A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity `omega` in a circular pathof radius R. A smooth groove AB of length `L(ltltR)` is made on the surface of the table. The groove makes an angle` theta` with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by tehparticle to reach the point B.

Explanation

Since the table rotates with the uniform angular velocity w. hence particle also rotates with the same angular velocity .
Since, centrifugal force acting on the particle is in a directiion paraller to `OA` , hence in component along the length of the grove` = m omega^(2) r cos theta `
If a be the acceleration of the particle along the groove, then
`a = omega^(2) r cos theta`

As the length of groove `ABL(lt lt R)` ,we can take the radius of rotation of the particle while moving on constant and equal to `r = R`
Hence the acceleration a of the particle along the groove will be constant then
`a = omega^(2) R cos theta`
Hence the time required by the particle to reach the end `B` of the groove, using
`L = ut +(1)/(2) at^(2)`, Here `u = 0, a = omega^(2) R cos theta`
`rAr t = sqrt((2L)/(omega^(2)R cos theta))`