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From a disc of radius R and mass M, a ci...

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

A

`(13 MR^(2))/(32)`

B

`(11 MR^(2))/(32)`

C

`(9 MR^(2))/(32)`

D

`(15MR^(2))/(32)`

Text Solution

Verified by Experts

The correct Answer is:
A
The mass of the removed part,
M. = `(M)/(pi R^(2)) xx pi ((R )/(2))^(2) = (M)/(4)`
The moment of inertia of the disc about the perpendicular axis passing through the centre,
` I = (MR^(2))/(2)`
The moment of inertia of the removed part about XX.
`I. = I_("cm") + M. d^(2)`
= `((M)/(4) ((R )/(2))^(2) )/(2) + (M)/(4) ((R )/(2))^(2)`
= `(MR^(2))/(32) + (MR^(2))/(16) = (3MR^(2))/(32)`
`therefore` The moment of inertia of the remaining part of the disc about XX. .
` I.. = I - I. = (MR^(2))/(2) -(3MR^(2))/(32) = (13 MR^(2))/(32)`
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Knowledge Check

  • The moment of inertia of a circular disc of mass m and radius r about an perpendicular axis passing through its centre is

    A
    `mr^(2)`
    B
    `(mr^(2))/(4)`
    C
    `(mr^(2))/(2)`
    D
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    A
    `4MR^2`
    B
    `40/9MR^2`
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    `10MR^2`
    D
    `37/9MR^2`

  • Moment of inertia of a circular ring of mass m and radius r about the normal axis passing through its centre is

    A
    `(mr^(2))/(4)`
    B
    `mr^(2)`
    C
    `(mr^(2))/(2)`
    D
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