Question 392: Centre of Mass and Rotational Motion

Question

M and R be the mass and radius of a disc. A small disc of radius $\mathrm{R} / 3$ is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis $A B$ passing through the centre $O$ and perpendicular to the plane of disc is $\frac{4}{x} \mathrm{MR}^2$. The value of $x$ is ____________.

Accepted Answer

Moment of Inertia of the Full Disc
:
The moment of inertia (M.I.) of the entire disc without any cavity is given by:
$    I_1 = \frac{1}{2} MR^2    $
Mass of the Removed Disc
:
The mass of the removed disc, which is of radius $ \frac{R}{3} $, is calculated as:
$    	ext{Mass of removed disc} = \left(\frac{M}{\pi R^2}ight) 	imes \left(\frac{R}{3}ight)^2 \pi = \frac{M}{9}    $
Moment of Inertia of the Removed Disc
:
The moment of inertia of the removed disc involves two parts: about its center and due to its position.
About its center
:
$      \frac{\frac{M}{9} \left(\frac{R}{3}ight)^2}{2} = \frac{MR^2}{162}      $
Due to its position (distance from O to the new center of the small disc)
:
The distance is $ \frac{2R}{3} $, so:
$      \frac{M}{9} 	imes \left(\frac{2R}{3}ight)^2 = \frac{4MR^2}{81}      $
Total M.I. of removed disc
:
$      I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18}      $
Moment of Inertia of the Remaining Part
:
Subtract the moment of inertia of the removed disc from the full disc:
$    I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4MR^2}{9}    $
Thus, the value of $x$ is $9$.

Centre of Mass and Rotational Motion

Choose the correct answer: