Question

One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure $1.In$ the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$. The finger rotates with an angular velocity ${\omega }_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2 ). The cocficieicient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$

The minimum value of ${\omega }_0$ below which the ring will drop down is

• A. $\sqrt{\frac{2g}{\mu (R-r)}}$
• B. $\sqrt{\frac{3g}{2\mu (R-r)}}$
• C. $\sqrt{\frac{g}{\mu (R-r)}}$
• D. $\sqrt{\frac{g}{2\mu (R-r)}}$

Answer 1

The correct option is C $\sqrt{\frac{g}{\mu (R-r)}}$


$N={\omega }^2(R-r)$

$f=mg$

$f\le \mu N$

For minimum value of ${\omega }_0$

${\tau }_N={\tau }_{mg}$

$\mu {\omega }^2(R-r)R=mgR$

${\omega }_0=\sqrt{\frac{g}{\mu (R-r)}}$