Rotation 1 Question 19

22. 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)

Show Answer

Answer:

Correct Answer: 22. (3)

Solution:

  1. $T=$ 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 .



NCERT Chapter Video Solution

Dual Pane