Question 330: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

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')
= $${1 \over 2}$$ $$ 	imes $$ $${M \over {16}}$$ $$ 	imes $$ $${\left( {{R \over 4}} ight)^2}$$ = $${{M{R^2}} \over {512}}$$
Moment of inertia of removed disc about  O,
= I
cm
+ mx
2
= $${{M{R^2}} \over {512}}$$ + $${M \over {16}}{\left( {{{3R} \over 4}} ight)^2}$$
= $${{19M{R^2}} \over {512}}$$
Moment of inertia of complete disc about point 'O',
I = $${{M{R^2}} \over 2}$$
$$	herefore\,\,\,$$ Moment of Inertia of remaining disc
= $${{M{R^2}} \over 2}$$ $$-$$ $${{19M{R^2}} \over {512}}$$
= $${{237M{R^2}} \over {512}}$$

Answer

$${{219\,M{R^2}} \over {256}}$$

Answer

$${{237\,M{R^2}} \over {512}}$$

Answer

$${{19\,M{R^2}} \over {512}}$$

Answer

$${{197\,M{R^2}} \over {256}}$$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: