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Find the moment of inertia of ring about...

Find the moment of inertia of ring about an axis passing through the centre and perpendicular to its plane.

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A uniform circular ring of radius R and mass M rotating is fixed axis passing through its centre, perpendicular to its plane of rotation with constant angular velocity `omega`.
Every mass element of this thing is at a distance R from the centre. Hence it rotates with `Romega` linear speed.
Kinetic energy of ring
`K=(1)/(2)Mv^(2)`
`K=(1)/(2)MR^(2)omega^(2)`
Equating this, with equation `K=(1)/(2)Iomega^(2)`,
`therefore I=MR^(2)`
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Knowledge Check

  • There is a thin rod of uniform cross-section of mass M and length L. If this rodis is bent at 90^(@) from the mid-point, then the moment of inertia about the axis passing through mid-point and perpendicular to the plane which includes both parts of rod is .........

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    B
    `(ML^(2))/(12)`
    C
    `(ML^(2))/(6)`
    D
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  • The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I_(0) . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is .............

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  • Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is …………

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    D
    `(4)/(3)Ml^(2)`

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