Question

A circular disc 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 $\dfrac{t}{4}$. Then the relation between the moment of inertia I$_X$ and I$_Y$ about their natural axis is-
(A) I$_Y$ = 64 I$_X$
(B) I$_Y$ = 32 I$_X$
(C) I$_Y$ = 16 I$_X$
(D) I$_Y$ = I$_X$

Answer 1

The moment of inertia of anybody gives a measure of distribution of mass in the body. It is the product of the mass and the distance from the axis. For our case, we may use the direct result of moment of inertia of a disc.
Formula used:
For a disc of mass M and radius R, the moment of inertia is given as:
$I =\dfrac{1}{2} M R^2$.

Complete answer:
In the given question, for the disc X and disc Y, the material used is the same. Therefore, the radius and the thickness will just change the mass of the given discs.
The mass of any body is the product of density and volume. When we assume that our disc is a very thin cylinder, we can directly say that the mass will be directly proportional to thickness and square of radius,
$M \propto t{R^2}$.
For I$_X$, we may write:
${I_X} \propto t{R^4}$
Similarly for I$_Y$ we may write:
\eqalign{ & {I_Y} \propto \left( {{t \over 4}} \right)\left( {{{(4R)}^4}} \right) \cr & \Rightarrow {I_Y} \propto 64t{R^4} \cr} The ratio of the two gives us,
I$_Y$ = 64 I$_X$

Therefore, the correct answer is option (A).

Additional Information:
The moment of inertia of uniform body is given as:
$I = \int {{r^2}} dm$.
For the case of a simple mass at some distance from the axis, the moment of inertia of that mass about that axis is just the product of mass and square of distance.

Note:
Moment of inertia for a disc can be found by considering the small mass element of the disc and multiplying the square of distance to it (from the axis) and then integrating it. The best way to perform this integration is by using polar coordinates. The key in this question lies in the difference in mass of the two discs X and Y because the radius is stated in the question itself.