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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

A)1

B)1/2

C)1/4

D)1/8
from Mht-cet 2017
in Physics
reopened | 89 views

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Detailed solution:

Moment of inertia of the rod about an axis passing through its centre and perpendicular to the length $$I_1={1\over12}ML^2$$

using $$Mk_1^2=I_1={1\over12}ML^2$$

we get radius of gyration  $$K_1={L\over\sqrt{12}}$$

Moment of inertia of the rod about an axis passing through its one end and perpendicular to the length  $$I_2={1\over3}ML^2$$

Using $$Mk_2^2=I_2={1\over3}ML^2$$

we get radius of gyration $$K_2={L\over\sqrt3}$$

Thus , the radius of gyaration $${k_1\over k_2}={L/\sqrt{12} \over L/\sqrt3}$$

⟹ $${k_1\over k_2}={1\over 2}$$

by Expert (11.5k points)