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Measuring position and velocity from a non inertial reference frame

  • #1

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?
 

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  • #2
PeroK
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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.
 
  • #3
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.
 

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  • #4
PeroK
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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?
 
  • #5
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?
 
  • #6
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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.
 
  • #7
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)
 
  • #8
PeroK
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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.
 
  • #9
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?
 
  • #10
PeroK
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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}##.
 

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