Question

Three thin rods each of length L and mass M are placed along the x, y and z axis such that one end of each rod is at origin. The moment of inertia of this system about z-axis is
A. $\dfrac{m{{l}^{2}}}{3}$
B.$\dfrac{2m{{l}^{2}}}{3}$
C. $m{{l}^{2}}$
D. $\dfrac{m{{l}^{2}}}{4}$

Answer 1

Hint: As a very first step, one could read the question well and hence understand the given system carefully and also note down the important points. Then you could find the moment of inertia along x and y axes easily. After that use the perpendicular axis to find that along the z-axis.

Complete step-by-step solution:
In the question, we are given three thin rods that are of lengths L and mass M and these rods are placed along x, y and z axes with one end of each of them placed at the origin. We are supposed to find the moment of inertia of this system about the z-axis.
As we are said that arrangement of the three rods is such that one end of each of them is at the origin, we could say that the moment of inertia of the system along x and y axes could be given by,
${{I}_{x}}={{I}_{y}}=\dfrac{m{{l}^{2}}}{3}$
Now, by using perpendicular axes theorem, we could get the moment of inertia along the z-axis, which would be the sum of the moment of inertias along x and y axes.
${{I}_{z}}={{I}_{x}}+{{I}_{y}}=\dfrac{m{{l}^{2}}}{3}+\dfrac{m{{l}^{2}}}{3}$
$\therefore {{I}_{z}}=\dfrac{2m{{l}^{2}}}{3}$
Therefore, we found the moment of inertia along the z-axis to be ${{I}_{z}}=\dfrac{2m{{l}^{2}}}{3}$.
Option B is correct.

Note: According to the perpendicular axes theorem, the moment of inertia of a planar lamina would be equal to the sum of the moments of inertia of the lamina about the two perpendicular axes. This lies in its own plane intersecting one another and the perpendicular axis is also passing through it.