Question

A circular protion of radius 10cm has been removed from a large disc
of radius 25cm. The centre of the hole is at a distance of 12 cm from
the centre of the original disc. If the mass of the remaining portion is
840 g, calculate its M.I. about an axis passing through the two centres.​

Answer 1

Let the mass of the original disc be M and that of the portion, which is removed from the disc to make the hole, be m. First we have to relate both M and m.

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Well, the areal density of both the disc and the portion should be same since a uniform disc is considered. So,

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$\dfrac {\textsf {Mass of the disc}}{\textsf {Area of the disc}}=\dfrac {\textsf {Mass of the removed portion}}{\textsf {Area of the removed portion}}\\\\\\\sf{\dfrac {M}{\pi(25)^2}=\dfrac {m}{\pi(10)^2}}\\\\\\\sf{m=\dfrac {4M}{25}}$

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Given that the mass of the remaining portion is 840 g, i.e.,

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$\sf{M-m=840\ g}\\\\\\\sf{M-\dfrac {4M}{25}=840\ g}\\\\\\\sf{\dfrac {21M}{25}=840\ g}\\\\\\\sf{M=1\ kg}$

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Then,

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$\sf{m=160\ g}$

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The moment of inertia of the remaining portion, passing through the center of the original disc perpendicular to the plane is,

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$\sf{I'=\dfrac {M(25)^2}{2}-\left[\dfrac {m(10)^2}{2}+m(12^2)\right]}\\\\\\\sf{I'=\dfrac {1000\times625}{2}-\left[\dfrac {160\times100}{2}+144\times160\right]}\\\\\\\sf{I'=281460\ g\ cm^2}$

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By theorem of perpendicular axes, the moment of inertia of the portion about the axis passing through the centers of hole and original disc is half of the above one, i.e., I'. Thus.,

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$\sf{\underline {\underline {I=140730\ g\ cm^2}}}$

Answer 2

140730 is the answer