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JEE Advanced · Physics · 8. Rotational Motion

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 \(g\).


Question :
The total kinetic energy of the ring is

  1. A Mω02R2
  2. B Mω02R-r2
  3. C 12Mω02R-r2
  4. D 32Mω02R-r2
Verified Solution

Answer & Solution

Correct Answer

(B) Mω02R-r2

Step-by-step Solution

Detailed explanation

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 rR
But we take Rω0 R-rω0
Then KE 22M R-r2  ω02
Then Mω02R-r2 is nearest possible option
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