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WBJEEParagraph: One twirls a circular ring (of mass MM and radius RR )...
Paragraph:
One twirls a circular ring (of mass MM and radius RR ) 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 rr. The finger rotates with an angular velocity ω0ω0. 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 μμ and the acceleration due to gravity is gg.
Question:
The total kinetic energy of the ring is
One twirls a circular ring (of mass MM and radius RR ) 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 rr. The finger rotates with an angular velocity ω0ω0. 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 μμ and the acceleration due to gravity is gg.
Question:
The total kinetic energy of the ring is
Solution:
VCMVCM of Ring = (R-r) ω0= (R−r) ω0
For no slipping
Rω1-(R-r)w0=rω0Rω1−(R−r)w0=rω0
Rω1-Rω0=0Rω1−Rω0=0
KE of ring =12M(R-r)2 ω20+12mR2ω20=12M(R−r)2 ω20+12mR2ω20
Which is not matching with any answer given if r≪Rr≪R
But we take Rω0 ≈(R-r)ω0Rω0 ≈(R−r)ω0
Then KE ≈22M (R-r)2 ω20KE ≈22M (R−r)2 ω20
Then Mω20(R-r)2Mω20(R−r)2 is nearest possible option
For no slipping
Rω1-(R-r)w0=rω0Rω1−(R−r)w0=rω0
Rω1-Rω0=0Rω1−Rω0=0
KE of ring =12M(R-r)2 ω20+12mR2ω20=12M(R−r)2 ω20+12mR2ω20
Which is not matching with any answer given if r≪Rr≪R
But we take Rω0 ≈(R-r)ω0Rω0 ≈(R−r)ω0
Then KE ≈22M (R-r)2 ω20KE ≈22M (R−r)2 ω20
Then Mω20(R-r)2Mω20(R−r)2 is nearest possible option
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