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# Help me understand! - Rotational Motion - JEE Main-3

A thin disc of mass M and radius R has mass per unit area $\sigma (r)=kr^{2}$ where

r is the distance from the center. Its moment of inertia abount an axis going through its centre of mass

and perpendicular to its plane is:

• Option 1)

$\frac{MR^{2}}{3}$

• Option 2)

$\frac{2MR^{2}}{3}$

• Option 3)

$\frac{MR^{2}}{6}$

• Option 4)

$\frac{MR^{2}}{2}$

Answers (1)
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Moment of inertia for continuous body -

$I= \int r^{2}dm$

- wherein

r is perpedicular distance of a particle of mass dm of rigid body from axis of rotation

$\sigma (r)=kr^{2}$

$dI=dmr^{2}$

$dm=\sigma \times 2\pi r dr$

$dm=kr^{2} \times 2\pi r dr$

$m=\int dm=2\pi k\int r^{3}dr=2\pi k(\frac{R^{4}}{4})$...............(1)

$dI=dmr^{2}$

$I=\int dI=2\pi k\int r^{2}\cdot r^{3}dr=2\pi k\int r^{5}dr = 2\pi k(\frac{R^{6}}{6})$................(2)

from (1) and (2)

$I=\frac{2MR^{2}}{3}$

Option 1)

$\frac{MR^{2}}{3}$

Option 2)

$\frac{2MR^{2}}{3}$

Option 3)

$\frac{MR^{2}}{6}$

Option 4)

$\frac{MR^{2}}{2}$

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