Q.(a) Derive the expression for moment of inertia of a rod about its centre and perpendicular to the rod. OR
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Start your 14-day free trial to unlock the full solution →Integrating x^2 dm along the rod's length from -L/2 to L/2 gives the standard result I = ML^2/12.
This question offers a choice between (a) deriving the moment of inertia of a rod about its centre perpendicular to the rod, and (b) explaining the variation of g with depth; part (a) is answered here.
Consider a uniform thin rod of mass M and length L, lying along the x-axis, with its centre (and the axis of rotation, perpendicular to the rod) at the origin.
The mass per unit length (linear mass density) is
lambda = M/L
Consider a small element of the rod of length dx at a distance x from the centre. Its mass is
dm = lambda dx = (M/L) dx
The moment of inertia of this element about the perpendicular axis through the centre is
dI = x^2 dm = x^2 (M/L) dx
Since the rod extends from x = -L/2 to x = +L/2, the total moment of inertia is obtained by integrating:
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