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I was thinking in this interesting problem:

Suppose we have this inverted pendulum:
enter image description here

But without this control force $F$ and the system would by loose from a height $h_0$, with initial velocity $0$ and with the pendulum angle equals $\theta$, and would do a free fall. What would be the equation of motion of this system?

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  • $\begingroup$ Hi, what have you done so far in this problem? $\endgroup$
    – JAlex
    Sep 14, 2021 at 17:55

3 Answers 3

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Lets think about it in the reference frame of the falling masses and ignore air resistance. By removing the control force $F$ and the force of gravity (by letting the whole system fall freely) there are no longer any forces left. That means that if $m$ has initial velocity 0 (with respect to $M$) it will continue to have velocity 0; i. e. it will not move with respect to $M$. From the non-falling perspective, that means that your whole system just accelerates downward uniformly and the angle $\theta$ does not change.

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Set cart displacement $x$ and rod angle $\theta$

  1. No external forces applied horizontally so the combined center of mass remains fixed in the horizontal location.

    $$ M x + m (x - \ell \sin \theta) = 0 $$

    or in terms of velocities

    $$ M \dot{x} + m ( \dot{x}- \ell \dot{\theta} \cos \theta) = 0 $$

  2. Conservation of energy applies which means

    $$ \underbrace{m \ell \cos \theta}_\text{PE} + \underbrace{ \frac{1}{2} m \dot{x}^2 + \frac{1}{2} (m \ell^2) \dot{\theta}^2}_\text{KE} + \underbrace{ \frac{1}{2} M \dot{x}^2}_\text{KE} = \text{(const)} $$

Use those two expressions in order to solve for $\dot{\theta}$ and $\dot{x}$ for each angle $\theta$.

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    $\begingroup$ I think that the center of mass x direction is ${\frac {m \left( x+l\sin \left( \theta \right) \right) +Mx}{m+M}}$ ? $\endgroup$
    – Eli
    Sep 14, 2021 at 19:39
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The position vector to the mass m is:

$$\mathbf R_m(t)=\left[ \begin {array}{c} x \left( t \right) +l\sin \left( \theta \left( t \right) \right) \\ l\cos \left( \theta \left( t \right) \right) \end {array} \right] $$

and to the mass M

$$\mathbf R_M(t)=\begin{bmatrix} x(t) \\ 0 \\ \end{bmatrix}$$

from here you can obtain the kinetic and potential energy and with EL the equation of motions

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