If I launch an object in a rotating reference frame, say at an angle of 90 degrees (see image) with a velocity of 5 m/s, the initial velocity of this object in the inertial reference frame will be, of course, 3 m/s in the 90 degrees direction. In the non-inertial frame, however, the direction of this initial velocity will inherit the linear velocity omega times the radius of the rotating body. The magnitude of the initial velocity in the non-inertial frame will also change, but I don't know how to calculate that. Or in other words, how do you convert the magnitude of initial velocity from the inertial frame to the non-inertial frame. enter image description here

Right now, I am thinking to do the following calculation - the velocity in the inertial frame plus the cross product of the angular velocity and the position vector.

enter image description here

Any advice would be appreciated.

  • $\begingroup$ where is the 5 m/s ? . please add coordinate systems initial and rotating system $\endgroup$
    – Eli
    Jul 5, 2021 at 6:59
  • $\begingroup$ @Eli sorry there, I have changed the value. $\endgroup$
    – Steven Oh
    Jul 5, 2021 at 9:21

1 Answer 1


If the initial velocity of 5 m/s is relative to the rotating frame then you have to add the rotational velocity vector (with magnitude $\omega r$) to find the velocity vector relative to the non-rotating frame.

On the other hand, if the initial velocity of 5 m/s is relative to the non-rotating frame then you have to subtract the rotational velocity vector to find the velocity vector relative to the rotating frame.

  • $\begingroup$ hey there thanks for your answer. Just to clarify, when I want to, for instance subtract the rotational vector from the initial velocity, would I do velocity vector - (angular velocity just magnitude * position vector) or would I do velocity vector -(angular velocity in vector * position vector) $\endgroup$
    – Steven Oh
    Jul 10, 2021 at 4:35
  • 1
    $\begingroup$ @StevenOh The rotational velocity vector has magnitude $r \omega$ and is at right angles to the radius vector - it is a tangential vector. In vector notation it is $\vec {\omega} \times \vec r$, but note that the “times” symbol here is the cross-product of the vectors, not ordinary multiplication. $\endgroup$
    – gandalf61
    Jul 10, 2021 at 7:46
  • $\begingroup$ thank you so much for your reply. Here, would 𝜔⃗ be [0, 0, w], where w is the magnitude of my angular velocity? $\endgroup$
    – Steven Oh
    Jul 10, 2021 at 10:02
  • 1
    $\begingroup$ @StevenOh If the disk is rotating clockwise when viewed from above then $\vec \omega = [0,0,-\omega]$. See en.wikipedia.org/wiki/Right-hand_rule. $\endgroup$
    – gandalf61
    Jul 10, 2021 at 13:54

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