Question:

The moment of inertia of a uniform circular disc of radius $ R $ and mass $ M $ about an axis touching the disc at its diameter and normal to the disc is

Updated On: Jan 23, 2024
  • $ MR^2 $
  • $ \frac{2}{5}MR^2 $
  • $ \frac{3}{2}MR^2 $
  • $ \frac{1}{2}MR^2 $
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The Correct Option is C

Solution and Explanation

The moment of inertia about an axis passing through centre of mass of disc and
perpendicular to its plane is
$I_{ CM }=\frac{1}{2} M R^{2}$
where $M$ is the mass of disc and $R$ its radius. According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc at its diameter and normal to the disc is
$ I =I_{ CM }+M R^{2} $
$=\frac{1}{2} M R^{2}+M R^{2} $
$=\frac{3}{2} M R^{2} $
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Concepts Used:

Moment of Inertia

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

Moment of inertia mainly depends on the following three factors:

  1. The density of the material
  2. Shape and size of the body
  3. Axis of rotation

Formula:

In general form, the moment of inertia can be expressed as, 

I = m × r²

Where, 

I = Moment of inertia. 

m = sum of the product of the mass. 

r = distance from the axis of the rotation. 

M¹ L² T° is the dimensional formula of the moment of inertia. 

The equation for moment of inertia is given by,

I = I = ∑mi ri²

Methods to calculate Moment of Inertia:

To calculate the moment of inertia, we use two important theorems-

  • Perpendicular axis theorem
  • Parallel axis theorem