Question

Let ${I_A}$ and ${I_B}$ be moments of inertia of a body about two axes $A$ and $B$ respectively. The axis $A$ passes through the center of mass of the body but $B$ does not
$\left( A \right)$ ${I_A} < {I_B}$
$\left( B \right)$ $If{\text{ }}{I_A} < {I_B}$
$\left( C \right)$ ${\text{If the axes are parallel, }}{I_A} < {I_B}$
$\left( D \right)$ ${\text{If the axes are not parallel, then }}{I_A} < {I_B}$

Answer 1

Hint: This question can be solved based on the parallel axis theorem. It states that the moment of inertia of a body regarding an axis parallel to the body passing through its center is adequate the sum of moment of inertia of the body regarding the axis passing through the center and products of mass of the body times the square of the distance between the two axes.

Formula used:
Parallel axis theorem,
$ \Rightarrow I = {I_c} + M{h^2}$
Where, $I$ ; Inertia of the body
${I_C}$; MOI about the center.
$M$; The mass of the body.
${h^2}$; The square of the distance between the two of the axes

Complete step by step solution:
Derivation for the parallel axis theorem
Let ${I_C}$ be the moment of inertia of an axis that is passing through the middle of mass and $I$ be the moment of inertia concerning the axis $A'B'$ at a distance$h$. Consider a particle of mass $m$ at a distance $r$ from the middle of gravity of the body. Then,
The Distance from $A'B' = r + h$
$ \Rightarrow I = \sum\limits_{}^{} {m{{\left( {r + h} \right)}^2}} $
$ \Rightarrow I = \sum\limits_{}^{} {m{{\left( {{r^2} + {h^2} + 2rh} \right)}^{}}} $
$ \Rightarrow I = \sum {m{r^2} + } \sum {m{h^2} + \sum {2rh} } $
$ \Rightarrow I = {I_C} + h2\sum {m + 2h\sum {mr} } $
$ \Rightarrow I = {I_C} + M{h^2} + 0$
$ \Rightarrow I = {I_C} + M{h^2}$
Therefore this is the derivation of the parallel axis theorem.
So from the question, if the axis is parallel then according to the parallel axis theorem,
Then ${I_A} < {I_B}$
And if the axis is not parallel then there might be some axis which is ${I_C}$which is parallel to $B$ the axis and perpendicular to $A$ the axis.
Therefore ${I_C}$ will be less than or more than ${I_A}$
Now if ${I_A} < {I_C}$then ${I_A} < {I_B}$
Else if-then \[{I_B}\]may be less than or greater than ${I_A}$.

Note: For any plane body the moment of inertia concerning any of its axes that are perpendicular to the plane is adequate the sum of the moment of inertia concerning any two perpendicular axes within the plane of the body that sees the primary axis within the plane.