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Math Help - [SOLVED] Moment of inertia?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Moment of inertia?

    A uniform disc has mass M and radius R. A circular hole of radius r is made in the disc with its centre at a distance x from the centre of the disc. Find the moment of inertia of the remaining portion of the disc about an axis passing through the centre of the hole and perpendicular to the plane of the disc.
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    A uniform disc has mass M and radius R. A circular hole of radius r is made in the disc with its centre at a distance x from the centre of the disc. Find the moment of inertia of the remaining portion of the disc about an axis passing through the centre of the hole and perpendicular to the plane of the disc.
    Parallel axis theorem plus thinking of hole as negative mass was the trick I used a loong time ago
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  3. #3
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    Just dotting the i's and crossing the t's

    Isomorphism is correct but just thought I'd add the explicit solution for absolute clarity and so this can get a SOLVED on it:

    Spoiler:


    Let the moment of inertia of about the centre of the large disc:
    the large (unpunctured) disc be I_A ;
    the small cut out disc be I_B;
    the punctured large disc be I_{AB}.

    Then we must have:

    I_{AB} + I_B = I_A

    where

    I_A = \frac{1}{2} M R^2

    and

    I_B = \frac{1}{2}m r^2 + m x^2 (Parallel axis theorem)

    so

    I_{AB} = I_A - I_B = \frac{1}{2}M R^2 - \frac{1}{2} m r^2 - m x^2

    but

    m = \frac{r^2}{R^2}M (same density and similarity of shape)

    so

    \color[rgb]{0,0,1}\boxed{I_{AB} = \frac{1}{2}\left(R^2 - \frac{r^4}{R^2} - \frac{2 r^2 x^2}{R^2} \right)M}

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