Question

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

Answer 1

Solution :

I=ml212+mR24I=ml212+mR24

or I=m4(l23+R2)I=m4(l23+R2) -------------(1)

Also m=πR2lpm=πR2lp

=> R2=mπlpR2=mπlp Put in equation (1)

For maxima & minima

dIdl=m4(2l3−mπl2ρ)=0dIdl=m4(2l3−mπl2ρ)=0

=> 2l3=mπl2ρ=>2l3=πR2lρπl2ρ2l3=mπl2ρ=>2l3=πR2lρπl2ρ

or 2l3=R2l2l3=R2l

=> l2R2=32l2R2=32

or l2R=32−−√