Derivation eqn motion particle rotating frame

[CMBR]

Homework Statement

So this isn't a homework problem but I don't know where else I am supposed to post for general help. I am basically trying to understand the derivation for the equation of motion of a particle in a rotating frame. See attachment for derivation and which steps I am stuck on.

Homework Equations

The Attempt at a Solution

From where the yellow dots are marked is where I get confused, since v_I=v_R+w\times r
why does: \dot{v}_I|_{I} = \dot{v}_R|_{I} + w\times v_I Theres 2 reasons this step is confusing me, if I am taking a time derivative of v_I=v_R+w\times r shouldn't I be using the product rule on the 2nd term in the RHS \dot{v}_I|_{I} = \dot{v}_R|_{I} + \frac{\mathrm{d} }{\mathrm{d} t}[w\times r]|_{I} ?

Secondly if that's just wrong the derivation does say to use (1.2.5) but then the 1st term on the RHS should be \dot{v}_I|_{R} instead of \dot{v}_R|_{I} no? But if that is the case then I don't see how to get to the step after that either... Really could use some help, thanks in advance.
[EDIT: I should specify that the I and R means inertial and rotational respectively]

 

[TSny]

Hello, CMBR. Welcome to PF!

From where the yellow dots are marked is where I get confused, since v_I=v_R+w\times r
why does: \dot{v}_I|_{I} = \dot{v}_R|_{I} + w\times v_I Theres 2 reasons this step is confusing me, if I am taking a time derivative of v_I=v_R+w\times r shouldn't I be using the product rule on the 2nd term in the RHS \dot{v}_I|_{I} = \dot{v}_R|_{I} + \frac{\mathrm{d} }{\mathrm{d} t}[w\times r]|_{I} ?

Yes, you would use the product rule. It must be that the text is assuming that the $\omega$ vector is time independent.

Secondly if that's just wrong the derivation does say to use (1.2.5) but then the 1st term on the RHS should be \dot{v}_I|_{R} instead of \dot{v}_R|_{I} no? But if that is the case then I don't see how to get to the step after that either... Really could use some help, thanks in advance.

You can derive the result this way, also. That is, you can use (1.2.5) at the beginning to write \dot{v}_I|_{I} = \dot{v}_I|_{R} + \omega\times v_I Then use (1.3.1) for the $v_{I}$'s on the right side.

The text approach is to first use (1.3.1) and then use the fact that the derivative of a sum is equal to the sum of the derivatives: \dot{v}_I|_{I} = \left. \frac{d}{dt}(v_R + \omega\times r) \right|_{I} = \dot{v}_R|_{I} + \omega\times v_I
Either approach will yield the same result.

Note $$\dot{v}_R|_{I} \equiv \left. \frac{d v_R}{dt} \right |_I$$ $$\dot{v}_I|_{R} \equiv \left. \frac{d v_I}{dt} \right |_R$$

etc.