Measuring position and velocity from a non inertial reference frame

In summary: An answer would be something of the form:##\vec{v_B} = (?,?,?)##Where the components depend on ##v_0## and...
  • #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
Like Tony Stark said:
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
PeroK said:
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
Like Tony Stark said:
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
PeroK said:
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
Like Tony Stark said:
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
PeroK said:
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
Like Tony Stark said:
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
PeroK said:
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
Like Tony Stark said:
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}##.
 

1. What is a non-inertial reference frame?

A non-inertial reference frame is a frame of reference that is accelerating or rotating, which causes the laws of motion to appear different than they would in an inertial reference frame. Examples of non-inertial reference frames include a car making a turn or a rotating planet.

2. How do you measure position and velocity from a non-inertial reference frame?

In order to measure position and velocity from a non-inertial reference frame, you must first determine the acceleration or rotation of the frame. This can be done through various methods such as using sensors or equations of motion. Once the acceleration or rotation is known, it can be accounted for in the measurements of position and velocity.

3. What are some common challenges when measuring position and velocity from a non-inertial reference frame?

One of the main challenges when measuring position and velocity from a non-inertial reference frame is accurately determining the acceleration or rotation of the frame. This can be difficult in situations where the frame is undergoing complex or changing movements. Another challenge is accounting for any external forces acting on the frame, such as friction or air resistance.

4. How does measuring from a non-inertial reference frame differ from an inertial reference frame?

In an inertial reference frame, the laws of motion are consistent and do not change. This means that measurements of position and velocity can be directly made without needing to account for any acceleration or rotation of the frame. In a non-inertial reference frame, these measurements must be adjusted to account for the changing motion of the frame.

5. What are some practical applications of measuring position and velocity from a non-inertial reference frame?

Measuring position and velocity from a non-inertial reference frame is important in a variety of fields, including aerospace engineering, navigation, and robotics. For example, in spacecraft navigation, it is necessary to account for the changing position and velocity of the spacecraft due to its rotation and orbit around a planet. In robotics, accurately measuring position and velocity from a non-inertial reference frame is crucial for precise movement and control of the robot.

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