Question

using perpendicular axis theorem calculate moment of inertia of a circular ring of radius R about its diameter

Answer 1

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A/c to perpendicular axis theorem,

$Ix + Iy = Iz$

$Where, \: Ix \: and \: Iy \: are \: the \: axis \: along \: the\\ \: plane \: of \: the \: ring \: and \: Iz \: is \: the \: axis \\ \: perpendicular \: to \: the \: plane \: and \: passing\\ \: through \: the \: centre.$

$We \: know \: Iz \: is \: M {R}^{2} \\ \\ where \: M \: is \: the \: mass \: of \: the \: ring \: and \: R \: is \: \\the \: radius \:$

$For \: a \: ring \: Ix = Iy = Id \\ where \: Id \: is \: the \: axis \: passing \: through \: the \\\: diameter \: of \: the \: ring$

$So, \: Id + Id = Iz = M {R}^{2} \\ \\2Id = M {R}^{2} \\ \\ Id = \frac{M {R}^{2}}{2}$

$Therefore, \: moment \: of \: inertia \: of \: a \: circular \: \\ring \: of \: radius \: R \: about \: its \: diameter \: is \: \: \frac{M {R}^{2}}{2} \:$

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