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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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