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

The moment of inertiaof a uniform cylinder of length l and radius R ab

1.	The moment of inertiaof a uniform cylinder of length l and radius R about its perpendicular bisector is l. What is the ratio (I)/(R ) such that moment of inertia is minimum ?
Answer» `sqrt((3)/(2))` `(sqrt(3))/(2)` 1 `(3)/(sqrt(2))` Solution :` I = (Ml^(2))/(12) + (MR^(2))/(4)` But M =` pi R^(2) L rho` [ `rho` = density of the material of the cylinder] or, `R^(2) = (M)/(pi rho l)` `therefore I = (Ml^(2))/(12) + (M^(2))/(4 pi rho l) "" or, (dl )/(dl) = (2 M l )/(12) - ( M^(2))/(4 pi rho ) ((1)/(l^(2)))` For the minimum value of `I`. `(dl)/(dl) = 0 ` Hence , `(2Ml)/(12) - (M^(2))/(4 pi rho) ((1)/(l^(2))) = 0 "" or, (I)/(R) = sqrt((3)/(2))`