Question

A wire of length l and mass m is bent in the form of rectangle ABCD with
AB/BC=2.The moment of inertia this wire frame about the side BC is=
​

Answer 1

first find mass and length of each part.

AB/BC = 2/1 so, AB = 2BC

perimeter = 2(AB + BC) = l

or, 2 × 3BC = l

or, BC = l/6 and AB = l/3

so, mass of AB = m/3 and mass of BC = m/6

here, ABCD is rectangular.

so, AB = CD = l/3 and BC = DA = l/6

we know, moment of inertia of rod about an axis passing through its end is 1/3 ML²

so, here M.I of AB about BC = 1/3 (m/3)(l/3)² = ml²/81

similarly, M.I of CD about BC = 1/3(m/3)(l/3)² = ml²/81

but here, DA is parallel to BC, so it can be assumed a particle placed r distance from the axis of rotation

so, M.I of DA about BC = (m/6)(l/3)² = ml²/54]

now, moment of inertia of frame about the side BC = moment of inertia of AB + moment of inertia of CD + moment of inertia of DA

= ml²/81 + ml²/81 + ml²/54

= (2 + 2 + 3)ml²/162

= 7ml²/162

hence, moment of inertia is 7ml²/162