Angular Momentum: Is Conservation Dependent on Net Torque?

[UrbanXrisis]

My book says that "Where there is no net torque on an object or system of particles, angular momentum is conserved."

Does this mean that when next torque does not equal zero, the angular momentum is not conserved?

 

[Galileo]

Yes, it goes both ways. The angular momentum will not change if and only if there is no torque on the system.
It's the same as for linear momentum. If there is no net force acting on the system, the total linear momentum is conserved and vice versa.
It can be easily seen from the equations:

$$\vec F = \frac{d}{dt}\vec P, \qquad \vec N = \frac{d}{dt}\vec L$$

If the left sides are not zero, the momenta are changing.

I`m talking here about the NET force, NET torque and TOTAL momenta. For a double star for example, the individual momenta of the stars are not conserved, but the total is.

EDIT: Just for clarity, the phrase:
"Where there is no net torque on an object or system of particles, angular momentum is conserved.",
does NOT mean the same as:
"If the torque is not zero, angular momentum is not conserved." It's a different statement that happens to be true.

 

[cepheid]

EDIT: Just for clarity, the phrase:
"Where there is no net torque on an object or system of particles, angular momentum is conserved.",
does NOT mean the same as:
"If the torque is not zero, angular momentum is not conserved." It's a different statement that happens to be true.

I don't really understand this...can you explain what you mean? Both results seem to follow directly from the equations, and I don't see how one could be true if the other weren't i.e. one is a restatement of the other in the same sense that some people argue that, given what we know now, Newton's first law is somewhat redundant in light of the second law...right?

 

[Galileo]

I don't really understand this...can you explain what you mean? Both results seem to follow directly from the equations, and I don't see how one could be true if the other weren't i.e. one is a restatement of the other in the same sense that some people argue that, given what we know now, Newton's first law is somewhat redundant in light of the second law...right?

I mean logically:

$$A \Rightarrow B$$
is not the same as
$$\neg A \Rightarrow \neg B$$
where $\neg$ denotes negation.

So: "No net torque => Angular momentum conserved"
is not the same as:
"Net torque => Angular momentum not conserved."

But yes, they both follow from the equations.

 

[xanthym]

My book says that "Where there is no net torque on an object or system of particles, angular momentum is conserved."

Does this mean that when next torque does not equal zero, the angular momentum is not conserved?

The 2 key questions for Angular Momentum Conservation are these: 1) What is the specific SYSTEM being considered?; and 2) Is the given torque from EXTERNAL sources (derived from elements not in the system) or from INTERNAL sources (derived from elements contained within the system). Definition of the SYSTEM being considered is prerequisite to applying Conservation of Angular Momentum. Conservation of Angular Momentum can be stated with the following:
Total Angular Momentum of a SYSTEM is conserved if that SYSTEM experiences only INTERNAL torques (derived from elements completely within the system) and has NO EXTERNAL torques (derived from elements outside the system) applied to it.
(If there are UNbalanced EXTERNAL torques applied to the system, Angular Momentum will not be conserved.)


~~

 

[cepheid]

I mean logically:

$$A \Rightarrow B$$
is not the same as
$$\neg A \Rightarrow \neg B$$
where $\neg$ denotes negation.

So: "No net torque => Angular momentum conserved"
is not the same as:
"Net torque => Angular momentum not conserved."

But yes, they both follow from the equations.

Yes I see you what you are saying, thanks. The fact that both are true is due to the physics. (the equations). Neither statement on its own (out of this context) implies that the other must be true based purely on logic. Makes sense now...

 

[Data]

Correct. Now, on the other hand

$$A \Rightarrow B$$

certainly implies

$$\neg B \Rightarrow \neg A$$

and visa-versa~