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  • Need clarity, kindly explain! From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centreis cut. What is the moment of inertia of the remaining part of the disc, about a perpendicular axis, passing through t

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?

  • Option 1)

    15MR2/32

  • Option 2)

    13MR2/32

  • Option 3)

    11MR2/32

  • Option 4)

    9MR2/32

 

Answers (1)

best_answer

 

Moment of inertia for disc -

I=frac{1}{2} MR^{2}

 

=\frac{MR^{2}}{2}-[\frac{3}{32}MR^{2}]

I_{remain}=\frac{13}{32}MR^{2} 

- wherein

About an axis perpendicular to the plane of disc & passing through its centre .

 

 I=I_{remain}+I_\frac{R}{2}

I_{remain}=I-I_\frac{R}{2}

=\frac{MR^{2}}{2}- \left[ \frac{ \frac{M}{4} \left( \frac{R}{2} \right )^{2} }{2} + \frac{M}{4} \left( \frac{R}{2} \right )^{2} \right ]

=\frac{MR^{2}}{2}-[\frac{MR^{2}}{32}+\frac{MR^{2}}{16}]

= \frac{MR^2}{2}-\frac{3MR^2}{32}

I_{remaining}=\frac{13MR^2}{32}

 


Option 1)

15MR2/32

This is incorrect option

Option 2)

13MR2/32

This is correct option

Option 3)

11MR2/32

This is incorrect option

Option 4)

9MR2/32

This is incorrect option

Posted by

prateek

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