Rotation Ques 24
- A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I _O$ and
$I _P$, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I _P}{I _O}$ to the nearest integer is
(2012)
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Answer:
Correct Answer: 24.(3)
Solution:
Formula:
Two Important Theorems On Moment Of Inertia :
- $T=$ Total portion
$R=$ Remaining portion and
$C=$ Cavity and let $\sigma=$ mass per unit area.
Then, $\quad m _T=\pi(2 R)^{2} \sigma=4 \pi R^{2} \sigma$
For $I _P$
$$ m _C=\pi(R)^{2} \sigma=\pi R^{2} \sigma $$
$$ \begin{aligned} I _R & =I _T-I _C \\ & =\frac{3}{2} m _T(2 R)^{2}-\frac{1}{2} m _C R^{2}+m _C r^{2} \\ & =\frac{3}{2}\left(4 \pi R^{2} \sigma\right)\left(4 R^{2}\right)-\frac{1}{2}\left(\pi R^{2} \sigma\right)+\left(\pi R^{2} \sigma\right)\left(5 R^{2}\right) \\ & =\left(18.5 \pi R^{4} \sigma\right) \end{aligned} $$
For $\boldsymbol{I} _{\boldsymbol{O}} I _R=I _T-I _C$
$=\frac{1}{2} m _T(2 R)^{2}-\frac{3}{2} m _C R^{2}$
$=\frac{1}{2}\left(4 \pi R^{2} \sigma\right)\left(4 R^{2}\right)-\frac{3}{2}\left(\pi R^{2} \sigma\right)\left(R^{2}\right)=6.5 \pi R^{4} \sigma$
$\therefore \quad \frac{I _P}{I _O}=\frac{18.5 \pi R^{4} \sigma}{6.5 \pi R^{4} \sigma}=2.846$
Therefore, the nearest integer is 3 .
BETA
