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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 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 :

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, $^{d m}=\sigma *(2 \pi x d x)=\frac{M}{\pi R^2} *(2 \pi x d x)$

$
\begin{aligned}
\Rightarrow d I & =x^2 d m \\
I & =\int d I=\int_0^R x^2 *\left(\frac{M}{\pi R^2} *(2 \pi x d x)\right)=\frac{2 M}{R^2} \int_0^R x^3 d x=\frac{M R^2}{2}
\end{aligned}
$
 

 

 

 

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Moment of inertia of a DISC

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