From a circular disc of radius R and mass 9M, | vedclass

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Question

(D) The moment of inertia of a circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its centre and perpendicular to its plan

Vedclass · Accepted Answer

$\frac{40}{9} M R^2$

Vedclass · Answer

(D) The moment of inertia of a circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M_{total} R^2$.For the original disc: $M_1 = 9M$ and $R_1 = R$. Thus,$I_1 = \frac{1}{2} (9M) R^2 = \frac{9}{2} M R^2$.For the removed disc: $M_2 = M$ and $R_2 = \frac{R}{3}$. Thus,$I_2 = \frac{1}{2} M (\frac{R}{3})^2 = \frac{1}{2} M (\frac{R^2}{9}) = \frac{1}{18} M R^2$.The moment of inertia of the remaining part is $I_{net} = I_1 - I_2$.$I_{net} = \frac{9}{2} M R^2 - \frac{1}{18} M R^2 = \frac{81 - 1}{18} M R^2 = \frac{80}{18} M R^2 = \frac{40}{9} M R^2$.

From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed concentrically. If the mass of the removed disc is $M$,what is the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre?

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