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A thin inelastic string is tightly wound on a cylinder. One of the ends of the string is attached to the above surface. Initially the whole system (string +cylinder) is held near the surface. When it is released from that position, find the acceleration of the centre of mass of the cylinder.

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When we are calculating the sum of the forces on the body we are not taking into account the frictional force between the cylinder and the thread. Why is it so?

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  • $\begingroup$ Short answer: because the problem does not specify the friction coefficient. Never make your life more complicated than it is. Long answer: for macroscopic objects in this problem neglecting the friction coefficient is a good approximation. It is very rare when it is significant in real life situations of this sort. $\endgroup$ – MsTais Dec 11 '18 at 21:50
  • $\begingroup$ Short answer #2: because friction is usually complicated, so, if neglecting it doesn't make things too wrong, then we do it. $\endgroup$ – FGSUZ Dec 11 '18 at 22:02
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The system that you are considering is the cylinder and the (massless?) string which is wound around the cylinder ie in contact with the cylinder.

The $T$ in your diagram is the external force exerted on the cylinder by the string and it is the tension in the string at the point where it just not touching the cylinder.

Below that point at each point of contact between the cylinder and the string there will be a frictional force on the cylinder due to the string and a Newton third law pair of a frictional force on the string due to the cylinder and these forces prevent slipping between the string and the cylinder. The frictional forces are internal forces and thus do not need to be considered in your calculation.

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Because there is no friction. No part of the string is rubbing against the cilinder at any time. All it does is stay tight to the cilinder until gradually coming loose.
You may have string rubbing against the string, you may have friction between cilinder and air, but not between cilinder and string.

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