Question

Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is ______.

Options

• 1

• `1/2`

• `1/4`

• `1/8`

Answer 1

Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is `underlinebb(1/2)`.

Explanation:

The equation for the moment of inertia of a thin rod whose axis is perpendicular to its length and passes through its middle is

I = `"ML"^2/12`                     ...(i)

In terms of radius of gyration, it can be written as

I = M`"K"_1^2`                  ...(ii)

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

`"MK"_1^2="ML"^2/12`

⇒ K₁ = `"L"/(2sqrt3)`          ...(iii)

When a rod's axis passes through one of its ends, its moment of inertia is given by

I = `"ML"^2/3`                   ...(iv)

In terms of radius of gyration, it can be written as

I = M`"K"_2^2`                 ...(v)

Comparing equation (iv) and (v), we get

`"ML"^2/3` = M`"K"_2^2`  

K₂ = `"L"/sqrt3`                ...(vi)

Again taking the ratio of K₁ and K₂ from Eqs. (iii) and (vi),

We get `"K"_1/"K"_2=("L"xxsqrt3)/(2sqrt3xx"L")=1/2`