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 l/r such that the moment of inertia is minimum ? (1) 3 2

Answer 1

Hey mate,

● Answer - l/r = √(3/2)

● Explanation-
Moment of inertia of cylinder along perpendicular bisector is given by-
I = ml^2/12 + mr^2/4

Mass of the cylinder is given by-
mass = volume × density
m = πr^2ld
r^2 = m / πld

Substituting this value-
I = m^2/4πdl + ml^2/12

For I to be minimum,
dI/dl = 0
-m^2/4πdl^2 + ml/12 = 0
m^2/4πdl^2 = ml/12
l^3 = 3m / 2πd
l^3 = 3πr^2ld / 2πd
l^2 = 3/2 r^2
l^2/r^2 = 3/2
l/r = √(3/2)

For I to be minimum, l/r = √(3/2)

Hope this is useful...