Answer:
The value is [tex]I = 0.0932 ML ^2[/tex]
Explanation:
From the question we are told that
The rotational inertia about one end is [tex]I_R = \frac{1}{3} ML^2[/tex]
The location of the axis of rotation considered is [tex]d = 0.4 L[/tex]
Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is [tex]0.4 M[/tex]
Generally the length of the rod from the its beginning to the axis of rotation consider is
[tex]k = 1 - 0.4 L = 0.6L[/tex]
Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is
[tex]m = 1- 0.4 M = 0.6 M[/tex]
Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is
[tex]I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
[tex]I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2[/tex]
Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is
[tex]I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is
[tex]I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
=> [tex]I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ][/tex]
=> [tex]I = 0.0932 ML ^2[/tex]
