Question

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point located 0.40 L from the end

Answer 1

Answer:

The value is  $I = 0.0932 ML ^2$  

Explanation:

From the question we are told that

  The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

   The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

      $k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    $m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     $I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

    $I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     $I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=>  $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  $I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=>   $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=>   $I = 0.0932 ML ^2$