Relationship between torque and angular momentum

I want to know how to deduce the equation $\vec{\tau}=\vec{\omega} \times \vec{L}$, where

$\vec{\tau}$ is the moment of force (also known as torque),

$\vec{L}$ is the angular momentum,

$\vec{\omega}$ is the angular velocity.

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This formula is not true in general. More fundamentally, torque is the time-derivative of angular momentum. Can you provide more context for how you encountered this equation? –  kleingordon Jul 30 '13 at 4:25

Here is the proof that I think you're looking for. As Ali remarks in his answer, the results holds true for a rigid body undergoing rotation with constant angular velocity.

Let $\vec r_i$ denote the position of some particle in a rigid body. Suppose this rigid body is undergoing rotation with angular velocity $\vec \omega$, then $$\dot {\vec r}_i = \vec \omega\times\vec r_i$$ See the appendix for a proof of this. By taking the derivative of both sides with respect to time and multiplying both sides by $m_i$, the mass of particle $i$, we obtain $$\dot {\vec p}_i = \omega\times \vec p_i$$ Now we simply note that if $\vec F_i$ denotes the net force on particle $i$, then Newton's Second Law gives $\vec F_i = \dot{\vec p_i}$ so that \begin{align} \vec\tau_i &= \vec r_i\times \vec F_i \\ &= \vec r_i\times\dot{\vec p_i} \\ &= \vec r_i\times(\vec\omega\times\vec p_i) \\ &= -\vec p_i\times(\vec r_i\times \vec\omega) - \vec\omega\times(\vec p_i\times\vec r_i) \\ &= \vec p_i\times(\vec\omega\times\vec r_i) + \vec\omega\times(\vec r_i\times\vec p_i) \\ &= \vec p_i\times \dot{\vec r}_i + \vec \omega\times \vec L_i \\ &= \vec\omega\times \vec L_i \end{align} This is basically the identity you were looking for. In the fourth equality, I used the so-called Jacobi identity. Now, by taking the sum over $i$, the result can readily be seen to also hold for the net torque $\tau$ on the body and the total angular momentum $\vec L$ of the body; $$\vec \tau = \vec\omega\times\vec L$$

Appendix. The motion of a rigid body undergoing rotation is generated by rotations. In other words, there is some time-dependent rotation $R(t)$ for which $$\vec r(t) = R(t) \vec r(0)$$ It follows that $$\dot{\vec r}(t) = \dot R(t) \vec r(0) = \dot R(t)R(t)^T\vec r(t) = \vec\omega(t)\times \vec r(t)$$ In the last step, I used the fact that $R(t)$ is an orthogonal matrix for each $t$ which implies that $\dot R R^T$ is antisymmetric. It follows that there exists some vector $\vec \omega$, which we call the angular velocity of the body, for which $\dot R R^T \vec A = \vec \omega\times\vec A$ for any $\vec A$.

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Hey josh, you like riddles: try this one –  JoeHobbit Jul 31 '13 at 20:43

As kleingordon and others pointed out, this equation is not true in general. But it can be true in a certain context. I will try to derive the conditions where it can be true.

We all know that:

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

But we want to have:

$$\vec{\tau}=\vec \omega \times\vec{L} \\ \Rightarrow \frac{d\vec L}{dt}=\vec \omega \times \vec L$$

Now, the case where the time derivative of a vector(in this case $\vec L$) is the cross product of a constant vector(in this case $\vec \omega$) with that vector is a well-known case. It's simply the case where that vector($\vec L$) is rotating around the constant vector($\vec \omega$) with constant angular velocity, which happens to be $|\vec \omega|$.

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This is wrong . The units of the equation itself are wrong . –  Dimensio1n0 Jul 30 '13 at 11:40
@Dimension10 No they are not! I think what you are missing is the fact that angles are dimensionless. –  Ali Jul 30 '13 at 11:41
Huh? $[\vec\omega] = \frac1{\operatorname{second}}$. $[\vec\alpha]=\frac1{\operatorname{second^2}}$. –  Dimensio1n0 Jul 30 '13 at 11:50
$[\omega] = [\frac{\theta}{\text{time}}]=\frac{1}{T}$ –  Ali Jul 30 '13 at 11:53
Yes, so? That doesn't refute the fact that angular acceeration is in $\frac1{\operatorname{second}^2}$ . –  Dimensio1n0 Jul 30 '13 at 11:54

I'm not sure where you have obtained this equation, but I'll try and show you why others are suggesting it's not correct.

You can try looking at the linear analogues of these values. In a linear system you are saying that:

$\text{Force} = \text{Velocity}\times \text{Momentum} = v \times mv = m v^2$

Which is somewhat different to Newton's law

$F = m a$

Where did you get this equation from and to what is it referring?

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In fact now I think about it, the fact that L=Iw means that the vectors will be in the same direction but scaled by I (rotational analogue of mass). Surely here the cross product will end up being 0 anyway? Are you meaning to use omega as angular velocity? You never explicitly said so, but it is the general convention. –  Folau Jul 30 '13 at 9:57
Hello Folau, "Rad" or angles in general are dimensionless; so the units do match. Also, you can use Latex commands to improve your answer. –  Ali Jul 30 '13 at 10:23
I'm at work, so I'm typing in somewhat of a hurry, but thanks :) I was certain I was missing something about the units. Is my second assumption solid or have I made another error? –  Folau Jul 30 '13 at 10:25
sure. I'm listening. –  Ali Jul 30 '13 at 12:53
Nope my bad, misread what you were saying :P I read it that you were saying that the angular velocity is constant, therefore making the angular acceleration 0, which is kind of trivial with respect to Newton II, but I misread how you interpreted the vectors. –  Folau Jul 30 '13 at 12:54

Torque is a rotational analouge of force.It plays the same role in rotational dynamics as force playes in linear motion.Torque is defined as the moment of force or turning effect of force about the given axis or point.It is measured as the cross product of position and force vector while as angular momentum is the rotational analouge of linear momentum.It is axial vector and is measured as the cross product of position and momentum vector.

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protected by Qmechanic♦Jul 23 '14 at 6:49

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