A horizontal disc of radius R rotates with a constant angular velocity `omega` about a stationary vertical axis passing through its edge. Along the circumference of the disc a particle of mass m moves with a velocity that is constant relative to the disc. At the moment when the particle is at the maximum distance from the rotation axis, the resultant of the inertial forces `F_(i n)` acting on the particle in the reference frame fixed to the disc turns into zero. Find:
(a) the acceleration `w^'` of the particle relative to the disc,
(b) the dependence of `F_(i n)` on the distance from the rotation axis.

Explanation

The disc OBAC is rotating with angular velocity `omega` about the axis `OO^'` passing through the edge point O. The equation of motion in rotating frame is,
`mvecw=vecF+momega^2vecR+2mvecvxxvecomega=vecF+vecF_(i n)`
where `vecF_(i n)` is the resultant inertial force (pseudo force) which is the vector sum of centrifugal and Coriolis forces.
(a) At A, `F_(i n)` vanishes. Thus `0=-2momega^2Rhatn+2mv^'omegahatn`
where `hatn` is the inward drawn unit vector to the centre from the point in question, here A.
Thus, `v^'=omegaR`
so, `w=(v^('^2))/(rho)=(v^('^2))/(R)=omega^2R`.
(b) At B `vecF_(i n)=momega^2OvecC+momega^2BvecC`
its magnitude is `m omega^2sqrt(4R^2-r^2)`, where `r=OB`.