Question 102: Centre of Mass and Rotational Motion

Question

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.

Accepted Answer

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 	imes 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 	imes L}{2} 	imes \left(\frac{L}{2\pi}ight)^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.

Answer

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

Answer

$ \frac{\lambda L^3}{16 \pi^2} $

Answer

$ \frac{\lambda L^3}{4 \pi^2} $

Answer

$ \frac{\lambda L^3}{12} $

Centre of Mass and Rotational Motion

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.

Choose the correct answer: