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The law that

$$\frac{d\vec{L}}{dt}= \vec{T}$$

where $\vec{T}$ is torque about a frame's origin $o$ and $\vec{L}$ is the angular momentum about that origin $o$.

Can this law be ultimately (always?) traced backed to Newton's Second Law ?

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It depends on how you define torque. For me, this would be the definition of torque, but it then requires a proof to show that $\mathbf{T}=\mathbf{r}\times\mathbf{F}$. Anyway, just write down $\sum \mathbf{r}\times\mathbf{p}$ and differentiate, applying the product rule to the cross products. Then apply Newton's second law. This is all for a Newtonian system of particles. In other contexts, e.g., relativity, it plays out differently. –  Ben Crowell Apr 20 '13 at 18:11
    
So you are asking if you can derive Euler's laws of rotational motion from Newton's laws? –  ja72 Apr 21 '13 at 6:52

3 Answers 3

I'll expand my comment into an answer.

I would take $\mathbf{T}=d\mathbf{L}/dt$ as the definition of torque, but it sounds like the OP takes $\mathbf{T}=\mathbf{r}\times\mathbf{F}$ as the definition. Either way, we need to prove that the two expressions are equivalent for a system of particles.

The total angular momentum is $$\mathbf{L}_{tot}=\sum \mathbf{r}_i\times \mathbf{p}_i$$ Differentiating with respect to time and applying the product rule, this becomes $$\sum \frac{d\mathbf{r}}{dt}_i\times \mathbf{p}_i+\sum \mathbf{r}_i\times \frac{d\mathbf{p}_i}{dt}$$ The first sum vanishes term by term. Applying Newton's second law to the second sum, we get $$\sum \mathbf{r}_i\times \mathbf{F}_i$$

The fact that there was a sum over multiple particles ended up being unimportant, because the manipulations were all term by term.

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How is this F=ma/(dp/dt) ? . This is what OP asked for . You have just derived the situation again . Your answer makes no sense –  user23432 Apr 21 '13 at 17:56
    
@Rishabh: By assuming $\mathbf{F}=d\mathbf{p}/dt$, I've proved $d\mathbf{L}/dt=\sum\mathbf{r}\times\mathbf{F}$. This is what the OP asked for. –  Ben Crowell Apr 21 '13 at 21:47

Torque is the rotational equivalent of force. The second law state, the sum of the forces $\sum F = ma$.

$\alpha$ is the rotational equivalent for acceleration, so the law would look like $\tau$ (torque) = $m \times \alpha$. The angular momentum would be $m \times \omega$. velocity over time is acceleration. $p/t = F$.

Therefore, angular momentum can be traced to torque, the rotational equivalent of force.

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Welcome to physics.SE! Please see our FAQ physics.stackexchange.com/faq#notation on how to notate math. I don't think your answer quite works, except as a heuristic or a mnemonic. It's true that there is a system of analogies between the linear stuff and the rotational stuff, but that doeesn't constitute a proof. –  Ben Crowell Apr 21 '13 at 0:20

No, it is a different law altogether .

Newton's law are true for point masses only

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Hi Ram. Welcome to Physics.SE. While writing answers, please make sure that you address the question specifically. Your answer looks very much a comment one way or the other... –  Waffle's Crazy Peanut Apr 21 '13 at 13:57
    
The question makes sense for a system of particles. –  Ben Crowell Apr 21 '13 at 16:54
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Misha Apr 23 '13 at 10:11

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