# Why use only centrifugal force in deriving formula of parabola formed by rotating liquid?

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?

• 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.
– Alv
Commented Sep 14, 2023 at 13:46
• @Alv please show the derivation from ground frame Commented Sep 14, 2023 at 13:50
• 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 $ma$. Commented Sep 14, 2023 at 14:04
• @Farcher , please share a link from where I can see the derivation. Commented Sep 14, 2023 at 14:08

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.