Question

If I₁ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I₂ is the moment of inertia of the ring formed by bending the rod about an axis perpendicular to the plane, the ratio of I₁ and I₂ is ____________.

Options

• 1 : 1

• `pi^2 : 3`

• `pi : 4`

• 3 : 5

Answer 1

If I₁ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I₂ is the moment of inertia of the ring formed by bending the rod about an axis perpendicular to the plane, the ratio of I₁ and I₂ is `pi^2 : 3`.

Explanation:

M.I. of thin rod I₁ `approx` ML² / 12    ......(i)

M.I. of ring I₂ = MR²    ..... (ii)

The rod is bend to form a ring, `"L" = 2pi"R"`

From equation (i) and equation (ii) we get,

`"I"_2/"I"_2 = "ML"^2/12 xx 1/"MR"^2`

`= ("M"(2 pi"R")^2)/12 xx 1/"MR"^2`

`= (4"M"pi^2"R"^2)/(12  "MR"^2)`

` = pi^3/3`