The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I . What is the ratio l / R such that the moment of inertia is minimum?

Step-by-step solution

m = ρ π R 2 l     . . . ( 1 ) d m = ρ π 2 R . d R l + R 2 d l 0 = ρ π 2 R l d R + R 2 d l 2 R l   d R = - R 2 d l     . . . ( 2 ) I = M R 2 4 + M l 2 12 I = M 12   3 R 2 + l 2 d I = M 12 6 R d R + 2 l   d l = 0     . . .   ( 3 ) Putting ( 1 ) ,   ( 2 ) in ( 3 ) 6 R - R 2 d l 2 R l + 2 l   d l = 0 - 3 R 2 l + 2 l d l = 0 - 3 R 2 + 2 l 2 = 0 2 l 2 = 3 R 2 l R =   3 2