Subjective Type

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.

Solution

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,
$$I=I_{remain}+I_{R/3}\\\implies I_{remain}=I-I_{R/3}$$
Now using the values we get
$$I_{remain}=\dfrac {MR^2}{2}-\left[\dfrac {\dfrac M4\left(\dfrac R3\right)^2}{2}+\dfrac M4\left(\dfrac R3\right)^2\right]\\\implies I_{remain}=\dfrac {MR^2}{2}-\left[\dfrac {MR^2}{72}+\dfrac {MR^2}{36}\right]\\\implies I_{remain}=\dfrac {36MR^2-MR^2-2MR^2}{72}\\\implies I_{remain}=\dfrac {33MR^2}{72}\\\implies I_{remain}=\dfrac {11MR^2}{24}$$
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.

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