JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 2)
A circular hole of radius $${R \over 4}$$ 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 :
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$${{219\,M{R^2}} \over {256}}$$
$${{237\,M{R^2}} \over {512}}$$
$${{19\,M{R^2}} \over {512}}$$
$${{197\,M{R^2}} \over {256}}$$
Explanation
Mass of removed disc = $${M \over {16}}$$ Radius of removed disc = $${R \over 4}$$
Moment of inertia of removed disc about it's own axis (O')
= $${1 \over 2}$$ $$ \times $$ $${M \over {16}}$$ $$ \times $$ $${\left( {{R \over 4}} \right)^2}$$ = $${{M{R^2}} \over {512}}$$
Moment of inertia of removed disc about O,
= Icm + mx2
= $${{M{R^2}} \over {512}}$$ + $${M \over {16}}{\left( {{{3R} \over 4}} \right)^2}$$
= $${{19M{R^2}} \over {512}}$$
Moment of inertia of complete disc about point 'O',
I = $${{M{R^2}} \over 2}$$
$$\therefore\,\,\,$$ Moment of Inertia of remaining disc
= $${{M{R^2}} \over 2}$$ $$-$$ $${{19M{R^2}} \over {512}}$$
= $${{237M{R^2}} \over {512}}$$
Moment of inertia of removed disc about it's own axis (O')
= $${1 \over 2}$$ $$ \times $$ $${M \over {16}}$$ $$ \times $$ $${\left( {{R \over 4}} \right)^2}$$ = $${{M{R^2}} \over {512}}$$
Moment of inertia of removed disc about O,
= Icm + mx2
= $${{M{R^2}} \over {512}}$$ + $${M \over {16}}{\left( {{{3R} \over 4}} \right)^2}$$
= $${{19M{R^2}} \over {512}}$$
Moment of inertia of complete disc about point 'O',
I = $${{M{R^2}} \over 2}$$
$$\therefore\,\,\,$$ Moment of Inertia of remaining disc
= $${{M{R^2}} \over 2}$$ $$-$$ $${{19M{R^2}} \over {512}}$$
= $${{237M{R^2}} \over {512}}$$
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