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Here, we are provided with a spoolhaving moment of inertia I=kmr², where k is a numerical factor.

I am in the confusion that how should we judge in which direction static friction would be applied,

I asked my teachers they told me that to use practical sense and said that since F is providing an anticlocwise torque friction must provide an clockwise torque

but I think this is not a valid justication as, in few cases friction acts in the same circular direction as of force

for example:- force applied at sphere at a distance of less than 2/5r from centre of massa

Can anyone please help me with this problem?

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2 Answers 2

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Since the force is lower than the center, the relative motion between the point of contact and the ground will be in the same sense due to both kinds of motion (translational as well as rotational), so friction can only be towards left to oppose relative motion.

The more interesting case is if the force was applied above the center: the torque would be clockwise but the translational motion would be towards the right. This case has been clearly explained in this answer.

Edit: The summary of the linked answer is that for any body, the net velocity of the bottommost point is towards the left relative to the ground, so friction always acts forward. In case of a ring ($I = MR^2$), if the force is applied at a distance $R$ above the center, the relative velocity of that point is zero, so no friction is needed.

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In the absence of friction the spool will move to the right, and the point of contact will be moving to the right even more quickly than the centre of mass due to the rotation of the spool. Therefore the friction force on the spool will be towards the left.

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