From a uniform circular thin disc of mass 9 M | vedclass

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(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through

Vedclass · Accepted Answer

$4 MR^{2}$

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(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is $I_{1} = \frac{1}{2} M_{total} R^{2} = \frac{1}{2} (9M) R^{2} = \frac{9 MR^{2}}{2}$.Since the mass is proportional to the area,the mass of the removed disc $m$ is given by $m = M_{total} 	imes \frac{\pi (R/3)^{2}}{\pi R^{2}} = 9M 	imes \frac{1}{9} = M$.The moment of inertia of the removed disc about the same axis is calculated using the parallel axis theorem: $I^{\prime} = I_{cm} + m d^{2}$,where $I_{cm} = \frac{1}{2} m (R/3)^{2}$ and $d = \frac{2R}{3}$.$I^{\prime} = \frac{1}{2} M (R/3)^{2} + M (2R/3)^{2} = \frac{MR^{2}}{18} + \frac{4MR^{2}}{9} = \frac{MR^{2} + 8MR^{2}}{18} = \frac{9MR^{2}}{18} = \frac{MR^{2}}{2}$.The moment of inertia of the remaining disc is $I_{remaining} = I_{1} - I^{\prime} = \frac{9 MR^{2}}{2} - \frac{MR^{2}}{2} = \frac{8 MR^{2}}{2} = 4 MR^{2}$.

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