JEE MAIN - Physics (2018 (Offline) - No. 19)
Seven identical circular planar disks, each of mass M and radius R are
welded symmetrically as shown. The moment of inertia of the arrangement
about the axis normal to the plane and passing through the point P is :
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$${{181} \over 2}M{R^2}$$
$${{55} \over 2}M{R^2}$$
$${{19} \over 2}M{R^2}$$
$${{73} \over 2}M{R^2}$$
Explanation
Moment of inertia of any disc form its center perpendicular to the plane of disc = $${1 \over 2}M{R^2}$$
Moment of inertia of any one of the outer disc about an axis passing through point O and perpendicular to the plane
$${I_1} = {1 \over 2}M{R^2} + M{\left( {2R} \right)^2}$$ = $${9 \over 2}M{R^2}$$
Moment of inertia of the entire system about point O
$${I_o} = {1 \over 2}M{R^2} + 6{I_1}$$
= $${1 \over 2}M{R^2} + 6 \times {9 \over 2}M{R^2}$$
= $${{55} \over 2}M{R^2}$$
By symmetry, point $O$ is center of mass of the complete configuration. The point $\mathrm{P}$ is located at a distance $d_p=3 R$ from the center of mass $\mathrm{O}$. The parallel axis theorem gives moment of inertia of the configuration about a normal axis through $\mathrm{P}$ as
$${I_P} = {I_O} + 7M{\left( {3R} \right)^2}$$
= $${{55} \over 2}M{R^2} + 63M{R^2}$$
= $${{181} \over 2}M{R^2}$$
Moment of inertia of any one of the outer disc about an axis passing through point O and perpendicular to the plane
$${I_1} = {1 \over 2}M{R^2} + M{\left( {2R} \right)^2}$$ = $${9 \over 2}M{R^2}$$
Moment of inertia of the entire system about point O
$${I_o} = {1 \over 2}M{R^2} + 6{I_1}$$
= $${1 \over 2}M{R^2} + 6 \times {9 \over 2}M{R^2}$$
= $${{55} \over 2}M{R^2}$$
By symmetry, point $O$ is center of mass of the complete configuration. The point $\mathrm{P}$ is located at a distance $d_p=3 R$ from the center of mass $\mathrm{O}$. The parallel axis theorem gives moment of inertia of the configuration about a normal axis through $\mathrm{P}$ as
$${I_P} = {I_O} + 7M{\left( {3R} \right)^2}$$
= $${{55} \over 2}M{R^2} + 63M{R^2}$$
= $${{181} \over 2}M{R^2}$$
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