Question

A uniform disc of radius $R$ has a round disc of radius $R/3$ cut as shown in fig. The mass of the remaining (shaded) portion of the disc equals $M$. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Answer 1

Given: A uniform disc of radius R has a round disc of radius R/3 cut as shown in fig. The mass of the remaining (shaded) portion of the disc equals M.

To find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Solution:

Moment of inertia of the disc,

$I=I_{remain}+I_{R/3}⟹I_{remain}=I-I_{R/3}$

Now using the values we get

$I_{remain}=\frac{M{R}^2}{2}-[\frac{\frac{M}{4}{\left(\frac{R}{3}\right)}^2}{2}+\frac{M}{4}{\left(\frac{R}{3}\right)}^2]⟹I_{remain}=\frac{M{R}^2}{2}-[\frac{M{R}^2}{72}+\frac{M{R}^2}{36}]⟹I_{remain}=\frac{36M{R}^2-M{R}^2-2M{R}^2}{72}⟹I_{remain}=\frac{33M{R}^2}{72}⟹I_{remain}=\frac{11M{R}^2}{24}$

is the the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.