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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,  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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