# The derivative of the unit velocity vector

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.

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.

References:

Previous stack post

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

• I want to only learn the derivation of coriolis force without the derivation of rotating vectors then go to this link drive.google.com/file/d/12D-Z-LrD2Itl8kiV8qvwez1aUZcLBG46/… Oct 16 '20 at 11:05
• Coincidently I also been studying the derivation of coriolis force since last week Oct 16 '20 at 11:05
• Shit sorry it requires velocity equals omega cross r so sorry Oct 16 '20 at 11:10

Since $$\vec v_{rel}$$ is a scalar multiple of $$\vec r$$ we have $$\hat v_{rel} = \hat r$$, so
$$\displaystyle \frac {d \hat v_{rel}}{dt} = \frac {d \hat r}{dt} = \vec \omega \times \hat r \\ \displaystyle \Rightarrow |\vec v_{rel}| \frac {d \hat v_{rel}}{dt} = |\vec v_{rel}| ( \vec \omega \times \hat r ) = \vec \omega \times (|\vec v_{rel}| \hat r) = \vec \omega \times \vec v_{rel}$$
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.
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 $$v$$
$$\frac{\rm d}{{\rm d}t} \vec{\rm vec} = v \frac{\rm d}{{\rm d}t} \hat{e} = v \left( \vec{\omega} \times \hat{e} \right) = \vec{\omega} \times \vec{\rm vec}$$