# Minimum speed needed for a gyroscope to precess

I was reading about gyroscopes and their precession. Based on the text, the angular speed of precession is: $\omega_{precession}=\dfrac{\tau}{L}$ But intuitively, if the wheel of the gyroscope is rotating with a very low angular speed, then the wheel won't precess, it will just fall. What is the minimum angular speed needed for the gyroscope to precess?

The purpose of this explanation is mainly to define the precession angular speed and spin speed as a spinning disc falls below $\theta = 45^\circ$ to the horizontal, having started at $\theta \approx 0^\circ$.

We assume a disk shape as in Feynman's Fig 20.5. The equation for precession angular frequency $$\omega_p = \frac{Mgr}{L} = \frac{Mgr_p\cos\theta}{I_s\omega_s} = \frac{Mgr_p\cos\theta}{\frac{1}{2}Mr_s^2\omega_s} \quad \tag 1$$

Where M is the mass of the disc, $\theta$ is the angle to the horizontal, g is the gravitational acceleration, $r_p, \omega_p, r_s, \omega_s$ are the precession radius, precession angular velocity, spin radius, spin angular velocity, respectively.

If we further assume that the shaft holding the disc has negligible mass and angular momentum as compared to the disc itself, then: $$\omega_p = \frac{2gr_p\cos\theta}{r_s^2\omega_s} \quad \tag 2$$

As stated at hyperphysics (bottom of page), this equation is only valid if $\omega_s >> \omega_p$. So assuming that, at $\theta = 0^\circ$, we set out to show the change in speed as $\theta$ changes $0^0 \to 45^\circ$ relative to the horizontal.

$$At \theta = 0^\circ,\qquad \omega_p = \frac{2gr_p}{r_s^2\omega_s} \quad \tag 3$$

$$At \theta = 45^\circ,\qquad \omega_{p1} = \frac{2gr_p\cos45^\circ}{r_s^2\omega_{s1}} \quad \tag 4$$

where 1 in the subscript indicates value at $\theta = 45^\circ$, provided, as already stated above, that $\omega_s >> \omega_p$.

We quickly see that if we set $\omega_{s1} = \omega_s$, then $\omega_{p1} \approx 0.707 \times \omega_p$. That is, for the same spin speed, the resulting precession angular speed at $\theta = 45^\circ$ us about 0.7 times that at $\theta = 0^\circ$, since precession radius $r_p$, which is measured in the horizontal direction, is now about 0.7 times of its $\theta = 0^\circ$ value.

However, if we start with spin angular speed $\omega_s$ at $\theta \approx 0^\circ$, then it will be less by the time the spinning disc sags to $\theta = 45^\circ$, and precession angular speed $\omega_{p1}$ will be higher than $\omega_p$, its value at $\theta = 0^\circ$.

To define the spin and precession angular speed changes, we turn to the principle of conservation of energy. Loss in kinetic and potential energy as the disc moves to a lower trajectory , as $\theta$ changes $0^\circ \to 45^\circ$ is:

$$\Delta K_s + \Delta K_p = \Delta E_p \qquad \tag 5$$ where $\Delta K_s, \Delta K_p, \Delta E_p are$ the spin kinetic energy, the precession kinetic energy, and the precession potential energy respectively. Equation 5 expands to: $$\frac{1}{2}(I_{s1}\omega_{s1}^2-I_{s}\omega_{s}^2) + \frac{1}{2}M(v_{p1}^2-v_{p}^2) = Mgr_p(\sin45^\circ-\sin0^\circ) \quad \tag 6$$ Further expanding, simplifying and multiplying by $2/M$: $$\frac{1}{2}r_{s1}^2\omega_{s1}^2-\frac{1}{2}r_{s}^2\omega_{s}^2 + r_{p1}^2\omega_{p1}^2 - r_{p}^2\omega_{p}^2= 2gr_p\sin45^\circ \quad \tag 7$$ We need to solve for only $\omega_{s1}and\omega_{p1}$, the spin and precession angular velocities at $\theta = 45^\circ$. Substituting the equation 4 expression for $\omega_{p1}$ into equation 7: $$\frac{1}{2}r_{s1}^2\omega_{s1}^2-\frac{1}{2}r_{s}^2\omega_{s}^2 + r_{p1}^2(\frac{2gr_p\cos45^\circ}{r_s^2\omega_{s1}})^2 - r_{p}^2\omega_{p}^2= 2gr_p\sin45^0 \quad \tag 8$$ What remains is to solve for $\omega_{s1}$, the only unknown in equation 8, and then substituting that value into equation 4 to evaluate $\omega_{p1}$.

• Just included the "\" before sin and cos too, and I must say it looks more beautiful. Thanks again @AccidentalFourierTransform. – Dlamini Jan 31 '18 at 4:34