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VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.2It means, that unless you apply an external net torque to a system, its angular momentum will not change in time.

siddhantsharan
 one year ago
Best ResponseYou've already chosen the best response.0^^ If the moment of inertia is constant. This statement is incorrect if : dw:1360225020292:dw

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.2@siddhantsharan I do not understand your point. Can you please elaborate?

siddhantsharan
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. What I was saying is basically a little thing that we always ignore, however it is still a possibility. Torque is defined as rate of change of angular momentum. Which is \[\tau = d(I \omega)/dt\] This is actually, \[I(d \omega /dt) + (dI/dt) \omega \] Using the product rule for differenciation. In your statement you have assumed that \[dI/dt = 0\] That is agreed, usually the case, However a small case two arises when the rate of change of angular mommentum is zero, i.e. Angular momentum is constant BUT there is still a net torque on the system. That is when, The system is a nonrelavistic sstem. Otherwise, The comment that I mentioned above may also be the case. I may chenge with time when mass is being perhaps ejected or it's distance from the axis of rotation is bieng chenged, etc.

siddhantsharan
 one year ago
Best ResponseYou've already chosen the best response.0Check the last line under "Proof for equivalance relations" here http://en.wikipedia.org/wiki/Torque @VincentLyon.Fr

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.2Reread my statement, I never assume that dI/dt BUT I assumed that the system was closed, as is usually the case when you quote "conservation of angular momentum". Of course, if the system is open and ejects matter, the angular momentum of the OPEN system will change without net torque applied. The initial question was : "What does it mean to say that angular momentum is conserved?" It implies the explanation of a general relation, not a particular one, where this would be the case with lots of adhoc conditions to meet the criteria.

siddhantsharan
 one year ago
Best ResponseYou've already chosen the best response.0Right you are. My bad. That is correct. I misunderstood. :)
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