Q53 / 54 · JEE Advanced 2017 (Paper 2) · Physics

Paragraph: 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 coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$. Figure
Figure

Question: The total kinetic energy of the ring is

  • A
    Mω02R2
  • Correct answer
    B
    Mω02R-r2
  • C
    12Mω02R-r2
  • D
    32Mω02R-r2

Solution

VCM of Ring = (R-r) ω0
For no slipping
Rω1-R-rw0=rω0
Rω1-Rω0=0
KE of ring =12MR-r2 ω02+12mR2ω02
Which is not matching with any answer given if r≪R
But we take Rω0 ≈R-rω0
Then KE ≈22M R-r2  ω02
Then Mω02R-r2 is nearest possible option

Hence, the correct answer is option: 'B'.

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