Question

The moment of inertia of a circular plate of uniform thickness and of radius R about an axis perpendicular to plane of plate and passing through its centre is I. A circular plate of radius $\dfrac{R}{2}$ is cut and removed from the initial plate. The moment of inertia of the plate after removing the part about an axis perpendicular to the plane of plate and passing through its centre is:

A. $\dfrac{I}{2}$
B. $\dfrac{I}{4}$
C. $\dfrac{I}{8}$
D. $\dfrac{{15I}}{{16}}$

Answer 1

Hint:Whenever a body is set into motion, it is observed that the body will not respond quickly to the rotation. This is due to the fact there is something called inertia which indicates the inability to adapt to the quick change in the state of the body. When this inertia is associated with rotation, we call it a moment of inertia.

Complete step-by-step answer:
The moment of inertia basically, gives us the distribution of mass of a body around its axis of rotation and it is the quantity that determines the torque required to produce an angular acceleration in the rotating body.
Consider a plate of radius R of uniform thickness as shown:

The moment of inertia of the plate of radius R about the axis passing through O,
${I_R} = \dfrac{{{M_R}{R^2}}}{2}$
Now, since the radius of the plate is decreased by half, the mass reduced will depend on:
$
  M \propto {R^2} \\
   \to \dfrac{M}{{{M_R}}} = {\left( {\dfrac{R}{{2R}}} \right)^2} \\
 $
The amount of mass that is removed –
$
  \dfrac{M}{{{M_R}}} = {\left( {\dfrac{R}{{2R}}} \right)^2} \\
   \to \dfrac{M}{{{M_R}}} = \dfrac{1}{4} \\
   \to M = \dfrac{{{M_R}}}{4} \\
 $
The moment of inertia of the removed portion:
$
  I = \dfrac{{M{R^2}}}{2} \\
  Substituting, \\
  I = \dfrac{{{M_R}{R^2}}}{{4 \times 2 \times 4}} \\
   \to I = \dfrac{{{M_R}{R^2}}}{{32}} \\
 $
The moment of inertia of the retained portion is equal to the difference between the moment of inertia of radius R and the moment of inertia of the removed mass.
\[
  {I_r} = {I_R} - I \\
  Substituting, \\
   \to {I_r} = {I_R} - I \\
   \to {I_r} = \dfrac{{{M_R}{R^2}}}{2} - \dfrac{{{M_R}{R^2}}}{{32}} \\
   \to {I_r} = \dfrac{{16{M_R}{R^2} - {M_R}{R^2}}}{{32}} \\
   \to {I_r} = \dfrac{{{M_R}{R^2}\left( {16 - 1} \right)}}{{32}} \\
   \to {I_r} = \dfrac{{15 \times {M_R}{R^2}}}{{32}} \\
   \to {I_r} = \dfrac{{15}}{{16}}\dfrac{{{M_R}{R^2}}}{2} \\
   \to {I_r} = \dfrac{{15}}{{16}}{I_R}\because {I_R} = \dfrac{{{M_R}{R^2}}}{2} \\
 \]
Hence, the new moment of inertia of the new plate is equal to $\dfrac{{15}}{{16}}I$.

Hence, the correct option is Option D.

Note: There are 2 ways of compressive stress failure in columns – Crushing and Buckling. Crushing occurs for shorter columns, also called struts and Buckling occurs in longer columns. Even though you may think that their classification of long and short columns is subjective, it depends on a number known as Slenderness Ratio, which is dependent on moment of inertia.
Slenderness ratio = $\dfrac{l}{k}$
where $l$= effective length of the column and $k$= radius of gyration.
The radius of gyration, $k = \sqrt {\dfrac{I}{A}} $
where $I$is the least moment of inertia of the cross-section of the column and A is the area of cross-section.