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Question 73 of 87

Q.(a) Derive the expression for moment of inertia of a rod about its centre and perpendicular to the rod. OR

(b) Explain the variation of Acceleration due to gravity
(g) with depth from the earth's surface.
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★
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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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