Inverted pendulum -- What force is applied at the center of mass?

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The discussion centers on understanding the forces acting on the center of mass of an inverted pendulum to maintain balance against gravitational torque. It explores the concept of torque and how adjusting the attachment point of the pendulum can redefine the torque experienced. The conversation distinguishes between fictitious inertial forces and real forces, such as compression or tension in the pendulum arm, which affect torque calculations. A visual representation of the forces and torques involved is requested, and a reference to relevant academic literature is provided. The thread also briefly mentions the historical significance of fuzzy logic in controlling inverted pendulums.
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Summary: I know that in order to keep inverted pendulum balanced when it leans forward, I should accelerate the bottom of it, but what is the force which is applied to center of mass of the pendulum with balances torque due to gravity? Is it inertial force? We assume that our inverted pendulum is massles rod with a center of mass at its top. As a Radius of rotation we denote "L"as angle between vertical and pendulum "a"

M*g*sin(a)*L= F*L
 
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fizzyfiz said:
what is the force which is applied to center of mass of the pendulum with balances torque due to gravity? Is it inertial force?
If you imagine that there is a torque due to gravity, you must imagine that there is an axis about which the torque applies. Let us choose the point where the pendulum is attached to the ground.

If you move the point where the pendulum is attached to the ground, you adjust the torque due to gravity by redefining your coordinate system. If you insist on viewing this as an applied force then yes, it would be a fictitious inertial force (the D'Alembert force) associated with the acceleration of the coordinate system.

If you prefer to stick to a single inertial frame then the relevant force is real and is, instead, the compression or tension force on the pendulum arm. When the pendulum's bottom end moves, this compression or tension force no longer produces zero torque about a reference axis at the original attachment point.
 
Thank you for reply! Could you draw the second case?
 
fizzyfiz said:
Thank you for reply! Could you draw the second case?
The star is the axis of rotation, the thin line is the downward force from gravity, the rectangle is where the support point was moved, the pendulum arm is the bold line and the circle is the pendulum bob. You can see that the compression force of arm on bob amounts to a clockwise torque while gravity amounts to a counter-clockwise torque.

My skills with mspaint are of questionable value.
inverted.png
 
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Check out this paper. It should be helpful with the equations, and the design.

https://link.springer.com/content/pdf/10.1007/0-387-29295-0_7.pdf
This thread triggered a memory. The inverted pendulum was the earliest and only? clear victory for fuzzy logic. So I searched for inverted pendulum fuzzy logic, and found many hits.
 
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