Calculate Particle Velocity/Accel in Referential w/ Motion & Coriolis Problem

[fluidistic]

Problem:
Calculate the velocity and acceleration from an inertial reference frame of a particle whose motion functions (in Cartesian's coordinates) are known from a moving referential. The motion of such a referential is in accelerated translation and rotation with respect to the inertial one. Identify the corresponding expression of tangential, centripetal and Coriolis's accelerations.

Homework Equations

Galileo's transformations... hmm not sure. I don't think so since I'm dealing with a non inertial reference frame.

The Attempt at a Solution

I'd like some guidance. I'm thinking of starting writing the motion of the particle the referential sees but I have a big confusion when it comes to the rotational part (is it a spin and an orbital motion?).
My other idea is to start to write down a similar relation to Galilean's transformation.
The translation from one frame to another involves an acceleration. I call it a(t)=\ddot f(t), v(t)=\dot f(t) and r(t)=f(t).
But I've no clue about the rotational part. Also big troubles with the translational part. I'd like some guidance.
Thanks.

 

[nickjer]

Please read this first: http://en.wikipedia.org/wiki/Rotating_reference_frame

 

[fluidistic]

Please read this first: http://en.wikipedia.org/wiki/Rotating_reference_frame

Thanks, very relevant. I will take time to read it and use it. I'll come back here if I need to.

 

[fluidistic]

I know it's been more than 1 year since I've posted but here I am.
In Wikipedia's article and using what they call the Transport theorem (replacing their "f" by \vec r (t)), I could show that \vec v (t)=\vec v _r (t)+ \vec \Omega \times \vec r; just like they did.
However I've a problem when it comes to find the acceleration.
If I set their "f" to \vec v (t) then I find \vec a (t)=\vec a _r (t) + \vec \Omega \times \vec v _r (t)+ \vec \Omega \times \vec \Omega \times \vec r (t) instead of \vec a (t)=\vec a _r (t) +2 \vec \Omega \times \vec v _r (t)+ \vec \Omega \times (\vec \Omega \times \vec r )- \frac{d \vec \Omega }{dt} \times \vec r (t).
I've no idea what I'm doing wrong.
By the way this problem seems really tough :/

 

[nickjer]

Use the relation:
\frac{d}{dt}\hat{\boldsymbol{u}} = \boldsymbol{\Omega \times \hat{u}}

I will do the first term for the derivative of the velocity (use the product rule):
\frac{d}{dt}\boldsymbol{v_r} = \frac{d}{dt}v(t)\hat{\boldsymbol{r}}<br /> = \frac{dv(t)}{dt}\hat{\boldsymbol{r}} + v(t) \frac{d\hat{\boldsymbol{r}}}{dt}<br /> = a(t)\hat{\boldsymbol{r}} + v(t) \boldsymbol{\Omega \times \hat{r}}<br /> = \boldsymbol{a_r} + \boldsymbol{\Omega \times v_r}

Then you would do the same for the last term in the velocity. You'd take its derivative and use the product rule.
\frac{d}{dt} \boldsymbol{\Omega \times r}

 

[fluidistic]

Thank you very much, I could demonstrate the formula in wikipedia thanks to you.
I could identify Euler acceleration, Coriolis acceleration and the centrifugal acceleration.
However in the question, they ask for the "tangential acceleration", do they mean the Euler acceleration by this term?

 

[nickjer]

Yep.

 

[fluidistic]

Ok thanks for all. Problem solved. :)