Question 668: Centre of Mass and Rotational Motion

Question

From a circular disc of radius R and mass 9M, a small disc of mass M and radius ${R \over 3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Accepted Answer

Mass of the disc = 9M
Mass of removed portion of disc = M
The moment of inertia of
the complete disc about
an axis passing through
its centre O and
perpendicular to its plane is ${I_1} = {9 \over 2}M{R^2}$
Now, the moment of inertia of the disc with
removed portion
${I_2} = {1 \over 2}M{\left( {{R \over 3}} \right)^2} = {1 \over {18}}M{R^2}$
Therefore, moment of inertia of the remaining portion of disc about O is
$I = {I_1} - {I_2} = {9 \over 2}M{R^2} - {{M{R^2}} \over {18}} = {{40M{R^2}} \over 9}$

Answer

${{40} \over 9}$ MR
2

Answer

MR
2

Answer

4MR
2

Answer

${4 \over 9}$ MR
2

Centre of Mass and Rotational Motion

From a circular disc of radius R and mass 9M, a small disc of mass M and radius {R \over 3} is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

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