A cabin is rotating in a circular path of radius R with a uniform...

In the rotating frame of the cabin, the particle experiences a centrifugal force directed radially outward from the center of rotation O.

At point A, the distance from the center O is $R$. The centrifugal force acting on the particle of mass $m$ is $F_c = m\omega^2 R$, directed along the extended line OA.

Since the length of the groove $L \ll R$, the centrifugal force remains approximately constant in magnitude and direction as the particle moves from A to B.

The groove AB is oriented at an angle $\theta$ with respect to the extended line OA. The component of the centrifugal force along the groove is:

$F_{\parallel} = m\omega^2 R \cos\theta$

The Coriolis force acts perpendicular to the velocity of the particle (perpendicular to the groove) and is balanced by the normal reaction from the smooth walls of the groove. Thus, it does not affect the motion along the groove.

The acceleration of the particle along the groove is:

$a = \dfrac{F_{\parallel}}{m} = \omega^2 R \cos\theta$

The particle is released from rest, so its initial velocity is $u = 0$. Using the kinematic equation $s = ut + \dfrac{1}{2}at^2$ for the distance $L$:

$L = 0 + \dfrac{1}{2}(\omega^2 R \cos\theta)t^2$

Solving for the time $t$ required to reach point B:

$t^2 = \dfrac{2L}{\omega^2 R \cos\theta}$

$t = \sqrt{\dfrac{2L}{\omega^2 R \cos\theta}}$