Question 559: Centre of Mass and Rotational Motion

Question

From a disc of mass '$M$' and radius '$R$', a circular hole of diameter '$R$' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

Accepted Answer

Moment of inertia of disc is given by
$\mathrm{I}_{	ext {disc }}=\mathrm{I}_{\mathrm{r}}+\mathrm{I}_{	ext {hole }}$ $\ldots .\left\{I_r=ight.$ M.I. of remaining part $\}$
$	herefore \quad \mathrm{I}_{\mathrm{r}}=\mathrm{I}_{	ext {disc }}-\mathrm{I}_{	ext {hole }}$ ..... (i)
$\mathrm{I}_{\mathrm{disc}}=\frac{\mathrm{MR}^2}{2}$ ..... (ii)
By parallel axes theorem we get,
$\mathrm{I}_{	ext {hole }}=\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}ight)^2}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}ight)^2ight]$
$\ldots .\left\{\begin{array}{l}\because \mathrm{M}_{	ext {hole }}=\frac{\mathrm{M}_{	ext {dise }}}{4} \ \because 	ext { the surface density is same }\end{array}ight\}$
$	herefore \quad \mathrm{I}_{	ext {hole }}=\left[\frac{\mathrm{MR}^2}{32}+\frac{\mathrm{MR}^2}{16}ight]$ ..... (iii)
Substituting eq (iii) and eq (ii) in eq (i) we get,
$\begin{aligned}
\mathrm{I}_{\mathrm{r}} & =\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{32}-\frac{\mathrm{MR}^2}{16} \
& =\mathrm{MR}^2\left[\frac{1}{2}-\frac{1}{32}-\frac{1}{16}ight] \
& =\frac{13}{32} \mathrm{MR}^2
\end{aligned}$

Answer

$\frac{13 \mathrm{MR}^2}{32}$

Answer

$\frac{11 \mathrm{MR}^2}{32}$

Answer

$\frac{9 \mathrm{MR}^2}{32}$

Answer

$\frac{7 \mathrm{MR}^2}{32}$

Centre of Mass and Rotational Motion

From a disc of mass 'M' and radius 'R', a circular hole of diameter 'R' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

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