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Consider the inverted pendulum shown below:

inverted pendulum

where $F$ is an external force, $B$ is the CoM of the rod, $A$ is the position of the joint where the rod attaches to the car. Angles increase in the counter-clockwise direction (the angle shown above is negative).

The forces applied on the rod are shown in the following free body diagram:

forces on rod

If we take Newton's law of rotational motion for the rod with respect to point $B$, we obtain the following equation:

$$ \tfrac{L}{2}(H\cos\theta - N\sin\theta) {}={} I\ddot{\theta}.\tag{1} $$

Question 1. My first question is what happens if we take Newton's law with respect to point $A$. Then we will have

$$ mg\sin\theta = I\ddot{\theta},\tag{2} $$

but this cannot be right because it seems that the dynamics of $\theta$ does not depend on $H$, therefore the external force $F$ seems not to affect the angle of the rod.

Question 2. My main motivation for the first question is the case where there is some friction at the joint which creates the torque

$$ T = -b\dot\theta,\tag{3} $$

where $T$ is a torque with respect to point $A$. How can we modify the equations of motion to accommodate such friction terms?

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Your inverted pendulum isn't in an inertial frame. You are looking at the pendulum in the frame accelerating with point A. Therefore, you have to consider the fictitious force $-ma$ acting on the rod as well. This is a uniform force, so you can consider it to act at the center of the pendulum. This is why if you calculate torque about the center of the pendulum you don't need to include it. So, you should have for torque about A $$\frac L2(mg\sin\theta-ma\cos\theta) = I\ddot{\theta}$$

I will leave your second question to you to think about based on this answer.

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    $\begingroup$ Thank's a log for your answer. What would be a good reference to read about non-inertial frames? I guess then the equation with friction becomes: $\frac{L}{2}(mg\sin \theta - ma \cos\theta) - b\dot{\theta} = I\ddot{\theta}$? $\endgroup$ – Pantelis Sopasakis Aug 18 at 19:34
  • $\begingroup$ @PantelisSopasakis That equation looks good! I think any mechanics book should cover non-inertial frames. Or I'm sure there are a lot of posted questions and answers here that could help. Also Wikipedia. $\endgroup$ – Aaron Stevens Aug 18 at 22:08

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