Question

The moment of inertia of a thin uniform circular disc about one of its diameter is l. Its moment of inertia about an axis tangent to it and perpendicular to its plane is
A. $\dfrac{{2l}}{3}$
B. $2l$
C. $\dfrac{l}{2}$
D. $6l$

Answer 1

Hint: Moment of inertia of a body about a given axis of rotation is defined as the sum of the product of masses of the particles and square of their distances from the axis of the rotation. It is not a fixed quantity; it depends upon the orientation and position of the axis of rotation of the body.

Complete step by step answer:
We know, the moment of inertia of a uniform circular disc of mass M and radius R about an axis passing through center is given by,
${{\rm{I}}_{centre}} = \dfrac{1}{2}{\rm{M}}{{\rm{R}}^2}$
From the perpendicular axis theorem,
moment of inertia of disc about a diameter,
${{\rm{I}}_{diametre}} = \dfrac{{{{\rm{I}}_{centre}}}}{2}$
According to the question, it is given that a moment of inertia of disc about diameter is I then

 ${\rm{I = }}\dfrac{{{{\rm{I}}_{centre}}}}{2} = \dfrac{1}{4}{\rm{M}}{{\rm{R}}^2}$
So, ${\rm{M}}{{\rm{R}}^2} = 4{\rm{I }}........{\rm{(1)}}$
Now, we find a moment of inertia about an axis perpendicular to its plane and passing through a point on its rim.
By applying the perpendicular axis theorem,
${\rm{I{'} = }}{{\rm{I}}_{centre}} + {\rm{M}}{{\rm{R}}^2}$
$ \Rightarrow {\rm{I' = }}\dfrac{1}{2}{\rm{M}}{{\rm{R}}^2} + {\rm{M}}{{\rm{R}}^2}$
$\therefore {\rm{I' = }}\dfrac{3}{2}{\rm{M}}{{\rm{R}}^2}$
Now put the value of $MR^2$ here we get,
${\rm{I' = 6l}}$

Hence the correct option (D).

Note:
According to the parallel axis theorem, the moment of inertia of a body about an axis in its plane and parallel to an axis passes through the center of mass of the body is equal to the moment of inertia about an axis passing through the center of mass of the body plus the product of the mass of the body and square of the distance between two axes. Also, the moment of inertia plays the same role in rotational motion as mass does in linear motion.