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What is the moment of inertia of a ring ...

What is the moment of inertia of a ring about a tangent to the circle of the ring?

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The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring.
The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem,

`I_("tangent")=I_(dia)+MR^(2)=(MR^(2))/(2)+MR^(2)=(3)/(2)MR^(2)`.
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(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR^(2)//5 , where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR^(2)//4 , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Define Moment of Inertia.

Knowledge Check

  • The moment of inertia of a solid sphere of mass M and radius R, about an axis through its centre, is (2)/(5)MR^(2) . The moment of inertia about an axis tangential to the surface of the sphere will be :

    A
    `(4)/(5)MR^(2)`
    B
    `(6)/(5)MR^(2)`
    C
    `(7)/(5)MR^(2)`
    D
    `MR^(2)`

  • The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameter is (1)/(4)MR^(2) . What is the moment of inertia of the disc about an axis passing through its centre and normal to the disc?

    A
    `MR^(2)`
    B
    `2MR^(2)`
    C
    `(3)/(2)MR^(2)`
    D
    `(1)/(2)MR^(2)`

  • If I_(1) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass, and I_(2) is the moment of inertia (about central axis) of the ring formed by bending the rod, then

    A
    `I_(1):I_(2)=1:1`
    B
    `I_(1):I_(2)=pi^(2):3`
    C
    `I_(1):I_(2)=pi:4`
    D
    `I_(1):I_(2)=3:5`

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