A spool with thread wound on it, of mass $m$ , rests on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to $I = γ m R^{2}$ , where $γ$ is a numerical factor and $R$ is the outside radius of the spool. The radius of the wound thread layer is equal to $r$ . The spool is pulled without sliding by the thread with a constant force $F$ directed at an angle $α$ to the horizontal as shown in the figure Find the projection of the acceleration vector of the spool axis on the $x - a x i s$ .

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$a_{x} = \frac{F {(\cos α - r / R)}^{2}}{m {(1 + γ)}^{2}}$

$a_{x} = \frac{F (\cos α - r / R)}{m (1 + γ)}$

$a_{x} = \frac{2 F (\cos α - r / R)}{m (1 + γ)}$

$a_{x} = \frac{2 F (\cos α - r / R)}{m (4 + γ)}$

answer is A.

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Detailed Solution

$I f a x b e t h e a c c e l e r a t i o n o f t h e s p o o l a l o n g x - a x i s, t h e n F \cos α - f = m a_{X} . . . . . (i) and f R - F r = I α . . . . . (ii) where α = \frac{a_{x}}{R} and I = γ m R^{2} A f t e r s i m p l i f y i n g a b o v e e q u a t i o n s, w e g e t$