The derivative of the unit velocity vector

Question

The set up:

An intertial frame Y-X used to describes trajectory of an insect on some rigid body using some relative vectors. Symbols: $ \vec{r_a}$ is is the vector connecting the origin to some point on the rigid body, $ \vec{r_b} $ is the vector connecting origin to the insect and $ \vec{r } $ is the vector connecting the reference to the insect. The relation between vectors:

$$ \vec{r_b} = \vec{r_a} + \vec{r } $$

In a video lecture about corollis force by professor M.S.Sivakumar, I don't get a formula at 19:12 which is used:

$$ |v_{rel} | \frac{ d \hat v_{rel} }{dt} = \vec{\omega} \times \vec{v_{rel}} $$

With,

$$ v_{rel} = \frac{ d|r| }{dt} \hat{r}$$

Where $ \hat{r} $ is a unit vector connecting the reference to the insect $|r|$ is the length of the whole vector connecting the reference to insect.

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In a previous post, I had it explained to me the relation about the time rate change of basis is related to the angular velocity by the equation $ \frac{d}{dt} \hat{u} = \omega \times \hat{u}$. However, I do not understand how that idea extends to this case as we are talking about the basis of velocity since $ \omega$ which was used initially was regarding the angular change of the position vectors.

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References:

Previous stack post

Lecture Series on Mechanics of Solids by Prof.M.S.Sivakumar, Department of Applied Mechanics, I.I.T.Madras.

Answer 1

Since $\hat{v}_{rel}$ is a unit vector $\dot{\hat{v}}_{rel} = \vec{\omega} \times \hat{v}_{rel}$. Multiplying by $|\vec{v}_{rel}|$ on both sides gives you that equation.

Answer 2

Since $\vec v_{rel}$ is a scalar multiple of $\vec r$ we have $\hat v_{rel} = \hat r$, so

$\displaystyle \frac {d \hat v_{rel}}{dt} = \frac {d \hat r}{dt} = \vec \omega \times \hat r \\ \displaystyle \Rightarrow |\vec v_{rel}| \frac {d \hat v_{rel}}{dt} = |\vec v_{rel}| ( \vec \omega \times \hat r ) = \vec \omega \times (|\vec v_{rel}| \hat r) = \vec \omega \times \vec v_{rel}$

Answer 3

If a rotating (constant) vector is decomposed into magnitude and direction $\vec{\rm vec} = v\, \hat{e} $ and the derivative of a unit vector is $\dot{\hat{e}} = \vec{\omega} \times \hat{e}$, then multiply both sides with $v$

$$ \frac{\rm d}{{\rm d}t} \vec{\rm vec} = v \frac{\rm d}{{\rm d}t} \hat{e} = v \left( \vec{\omega} \times \hat{e} \right) = \vec{\omega} \times \vec{\rm vec}$$