When t = 0, 0 = 0 and hence the first member is zero and C, = 0. Whence

0 = B sin t VA

which is of the same form as the equation of a simple harmonic motion of rotation, (48),

0 = T) sin (t T )

Therefore, the undamped motion of the gyro-axle of a pendulous gyroscope back and forth across the meridian is approximately

simple harmonic and of the period T = 2A . Replacing the value of A in this equation

T =   2 ,r   h

/mglwe cos A 2 " mglwe cos A   (88)

~l   h,

(Among other things, this equation shows that the period of the gyro-axle of a pendulous gyroscope back and forth across the meridian is increased by diminishing the distance between the point of support and the center of gravity. This fact is utilized in the design of various gyro-apparatus.

69. The Torque with Which the Second Frame of a Gyroscope Resists Angular Deflection. - Consider a gyro-wheel G of spinvelocity w, and moment of inertia KS with respect to the spin-axis, mounted in two coplanar frames A and B, Fig. 89. Let the axis pp' pass through the center of gravity of the wheel and inner frame.