An inextensible thread is wound round a cylinder on mass M, the cylinder is placed over a rough horizontal surface and the thread is passed over massless and frictionless pulley such that the part of the thread between cylinder and pulley is horizontal as shown in figure.When a block of mass m is attached with free end of the thread and system is released, it is observed that friction just prevents slipping of the cylinder over the horizontal surface. If Mm=4, then find minimum coefficient of friction between horizontal surface and cylinder.

As the cylinder does not slip over the horizontal surface, the friction will be static in nature. Let us assume it is acting in forward direction. The instantaneous axis of rotation will pass through the lowest point P. Let angular acceleration of the cylinder is α, (clockwise) then acceleration of its centre will be equal to Rα horizontally rightwards and that of block will be 2Rα vertically downward. We can apply torque equation about ICR. Writing torque equation about P,where IP is moment of inertial of the cylinder about PT.2R=3MR22.α⇒T=34MRα          …(i)Now considering F.B.DsForce equation of the block,mg−T=m(2Rα)                  …(ii)From Equations (i) and (ii)  Rα=mg2m+34M          …(iii)Force equation for cylinder, T+f=M.(Rα)                …(iv)From (i), (iii) and (iv) we get, f=Mmg(8m+3M)As friction is static nature, f≤μN⇒Mmg(8m+3M)≤μMgor μ≥m(8m+3M)=18+3Mm=1(8+3×4)=120=0.05