Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through center 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

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 center 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 (A) 1 (B) (1/2) (C) (1/4) (D) (1/8)

Tardigrade Admin · Accepted Answer

Moment of inertia of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is

I = \frac{M L ^{2}}{12} an d I = M K_{1}^{2}

(where

K_{1}

radius of gravitation)

\Rightarrow K_{1} = \frac{L}{2 3}

..... (i)
When axis of rotation of rod is passing through one end of rod, then

I = M K_{2}^{2} = \frac{M L ^{2}}{3} \Rightarrow K_{2} = \frac{L}{3}

..... (ii)
Taking ratios of (i) and (ii) we get

\frac{K _{1}}{K _{2}} = \frac{L}{2 3} \times \frac{3}{L} = \frac{1}{2}

\Rightarrow \frac{K _{1}}{K _{2}} = \frac{1}{2}

$1$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

1

21​

41​

81​

$1$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$