Report Issue

JEE Advance 2017
Paper-2
Question
Paragraph - 2
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 (Figure2). The coefficient 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 -
Select the correct option:
A
$\sqrt{\frac{3 g}{2 \mu(R-r)}}$
B
$\sqrt{\frac{\mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$
C
$\sqrt{\frac{g}{2 \mu(R-r)}}$
D
$\sqrt{\frac{2 g}{\mu(R-r)}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Advance
Physics
Easy
Start Preparing for JEE with Competishun