## Discussion of 347(7)

Many thanks! The method is to equate the lab frame and moving frame forces using the expression for acceleration in spherical polar coordinates, used to represent the moving frame. For clarity one should replace r on the left hjand side by R, so g = -MG/ R squared. A point r bold inside the gyro is considered, this can be the centre of mass. So the gyro has mass m and its centre of mass moves according to

F bold = mg bold = m d2 r bold / dt squared = -mM e sub r / R squared

In the spherical polar coordinate system

r bold = r e sub r

The moving frame of the gyro is represented by spherical polar coordinates, thus giving Eq. (7). The method is exactly the same as planar orbital theory, but the plane polar coordinates are replaced by spherical polar coordinates. The force between m and the earth of mass M is given by Eq. (11). So instead of the sun of mass M attracting a planet of mass m, the earth of mass M attracts the gyro of mass m. Instead of falling directly into M, the gyro is counterbalanced by the centrifugal force (12). The motion of the centre of mass of the gyro is then given by Eq. (19). In planar orbital theory:

F bold = -mMG / r squared e sub r = m dt r bold / dt squared

= ( r double dot – r theta dot squared e sub r

In both cases F (Cartesian) = F (moving coordinate system).

To: EMyrone@aol.com
Sent: 13/01/2017 15:42:47 GMT Standard Time
Subj: Re: 347(7): Solution of the Gyro with Additional Torque