Inverse Dynamics - 2 Rigid Bodies connected with hinge

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SUMMARY

This discussion focuses on simulating inverse dynamics for two rigid bodies connected by a hinge joint. The user initially calculated torque using the formula \(\tau=I\alpha\) for a single rod and then adapted it to \(\tau=(I_{COM}+Mh^2)\alpha\) when a hinge joint was introduced. The main challenge arises when adding a second rod, as the user seeks to understand the interaction between the two rods and how to calculate the torque on the rod connected to the ground. The user also notes that gravity has been disabled for this simulation.

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  • Understanding of inverse dynamics and torque calculations
  • Familiarity with the parallel axis theorem
  • Knowledge of rigid body dynamics
  • Basic principles of Lagrangian dynamics
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This discussion is beneficial for mechanical engineers, robotics researchers, and anyone involved in simulating physical systems with multiple rigid bodies.

leonj
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I am trying to simulate and animate inverse dynamics. So I need to calculate torques. I started out by just taking a rod which rotates about its COM, applied a torque and then just used [itex]\tau=I\alpha[/itex] to make sure I get back the same torque I applied. Next, I attached the rod to the ground with a hinge joint. This time to get the torque I used the parallel axis theorem to get [itex]\tau=(I_{COM}+Mh^2)\alpha[/itex]. So my next step is to attach a second rod to the end of the one attached to the ground. What I am stuck on now is I am not sure how the two rods interact.

For example how do I calculate the torque on the rod connected to the ground? When it is just one rod I used the aforementioned [itex]\tau=(I_{COM}+Mh^2)\alpha[/itex], but now with the other rod attached to it, what else do I need to take into account? Is there a negative torque on the opposite side?

Note: For now I have disabled gravity.

I would appreciate any insight. Thanks
 
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