Is there a proof about angular momentum conservation?

[LCSphysicist]

Homework Statement

Angular momentum and it conservation.

Relevant Equations

L = rp

Angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved).
There is a proof about this conservation?

 

[haruspex]

Consider two points in the system interacting, i.e. exerting a force on each other. If we accept the rule that action and reaction are equal and opposite then these forces are equal and opposite, $\vec F$ and $-\vec F$, and lie in the same line.
If the forces exist for time dt then the points impart equal and opposite momenta, $\vec Fdt$ and $-\vec Fdt$.
If we consider all this in respect of some axis O, and a vector from there to the line of action of the forces is $\vec r$, then the angular momenta they impart on each other are $\vec r\times\vec Fdt$ and $-\vec r\times\vec Fdt$.
Hence the sum of the imparted angular momenta is zero.

 

[ORF]

Hi,

A more abstract (and fancy) proof is given by Noether's theorem: if your system is invariant under rotation, angular momentum is conserved

https://en.wikipedia.org/wiki/Noether's_theorem#Applications

Regards,
ORF

 

[LCSphysicist]

Hi,

A more abstract (and fancy) proof is given by Noether's theorem: if your system is invariant under rotation, angular momentum is conserved

https://en.wikipedia.org/wiki/Noether's_theorem#Applications

Regards,
ORF

Wow, this was scarry, i was just reading this in French Newtonian mechanics right now about this, like ten minutes ago kkkk unfortunately I don't understand much about these concepts of symmetry. but i will save it thx

 

[Delta2]

Consider two points in the system interacting, i.e. exerting a force on each other. If we accept the rule that action and reaction are equal and opposite then these forces are equal and opposite, $\vec F$ and $-\vec F$, and lie in the same line.
If the forces exist for time dt then the points impart equal and opposite momenta, $\vec Fdt$ and $-\vec Fdt$.
If we consider all this in respect of some axis O, and a vector from there to the line of action of the forces is $\vec r$, then the angular momenta they impart on each other are $\vec r\times\vec Fdt$ and $-\vec r\times\vec Fdt$.
Hence the sum of the imparted angular momenta is zero.

You are using a theorem of classical Newtonian mechanics here, namely that the sum of torques $\sum \vec{T_i}=\frac{d\vec{L}}{dt}$ equals the rate of change of angular momentum L. It is a theorem in the sense that it can be proved from Newton's laws.

 

[ehild]

You are using a theorem of classical Newtonian mechanics here, namely that the sum of torques $\sum \vec{T_i}=\frac{d\vec{L}}{dt}$ equals the rate of change of angular momentum L. It is a theorem in the sense that it can be proved from Newton's laws.

Start from Newton's second law $\vec F = m \vec a$.
By definition, the angular momentum is $\vec L = \vec r \times\vec p= m \vec r \times\vec v$. (1)
(x means cross product)
$\vec r$ is the position vector, $\vec v$ is the velocity $\vec v = \dot {\vec r}$
$\vec a$ is the acceleration,$\dot{\vec v}= \vec a$.
The force is $\vec F =m\vec a$ .
The torque is $\vec\tau=\vec r \times \vec F$
Take the time derivative of eq. (1)
$\dot{\vec L }= m \left (\dot{\vec r}\times \vec v + \vec r \times \dot{\vec v}\right)$
$\dot{\vec r}\times \vec v ==0$
so
$\dot{\vec L }= m \vec r \times \dot{\vec v}= m \vec r \times \vec a = \vec r \times\vec F= \vec \tau$