To complete this reflection, determine the relationship between $\mathit{\omega }$and $\mathit{\Omega }$for the case of pure precession, but with the spin axis at an arbitrary angle $\mathit{\theta }$to the vertical (figure) $\mathit{\theta }\mathbf{=}\mathbf{90}\mathbf{°}$is the case of horizontal precession we treated). If you have the opportunity, see whether this relationship holds for a real gyroscope.

Angular speed is the rate at which the central angle of a spinning body varies over time.

The rate of change of angular displacement is known as angular speed.

The force exerted by the support on the gyroscope is in upward direction, and Earth pull down with a force $Mg$through the center of mass as shown in the below figure. Since, the center of mass stays at the same height all the time in the case of precession, the vertical component of the net force must be zero. Such that

${\overrightarrow{F}}_{net}=F_N-Mg=0$

Thus, the vertical force on the gyroscope by the support is

$F_N=Mg$

The given situation is as shown in the following figure and also it shows the direction of torque is into the page.

The magnitude of translational angular momentum is constant, and it always points vertically upward. The direction of rotating angular momentum, on the other hand, is constantly shifting as the gyroscope processes. The rate at which the rotational angular momentum vector changes must be calculated.

The rate of change of rotational angular momentum of the gyroscope is

$\left|\frac{d{L}_{rot}}{dt}\right|=\Omega L_{rot}$

Here, the precession angular speed is $\Omega$and the rotational angular momentum of the gyroscope is $L_{rot}$.

The net torque acting on the gyroscope about the center of mass is

$$\begin{gathered} \overrightarrow{\tau }=\left(R\sin \theta \right)F_N \\ =R{F}_N\sin \theta \\ =R\left(Mg\right)\sin \theta \left(1\right) \end{gathered}$$

The rate of change in angular momentum of a multiple particle system is equal to the net torque applied on the system, according to the angular momentum principle. as a result.

$\left|\frac{d{L}_{rot}}{dt}\right|=\Omega {\overrightarrow{L}}_{rot}={\tau }_{CM}.....\left(2\right)$

When the spin axis made by the angle $\theta$with the vertical, the rotational angular momentum of the gyroscope can be expressed as

${\overrightarrow{L}}_{rot}=\sin \theta$ (since, $L_{rot}=I\omega$)

Substituting the above equation in the equation (2),

${\tau }_{CM}=\Omega I\omega \sin \theta .$

Thus, the precession angular speed of the gyroscope can be expressed as

$\Omega =\frac{{\tau }_{CM}}{I\omega \sin \theta }.$

Substituting the equation (1) in the above equation

$$\begin{gathered} \Omega =\frac{{\tau }_{CM}}{I\omega \sin \theta }. \\ =\frac{RMg\sin \theta }{I\omega \sin \theta } \\ =\frac{RMg}{I\omega } \end{gathered}$$

Hence, the precession angular speed of the gyroscope is $\frac{RMg}{I\omega }$.