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?

Question

Aicharya · Accepted Answer

m = &#961; &#960; R 2 l &#160; &#160; . . . ( 1 ) d m = &#961; &#960; 2 R . d R l + R 2 d l 0 = &#961; &#960; 2 R l d R + R 2 d l 2 R l &#160; d R = - R 2 d l &#160; &#160; . . . ( 2 ) I = M R 2 4 + M l 2 12 I = M 12 &#160; 3 R 2 + l 2 d I = M 12 6 R d R + 2 l &#160; d l = 0 &#160; &#160; . . . &#160; ( 3 ) Putting ( 1 ) , &#160; ( 2 ) in ( 3 ) 6 R - R 2 d l 2 R l + 2 l &#160; 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 = &#160; 3 2

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I .…

Solution