- #1
TOAsh2004
- 10
- 5
- TL;DR Summary
- How does the velocity transform under two successive rotations? Is the velocity relative to one rotating frame invariant under a second rotating around the same origin?
Good evening,
I was wondering about how velocities transform when two successive rotations are applied. In other words, how is the transformation law between two frames which are rotating relative to another.
Lets say some particle is moving with a velocity v in an inertial frame S. If we go from this inertial frame to one rotating with an angular frequency ω around the origin, the velocity v in the non rotating system is related to the velocity ##\vec v^{'}## in the rotating frame by ##S^{'}## by $$\vec v = \vec v^{'}+\vec ω \times \vec r^{'} \qquad (1)$$ Now, we move to another frame ##S^{''}## which is rotating relative to ##S^{'}## around the same origin with an angular velocity Ω. My question is now: How does the expression for v above look in this frame. I know that for the velocities one can derive the expression (1) by writing the position vector in each of the bases and then compare the total time derivatives. Consequently, the velocities should be related by the same equation as above, like $$\vec v^{'} = \vec v^{''}+\vec Ω \times \vec r^{''}$$ But I am not sure though. Is it right to just express the crossproduct in (1) in the new frame like $$ D \vec ω \times \vec r^{'}= \vec ω^{''} \times \vec r^{''} $$ with D being the rotation matrix between ##S^{'}## and ##S^{''}##? One would then arrive at $$\vec v = \vec v^{''}+\vec Ω_{tot} \times \vec r^{''} $$ with ##\vec Ω_{tot}=\vec Ω+\vec ω^{''}##
Further, I'd like to ask about the relative velocity of the particle to the rotating frame ##S^{'}##. Is this something like $$\vec v_{S,S^{'}} = \vec v-\vec ω \times \vec r^{'} \qquad (2)$$ And if we transform this quantity into the rotating frame ##S^{''}##, does it stay invariant? Like the analogue case of linear relative velocities that stay invariant under linear velocity transformations. How would the crossproduct term transform here? Thanks in advance for your answers.
I was wondering about how velocities transform when two successive rotations are applied. In other words, how is the transformation law between two frames which are rotating relative to another.
Lets say some particle is moving with a velocity v in an inertial frame S. If we go from this inertial frame to one rotating with an angular frequency ω around the origin, the velocity v in the non rotating system is related to the velocity ##\vec v^{'}## in the rotating frame by ##S^{'}## by $$\vec v = \vec v^{'}+\vec ω \times \vec r^{'} \qquad (1)$$ Now, we move to another frame ##S^{''}## which is rotating relative to ##S^{'}## around the same origin with an angular velocity Ω. My question is now: How does the expression for v above look in this frame. I know that for the velocities one can derive the expression (1) by writing the position vector in each of the bases and then compare the total time derivatives. Consequently, the velocities should be related by the same equation as above, like $$\vec v^{'} = \vec v^{''}+\vec Ω \times \vec r^{''}$$ But I am not sure though. Is it right to just express the crossproduct in (1) in the new frame like $$ D \vec ω \times \vec r^{'}= \vec ω^{''} \times \vec r^{''} $$ with D being the rotation matrix between ##S^{'}## and ##S^{''}##? One would then arrive at $$\vec v = \vec v^{''}+\vec Ω_{tot} \times \vec r^{''} $$ with ##\vec Ω_{tot}=\vec Ω+\vec ω^{''}##
Further, I'd like to ask about the relative velocity of the particle to the rotating frame ##S^{'}##. Is this something like $$\vec v_{S,S^{'}} = \vec v-\vec ω \times \vec r^{'} \qquad (2)$$ And if we transform this quantity into the rotating frame ##S^{''}##, does it stay invariant? Like the analogue case of linear relative velocities that stay invariant under linear velocity transformations. How would the crossproduct term transform here? Thanks in advance for your answers.