Question 387: Centre of Mass and Rotational Motion

Question

Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is $\_\_\_\_$ .

Accepted Answer

Mass of full disc is M
Radius of full disc is $r$
Surface Mass Density $=\sigma=\frac{\mathrm{M}}{\pi \mathrm{r}^2}$
Two identical small discs of radius $r^{\prime}=\frac{r}{4}$ are removed.
Mass of each small disc (m):
$$ \mathrm{m}=\sigma 	imes \pi\left(\mathrm{r}^{\prime}ight)^2=\frac{\mathrm{M}}{\pi \mathrm{r}^2} 	imes \pi\left(\frac{\mathrm{r}}{4}ight)^2=\frac{\mathrm{M}}{16} $$
The centre of each small disc is at a distance $\mathrm{d}=\frac{3}{4} \mathrm{r}$ from the central point. Since axis A passes through the centre, the distance of each centre from axis A is $\mathrm{d}=\frac{3}{4} \mathrm{r}$.
The moment of inertia of full disc $\left(\mathrm{I}_{	ext {total }}ight)$ :
$$ \mathrm{I}_{	ext {total }}=\frac{1}{2} \mathrm{Mr}^2 $$
The moment of inertia of small disc about its own centre ( $\mathrm{I}_{\mathrm{cm}}$ ):
$$ \mathrm{I}_{\mathrm{cm}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{r}^{\prime}ight)^2=\frac{1}{2}\left(\frac{\mathrm{M}}{16}ight)\left(\frac{\mathrm{r}}{4}ight)^2=\frac{\mathrm{Mr}^2}{512} $$
The moment of inertia of small disc about axis A ( $\mathrm{I}_{	ext {removed }}$ ) using the parallel axis theorem :
$$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2 $$
$$\Rightarrow $$ $\mathrm{I}_{	ext {removed }}=\frac{\mathrm{Mr}^2}{512}+\left(\frac{\mathrm{M}}{16}ight)\left(\frac{3 \mathrm{r}}{4}ight)^2=\frac{\mathrm{Mr}^2}{512}+\frac{9 \mathrm{Mr}^2}{256}=\frac{\mathrm{Mr}^2+18 \mathrm{Mr}^2}{512}=\frac{19}{512} \mathrm{Mr}^2$
The moment of inertia of rest of the system is ( $I_{	ext {rem }}$ )
$$ \mathrm{I}_{	ext {rem }}=\mathrm{I}_{	ext {total }}-2\left(\mathrm{I}_{	ext {removed }}ight) $$
$$\Rightarrow $$ $I_{\mathrm{rem}}=\frac{1}{2} \mathrm{Mr}^2-2\left(\frac{19}{512} \mathrm{Mr}^2ight)=\frac{1}{2} \mathrm{Mr}^2-\frac{19}{256} \mathrm{Mr}^2$
$$\Rightarrow $$ $\mathrm{I}_{\mathrm{rem}}=\left(\frac{128-19}{256}ight) \mathrm{Mr}^2=\frac{109}{256} \mathrm{Mr}^2=\frac{\mathrm{x}}{256} \mathrm{Mr}^2$
$$ \Rightarrow x=109 $$
Therefore, the value of x is 109 . Hence, the correct answer is $\mathbf{1 0 9}$.

Centre of Mass and Rotational Motion

Suppose there is a uniform circular disc of mass M \mathrm{~kg} and radius r \mathrm{~m} shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by \frac{x}{256} M r^2. The value of x is \_\_\_\_ .

Integer / Numeric Answer Type