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Moment Of Inertia Of A Disc - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Moment of inertia of a DISC is considered one of the most asked concept.
13 Questions around this concept.

Solve by difficulty

The moment of inertia of a uniform semicircular disc of mass M and radius $r$ about a line perpendicular to the plane of the disc through the center is :d

A

$Mr^2$

B

$\frac{1}{2}Mr^{2}$

C

$\frac{1}{4}Mr^{2}$

D

$\frac{2}{5}Mr^{2}$

A circular disc X of radius R is made from an iron plate of thickness $t$ and another disc Y of radius $4R$ is made from an iron plate of thickness $t/4$ . Then the relation between the moment of inertia. $I_{X}$ and $I_{Y}$ is :

A

$I_{Y}= 32I_{X}$

B

$I_{Y}= 16I_{X}$

C

$I_{Y}= I_{X}$

D

$I_{Y}= 64I_{X}$

Concepts Covered - 1

Moment of inertia of a DISC

Let I=Moment of inertia of a DISC about an axis through its centre and perpendicular to its plane

To calculate I

Consider a circular disc of mass M, radius R and centre O.

And mass per unit area = $\sigma = \frac{M}{\pi R^2}$

Take an elementary ring of mass dm of radius x as shown in figure

So, $dm=\sigma *(2\pi xdx)=\frac{M}{\pi R^2}*(2\pi xdx)$

$\Rightarrow dI= x^2dm$

$I=\int dI=\int_{0}^{R}x^2*(\frac{M}{\pi R^2}*(2\pi xdx))=\frac{2M}{R^2}\int_{0}^{R}x^3dx=\frac{MR^2}{2}$

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Moment of inertia of a DISC

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