JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 3)
A uniform circular disc of radius ' $\mathrm{R}^{\prime}$ and mass ' $\mathrm{M}^{\prime}$ is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius $R / 2$ is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
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Explanation
_22nd_January_Morning_Shift_en_3_2.png)
Mass of removed disc, $$m = {M \over {\pi {R^2}}} \times \pi {\left( {{R \over 2}} \right)^2}$$
$$ \Rightarrow m = {M \over 4}$$
$$I = {I_1} - {I_2}$$
where, I$_1$ = MOI of whole disc about perpendicular axis through O
I$_2$ = MOI of removed dis about $$ \bot $$ axis through O
$$ \Rightarrow I = {{M{R^2}} \over 2} - \left[ {{{{M \over 4}{{\left( {{R \over 2}} \right)}^2}} \over 2} + {M \over 4}{{\left( {{R \over 2}} \right)}^2}} \right]$$ (using parallel axis theorem)
$$ \Rightarrow I = {{M{R^2}} \over 2} - \left( {{{M{R^2}} \over {32}} + {{M{R^2}} \over {16}}} \right)$$
$$ \Rightarrow I = M{R^2}\left( {{1 \over 2} - {1 \over {32}} - {1 \over {16}}} \right) = {{13} \over {32}}M{R^2}$$
$$ \Rightarrow I = {{13} \over {32}}M{R^2}$$
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