# Initial velocity of an object in the inertial frame and the rotating frame (non-inertial)

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.

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.

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

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.
• @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. Jul 10, 2021 at 7:46
• @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. Jul 10, 2021 at 13:54