KE of off-center impacts in rotating and translating rigid bodies

  • Thread starter dpettig
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I'm trying to understand basic principles of ancient thrown weaponry. Let's say we have something like a bar with a known inertia tensor that is thrown from one end such that it is both rotating and translating. If it strikes something along either side of its center of mass (an off-center impact), how can we understand the energy available in the impact? For example if it strikes with its long axis roughly perpendicular to the vector of translation (see sketch), on one side (the "advancing" side relative to the combined rotation and translation vectors) the kinetic energy in the impact will be much higher than on the "receding" side. But even here not all the KE available in the body will be transferred to the impacted object (which for the sake of simplicity can be inelastic). I hope it's as simple as including the distance (r) from CM somewhere in the equation? Please forgive the terrible sketch.
 

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  • #2
A.T.
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... (which for the sake of simplicity can be inelastic). ...
Working in the rest frame of the common center of mass: For completely inelastic collisions (objects stick together), there is no linear KE after the collision. The remaining rotational KE follows from the conserved angular momentum.
 
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My apologies, I forgot the difference between elastic and inelastic collisions. This would be of the latter form, but all I care about is the KE available in the impact, so the specifics of the collision can be simplified as much as possible.
 
  • #4
jbriggs444
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Working in the rest frame of the common center of mass: For completely inelastic collisions (objects stick together), there is no linear KE after the collision. The remaining rotational KE follows from the conserved angular momentum.
I think that this over-idealizes the situation.

[I've cast the problem as two men rather than a man and a weapon. It is easier to refer to body parts and is pleasantly symmetric]

In a completely inelastic collision of one man's hand on another man's hand, the hands will come to relative rest, but not necessarily the men. One can calculate the impulse that is required to deliver enough combined linear and angular velocity change on each man so that the two hands each come to relative rest. One can then calculate the resulting change in linear and rotational energy for each man.
 
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  • #5
osilmag
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The change in momentum from the impact would be split between linear and angular. You would have to find the final linear and angular velocity to get the appropriate amount of linear and rotational energy. Good luck finding that. 🤷‍♂️
 
  • #6
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Thanks osilmag. Right, this is challenging, hence why I've read everything including a book on off-center impacts and haven't found a simple solution. But I am not a physicist by training, so I am suspicious that I've already come across the solution and didn't recognize it.
 

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