One hollow and one solid cylinder of the same radius are rolling down an inclined plane. Which of them will reach the end of the plane first?

Explanation

mgcos∅ will get balanced by the normal force For rolling without slipping (pure rolling), we have a = àR ... (1) I am using symbol à for angular acceleration Two forces are acting, f and mgsin∅ submission of F =ma mg sin∅ -f= ma ... (2) (Here f is the friction force, a = linear acceleration à = angular acceleration) As we know, T= là (T=torque, I=inertia) from eqn(1)....à=a/R T=I(a/R) We know torque equals force into the radius But we also know the force of friction and radius are also perpendicular so T = f×R f×R = I ( a/R ) f =I ( a/R² ).......(3) From eqn (2) and (3), mgsin∅ - I ( a/R² ) =ma a = (mgsin∅) / (1/R²+m) a = (mgR²sin∅) / I+mR² m, R and ∅ are constant So, a is proportional to 1/I (I=inertia) The body whose inertia will less reach the bottom first We know, Inertia of solid cylinder = ½ mR² Inertia of hollow cylinder= ½ m ( R2² +R1²) I (solid) < I (hollow) So solid cylinder will have greater acceleration So, it will reach the end first