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?

  • 3
    $\begingroup$ Why would an observer, outside the merry-go-round, in an inertial frame of reference, need to invoke a fictitious force. $\endgroup$
    – DJohnM
    Apr 20, 2014 at 14:32
  • $\begingroup$ @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? $\endgroup$
    – Francisco
    Apr 20, 2014 at 15:10

1 Answer 1


WRT the second paragraph:

The inertial observer will see an object (the walker) moving in a fixed radius circle with a lower tangential (and thus lower angular) velocity, than was seen when the walker was not walking...

This in turn would reduce the centripetal force required to keep the walker moving in his slower, fixed radius circle, and this would be reflected in the lower radial inward friction force between the walker's shoes and the floor of the merry-go-round. The walker, who, while not walking, had been leaning towards the center of the circle, could walk more upright.

This leads to a novel way of commuting in a large, wheel-type space habitat. The commuter faces against the rotation of the wheel and takes enough running steps to cancel the wheel's tangential velocity. Then he simply lifts his feet and floats in space until his destination comes along,

  • $\begingroup$ 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? $\endgroup$
    – Francisco
    Apr 20, 2014 at 18:19
  • $\begingroup$ it will not be present in the inertial frame of reference... $\endgroup$
    – DJohnM
    Apr 20, 2014 at 18:47
  • $\begingroup$ 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? $\endgroup$
    – Francisco
    Apr 20, 2014 at 18:52

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