Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 ?

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
Welcome to physics.SE! Please see our FAQ 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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.