Question 11.15: Calculate the precession rate ωp for a toy top like that in ...
Calculate the precession rate \omega_p for a toy top like that in Fig. 11.19 if m = 0.5 kg, A(=I_x=I_y) = 12(10^{-4}) kg·m², C (= I_z) = 4.5(10^{-4}) kg·m², and d = 0.05 m.
For an inclination of, say, 60°, (A C) cos θ > 0, so that Eq. (11.98) requires \left.\omega_s\right)_{\min } = 407.01 rpm. Let us choose the spin rate to be \omega_s = 1000 rpm = 104.7 rad/s. Then, from Eq. (11.97), the steady precession rate as a function of the inclination θ is given by either one of the following formulas:
\omega_p=\frac{C}{2(A-C) \cos \theta}\left(\omega_s \pm \sqrt{\omega_s^2-\frac{4 m g d(A-C) \cos \theta}{C^2}}\right) (11.97)
\left.\omega_s\right)_{\min }=\frac{2}{C} \sqrt{m g d(A-C) \cos \theta} \text { if }(A-C) \cos \theta>0 (11.98)
\omega_p=31.42 \frac{1+\sqrt{1-0.3312 \cos \theta}}{\cos \theta} \text { and } \omega_p=31.42 \frac{1-\sqrt{1-0.3312 \cos \theta}}{\cos \theta} (a)
These are plotted in Fig. 11.20. For θ = 60°, the high-energy precession rate is 1148.1 rpm, which exceeds the spin rate, whereas the low-energy precession rate is a leisurely 51.93 rpm.
