Question

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

Answer 1

moment of inertia of cylinder of radius R about an axis passing through perpendicular bisector of its length is given by,$i=\frac{ml^2}{12}+m\frac{R^2}{4}$

let density of cylinder is d

then, mass, m = πR²ld

R² = m/πld .......(1)

so, $i=\frac{ml^2}{12}+m\frac{m}{4\pi ld}$

differentiating i with respect to l,

$\frac{di}{dl}=\frac{ml}{6}-\frac{m^2}{4\pi l^2d}$

for Maxima and minima ,

di/dl = 0 => ml/6 - m²/4πl²d = 0

or, l/6 = m/4πl²d

or, 2/3 = πR²ld/4πl³d = R²/l² [ from equation (1) ]

l²/R² = 3/2 => l/R = √(3/2)

hence, answer is √{3/2}.