wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 ω0. 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 cofficient of friction between the ring and the finger is μ and the acceleration due to gravity is g


A
Mω20(Rr)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Mω20R2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
32Mo20(Rr)2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12Mc20(Rr)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B Mω20R2


k=12Mv2c+12Icω2

=12Mω20(Rr)2+12MR2ω20

=12Mω20R2(1rR)2+12Mω20R2

r<<R

rR0

k=12Mω20R2+12Mω20R2

=Mω20R2

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rubbing It In: The Basics of Friction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon