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Derive the expresssion for moment of inertia of a rod about its centre and perpendicular to the rod. |
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Answer» Solution :Let US consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an AXIS that passes through the CENTER of mass and perpendicular to the rod: First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dl) of this mass (dm) about the axis is `dI = (dm) x^2`  As the mass isuniformly distributed , , the mass per unit length`(lamda)` of the rod is `lamda = M/l` The (dm) massof the infinitesimally small lengthas , `dm= lamda dx = M/l dx` The moment of inertia (I) o the entire rod can be found by integrating dI `I= INT dI = int (dm) x^2 in ( M/l dx) x^2` ` I = M/l int x^2 dx` As the mass is distributed on either side of the origin the limits for integration are taken from to `- l//2 " to " l//2` ` I = m/l int_(-l//2)^(l//2) x^2 dx= M/l [(x^3)/(3)]^(l//2)` ` I = M/l [ (l^3)/(24) - (- (l^3)/(24) )] = M/l [ (l^3)/(24) + (l^3)/(24) ]` ` I = M/l [2( (l^3)/(24) ) ]` ` I = 1/12 ml^2` |
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