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.

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)
$$ \Rightarrow I = {{M{R^2}} \over 2} - \left( {{{M{R^2}} \over {32}} + {{M{R^2}} \over {16}}} ight)$$
$$ \Rightarrow I = M{R^2}\left( {{1 \over 2} - {1 \over {32}} - {1 \over {16}}} ight) = {{13} \over {32}}M{R^2}$$
$$ \Rightarrow I = {{13} \over {32}}M{R^2}$$

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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