Question 5.6: Study the stability of the stationary precession (θ = θ0) of......

Study the stability of the stationary precession (\theta = \theta_{0} ) of a spherical pendulum.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

From (5.67) one identifies

E = T + V =\frac{ml^{2}}{2}\dot{\theta}^{2}+\frac{p^{2}_{\phi }}{2ml^{2} \sin^{2}\theta }+mgl \left(1- \cos \theta\right).  (5.67)

V_{eff}\left(\theta\right)=mgl \left(1- \cos \theta\right)+ \frac{p^{2}_{\phi }}{2ml^{2} \sin^{2}\theta }.   (5.73)

The stationary motion \theta = \theta_{0} requires

\left(\frac{dV_{eff}}{d\theta}\right)_{\theta = \theta_{0}}=mgl \sin\theta_{0}- \frac{p^{2}_{\phi }\cos \theta_{0}}{ml^{2}\sin^{3}\theta_{0}} =0 \Longrightarrow p^{2}_{\phi }=m^{2}gl^{3} \frac{\sin^{4}\theta_{0}}{\cos \theta_{0}}.   (5.74)

Making use of p_{\phi }= ml^{2}\sin^{2}\theta_{0}\dot{\phi } there results \cos \theta_{0}={g}/{\left(l \dot{\phi }^{2}\right)} as long as \left|\dot{\phi }\right| \gt \left({g}/{l}\right)^{{1}/{2}}, the same restriction found in Example 5.4. On the other hand,

k^{\left(0\right)}=\left(\frac{d^{2}V_{eff}}{d\theta^{2}}\right)_{\theta = \theta_{0}}=mgl \cos \theta_{0}+ \frac{p^{2}_{\phi }}{ml^{2}\sin^{2}\theta_{0}}+3 \frac{p^{2}_{\phi }\cos^{2} \theta_{0}}{ml^{2}\sin^{4}\theta_{0}}

= \frac{mgl}{\cos \theta_{0}}\left(1+3\cos^{2} \theta_{0}\right)\gt 0 ,   (5.75)

where we have used (5.74). The precession is stable and

ω = \left(\frac{k^{\left(0\right)}}{\alpha ^{\left(0\right)}}\right)^{{1}/{2}}=\left(\frac{k^{\left(0\right)}}{ml^{2}}\right)^{{1}/{2}}=\left(\frac{g}{l}\frac{1+3\cos^{2} \theta_{0}}{\cos \theta_{0}}\right)^{{1}/{2}}  (5.76)

is the frequency of the small oscillations about the stationary motion.

Related Answered Questions