Answer:
[tex]I=1.84\ kg.m^2[/tex]
Explanation:
It is given that,
Mass of each mass, m = 2 kg
Side of the square frame, l = 0.68 m
It is clearly written in the question that the four point masses positioned at the corners of an essentially massless rigid frame if the frame rotates around one side like a door rotates on its hinges. Let ABCD are the vertices of square and AB acts like a door.
The moment of inertia for a set of point masses is given by :
[tex]I=\sum n_im_ir_i^2[/tex]
So,
[tex]I=I_A+I_B+I_C+I_D[/tex]
The sum of inertia at A and B is 0 because l = 0
[tex]I=0+0+ml^2+ml^2[/tex]
[tex]I=2ml^2[/tex]
[tex]I=2\times 2\ kg\times (0.68\ m)^2[/tex]
[tex]I=1.84\ kg.m^2[/tex]
So,the moment of inertia for a frame of four point masses positioned at the corners is [tex]1.84\ kg.m^2[/tex]. Hence, this is the required solution.
