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

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Homework Help Overview

The discussion revolves around a physics problem involving a hollow rod and a point mass sliding within it, focusing on angular and linear motion after an impact. The problem is set in the context of rotational dynamics and conservation laws, specifically addressing the effects of an elastic collision and subsequent motion of the mass within the rod.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore conservation of angular momentum and kinetic energy to derive expressions for velocities and angular speeds after impacts.
  • Some participants question the definitions and assumptions made regarding impulse and the components of velocity.
  • There are discussions about the implications of the mass's motion along radial and tangential directions, and how these affect the overall momentum and energy equations.
  • Several participants express uncertainty about specific calculations and the correct interpretation of the problem's requirements.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections to each other's reasoning. Some have offered guidance on the correct application of conservation laws, while others are still seeking clarification on specific aspects of the problem. Multiple interpretations of the equations and assumptions are being explored without a clear consensus yet.

Contextual Notes

Participants note that the problem may involve complexities due to the elastic nature of the collision and the constraints on the rod's motion. There is also mention of potential confusion regarding the impulse of the reaction forces and the definitions of the velocities involved.

  • #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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