Question about Formulae for Motion in a Rotating Reference Frame

[Master1022]

Homework Statement

Particle P moves in a circular groove with radius $a$ which has been cut into a square plate with sides of length $l$. The plate rotates about its corner $O$ with with angular velocity $\omega \hat k$ and angular acceleration $\dot \omega \hat k$. Particle P rotates about the centre of the circular groove (which coincides with the centre of the square plate) with with angular velocity $\alpha \hat k$ and angular acceleration $\dot \alpha \hat k$

Relevant Equations

$\ddot r = \ddot R + \ddot \rho + \omega \times \omega \times \rho + 2 \omega \times \rho$

Hi,

I am reading the following question: "Particle P moves in a circular groove with radius $a$ which has been cut into a square plate with sides of length $l$. The plate rotates about its corner $O$ with with angular velocity $\omega \hat k$ and angular acceleration $\dot \omega \hat k$. Particle P rotates about the centre of the circular groove (which coincides with the centre of the square plate) with with angular velocity $\alpha \hat k$ and angular acceleration $\dot \alpha \hat k$"

I am just trying to understand what the terms in these equations for velocity and acceleration in a rotating reference frame mean. $\vec R$ is defined to be the vector from some origin $O$ to some point $o$, $\vec \omega$ is defined to be the angular velocity of $o$ about O, and $\rho$ is defined to to be a vector from $o$ to some point P. $\vec r$ is defined to point from O to P.

My attempt:
First to start with the velocity equation (generic formula, terms aren't related to problem)
$\dot r_p = \dot R + \dot \rho + (\omega \times \rho)$

This is what I think the terms are:
- $\dot R$: this will include the $\omega_{O} \times r_{O/A} = \omega \hat k \times \frac{l \sqrt 2}{2} \hat j = \frac{- \omega l \sqrt 2}{2} \hat i$ term
- $\dot \rho$: this will include the $\omega_{P/o} \times r_{P/o} = \alpha \hat k \times a \hat i = \alpha a \hat j$
- $\omega \times \rho$ term which will equal $\omega_{O} \times r_{P/o} = \omega \hat k \times a \hat i = \omega a \hat j$

I can understand what the first two terms are, but what is the third term? Originally I thought that $\dot \rho$ would correspond to changes in the radius (which is fixed in this case and thus would be 0) and the $\omega \times \rho$ would be what I have denoted $\dot \rho$ as.

Now to consider the acceleration equation (once again the symbols are just generic and bear no relation to variable names in the problem)
$\ddot r_p = \ddot R + \ddot \rho + (\omega \times \omega \times \rho) + 2 \omega \times \dot \rho + (\dot \omega \times \rho)$

- Coriolis term: $2 \omega \times \dot \rho = 2 \omega_{O} \times \dot \rho$ where $\dot \rho$ was found above
- The $(\omega \times \omega \times \rho)$ term. At first I thought this was centripetal of P relative to o, but the answer seems to denote it as a centripetal term relative to O: $\omega_{O} \times \omega_{O} \times a$ (have dropped the vector directions for simplicity).
- The $\ddot \rho$ term which includes the centripetal acceleration of P about o $\alpha ^ 2 a$ and the tangential acceleration of P about o $a \dot \alpha$.
- $\ddot R$: which will include the centripetal acceleration of o around O and the tangential term $\omega_{O} ^ 2 \frac{l \sqrt 2}{2}$
- $\dot \omega \times \rho = \dot \omega_{O} \times a$

Are those correct interpretations of those terms?

Any help would be greatly appreciated

 

[PeterDonis]

I am just trying to understand what the terms in these equations for velocity and acceleration in a rotating reference frame mean.

Is the problem even using a rotating frame? It looks to me like they are using a non-rotating frame, and the angular velocities and accelerations given are relative to that non-rotating frame.

 

[kuruman]

To me it looks like the particle is constrained to move in a circular groove cut into a square plate of side $l$. The square plate has angular acceleration about one of its corners. So in addition to the contact forces exerted by the groove, the particle is subjected to the inertial forces in the plate's rotating frame.

The particle's angular velocity $\alpha$ (poor choice of symbol if you ask me) and angular acceleration $\dot{\alpha}$ are given relative to the center of the groove which is at distance $\frac{l}{\sqrt{2}}$ from the axis of rotation of the plate.