Measuring position and velocity from a non inertial reference frame

[Like Tony Stark]

Homework Statement:

A particle moves with constant velocity $v_0$ in the $y$ direction with respect to an inertial system $A(x;y;z)$ as depicted in the picture. There is another system $B(x′;y′;z′)$, which is not inertial and rotates with constant angular velocity ω.
Determine
I) $r(t)$ from the perspective of $B$ and $A$
II) the velocity measured from $B$, i.e., the velocity relative to $B$
III) the acceleration measured from $B$, i.e., the acceleration relative to $B$

Relevant Equations:

Relative motion equations for non inertial reference frame

I) For $A$, the positition is $\vec r=(0;V_0 . t;0)$.
For $B$, we have $\vec r_A=\vec r_B + \vec r_{A/B}$, but $\vec r_{A/B}$ is equal to zero because they have the same origin, so the position measured from $A$ is equal to the position measured from $B$

II) For $A$, velocity is equal to $V_0$.
For $B$ we know that $\vec V_A = \vec V_B +\vec \omega \times \vec V_{rel}$. $\vec V_A$ is equal to $(0;V_0;0)$; $\vec V_B=0$, since the origin doesn't move; $\vec \omega$ is equal to $(0;0;\theta /t)$, $\vec r$ is the vector calculated previously and then I solve for $\vec V_{rel}$.

III) For acceleration we just have to replace the values in the formula and solve for $\vec a_{rel}$

Is this ok?

 

[PeroK]

Homework Statement: A particle moves with constant velocity $v_0$ in the $y$ direction with respect to an inertial system $A(x;y;z)$ as depicted in the picture. There is another system $B(x′;y′;z′)$, which is not inertial and rotates with constant angular velocity ω.
Determine
I) $r(t)$ from the perspective of $B$ and $A$
II) the velocity measured from $B$, i.e., the velocity relative to $B$
III) the acceleration measured from $B$, i.e., the acceleration relative to $B$
Homework Equations: Relative motion equations for non inertial reference frame

I) For $A$, the positition is $\vec r=(0;V_0 . t;0)$.
For $B$, we have $\vec r_A=\vec r_B + \vec r_{A/B}$, but $\vec r_{A/B}$ is equal to zero because they have the same origin, so the position measured from $A$ is equal to the position measured from $B$

How can that be? In any case, you have to calculate $r_B(t)$. Can you do that?

Hint: the axes of frames A and B are rotating wrt each other.

 

[Like Tony Stark]

How can that be? In any case, you have to calculate $r_B(t)$. Can you do that?

Hint: the axes of frames A and B are rotating wrt each other.

I can't see why it's incorrect. I mean, I also think that the positions must be different since they're measured from different systems, but if I go to the equations I find out that $\vec r_{A/B}$ is the vector that goes from the origin of the fixed system to the origin of the rotating system, and in the picture I see that both systems share the same origin, and it's always like that since $B$ rotates but its origin doesn't move.

 

[PeroK]

I can't see why it's incorrect. I mean, I also think that the positions must be different since they're measured from different systems, but if I go to the equations I find out that $\vec r_{A/B}$ is the vector that goes from the origin of the fixed system to the origin of the rotating system, and in the picture I see that both systems share the same origin, and it's always like that since $B$ rotates but its origin doesn't move.

If an object is at rest in frame A, is it at rest in frame B?

 

[Like Tony Stark]

If an object is at rest in frame A, is it at rest in frame B?

No, because of the angular velocity. I can notice that, from frame $B$, the particle moves so the position vector is changing. I understand that. But my doubt is about the equation: what's wrong with saying that the vector from one system to another is zero?

 

[PeroK]

No, because of the angular velocity. I can notice that, from frame $B$, the particle moves so the position vector is changing. I understand that. But my doubt is about the equation: what's wrong with saying that the vector from one system to another is zero?

That's only valid for translations. Rotations entail a different transformation between coordinates.

 

[Like Tony Stark]

That's only valid for translations. Rotations entail a different transformation between coordinates.

Ok. But apart from that, is the solution for velocity and acceleration ok? (Obviously, I have to change the values for position, but without considering that mistake)

 

[PeroK]

Ok. But apart from that, is the solution for velocity and acceleration ok? (Obviously, I have to change the values for position, but without considering that mistake)

I don't see a clear solution for those, but I'm not convinced that you understand the rotation transformation from what you've written.

 

[Like Tony Stark]

I don't see a clear solution for those, but I'm not convinced that you understand the rotation transformation from what you've written.

I've changed basis. I wrote the position of the particle in terms of the basis vectors of $B$. Now I have to replace this new values in the $\vec r$ part of the equation of velocity and acceleration. Right?

 

[PeroK]

I've changed basis. I wrote the position of the particle in terms of the basis vectors of $B$. Now I have to replace this new values in the $\vec r$ part of the equation of velocity and acceleration. Right?

An answer would be something of the form:

$\vec{v_B} = (?,?,?)$

Where the components depend on $v_0$ and $\vec{\omega}$.