"A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is."

Question

Accepted Answer

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First, relate the length of the rod to the circumference of the ring:

$ L = 2\pi R $

The mass of the ring, denoted by $ M $, is the product of the linear mass density and the length of the rod:

$ M = \lambda \times L $

The moment of inertia of a ring about a diameter is given by the formula:

$ I = \frac{MR^2}{2} $

Substituting the expression for mass $ M $ and using the relation for $ R $ derived from the circumference:

$ I = \frac{\lambda \times L}{2} \times \left(\frac{L}{2\pi}\right)^2 $

Simplifying the expression, we arrive at:

$ I = \frac{\lambda L^3}{8\pi^2} $

This is the moment of inertia of the ring about any of its diameters.

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A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.