QJEEAdv17Ph2eaa: JEE Advanced 2017 Physics - Class 11 JEEAdv17Ph2eaa Solution: The minimum value of 0 below which the ring will drop down is

Let the ring makes an angle $\alpha$ with the horizontal and the finger makes an angle $\beta$ with the vertical (see figure). The forces acting on the ring are its weight mg at the centre C, normal reaction N and the frictional force f at the contact point P. Resolve the forces in the horizontal and the vertical directions. The centre of mass C rotates with an angular velocity $\omega$ ₀ in a horizontal circle of radius (R $-$ r) cos $\alpha$ centred at O. Apply Newton's second law in the horizontal and the vertical directions to get

$f\cos \beta - N\sin \beta - Mg = 0$ ..... (1)

$N\cos \beta + f\sin \beta = M\omega _0^2(R - r)\cos \alpha$ .... (2)


The minimum value of $\omega$ ₀ occurs when the frictional force attains its limiting value i.e.,

$f = \mu N$ ....... (3)

Eliminate f and N from equations (1) - (3) to get

${\omega _{0,\,\min }} = {\left[ {{{(\cos \beta + \mu \sin \beta )g} \over {(\mu \cos \beta - \sin \beta )(R - r)\cos \alpha }}} \right]^{1/2}}$

$\approx {\left[ {{g \over {\mu (R - r)}}} \right]^{1/2}}$ ( $\because$ $\alpha$ = 0 and $\beta$ = 0).