When a fluid-filled container containing an ideal liquid is rotated about its central axis with angular velocity $\omega$, the surface of the liquid forms a paraboloid shape. Now when we derive the equation of the parabola formed by taking a cross section through the midpoint of the paraboloid, using the standard technique used for deriving this formula: $$y=\frac{\omega^2·x^2}{2g},$$ assuming the vertex of the parabola to be the origin, we assume a centrifugal force $$F_\text{c}= \mathrm{d}m · (\omega)^2 · x$$ to be acting on the liquid particles situated along the parabola. My teachers said we assume a centrifugal force because we are viewing from the frame of the container.

But, then why cannot the same formula be derived from the ground frame, by taking a centripetal force instead of a centrifugal force?

  • $\begingroup$ It can derived from any frame as long as you are taking everything into account. In ground frame, centripetal force required for the rotational has the same magnitude as that of centrifugal in container's frame. $\endgroup$
    – Alv
    Commented Sep 14, 2023 at 13:46
  • $\begingroup$ @Alv please show the derivation from ground frame $\endgroup$ Commented Sep 14, 2023 at 13:50
  • $\begingroup$ The derivation is essentially the same in that in the rotating frame there is a centrifugal force term $-ma$ on the force side of N2L with the other side being equal to zero. In the inertial frame the $-ma$ term is moved to the other side and becomes $ma$. $\endgroup$
    – Farcher
    Commented Sep 14, 2023 at 14:04
  • $\begingroup$ @Farcher , please share a link from where I can see the derivation. $\endgroup$ Commented Sep 14, 2023 at 14:08

1 Answer 1


Yes, of cause you can do it from the ground frame. You need a centripetal force to keep the fluid particles on a circle. So on the particle you have the weight $mg$ and the force $m·\omega^2·r$ perpendicular to the slope; together they give the centripetal force.

  • $\begingroup$ can you please share a link, where I can see the derivation done in that manner, I am unable to get the same result on my own. $\endgroup$ Commented Sep 14, 2023 at 15:04
  • $\begingroup$ I can't see almost no difference in the two calculations, so maybe you post what you have calculated to get the parabola with centrifugal force. $\endgroup$
    – trula
    Commented Sep 14, 2023 at 15:27
  • $\begingroup$ since the direction of the centripetal force is opposite the direction of the centrifugal force, I am getting the formula upside down, relative to the formula derived by considering a centrifugal force. $\endgroup$ Commented Sep 14, 2023 at 15:43
  • $\begingroup$ I don't understand your "upside down" did you sketch the two acting forces mg down and the force of the surface perpendicular to it? $\endgroup$
    – trula
    Commented Sep 14, 2023 at 20:32
  • $\begingroup$ by upside down I mean the equation is coming as y=2g/(omega)^2.x^2 $\endgroup$ Commented Sep 15, 2023 at 7:29

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