Solving the First Equation: v1, v2, & ωR

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SUMMARY

The discussion centers on the dynamics of a ring and a bug moving along its circumference, specifically addressing the relationship between their velocities (v1 and v2) and angular velocity (ωR). It is established that v1, the velocity of the ring's mass center, and v2, the velocity of the bug, must be equal and opposite to maintain zero total linear momentum. The confusion arises from the direction of rotation, where the ring rotates clockwise while the line joining the mass centers rotates anticlockwise, ensuring total angular momentum remains zero. The conversation emphasizes the importance of visualizing the system's motion and understanding the frames of reference for momentum conservation.

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Gourab_chill
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Homework Statement
I have posted the question below. My question is how to form the equations to solve the question below.
Relevant Equations
v=ωr
Capture.PNG


I've marked correct answers above. Have a look at the solutions:
Capture1.PNG

How is the first equation justified? Shouldn't v2 and ωR be of opposite signs? What is v1? And how is it equal to v2? My biggest problem is the source of v1 since the ring is not having vertical displacement, then what is v1?
 
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Gourab_chill said:
Shouldn't v2 and ωR be of opposite signs?
Don't confuse the rotation of the line joining the mass centres, which will be anticlockwise, with the rotation of the ring about its own centre, which will therefore be clockwise so that total angular momentum remains zero.
ω is the rate of the latter.
Gourab_chill said:
What is v1? And how is it equal to v2?
v1 is the velocity of the ring's mass centre in the ground frame. v2 is the velocity of the bug in the ground frame. These must be equal and opposite so that total linear momentum remains zero.
 
haruspex said:
Don't confuse the rotation of the line joining the mass centres, which will be anticlockwise, with the rotation of the ring about its own centre, which will therefore be clockwise so that total angular momentum remains zero.
ω is the rate of the latter.

So, the first equation is for the rotation for line joining the center of masses. Okay. But, if the bug moves upwards (let's say) then the ring will rotate clockwise and so the direction of v1 and ωR would be same, right? Wouldn't they be of opposite signs? You did say the rotation will be anti-clockwise, am I mistaking somewhere or did you consider the bug to move down?
haruspex said:
v1 is the velocity of the ring's mass centre in the ground frame. v2 is the velocity of the bug in the ground frame. These must be equal and opposite so that total linear momentum remains zero.

So the ring does have some motion along the vertical plane right? It's kinda hard to visualize though. Can I conclude that the ring will keep bouncing all the time?
 
Gourab_chill said:
So, the first equation is for the rotation for line joining the center of masses.
No, I wrote "the latter", i.e. the rotation of the ring about its own centre.

Yes, it is hard to visualise. Part of the difficulty is that we are only concerned with the instantaneous velocities at the starting position. It will soon look rather different. To minimise this problem, take the part of the circle where the bug is to be a straight line.
Try drawing a diagram showing:
  • The initial positions of mass centre, ring centre and bug
  • Positions of ring centre, bug, and the point X on the ring where the bug started, after a short time t.
The mass centre does not move.
The ring centre will have moved down v1t, X will have moved further down, and the bug will have moved up v2t.
 
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It's a fiddly little thing. You need to be careful to think about in which frames of reference angular momentum and/or linear momentum are conserved. They're both conserved in the lab frame, and also in the CM frame of the whole system.

Once you've got that, it's more a matter of just being careful with the algebra in finding the correct relative velocities.
 
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