Question

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is \[{I_o}\]. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is:
A. \[{I_o} + \dfrac{{M{L^2}}}{4}\]
B. \[{I_o} + 2M{L^2}\]
C. \[{I_o} + M{L^2}\]
D. \[{I_o} + \dfrac{{M{L^2}}}{2}\]

Answer 1

Hint:To calculate the moment of inertia about an axis through an object, two theorems are used. One is the parallel axis theorem and another is the perpendicular axis theorem. Both are used depending upon the situations given in the question. In this problem, the parallel axis theorem is used for the calculation of the required moment of inertia about the given axis of rotation of the rod.

Formula Used:
Theorem of parallel axis,
\[I = {I_C} + M{h^2}\]
Where, \[I\]= Moment of inertia about the axis parallel to the axis passing through the centre of the body.
\[{I_C}\]= Moment of inertia about the axis passing through the centre of the body.
M = mass of the body
h = distance between the two axes

Complete step by step solution:
Given: mass of the rod = M, length of the rod = L, distance between the two axes = L/2. Moment of inertia about the centre of the rod, \[{I_C}\] = \[{I_o}\]
Let M be the mass of the uniform light rod. It is given that the moment of inertia of the rod about an axis perpendicular to its length and passing through the mid-point of that length is \[{I_o}\].

From the theorem of the parallel axis, the moment of inertia of an object about an axis parallel to that object passing through its centre is equal to the sum of the moment of inertia of the object about the axis passing through its midpoint and the product of the mass of the object times the square of the distance of between the two axes i.e.
\[I = {I_C} + M{h^2}\]-----(1)
We have to calculate the moment of inertia about an axis passing through one of the ends of the rod. So, the distance between this axis and the axis passing through the centre is equal to L/2. So,
\[I = {I_o} + M{\left( {\dfrac{L}{2}} \right)^2}\]---(2)
\[\therefore I = {I_o} + \dfrac{{M{L^2}}}{4}\]

Hence option (A) is the correct option.

Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. Also, the moment of inertia of a rod about its centre and perpendicular to length is \[\dfrac{{M{L^2}}}{{12}}\].