Rod on a Plane: Calculating Angular and Linear Motion After Impact"

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SUMMARY

The discussion centers on the dynamics of a hollow rod with mass M and length R, free to rotate around the z-axis, and a point mass m sliding inside it. After an elastic collision with a ball of mass m0, the participants calculated the speeds and angular velocities post-impact using conservation of angular momentum and kinetic energy principles. Key equations derived include the final speed of the ball v1 and the angular speed of the rod ωi, with results showing that the impulse of the constraint reaction J is directly related to the velocities involved. The discussion also explores the motion of the center of mass and the implications of inelastic collisions.

PREREQUISITES
  • Understanding of angular momentum conservation principles
  • Familiarity with kinetic energy conservation in rotational systems
  • Knowledge of impulse and momentum concepts
  • Basic mechanics involving elastic and inelastic collisions
NEXT STEPS
  • Study the conservation of angular momentum in complex systems
  • Learn about the dynamics of inelastic collisions and their effects on motion
  • Explore the implications of rotational kinetic energy in mechanical systems
  • Investigate the motion of the center of mass in multi-body systems
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in advanced dynamics and collision analysis will benefit from this discussion.

  • #31
Delta2 said:
This looks good

You have to take the angular momentum around the CM of the system. And also the total moment of inertia around the CM of the system. I can't understand what the term ##I_{TOT}V_{CM_y}(R-x_{CM})## is about?
The terms is an error, i tought that ## R - x_{CM} ## was the radius of the circumference along which the system rotates.
I'm not understanding perfectly what you mean (probably because I'm a little dumb) but if i fix that term i got:
## -m_0 v_0 R = I_{TOT} V_{CM_y} (x_{CM}) + I_{TOT} \omega_{CM} ##
is it right?
 
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  • #32
To be honest i am not sure myself either at this point, i think the correct equation is $$-m_0v_0(R-x_{CM})=I_{TOT}\omega_{CM}$$, where ##I_{TOT}## the moment of inertia of the system around the CM.
 
  • #33
Delta2 said:
To be honest i am not sure myself either at this point, i think the correct equation is $$-m_0v_0(R-x_{CM})=I_{TOT}\omega_{CM}$$, where ##I_{TOT}## the moment of inertia of the system around the CM.
Ummm ok.
Is the total moment of inertia that I found in post 22 correct?
 
  • #34
Yes it is correct.
 
  • #35
Delta2 said:
Yes it is correct.
Ok thanks for your patience, but this problem was difficult as f*ck.
I hope this year the exam will be a little easier
 
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  • #36
Yes i agree, quite hard problem, testing to the extreme the understanding of concepts and principles.
Is it college level or university level exams this is from?
 
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  • #37
Delta2 said:
If we don't make this assumption, then a mini paradox will appear if we do the balance of the kinetic energy when the particle returns to A. Somehow the total kinetic energy will be increased (in point A , with regards to point B after the inelastic collision), and we won't be able to infer where this surplus "ghost" energy has come from.
Ah yes, it wouldn’t make it back to A if it were not elastic.
 
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  • #38
haruspex said:
Ah yes, it wouldn’t make it back to A if it were not elastic.
Thanks for resolving the paradox, if i tell you that i spend part of last night before sleep thinking where does the surplus of kinetic energy comes from you ll laugh with me won't you?
 
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  • #39
Delta2 said:
Yes i agree, quite hard problem, testing to the extreme the understanding of concepts and principles.
Is it college level or university level exams this is from?
University.
 
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