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A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Option 1)
$\frac{219 MR^{2}}{256}$

Option 2)
$\frac{237MR^{2}}{512}$

Option 3)
$\frac{19MR^{2}}{512}$

Option 4)
$\frac{197MR^{2}}{256}$

Answers (1)

best_answer

As we learnt in

Paraller Axis Theorem -

$I_{b\: b'}=I_{a\: a'}+mR^{2}$

- wherein

$b\: b'$ is axis parallel to $a\: a'$ & $a\: a'$ an axis passing through centre of mass.

Moment of inertia of complete disc about O point.

$\sigma_{1}=\frac{m}{A}=\frac{M}{\pi R^{2}}$

$\sigma_{2}=\frac{m}{A'}=\frac{M'}{\frac{\pi R^{2}}{16}}$

$\sigma_{2}=\frac{16M'}{\pi R^{2}}$

$\sigma_{1}=\sigma_{2}=\frac{M}{\pi R^{2}}=\frac{16M'}{\pi R^{2}}$

$M'=\frac{M}{16}$

M.O.I. of removed disc about its axis

$I=\frac{1}{2}\frac{M}{16}\times\left(\frac{R}{4} \right )^{2}=\frac{MR^{2}}{512}$

M.O.I. of removed disc about O

$I_{removed}=I_{cm}+mx^{2}$

Where $x=\frac{3R}{4}, I_{removed}=\frac{MR^{2}}{16}+\frac{M}{16}\left(\frac{3R}{4} \right )^{2}=\frac{19MR^{2}}{512}$

$\therefore\ \; I_{rem}=\frac{1}{2}MR^{2}-\frac{19MR^{2}}{512}=\frac{237MR^{2}}{512}$

Option 1)

$\frac{219 MR^{2}}{256}$

This is an incorrect option.

Option 2)

$\frac{237MR^{2}}{512}$

This is the correct option.

Option 3)

$\frac{19MR^{2}}{512}$

This is an incorrect option.

Option 4)

$\frac{197MR^{2}}{256}$

This is an incorrect option.

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