Four small objects, each of mass m, are fixed | vedclass

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(A) Let the rectangular frame have corners $A, B, C,$ and $D$ such that side $AD = BC = a$ and side $AB = CD = b$. The masses $m$ are placed at each c

Vedclass · Accepted Answer

$2 ma^{2}$

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(A) Let the rectangular frame have corners $A, B, C,$ and $D$ such that side $AD = BC = a$ and side $AB = CD = b$. The masses $m$ are placed at each corner.The axis of rotation passes along the side of length $b$ (let this be side $CD$).The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of the $i$-th mass from the axis of rotation.$1$. The masses at $C$ and $D$ lie on the axis of rotation, so their perpendicular distances are $r_C = 0$ and $r_D = 0$. Thus, their contribution to the moment of inertia is $m(0)^2 + m(0)^2 = 0$.$2$. The masses at $A$ and $B$ are at a perpendicular distance $a$ from the axis $CD$. Thus, their contribution is $m(a)^2 + m(a)^2 = 2ma^2$.Therefore, the total moment of inertia of the system about the axis $CD$ is $I = 0 + 2ma^2 = 2ma^2$.

Four small objects, each of mass $m$, are fixed at the corners of a rectangular wire-frame of negligible mass with sides $a$ and $b$ $(a > b)$. 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:

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