Question 128: Centre of Mass and Rotational Motion

Question

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.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m6qnsgoo/7758b425-b0e8-44d9-9b4b-31ea7c179344/1e45af80-e310-11ef-8161-69fc95b6cece/file-jaoe38c1m6qnsgop.png?format=png)

Accepted Answer

Mass of removed disc, $m = {M \over {\pi {R^2}}} 	imes \pi {\left( {{R \over 2}} ight)^2}$
$ \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}} ight)}^2}} \over 2} + {M \over 4}{{\left( {{R \over 2}} ight)}^2}} ight]$ (using parallel axis theorem)
$ \Rightarrow I = {{M{R^2}} \over 2} - \left( {{{M{R^2}} \over {32}} + {{M{R^2}} \over {16}}} ight)$
$ \Rightarrow I = M{R^2}\left( {{1 \over 2} - {1 \over {32}} - {1 \over {16}}} ight) = {{13} \over {32}}M{R^2}$
$ \Rightarrow I = {{13} \over {32}}M{R^2}$
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m7si91hr/312c0570-a7e6-4ad8-8435-ee259f196000/3d31e5f0-f7e0-11ef-8e83-4d7d1d05a0d3/file-jaoe38c1m7si91hs.png?format=png)

Answer

$\frac{17}{32} \mathrm{MR}^2$

Answer

$\frac{13}{32} \mathrm{MR}^2$

Answer

$\frac{9}{32} \mathrm{MR}^2$

Answer

$\frac{7}{32} \mathrm{MR}^2$

Centre of Mass and Rotational Motion

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