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Derivation eqn motion particle rotating frame

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



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

    Secondly if thats just wrong the derivation does say to use (1.2.5) but then the 1st term on the RHS should be [itex]\dot{v}_I|_{R}[/itex] instead of [itex]\dot{v}_R|_{I}[/itex] 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]
     

    Attached Files:

  2. jcsd
  3. Jul 18, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, CMBR. Welcome to PF!

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

    You can derive the result this way, also. That is, you can use (1.2.5) at the beginning to write [tex]\dot{v}_I|_{I} = \dot{v}_I|_{R} + \omega\times v_I[/tex] 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: [tex]\dot{v}_I|_{I} = \left. \frac{d}{dt}(v_R + \omega\times r) \right|_{I} = \dot{v}_R|_{I} + \omega\times v_I[/tex]
    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.
     
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