[Physics] Derivative of angular momentum of rigid body

angular momentumrigid-body-dynamicsrotationrotational-dynamics

I found this equation that describes the change in angular momentum $\vec{L}$ of a rigid body rotating about a fixed point $O$. $I_o$ is the moment of inertia of the body with respect to the axis of rotation and I marked with $⊥$ the component of the angular momentum perpendicular to the axis of rotation.

$$\vec{L_o} = I_o \vec{ω} + \vec{L_o}_⊥ \implies
\frac{d\vec{L_o}}{dt}= I_o \frac{d\vec{ω}}{dt}+ \frac{1}{ω}
\frac{dω}{dt} \vec{L_o}_⊥ + \vec{ω} \times \vec{L_o}_⊥ =

In the derivation the following was used $ \vec{L_o}_⊥ = \vec{\omega}A \hat{L_o}_⊥ $ with $A$ some constant depending on the distribution of mass.

How does the middle term $\frac{1}{ω} \frac{dω}{dt} \vec{L_o}_⊥$ come out and what does it means?

It is a part of the derivative of $\vec{L_o}_⊥$ but I don't understand how and why it is there.

Best Answer

As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity.
You can also derive this. formula like this:
$$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega} \frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } \times \vec{L_o}_⊥ $$.

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