Question

A circular portion of radius 4cm has been removed from the centre of large circular disc of radius 16cm. If the mass of the remaining portion is 180g, what is its moment of inertia about an axis passing through its centre and perpendicular to its plane

Answer 1

Radius of the original disc, $\sf{R=16\ cm.}$

Radius of the removed circular portion, $\sf{r=4\ cm.}$

Let mass of the original disc be $\sf{M}$ in grams, so that mass of the removed portion will be $\sf{M-180}$ grams.

Assume the disc to be uniform, so the areal density of the disc will be same everywhere in it.

Therefore, density of the original disc is equal to that of the removed portion.

$\longrightarrow\sf{\dfrac{M}{\pi R^2}=\dfrac{M-180}{\pi r^2}}$

$\longrightarrow\sf{\dfrac{M}{\pi(16)^2}=\dfrac{M-180}{\pi(4)^2}}$

$\longrightarrow\sf{\dfrac{M-180}{M}=\dfrac{4^2}{16^2}}$

$\longrightarrow\sf{\dfrac{M-180}{M}=\dfrac{16-15}{16}}$

By rule of dividendo,

$\longrightarrow\sf{\dfrac{M}{180}=\dfrac{16}{15}}$

$\longrightarrow\sf{M=192\ g}$

Moment of inertia of the original disc is,

$\longrightarrow\sf{I_1=\dfrac{1}{2}\,MR^2}$

$\longrightarrow\sf{I_1=\dfrac{1}{2}\times192\times16^2}$

$\longrightarrow\sf{I_1=24576\ g\,cm^2}$

Moment of inertia of the removed portion is,

$\longrightarrow\sf{I_2=\dfrac{1}{2}\,(M-180)r^2}$

$\longrightarrow\sf{I_2=\dfrac{1}{2}\times(192-180)\times4^2}$

$\longrightarrow\sf{I_2=96\ g\,cm^2}$

Hence moment of inertia of the remaining portion is,

$\longrightarrow\sf{I=I_1-I_2}$

$\longrightarrow\sf{I=24576-96}$

$\longrightarrow\underline{\underline{\sf{I=24480\ g\,cm^2}}}$