Relating Angular Momentum and Torque Momentum

Torque is a rotational analogue of force. It plays the same role in rotational dynamics as force plays 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 analogue of linear momentum. It is axial vector and is measured as the cross product of position and momentum vector.

Let $\vec{r}$ denote the position of some particle in a rigid body. Suppose this rigid body is undergoing rotation with angular velocity $\vec{\omega},$ then $$\dot{\vec{r_i}}=\vec{\omega_i} \times \vec{r_i}.$$

See the appendix for a proof of this. By taking the derivative of both sides with respect to time and multiplying both sides by $m$, the mass of particle, we obtain $$\dot{\vec{p}}=\vec{\omega}\times \vec{p}.$$

Now we simply note that if $\vec{F}$ denotes the net force on particle ii, then Newton's Second Law gives $\vec{F} = \vec{p}$ so that $$\tau = r \times F = r \times p = r \times (\omega \times p) = -p \times (r \times \omega) - \omega \times (p \times r) = p \times (\omega \times r) + (\omega \times (r \times p)) = p \times r + \omega \times L = \omega \times L$$ $$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \vec{p} = \vec{r} \times (\vec{\omega} \times \vec{p}) = - \vec{p} \times (\vec{r} \times \vec{\omega})$$

$-ω→×(p→i×r→i)=p→i×(ω→×r→i)+ω→×(r→i×p→i)=p→i×r→˙i+ω→×L→i=ω→×L→i$ 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 ii, the result can readily be seen to also hold for the net torque ττ on the body and the total angular momentum L⃗ L→ of the body; $τ⃗ =ω⃗ ×L⃗$ $τ→=ω→×L→$ Appendix. The motion of a rigid body undergoing rotation is generated by rotations. In other words, there is some time-dependent rotation R(t)R(t) for which $r⃗ (t)=R(t)r⃗ (0)$ $r→(t)=R(t)r→(0)$ It follows that $r⃗ ˙(t)=R˙(t)r⃗ (0)=R˙(t)R(t)Tr⃗ (t)=ω⃗ (t)×r⃗ (t)$ $r→˙(t)=R˙(t)r→(0)=R˙(t)R(t)Tr→(t)=ω→(t)×r→(t)$ In the last step, I used the fact that R(t)R(t) is an orthogonal matrix for each tt which implies that $R˙RTR˙RT$ is antisymmetric. It follows that there exists some vector ω⃗ ω→, which we call the angular velocity of the body, for which $R˙RTA⃗ =ω⃗ ×A⃗ R˙RTA→=ω→×A→$for any $A⃗ A→.$