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Question

The frictional force on the ring (as shown in figure) is


A
F3 towards left
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B
2F3 towards left
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C
F3 towards right
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D
Zero
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Solution

The correct option is D Zero
Let frictional force (f), be acting opposite to applied force F.


Applying Fnet=maCOM
Ff=maCOM ...(1)
From equation of torque about centre of mass of the ring,
τnet, COM=ICOMα
FR+fR=mR2α ...(2)
[Both torques tend to produce rotation in clockwise sense, hence will add up.]

α=aCOM/R (in case of pure rolling)
(F+f)R=mR2×aCOMR
Or, F+f=maCOM ...(3)
From Eq (1) and (3), we get
f=0
Hence no frictional force will be acting on the ring.

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