Uniform Circular Motion

AI Thread Summary
In uniform circular motion, as an object's mass decreases due to evaporation, its velocity is expected to increase to maintain centripetal force, while the angular momentum is debated. Some argue that angular momentum changes because the evaporating mass carries momentum away, contradicting the book's assertion that it remains constant. The direction of gas escape is crucial; if it escapes uniformly, it will affect angular momentum proportionally to mass loss. However, if the gas escapes in a preferred direction, it could either increase or decrease angular momentum, depending on various factors. The discussion highlights the need for clarity in the problem's wording regarding the escape direction of the gas.
Nugso
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Homework Statement



An object is in uniform circular motion. As the time passes by, its mass decreases. ( It has Co2 in it, and it evaporates) What will happen to the object's velocity and angular momentum?



Homework Equations



F = mV²/r, L = mvr


The Attempt at a Solution



Well, I'm trying to figure out if the centripetal force will change. According to the first formula, it seems like it will. ( As the mass of the object decreases). But I do think that maybe the object will increase in order to balance so that the centripetal force won't change.


The answer is = V increases, L does not change. Even if the V increases, it'll increase to balance the decrease of mass(Actually, V² will increase proportionally to M, so in all cases angular moment mvr, should decrease)

Can anybody help me with this?
 
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I do not think the answer where the angular momentum of the mass is constant is correct. The mass evaporates alright, but the vapor carries angular momentum away with it. You cannot just assume that does not happen, that would be nonphysical, unless you can describe some plausible mechanism preventing that.
 
voko said:
I do not think the answer where the angular momentum of the mass is constant is correct. The mass evaporates alright, but the vapor carries angular momentum away with it. You cannot just assume that does not happen, that would be nonphysical, unless you can describe some plausible mechanism preventing that.

Thanks voko. I kind of think the same. In all cases I've thought of so far, angular moment changes. Well, guess the answer given by the book is incorrect.
 
The problem is very badly worded. For instance, it is not clear what the object is rotating about. If that is its center of mass, and the axis of rotation do not change, then it is possible for vapors to escape axially, without affecting the angular momentum of the object.

In any other configuration, that seems impossible.
 
In this situation, much depends on the possible direction(s) for the gas to escape. If the gas can escape in all possible directions equally, then it will on average take away the angular momentum proportional to its mass. If any direction is preferred, then it may increase or decrease the angular momentum of the rotating body, but the rate of the increase or decrease will depend on many factors.
 
voko said:
In this situation, much depends on the possible direction(s) for the gas to escape. If the gas can escape in all possible directions equally, then it will on average take away the angular momentum proportional to its mass. If any direction is preferred, then it may increase or decrease the angular momentum of the rotating body, but the rate of the increase or decrease will depend on many factors.

Oh. I beg your pardon. I did not think that it was important. It says that the gas goes perpendicular to horizontal plane. By the way, how's that important at all?
 
Nugso said:
Oh. I beg your pardon. I did not think that it was important. It says that the gas goes perpendicular to horizontal plane. By the way, how's that important at all?

Still somewhat unclear. If the gas escapes vertically relatively to the rotating mass, then relatively to the fixed center of rotation it will still have the horizontal velocity equal to that of the rotating mass.

If, however, the gas escapes vertically relatively to the fixed center of rotation, then it will have zero horizontal velocity. But that means that, relatively to the rotating mass, it will have to escape at a certain angle and with a very specific escape velocity, which is possible, but requires careful engineering.
 
voko said:
Still somewhat unclear. If the gas escapes vertically relatively to the rotating mass, then relatively to the fixed center of rotation it will still have the horizontal velocity equal to that of the rotating mass.

If, however, the gas escapes vertically relatively to the fixed center of rotation, then it will have zero horizontal velocity. But that means that, relatively to the rotating mass, it will have to escape at a certain angle and with a very specific escape velocity, which is possible, but requires careful engineering.

Thank you very much voko. I hope it's the book's mistake, not mine.
 
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