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A rod is of mass M = 3kg and length 2a (...

A rod is of mass M = 3kg and length 2a (a = 2m). Find moment of inertia about an axis through the centre of the rod and perpendicular to the rod. (b) parallel to the rod and distant d = 2m from it?

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Let the rod be divided into elements of length et, each element being approximately a particle. (a) For a typical element, `mass=M/(2a)dx` moment of inertia about YY.`=(M/(2a)dx)x^2` Therefore `I_(yy)` , the moment of inertia of the rod about yr is given by `I_(yy)=(Ma)/(2a)int_(-a)^(a)x^2dx=1/3Ma^2=4kgm^2` (b) In this case every element of the rod is the same distanced from the axis XY. The moment of inertia of an element about XY `(M/(2a)dx)(d^2)` Therefore the moment of inertia of the rod about `I_(XY)=int_(0)^(2a)(M/(2a))dx(d^2)=Md^2=12kgm^2`.
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