How Can the Stability of a Kapitza Pendulum Be Demonstrated?

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Homework Help Overview

The discussion revolves around the stability of a Kapitza pendulum, particularly under the condition where the amplitude is significantly larger than the gravitational acceleration. Participants are exploring the implications of this condition on the pendulum's motion and stability.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the conditions under which the pendulum remains stable, particularly regarding the parameter ##T_d## and its implications. There are attempts to understand the mechanics involved and the potential issues with the problem setup.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the conditions necessary for stability. Some have suggested that the problem may be poorly defined, while others have noted that the stability can be analyzed through advanced theories like KAM theory, indicating a lack of consensus on the basic understanding.

Contextual Notes

There appears to be uncertainty regarding the specific conditions for ##T_d##, with some participants noting that no conditions were provided in the original problem statement. This lack of information may be contributing to the confusion surrounding the stability of the pendulum.

HansBu
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Homework Statement
I am having confusion with regards to the proof of my problem. This involves a pendulum with harmonically driven pivot and the task is to show that the pendulum is stable in the inverted position when the amplitude of the driving acceleration is sufficiently high. For reference, consider the problem below.

> Consider a pendulum with harmonically driven pivot. The equation of motion is
$$\frac{d^2\theta}{dt^2}=-\frac{g+a_d(t)}{L}\sin\theta$$
where $$a_d(t)=A_0\sin(2\pi t/T_d)$$ is the time-varying acceleration of the pivot. Show that when the amplitude of the driving acceleration is sufficiently high $$A_0\gg g$$ the pendulum is stable in the inverted position i.e., if $$\theta(t=0)\approx180°$$.
Relevant Equations
$$\frac{d^2\theta}{dt^2}=-\frac{g+a_d(t)}{L}\sin\theta$$
where $$a_d(t)=A_0\sin(2\pi t/T_d)$$ is the time-varying acceleration of the pivot.
I understand that when $$A_0 \gg g$$, the g term in the equation of motion can be dropped. The equation of motion then becomes
$$\frac{d^2\theta}{dt^2}=-\frac{a_d(t)}{L}\sin\theta$$

But how can I show that the pendulum is stable for such case? I am totally clueless.
 
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Is there any condition on ##T_d##? Say ##T_d=\infty## the inverted pendulum is unstable.
 
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looks like an ill - understood Kapitza's pendulum
 
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mitochan said:
Is there any condition on ##T_d##? Say ##T_d=\infty## the inverted pendulum is unstable.
Hi, mitochan! There were no specified conditions for ##T_d## given in the problem.
 
wrobel said:
looks like an ill - understood Kapitza's pendulum
It is quite consistent with the mechanics of Kapitza's pendulum. I presume that something is wrong in the problem, right?
 
I do not know.

Anyway stability of the Kapitza pendulum is proved by means of KAM theory. All this field is far beyond undergraduate courses.
 

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