Undergrad What Happens to Rotational Velocity When AC Part is Dropped?

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When the AC part of the rod is dropped, the rotational velocity of the remaining rod will change due to the loss of mass and the shift in mass distribution. The initial angular momentum of the entire rod must be recalculated by subtracting the angular momentum of the dropped section. The challenge lies in determining the angular momentum of the dropped part accurately. The remaining rod will rotate with a different velocity, influenced by its new mass configuration. For further assistance, users are advised to repost their questions in the Homework Help forums and include their attempts at solutions.
Anisur Rahman
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TL;DR
Need help in this question.
In figure AO = OB = 4m. C is the midpoint of AO. The rod rotates with a velocity of 5 rad/s about the axis PQ. What will be the rotational velocity of the remaining rod if AC part is suddenly dropped from the rod? Assume that the mass of this uniform rod is M.
Screenshot 2024-06-05 221032.png
 
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What do you think it will happen?
You will have less mass rotating, but not in a concentric way.
 
Lnewqban said:
What do you think it will happen?
You will have less mass rotating, but not in a concentric way.
To calculate the angular momentum of the remaining rod, I think to substract the angular momentum of the dropped part from primary angular momentum. But I am confused about to calculate the angular momentum of the dropped part. How can I calculate that?
 
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Anisur Rahman said:
TL;DR Summary: Need help in this question.

In figure AO = OB = 4m. C is the midpoint of AO. The rod rotates with a velocity of 5 rad/s about the axis PQ. What will be the rotational velocity of the remaining rod if AC part is suddenly dropped from the rod? Assume that the mass of this uniform rod is M.
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Please repost your question in the Homework Help forums and always show your Attempt at the Solution when starting new schoolwork threads. Thank you.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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