Equation of motion of a classic inverted pendulum in free fall

Question

I was thinking in this interesting problem:

Suppose we have this inverted pendulum:

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?

Answer 1

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.

Answer 2

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 $\theta$.

Answer 3

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