Question

A thin uniform rod rotating about an axis passing through its centre and perpendicular to its length, moment of inertia is ${I_o}$. $M.I.$ of rod about an axis through one end and perpendicular to the rod will be:
A) $\dfrac{{{I_o}}}{2}$.
B) $3{I_o}$.
C) $5{I_o}$.
D) $4{I_o}$.

Answer 1

The parallel axis theorem can be used to know the answer for this problem. We can apply the parallel axis theorem only when the moment of inertia about the axis of the center of mass is given and also the distance between the center of mass and the new position where the moment of inertia is to be calculated is given.

Formula used: The formula for parallel axis theorem is given by $I = {I_{cm}} + m{d^2}$, where $I$ is moment of inertia about new position ${I_{cm}}$ is the moment of inertia about center of mass $d$ is the distance between the new moment of inertia and moment of inertia about center of mass.

Complete step-by-step answer:
It is given the moment of inertia about the center of mass if given as ${I_o}$ and it is asked to find the moment of inertia about one end and is perpendicular to the rod.
The moment of inertia of the rod at the center of the rod and perpendicular to the rod is ${I_o} = \dfrac{1}{{12}} \cdot M{L^2}$ where M is mass of the rod and L is the length of the rod.

We have to find the moment of inertia of the rod about one of the ends and perpendicular to the rod can be calculated by the applying parallel axis theorem.
$I = {I_o} + M{d^2}$here the $I$ is moment of inertia that is asked in problem $M$ is the mass of the rod also$d$ is distance between the two moment of inertia about the two axes.

As any end from the centre of the rod is $d = \dfrac{L}{2}$ and ${I_o}$is equal to ${I_o} = \dfrac{1}{{12}} \cdot M{L^2}$. Let us put the value of ${I_o}$ and $d$ in the parallel axis theorem,
$I = {I_o} + M{d^2} \\\ \Rightarrow I = \dfrac{{M{L^2}}}{{12}} + M{\left( {\dfrac{L}{2}} \right)^2} \\\ \Rightarrow I = \dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{4} \\\ \Rightarrow I = \dfrac{{M{L^2} + 3M{L^2}}}{{12}} \\\ \Rightarrow I = \dfrac{{4M{L^2}}}{{12}} \\\ \Rightarrow I = \dfrac{{M{L^2}}}{3} \\\$
So the moment of inertia about the new axis is$I = \dfrac{{M{L^2}}}{3}$. Let us express a new moment of inertia in terms of ${I_o}$.
$I = \dfrac{{M{L^2}}}{3} \\\ \Rightarrow I = \dfrac{4}{4}\left( {\dfrac{{M{L^2}}}{3}} \right) \\\ \Rightarrow I = \dfrac{{4M{l^2}}}{{12}} \\\ \Rightarrow I = 4{I_o} \\\$ (Multiply 4 in denominator and numerator.)
So the correct answer for this problem is $I = 4{I_o}$.

So the correct answer is option D.

Note: It is important for students to remember the moment of inertia of some of the standard shapes like rod, rectangular sheet, cuboid, disc, conical shape and sphere. It is advisable to learn the parallel axis theorem and also perpendicular axis theorem as these theorems help the students to find moments of inertia about different axes.