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8 January, 17:23

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

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  1. 8 January, 17:28
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    I = 2 MR²

    Explanation:

    Given that

    Radius of the hollow ring (hoop) = R

    The mass of the hoop = M

    We know that mass moment of inertia of a hoop about its center is given as

    Io = M R²

    By using theorem, mass moment of inertia at distance d from center is given as

    I = Io + m d²

    Here, M = m, d = R

    Now by putting the values in the above equation we get

    I = M R² + M R²

    I = 2 MR²

    Therefore the mass moment of inertia will be 2 M R².
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