A disc of mass m1 radius r (with head thread wrapped on its periphery) is released from rest in the figure shown. If the cylinder of radius R is on the verge of slipping as well as toppling then coefficient of friction between cylinder and table surface is (assuming pulley to be smooth and thread is not slipping on disc)
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answer is 1.
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Detailed Solution
m1g−T=m1a−−−1⇒τ=Iα⇒Tr=m1r22ar⇒T=m1a2−−−2 ∵I=m1r22 and α=arsolving 1 and 2 we get m1g−T=m1a⇒m1g−m1a2=m1a⇒a=2g3;T=m1g3Also, T=μm2g ⇒ μ=m13m2Balancing torque about A, we getTR=m2gRm1g3R=m2gR⇒m1=3m2 ⇒μ=m13m2=1