Moment of inertia of a uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius R is

Step-by-step solution

Moment of inertial of solid cylinder, about the axis passing through its centre of mass and perpendicular to its plane I_ X Y =M ( l^2 12 + R^2 4 ) From theorem of parallel axis, moment of inertial of cylinder ahout an axis passing through its edge and perpendicular to it plane. aligned I_ A B & =I_ X Y +M ( l 2 )^2 I_ A B & =M ( l^2 12 + R^2 4 )+M l^2 4 & =M ( l^2 3 + R^2 4 ) aligned But l=6 R aligned I_ A B & =M [ (6 R)^2 3 + R^2 4 ] & =M [12 R^2+ R^2 4 ] & =M ( 49 R^2 4 )= 49 M R^2 4 aligned