# Coriolis Force in a Merry go Round

Suppose there is a person standing in a Merry go Round, which is rotating at a constant angular velocity $\vec \omega$. He experiences, of course, a centripetal acceleration $\vec a_{cen}$ and has some tangental velocity $\vec v_{tan}$.

This person now starts to walk in a circular path against the direction of rotation with a velocity $\vec v$ with respect to the Merry go Round. From an inertial frame of reference outside the Merry go Round, an observer should notice a Coriolis acceleration pointing radially outwards, given by $\vec a_{coriolis}=2 (\vec \omega \times \vec v)$.

If $2|\vec v|=|\vec v_{tan}|$ then the Coriolis acceleration would be equal and opposite to the centripetal acceleration and the observer in the inertial frame of reference would see this person standing still.

If $2|\vec v|>|\vec v_{tan}|$ then the Coriolis acceleration is greater than the centripel acceleration and the person would move in a spiral with an increasing radius against the direction of rotation of the Merry go Round, as seen from the inertial frame of reference.

Is this reasoning correct?

• Why would an observer, outside the merry-go-round, in an inertial frame of reference, need to invoke a fictitious force. Commented Apr 20, 2014 at 14:32
• @User58220 I thought an observer in an inertial frame of reference would introduce the Coriolis acceleration to explain what happens in another frame of reference that's moving and rotating. This is, if $O$ represents the non-inertial frame of reference and $O'$ the inertial frame of reference, a particle $P$ seen from $O'$ would have and acceleration $\vec a_{P|O'} = \vec a_{O|O'} + \vec a_{P|O} + \vec \alpha \times \vec{OP} + \vec \omega \times (\vec \omega \times \vec{OP}) + 2(\vec \omega \times \vec v_{P|O})$. Is this correct? Commented Apr 20, 2014 at 15:10

• So what you say is that the centripetal acceleration would be $\vec a = \frac{(\vec v_{tan}- \vec v)^2}{r}$ after he starts walking. But will the Coriolis acceleration be present/have any effect on the motion, as seen from the inertial frame of reference? Commented Apr 20, 2014 at 18:19
• Could this be because in the inertial frame of reference the walker would not be rotating, thus the term $2(\omega \times v)$ is ignored? Commented Apr 20, 2014 at 18:52