(a) The cylinder rotates about the point of contact. Hence, the mechanical energy of the cylinder will be conserved i.e.∴(PE+KE)a​=(PE+KE)b​∴mgR+0=mgRcosθ+21​Iω2+21​mv2 but ω=v/R (no slipping at point of contact)  and I=21​mR2​Therefore,mgR=mgRcosθ+21​(21​mR2)(R2v2​)+21​mv2or 43​v2=gR(1−cosθ)v2=34​gR(1−cosθ)Rv2​=34​g(1−cosθ)​At the time of leaving contact, normal reaction N=0 and θ=θc​ hence,mgcosθ=Rmv2​ or Rv2​=gcosθFrom Eqs. (i) and (ii),34​g(1−cosθc​)=gcosθc​ or 47​cosθc​=1 or cosθc​=4/7 or θc​=cos−1​(b) v=34​gR(1−cosθ)​At the time of losing contactcosθ∴v​=cosθc​=4/7=34​gR(1−74​)​ or v=74​gR​​Therefore, speed of CM of cylinder just before losing contact is 74​gR​(c) At the moment, when cylinder loses contactv=74​gR​Therefore, rotational kinetic energy, KR​=21​Iω2OrKR​=21​(21​mR2)R2v2​=41​mv2=41​m(74​gR)or KR​=7mgR​or KR​=7I0​​Now, once the cylinder loses its contact, N=0, i.e the frictional force, which is responsible for its rotation, also vanishes. Hence, its rotational kinetic energy now becomes constant, while its translational kinetic energy increases. Applying conservation of energy at (a) and (c). Decrease in gravitational PE= Gain in rotational KE+ translational KE∴ Translational KE(KT​)= Decrease in gravitational PE−KR​ or KT​=(mgR)−7mgR​=76​mgR∴KR​KT​​=7mgR​76​mgR​ or KR​KT​​=6​