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From a disc of radius r(1), a concentric...

From a disc of radius `r_(1)`, a concentric disc of radius `r_(2)` is removed . The mass of the remaining portion is `m`. Find the `M.I.` of the remaining about an axis passing through the center of mass and perpendicular to the plane.

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To find the moment of inertia (M.I.) of the remaining portion of the disc about an axis passing through the center of mass and perpendicular to the plane, we can follow these steps: ### Step 1: Define the Masses Let: - \( m_A \) = mass of the original disc of radius \( r_1 \) - \( m_B \) = mass of the removed concentric disc of radius \( r_2 \) - \( m \) = mass of the remaining portion, which is given. ...
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Knowledge Check

  • The radius of gyration of a uniform disc of radius R, about an axis passing through a point R/2 away from the centre of disc, and perpendicular to the plane of disc is:

    A
    `sqrt(1/2)R`
    B
    `sqrt2R`
    C
    `sqrt3/2R`
    D
    `sqrt3/4R`

  • Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass and perpendicular to its plane is :

    A
    `(MR^(2))/(2)-M((4R)/(3pi))^(2)`
    B
    `(MR^(2))/(2)-M(sqrt(2)(4R)/(3pi))^(2)`
    C
    `(MR^(2))/(2)+M((4R)/(3pi))^(2)`
    D
    `(MR^(2))/(2)+M(sqrt(2)(4R)/(3pi))^(2)`

  • Out of a disc of mass M and radius R a concentric disc of mass m and radius r is removed. The M.I. of the remaining part about the symmetric axis will be :

    A
    `(M-m)(R+r)^(2)//2`
    B
    `(M-m)(R^(2)-r^(2))//2`
    C
    `(M-m)(R-r)^(2)//2`
    D
    `(M-m)(R^(2)+r^(2))//2`

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