Solveeit Logo
Login

Question

Physics Question on System of Particles & Rotational Motion

Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides aa and b(a>b)b (a > b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is

A

2 ma22\,ma^2

B

4 ma24\,ma^2

C

2 m(a2+b2)2\,m(a^2 + b^2)

D

2 m(a2−b2)2\,m(a^2 - b^2)

Answer

2 ma22\,ma^2

Explanation

Solution

The given situation can be shown as

Moment of inertia of the system about side of length bb say CDC D is
== M.I. of mass at AA about CD+M.I.C D+M . I . of mass at BB about CD+M.ICD + M . I. of mass at CC about CDCD ++ M.I. of mass ot DD about CDC D
=m(a)2+m(a)2+m(0)2+m(0)2=m(a)^{2}+m(a)^{2}+m(0)^{2}+m(0)^{2}
=2 ma2=2\, m a^{2}