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A man, standing on a rotating disc with angular frequency w starting from the center, moves radially outwards with a velocity of 6 meters per second. If an inertial frame of reference is placed at the center of the disc is it correct to say that the person has a velocity of 6m/s in direction of i' or that the displacement is 6m/s*time in direction of i' . Where i' is a unit vector pointing radially outwards in the rotating frame of reference and is a function of time in the inertial frame of reference.

Both descriptions seem the same to me but if we use the second description and the fact that velocity is derivative of the displacement wrt time, there is an other component for the velocity in direction of j', the unit vector perpendicular to i'. I can't get why that exists.

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You are mixing up "inertial frame" and "rotating frame" in my opinion (based on a reading of your statements). If you fix the i' axis to the rotating platform it is no longer inertial! You are now in an accelerated frame. It should be fairly obvious that in the non-moving frame of the earth (there are a lot of assumptions in that last statement) the person is moving along a type of spiral so obtaining a j component should not be a surprise. The problem is that the definitions of velocity and acceleration are modified in the non-inertial frame. The modification can be described by an additional term in the time derivative.

The operator d[]/dt needs to be changed to d[]/dt + omega (X) []

where omega is the vector form of rotation, (X) the cross product and [] means insert vector here. In the two term derivative d[]/dt is the rate in the rotating frame. This is sometimes referred to as fictitious force.

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  • $\begingroup$ actually there are two frames of reference one is rotating and the other is fixed i' and j' are in the rotating and i and j are fixed $\endgroup$ – M.Hamza Ali May 25 '18 at 21:39
  • $\begingroup$ Well, which one are you in when you take the derivative? Use the correct differentiation formula and it should work. $\endgroup$ – ggcg May 25 '18 at 21:57
  • $\begingroup$ how is the definition of velocity and acceleration different in non inertial frames of reference? $\endgroup$ – M.Hamza Ali May 29 '18 at 16:37
  • $\begingroup$ That is described above. These quantities are defined relative to an observer in each frame. You can find reference and description in Goldstien, and other texts. $\endgroup$ – ggcg May 29 '18 at 19:51
  • $\begingroup$ @M.HamzaAli Don think that the derivative should be the same because x and t are scalars. x is a vector and its component mix under rotation. One of the derivative you are taking is in a rotation frame the omega cross term basically takes the derivative of that time dependent rotation, from one frame to the other. $\endgroup$ – ggcg May 30 '18 at 13:26

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