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State parallel -axes theorem of moment o...

State parallel -axes theorem of moment of inertia. The moment of inertia of a circular ring about its diameter is `(1)/(2) MR^(2)`. Calculate the moment of inertia of the ring about its tangent lying in its plane.

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Given, moment of inertia of this circular ring about its diameter AB, `I_(cm) = (1)/(2) MR^(2)` Fig. From the parallel-axes theorem the M.I. of the ring about the tangent CD, lying in its plane is,
`I =I_("CM")+MR^(2)`
`=(1)/(2)MR^(2)+MR^(2)=(3)/(2)MR^(2)`
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Knowledge Check

  • Moment of inertia of a circular wire of mass m and radius r about its diameter is

    A
    `(1)/(2)mr^(2)`
    B
    `(1)/(4) mr^(2)`
    C
    `mr^(2)`
    D
    `2 mr^(2)`

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