Rotating Frames & Centrifugal Force

On a spinning ride you feel flung outward, yet nothing outside you is pushing. That feeling has a name, and a precise place in physics.

A box on a turntable#

Put a heavy box on the rough floor of a closed cabin rotating steadily about a vertical axis with angular velocity ω. The box sits at distance $r$ from the axis and doesn’t slip.

From the ground (inertial): the box goes round a circle, so it needs $m\omega^2 r$ toward the axis. The forces on it are its weight, the floor’s normal force and friction; only friction is horizontal, so $f = m\omega^2 r$, pointing inward.

From inside the cabin: the box is at rest. Newton’s first law says the forces must cancel, but friction still points inward. To make the books balance we must add a pseudo force of size $m\omega^2 r$, pointing radially outward. Then $f = m\omega^2 r$ again: the same physics, described from a different seat.

Centrifugal force is enough for bodies at rest in the rotating frame. A body that moves within the rotating frame also needs the Coriolis force, which is perpendicular both to its velocity (in that frame) and to the rotation axis. It explains why large weather systems swirl, and why a ball rolled across a spinning platform appears to curve.

Earth is a rotating frame#

Earth spins once a day, $\omega = 2\pi/(24\ \text{h}) = 7.27\times10^{-5}$ rad/s. A point at colatitude θ (angle from the north pole) circles the axis on a radius $R\sin\theta$. Working from Earth’s frame, a mass hanging on a string feels gravity $mg$ toward Earth’s centre and centrifugal force $m\omega^2R\sin\theta$ away from the axis. Their resultant is the apparent weight $mg'$:

Since $2g > \omega^2R$, $g' < g$ everywhere except at the poles. The direction of $g'$ is the apparent vertical: plumb lines hang along it, walls are built along it, and still water lies perpendicular to it. At the equator the two forces are exactly opposite: