A spool of wire of M and radius 2R is unwound under a constant force F. Assuming the spool is a uniform, solid cylinder that does not slip.

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The force of friction is to the left and equal in magnitude to $\frac{F}{3}$

The force of friction is to the right and equal in magnitude to $\frac{4 F}{3}$

The force of friction is to the right equal in magnitude to $\frac{F}{3}$

The acceleration of the centre of mass is $\frac{4 F}{3 M}$

answer is A, B.

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Detailed Solution

$a_{c m} = 2 R α$

From Newton’s second law,

$F + f = M a_{c m}$ …………………….(1)

The net torque about centre,

$F (2 R) - f (2 R) = I α = \frac{M {(2 R)}^{2}}{2} α$

$F - f = \frac{M (2 R)}{2} α = \frac{M}{2} a_{c m}$ …………………….(2)

$(1) + (2),$ $2 F = \frac{3 M}{2} a_{C M} \Rightarrow F = \frac{3 M a_{c m}}{4}$

$a_{C M} = \frac{4 F}{3 M}$

From (2), $f = F - \frac{M}{2} a_{C M}$

$= F - \frac{M}{2} (\frac{4 F}{3 M}) = \frac{F}{3}$

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